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ADVANCEDMANAGEMENTACCOUNTINGConcepts & Theory (Updated Till Nov-12)

CA B L Chakravarti(B.Sc, M.Sc, FCA, CPA, CAIIB)

Gurukul ComU-158 Vats Complex

merce Academy Laxmi nager Delhi - 92 ,9910993376 @gmail.com

+91 -9810993376 E-mai : gurukulca

CA B L Chakravarti is a Fellow Member of Institute of Chartered Accountants of India.

15 years. Mr. Chakravarti is a Science Graduate of 1995 from B.B.C Degree College, Jhansi and scored more than 70% in almost all the subjects. Mr. Chakravarti became topper of 1994 by securing 96% in an individual Subject

He hascompleted Chartered Accountancy in 2002 and having teaching experience of more than

He Completed his M.Sc in Mathematics.

After Completion of Chartered Accountancy, Mr. Chakravarti joined Punjab National Bank as

Financial Analyst and represented Credit Risk Management Committee at Zonal Office &

Head Office. His last Job was in the capacity of Chief Finance Officer in a Leading IT

Company. He has more than 10 years experience in Corporate Finance, Risk Analysis

and Re-Structuring of Loans.

Mr. Chakravarti has fondness of teaching and started teaching from college days. Mr.

During 1997 to 2004, all the XII-ISC Topper in Accountancy of Jhansi was Mr. Chakravarti's Students. Mr. Chakravarti started classes for CA Foundation in 1999 and Most of the Students of CA Foundation secured more than 80%-90% in Accountancy.

Chakravarti's specific teaching style and focus on concepts makes complexchapters easy.

150/-

Exclusive Costing Classes for

CA/CMA/CS By: CA B L Chakravarti

(B.Sc, M.Sc, FCA, CPA, CAIIB)

Special Features: 1) Best Concept with Practical approach 2) Free Trial Class Available 3) Best Notes specifically designed

a) Which Covers ICAI Study Material / Practice Manual/ Compiler b) Which Covers Scanner/ Theory also

4) After completion of Syllabus, Revision through Model Test Paper. 5) Air-Conditioned Class Room 6) Live Back-Up of Classes

7) SMS Your Name and Batch Required at 9910993376

K K Klasses (We Not Only Teach Rather We Prepare for Examinations also)

Refundable Registration: Bring two photograph & Rs. 1000

8459423435, 9910993376

1/11, Lalita Park, Behind Mahindra Show Room, Delhi-92

Exclusive Costing Classes for

CA Final AMA/ IPCC Cost FM CMA Final: Cost FM (2012 Syllabus)

CMA Gr I: Cost FM (2012 Syllabus)

CMA Gr II: Cost MA (2012 Syllabus)

CMA Gr II: Cost MA (2008 Syllabus)

Course Days Timing Tenure Start Fees

CA Final AMA

CMA Final

TTSS 7 to 10.30 am 45 Lectures

Call or SMS 9910993376 ₹ 10,000

MWF 7 to 10.30 am

IPCC Cost FM

CMA Gr II Cost MA (2008 Syllabus)

TTSS 11 to 2 pm 50 Lectures

Call or SMS 9910993376

₹ 5,000

MWF 11 to 2 pm ₹ 5,000

CMA Gr I Cost FM

CMA Gr II Cost MA

(2012 Syllabus)

TTSS 11 to 2 pm 40

Lectures Call or SMS

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₹ 5,000

MWF 11 to 2 pm ₹ 5,000

1. New Batches starts in First Week of Oct, Dec, April & Jun

2. Completion of syllabus within tenure.

3. In Other Coaching, Most of the students do not revise the syllabus

and before examination, got frustrated or afraid from Costing.

4. At KK Klasses, We ensure your 100% revision through Model Test

Paper and after revision, Student get relaxed from Costing Paper.

Advanced Management Accounting Theory & Concepts for CA Final

Updated Till November 2012

By: CA B L Chakravarti (B.Sc, M.Sc, FCA, CPA, CAIIB)

Gurukul Commerce Academy U-158, Vats Complex, Laxmi Nagar,

Delhi-110092

Acknowledgement

This Book is Dedictated to

My mother

Thanks Maa…..

For everything

Price: ₹ 150/- No part of this book may be reproduced or utilized in any form or by any means of graphical, electronic or mechanical including photocopying, recording taping or information retrieval system or reproduced on any disc, tape perforated media or any other storage device etc. and retrieval system, without the written permission from the Author.

Preface

This is the first edition of Advance Management Accounting – Concepts & Theory. I feel a

great feeling of achievement and enthusiasm in presenting first edition of my book. This

book has primarily been written with the objective for meeting the needs and interest of

C.A. final students relating to theory aspect of Advance Management Accounting which

covers almost 20% - 25% marks of the CA Final – Advance Management Accounting Paper.

This book covers concepts and theory of all the chapters of CA Final – Advance

Management Accounting Paper. Every Chapter has been dealt precisely and to the point in

a simple and easy language. This Book Along with Past Year Questions, Question of

Revision Test Papers also covered. The content have been so prepared with modification

so as to retain the basics of the subject and enable students and followers to achieve better

results

I have devoted my sincere efforts in preparing this book. However Inspite of my best

efforts, I am aware of possible errors and omissions that escaped my notice. I shall be

extremely thankful for your valuable suggestions, criticism, and observations for further

improvement of this book. I am reachable at [email protected]

For guidance and practical problems and concepts, you can attend our face to face classes

at Laxmi Nagar or call at 9910993376 & 9810993376.

31st March 2013 CA. B L Chakravarti

mailto:[email protected]

Syllabus

Paper 5: Advanced Management Accounting

(One paper – Three hours – 100 marks)

Level of Knowledge: Advanced knowledge

Objective:

To apply various management accounting techniques to all types of organizations for planning,

decision making and control purposes in practical situations.

To develop ability to apply quantitative techniques to business problems

1. Cost Management

a) Developments in the business environment; just in time; manufacturing resources

planning; (MRP); automated manufacturing; synchronous manufacturing and back flush

systems to reflect the importance of accurate bills of material and routings; world class

manufacturing; total quality management.

b) Activity based approaches to management and cost analysis

c) Analysis of common costs in manufacturing and service industry

d) Techniques for profit improvement, cost reduction, and value analysis

e) Throughput accounting

f) Target costing; cost ascertainment and pricing of products and services

g) Life cycle costing

h) Shut down and divestment.

2. Cost Volume Profit Analysis

a) Relevant cost

b) Product sales pricing and mix

c) Limiting factors

d) Multiple scarce resource problems

e) Decisions about alternatives such as make or buy, selection of products, etc.

3. Pricing Decisions

a) Pricing of a finished product

b) Theory of price

c) Pricing policy

d) Principles of product pricing

e) New product pricing

f) Pricing strategies

g) Pricing of services

h) Pareto analysis

4. Budgets and Budgetary Control

The budget manual, Preparation and monitoring procedures, Budget variances, Flexible budgets,

Preparation of functional budget for operating and non-operating functions, Cash budgets, Capital

expenditure budget, Master budget, Principal budget factors.

5. Standard Costing and Variance Analysis

Types of standards and sources of standard cost information; evolution ofstandards, continuous-

improvement; keeping standards meaningful and relevant; variance analysis; disposal of

variances.

a) Investigation and interpretation of variances and their inter relationship

b) Behavioural considerations.

6. Transfer pricing

a) Objectives of transfer pricing

b) Methods of transfer pricing

c) Conflict between a division and a company

d) Multi-national transfer pricing.

7. Cost Management in Service Sector

8. Uniform Costing and Inter firm comparison

9. Profitability analysis - Product wise/ segment wise / customer wise

10. Financial Decision Modeling

a) Linear Programming

b) Network analysis - PERT/CPM, resource allocation and resource leveling

c) Transportation problems

d) Assignment problems

e) Simulation

f) Learning Curve Theory

g) Time Series Fore casting

h) Sampling and Test of Hypothesis.

Index

Page No.

November 2012 Question Papers:

Chapters:

01. Cost Concept & CVP Analysis in Decision Making 1.1 to 1.16

02. Transfer Pricing 2.1 to 2.04

03. Pricing Decision 3.1 to 3.14

04. Budget & Budgetary Control 4.1 to 4.10

05. Standard Costing 5.1 to 5.14

06. Costing of by Service Sector 6.1 to 6.06

07. Development in Business Environment 7.1 to 7.56

a) Total Quality Management

b) Activity Based Management

c) Target Costing

d) Life Cycle Costing

e) Value Chain Analysis

f) Cost Control & Cost Reduction

g) Computer Aided Manufacturing

h) Just In Time

i) Manufacturing Resource Planning

j) Synchronous Manufacturing

k) Business Process Re-Engineering

l) Throughput Accounting

08. Uniform Costing & Inter Firm Comparison 8.1 to 8.06

09. Cost Sheet, Profitability & Responsibility Accounting 9.1 to 9.14

10. Linear Programming 10.1 to 10.10

11. Transportation Problems 11.1 to 11.06

12. Assignment Problems 12.1 to 12.04

13. CPM & PERT 13.1 to 13.14

14. Simulation 14.1 to 14.06

15. Learning Curve 15.1 to 15.08

November 2012 Question Papers PAPER – 5: ADVANCED MANAGEMENT ACCOUNTING

Question No.1 is compulsory.

Answer any five out of the remaining six questions

Working notes should form part of the answer

No statistical or other table is to be distributed along with this paper

Question 1 (a) If Moonlite Limited operates its plant at normal capacity it produces 200000 units from the plant

“Meghdoot”. The unit cost of manufacturing at normal capacity is as under: ₹

Direct Material 65 Direct Labour 30 Variable Overhead 33 Fixed Overhead 7

135 Direct labour cost represents the compensation to highly-skilled workers, who are permanent employees of the company. The company cannot afford to lose them. One labour hour is required to complete one unit of the product.

The company sells its product for ₹ 200 per unit with variable selling expenses of ₹ 16 per unit. The company estimates that due to economic down turn. It will not be able to operate the plant at the normal capacity, at least during the next year. It is evaluating the feasibility of shutting down the plant temporarily for one year.

If it shuts down the plant, the fixed manufacturing overhead will be reduced to ₹ 125000. The overhead costs are incurred at a uniform rate throughout the year. It is also estimated that the additional cost of shutting down will be ₹ 50000 and the cost of re-opening will be ₹ 1,00,000.

Required: Calculate the minimum level of production at which it will be economically beneficial to continue to operate the plant next year if 50% of the labour hours can be utilized in another activity, which is expected to contribute at the rate of ₹ 40 per labour hours. The additional activity will relate to a job which will be off-loaded by a sister company only if the company decides to shut down the plant. (Assume that the cost structure will remain unchanged next year. Ignore income tax and time value of money.) (5 Marks)

(b) An investor is interested in investing Rs. 15,00,000 in a portfolio of investments. The investment choices

and expected rates of return on each one of them are: The investor

wants at least 40% of his investment in Government Bonds. Because of the higher perceived risk of the two shares, he has specified that the combined investment in these two shares not to exceed Rs. 2,60,000. The investor has also specified that at least 25% of the investment should be in the money market fund and that the amount of money invested in shares should not exceed the amount invested in Mutual Funds. His final investment condition is that the amount invested in mutual fund 'XY' should be no more than the amount invested in mutual fund 'HN'. The problem is to decide the amount of money to invest in each alternative so as to obtain the highest annual total return.

Required: Formulate the above as a linear programming problem. (5 Marks)

Investment Projected Rate of Return Mutual Fund XY 15% Mutual Fund HN 9% Money Market Fund 8% Government Bond 8.75% Share P 17% Share Q 18%

(c) PQR Limited sells two versions: Deluxe and Premium of its only product GoGo Juicer. The GoGo Juicer

uses patented technology to extract the last drop of juice from most fruits. The ‘Premium’ version can handle larger fruit and has more option relative to the ‘Deluxe’ version. The following table provides the financial results of the most recent years of operations: Partiulars Deluxe

90000 units Premium

10,000 units Total

1,00,000 Units Revenue 63,00,000 9,00,000 72,00,000 Material Cost 10,80,000 2,50,000 13,30,000 Direct Labour Cost 14,40,000 1,60,000 16,00,000 Contribution Margin 37,80,000 4,90,000 42,70,000 Allocated Fixed Manufacturing Overhead 34,20,000 3,80,000 38,00,000 Allocated fixed Selling & administrative Overheads

2,51,563 35,937 2,87,500

Profit Margin 1,08,437 74,063 1,82,500 Profit Margin per unit 1.2048 7.4063 Labour cost is ₹ 16 per hour and each require one hour of labour. The company current allocates all fixed manufacturing overheads, using labour hours as the allocation basis. It allocates fixed selling and administrative overheads; using revenue is the allocation base.

Although the profit margin per unit of ‘Deluxe’ Juicer is rather low. PQR limited believes that it is important to keep this model in the product mix. However PQR can tailor its promotion and sales strategies to improve the sales mix to 16:4 ratio from the current 9:1 ratio of ‘Deluxe’ to ‘Premium’ juicers with total volume staying at 1,00,000 units.

PQR Limited finds that ₹ 1,1 milion of the ₹ 3.8 milion of fixed manufacturing overheads pertains to batch related activities such as scheduling production runs. Similarly, ₹ 1,15,000 is the amount of administrative overheads out of the ₹ 2,87,500 of selling and administrative overheads.

It is found that the ‘Premium’ juicer is produced in smaller batches (250 units per batch) than that of ‘Deluxe” juicer (500 units per batch). Similarly it takes 10 sales visits to sell 1,000 units of the ‘Deluxe’ juicers, while it takes 25 visits to sell 1,000 units of ‘Premium’ juicer.

Required:

i) Prepare a profitability statement based on the proposed sales mix, using the most appropriate basis of allocating fixed overheads. (In absence of an appropriate basis, do not allocate overheads to products).

ii) Advise the company on whether it should go ahead with the propose change in sales mix. (10 Marks)

Question 2

(a) PEX is a manufacturing company of which division PQR manufactures a single standardized product. Some of the output is sold externa lly whilst the remainder is transferred to division RPQ where it is a sub assembly in the manufacture of that division's product. PQR has the capacity (annual) to produce 30,000 units of the product. The unit costs of division PQR's product are as under:

₹ Direct Material 40 Direct Labour 20 Direct Expenses 20 Variable manufacturing overheads 20 Fixed manufacturing overheads 40 Sells and packaging expenses- variable 10 Total 150

Annually 20,000 units of the product are sold externally at the standard price of ₹ 300 per unit.

In addition to the external sales, 10,000 units are transferred annually to division RPQ at an internal transfer price of ₹ 290 per unit. This transfer price is obtained by deducting variable selling and packing expenses from the external price since those expenses are not incurred for internal transfers.

Division RPQ incorporates the transferred- in goods into a more advanced product. The unit costs of this product are as follows:.

₹ Transferred- in - item (from division PQR) 290 Direct Material and components 230 Direct Labour 30 Variable overheads 120 Fixed overheads 120 Selling and packing expenses- variable 10 Total 800

Division RPQ's manager disagrees with the basis used to set the transfer price. He argues that the transfers should be made at variable cost plus an agreed (minimal) mark up because his division is taking output that division PQR would be unable to sell at the price of ₹ 300.

Partly because of this disagreement, a study of the relationship between selling price and demand has recently been carried out for each division by the company's sales director. The study has brought out the following demand schedule:

Division PQR Selling price (₹) 200 300 400 Demand (units) 30,000 20,000 10,000

Division RPQ

Selling price (₹) 800 900 1,000 Demand (units) 14,400 10,000 5,600

The manager of the division RPQ claims that this study supports his case. He suggests that a transfer price of ₹ 120 would give division PQR a reasonable contribution to its fixed overheads while allowing division RPQ to earn a reasonable profit. He also believes that it would lead to an increase of output and an improvement in the overall level of company profits. Required:

(i) Calculate the effect of the transfer price of ₹ 290 per unit on company's operating profit. Calculate the optimal product mix.

(ii) Advise the company on whether the transfer price should be revised to ₹ 120 per unit. (11 Marks)

(b) What do you mean by Degeneracy in transportation problem? How this can be solved? (5 Marks) Question 3

(a) Sunglow Limited manufactures and sells a single product. From the records of the company the following information is available for November 2012:

Direct Material Unit ₹ X 8 320 Y 24 1680 Z 16 400 2400 Direct Wages (₹ 40 per hour) 1600 Variable Overhead (25% of Direct Wages) 400 Fixed overhead (based on budgeted production 600 of 10000 units of the final product per month) 5000

The budgeted selling price is ₹ 700 each and the budgeted sales for the month were 14000 units The following were the transactions for the month

Material Units Purchased Price P.U Issued Unit X 44000 42 82400 Y 140000 71 246400 Z 60000 24 164000

Direct Wages: ₹ 90,00,000 (3,98,000 hours) Overhead: Variable ₹ 2,00,000 Fixed ₹ 3,00,000 Production 11000 units Sales 9,000 units at ₹ 700 each and 3500 units at ₹ 750 each. Required: Calculate

(i) Material Price Variance; (ii) Material Mix Variance; (iii) Labour rate variance; (iv) Labour efficiency variance; (v) Variable Overhead efficiency variance and (vi) Fixed Overhead efficiency variance.

(9 Marks) (b) An international tourist company deals with numerous personal callers each day and prides itself on its

level of service. The time to deal with each caller depends on the client's requirements which range from, say, a request for a brochure to booking a round-the- world cruise. If a client has to wait for more than10 minutes for attention, it is company's policy for the manager to see him personally and to give him a holiday voucher worth Rs. 15.

The company's observations have shown that the time taken to deal with clients and the arrival pattern

of their calls follow the following distribution pattern:

Time to deal with clients Minutes 2 4 6 10 14 20 30 Probability 0.05 0.10 0.15 0.30 0.25 0.10 0.05

Time between call arrivals

Minutes 1 8 15 25 Probability 0.2 0.4 0.3 0.1

Required: (i) Describe how you would simulate the operation of the travel agency based on the use of random

number tables; (ii) Simulate the arrival and serving of 12 clients and show the number of clients who receive a

voucher (use line 1 of the random numbers below to derive the arrival pattern and line 2 for serving times); and

(iii) Calculate the weekly cost of vouchers; assuming the proportion of clients receiving vouchers derived from ( ii ) applies throughout a week of 75 operating hours.

Random Numbers

Line 1 03 47 43 73 86 36 96 47 36 61 46 98 Line 2 63 71 62 33 26 16 80 45 60 11 14 10

(7 Marks)

Question 4 (a) A production supervisor is considering how he should assign five jobs that are to be performed to five

operators. He wants to assign the jobs to the operators in such a manner that the aggregate cost to perform the job is the least. He has the following information about the wages paid to the operators for performing these jobs:

Jobs

Operators 1 2 3 4 5

A 10 3 3 2 8 B 9 7 8 2 7 C 7 5 6 2 4 D 3 5 8 2 4 E 9 10 9 6 10

Required: Assign the job to the operators so that the aggregate cost is least. (8 Marks)

(b) Discuss the characteristic of Zero Base Budgeting. (4 Marks)

(c) What are the essential requisites for the installation of Uniform costing system? (4 Marks) Question 5

(a) The Board of Directors of XY Company Limited are considering a new type of handy sewing machine which their R & D department has developed. The expenditure so far on research has been ₹ 95,000 and a consultant’s report has been prepared at a cost of ₹ 22,500. The report provides the following information:

Cost of production per unit: Material ₹ 45.00 Labour ₹ 75.00 Fixed Overhead ₹ 20.00 (Based on Company’s normal allocation rates) Total ₹ 140.00

Anticipated additional Fixed Cost: Rent for additional space ₹ 1,25,000 per annum Other additional fixed costs ₹ 70,000 per annum A new machine will be built with the available facilities with a cost of ₹ 1,10,000 (Material ₹ 90,000 and Labour ₹ 20,000). The materials are readily available in stores which are regularly used. However, these are to be replenished immediately. The price of these materials have since been increased by 50%. Scrap value of the machine at the end of the 10th year is estimated at ₹ 20,000. The product scraps generated can be disposed off at the end of year 10 for a price of ₹ 1,43,000.

Year 1-5 Year 6-10 Demand (Unit) Probability Demand (Unit) Probability

40,000 0.15 24,000 0.30 20,000 0.60 16,000 0.50 12,000 0.25 4,000 0.20

It is estimated that the commercial life of the machine will be no longer than 10 years and the after tax cost of capital is 10%. The full cost of the machine will be depreciated on straight line basis, which is allowed for computing the taxable income, over a period of 10 yea₹ Tax rate is 30% DCF factors at 10%: 1-5 year (cumulative) 3.79 6-10 year (cumulative) 2.355 10th year 0.386

Required: Compute minimum selling price for the handy sewing machine. (12 Marks) (b) What are the distinctive features of learning curve theory in manufacturing environment? Explain the

learning curve ratio. (4 Marks)

Question 6 (a) XY Hotel has 40 bed rooms with a maximum occupancy of 490 sleeper nights per week. Average

occupancy is 60% throughout the year. Meals provided to guests have been costs and the average food cost per person per day is as follows:

₹ Breakfast 72.00 Lunch 220.00 Dinner 268.00 560.00

Direct wages and staff meals per week are as under: ₹

Housekeeping 39,040.00 Restaurant and kitchen 68,600.00 General 35,200.00

Direct expenses per annum are ₹ 9,15,200 for house keeping and ₹ 10,40,000 for restaurant. Indirect expenses amount to ₹ 68,22,400, which should be apportioned on the basis of floor area. The floor areas are as fol lows:

Sq. Mt. Bed rooms 3,600 Restaurant 1,200 Service Area 600

A net profit of 10% must be made on the restaurant taking and also on accommodation takings. Required: Calculate what inclusive term per person should be charged per day and also show the split between meals and accommodation charges. (7 Marks)

(b) In the context of Activity Based Costing System, Explain the following statement:

“Strategic cost analysis should exploit internal linkages” (4 Marks)

(c) Write difference between – PERT and CPM (5 Marks)

Question 7

(a) What is Target Costing? It is said that target costing fosters team work within the organization. Explain how target costing creates an environment in which team work fosters. (4 Marks)

(b) What qualitative factors should be considered in an decision to outsource manufacturing of a product? (4 Marks)

(c) “Sunk cost is irrelevant in decision-making, but irrelevant costs are not sunk costs”. Explain with example. (4 Marks)

(d) Write Short Notes on the characteristic of the dual problem (4 Marks)

(e) Brief the principles associated with synchronous manufacturing (4 Marks)

10

CA. B L Chakravarti (B.Sc, M.Sc, FCA, CPA, CAIIB) 9910993376 & 9810993376 Page 10.1

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Linear Programming

Past Examinations Coverage:

Year M-13 N-12 M-12 N-11 M-11 N-10 N-10* M-10 M-10* N-09

Q No. - 7e+1b 4 (b) 5 (b) 5 (b) 3 (b) 1d+3 a 5 (b)

Marks - 4#+5 8 10 5 6 5+7 6

Year N-09* J-09 J-09* N-08 N-08* M-08 N-07 M-07 N-06 M-06

Q No. 4 (b) 6 (a) 6 (b) 5 (c) 5 (a) 4a+6b 5 (c)

Marks 4 6 9 11 8 7+7# 8

Year N-05 M-05 N-04 M-04 N-03 M-03 N-02 M-02 N-01 M-01

Q No. 6 (a) 6 (b) 3 (b) 1b+5c 5 (b) 1b+3b 6 (c) 1 (b) 1 (b) 1 (b)

Marks 11 9 8 4#+8 8 4#+3# 8 10 10 10

Year N-00 M-00 N-99 M-99 N-98 M-98 N-97 M-97 N-96 M-96

Q No. 1b+7(iv) 1 (b) 1 (b) 1 (b) 1 (c) 1c+6b 1 (b) 1 (b) 1 (b) 1b+7b

Marks 10+5# 10 10 10 10 10+5# 10 10 10 10+5#

* means questions from old syllabus

# means theory questions

Learning Objectives: Understanding the feature of L.P.P.

Formulation of L.P.P.

Graphical Method

Simplex Method

Degeneracy

Multiple Optimal Solution

Limitations of L.P.P.

Applications of L.P.P.

Advance Management Accounting

1. Email: [email protected] Page 10.2

Linear programming Linear programming is a mathematical technique for determining the optimal allocation of resources and achieving the specified objective when there are alternative uses of the resources like money, manpower, materials, machines and other facilities. The objective in resource allocation may be either cost minimization or profit maximization. The adjective ‘linear’, is to be particularly noted here. It implies that all the limitations or constraints and the objective must be expressed as linear functions. Although the technique of linear programming is applicable to all such problems, there are some more easy methods for solving specific categories of problems. This has led to the following categories of the linear programming problems:

(i) General Linear Programming Problems (ii) Transportation Problems. (iii) Assignment Problems.

Formulation of LP The practical steps involved in the formulation of Linear Programming problem are as follows: Step 1: Identify the Decision Variables of interest to the decision maker and express them as X1, X2, X3… Step 2: Ascertain the Objective of the decision maker whether he wants to Minimize or to Maximize. Step 3: Ascertain the cost (in case of Minimization problem) OR the Profit (in case of Maximization problem) per unit of each of the decision variables. Step 4: Ascertain the constraints representing the maximum availability or minimum commitment or equality and represent them “as less than or equal to” (≤) type inequality or “greater than or equal to” (≥) type inequality or “equal to” (=) type equality respectively. Step 5: Put “non-negative restriction” as under. Step 6: Now formulate the LP problem Objective Function

Minimize (or Maximize) Z = C1X1 + C2X2 + C3X3 +………… CnXn Subject to Constraints: a11X1 + a12X2 + a13X3 + ………. a1nXn < b1 (Maximum Availability)

a21X1 + a22X2 + a23X3 + ………. a2nXn > b2 (Minimum Availability) a31X1 + a32X2 + a33X3 + ………. a3nXn = b3 (Equity) .. .. am1X1 + am2X2 + am3X3 + ………. amnXn ≤ bm

and X1; X2; X3 ………. Xn ≥ 0 (Non-negative restriction)

Where

(i) X1; X2; X3 ………. Xn are the variables whose values we wish to determine and are called the decision or structural variables

(ii) C1; C2; C3 ………… Cn; represents the contribution (profit or cost) to the objective function. (iii) aij (I = 1,2,3,… m ; j = 1,2,3,… n) are referred to as the technological or substation coefficients. (iv) b1; b2; b3 ………… bm; is the constants representing the requirements or availability of the

constraints..

Linear Programming


Graphical Method: When only two basic variables are involved in the calculations graphical solutions are handy which consists the following steps:

1. Formulate the problem i.e. define the objective function and constraints. 2. Represent each constraint by a straight line as if the inequality were equality. 3. Identify the feasible solution region. 4. The optimal solution lies at corners of the solution region. Compute the profit (or cost) at each corner

point by substituting the points co-ordinates in the objective function and find out which corner point yields maximization (or minimization) value.

The practical steps involved in solving the Linear Programming problem by Graphical Method are as follows: Step 1: Consider each inequality constraint as equation. Step 2: Take one variable (say X) in a given equation equal to zero and find the value of other variable (say Y) by solving that equation to get one co-ordinate (say (0, Y) for that equation. Step 3: Take the second variable (say Y) in a given equation equal to zero and find the value of other variable (say X) by solving that equation to get one co-ordinate (say X,0) for that equation. Step 4: Plot both the co-ordinate so obtained (i.e. (0,Y) and (X,0) on the graph and join them by a straight line. This straight line shows that any point on that line satisfies the equality and any point below or above that line shows inequality. Shade the feasible reason which may be either convex to the origin in case of less than type of inequality (<) or opposite to the origin in case of more than type of inequality (>). Step 5: Repeat steps 2 to 4 for other constraints. Step 6: Find the common shaded feasible region and mark the co-ordinates of the corner points (vertices). Step 7: Put the co-ordinates of each of such vertex in the Objective Function (Z). Choose that vertex which achieves the most optimal solution (i.e. in the case of maximization, the vertex that gives the maximum value of Z, & in case of minimization, the vertex that gives the minimum value of Z. Vertex Co-ordinates of Vertex Value of Objective Function (Z) 1 (X1, Y1) Z1 = … 2 (X2, Y2) Z2 = … 3 (X3, Y3) Z3 = … . . . . Advantage of Graphical Method:

1. This is simple pictorial method which is very easy to follow. 2. It is easy to see from the graph how a slight alteration to one or more of the constraints may affect the

situation. Disadvantage of Graphical Method:

1. If results are to be read from a graph, a small degree of accuracy is lost. 2. Too many constraints lead to an overcomplicated graph. 3. The graphical method is of no practical value when the number of variable exceeds three.



Simplex Method: Step 1: Add non-negative slack variables (Say S1, S2 .. ) in each constraints to convert inequalities into equation. A slack variable indicates under – utilization of capacity of the constraints; hence, its contribution to objective function is assumed to be zero (or negative) and thus called Non-basic variables.. Step 2: Assuming the value of each Non-basic variable equal to “zero” (i.e. assuming that no performance has taken place), calculate the value of each Basic variable from the equation. Step 3: Fine the value of “Z” by putting the value of each basic and non-basic variable in the objective function.

Step 4: Draw the initial Simplex Table as follows

Cj Contribution

Basic Variable

Qty Replacement Ratio

X1 X2 X3 S1 S2 S3 C1 S1 C2 S2 C3 S3 Total Contribution Zj

Opportunity Cost Cj-Zj Note: The value in the Zj row for X1 variable is computed by the formula = C1X1+C2X1+C3X1 . The values in the Zj

row in the column for other variables are computed by the same formula as stated.

Step 5: Key Column – Mark the column having maximum positive value in (Cj-Zj) Row. It represents the

opportunity cost or loss of not introducing one unity of the variable of that column. This column is also known as

Key Column. This column indicates the selection of incoming variable in the next Simplex Table.

Step 6: Replacement Ratio:- Find out the replacement ratio (also known as key ratio) by dividing value in the

quantity column of each row by corresponding key column value. Replacement ratio represents how much

quantum of variable can be produced based on that row taking the Key column value.

Step 7: Key Row: Mark the row having minimum non-negative replacement ratio by Sign. Minimum

replacement ratio ensures that no basic variable will ever be negative. This can be verified by putting higher

replacement ratio in all constraints. This row is known as Key Row. This row indicates the selection of outgoing

variable from the current simplex table.

Step 8: Key Element: Encircle the element at the intersection of Key Row and Key Column. This value is known

as Key Element or Pivot Element.

Step 9: Replace the outgoing variable by the incoming variable with its contribution per unit i.e. C1 in the first

Column for Contribution per unit.

Step 10: Calculate the new values of Key Row

New value of key row = Old Value of Key Row/ Key Element.

Step 11: Calculate the new value of other rows (Non-key rows)

A. Old Value

B. New Values of Key Row

C. Key Column Element

D. Product of B & C

E. New Values (A-D)

Step 12: Draw another Simplex Table based on new values of each row.

Linear Programming


Step 13: Repeat Steps 4 to 12 till all values in Cj-Zj row becomes zero or negative which indicates that any

further introduction of additional unit of the variable having negative value in Cj-Zj row will have negative

contribution towards Objective Function (no effect in case of Zero in Cj-Zj row). This is also known as condition

for optimal solution.

Step 14: Find optimal Solutions from the Final Simplex Table i.e. the value of Z and quantity of non-basic

variables.

Value of Z = Ci (Qty of Xi) + Cj (Qty of Xj)

Important Points:

1. It may be desired to convert a maximization problem into a minimization one and vice versa. Mathematically, this can be accomplished by reversing signs though of just the objective function.

2. Inequalities in the wrong direction: Consider the problem:

Maximize Z = x1 + 5x2 Subject to 3x1 + 4x2 ≤ 6 ... (i)

x1 + 3x2 ≥ 2 ... (ii) x1, x2 ≥ 0

[Whether to introduce slack or surplus or artificial variables depends on the type of inequality and has nothing to do with type of the problem i.e., maximization or minimization]. The 2nd inequality is in the wrong direction. Upon introducing the "surplus" variable;

x1 + 3x2 – S2 = 2

If S2 is taken in the initial solution it would be – ve when x1 and x2 are zero. To circumvent this, an artificial variable is also introduced in this inequality. The problem becomes:

Maximize Z = x1 + 5x2 - MA2 Subject to 3x1 + 4x2 + S1 = 6

x1 + 3x2 – S2 +A2 = 2 x1 , x2, S1, S2, A2 > 0

(Note that in maximization problems M always has -ve sign and in minimization problems M always has a + ve sign in the objective function). The initial solution consists of S1 and A2. Several examples on inequalities in the wrong direction follow. Surplus variables can never come in initial solution.

3. Any linear programming problem can be re-formulated into what is known as its dual. Any of the primal (the original) or the dual may be selected for iterating by the simplex method. The selection is made on the basis of computational burden. Also the dual provides interesting insights into the methodology of the LP solution. This matter is discussed in a following section at a greater length.

4. If two or more variables share the maximum positive co-efficient in the net evaluation row any one may be chosen for introduction for the new solution arbitrarily, viz., in Z = 2x1 + 2x2 + x3 it matters little if x1 or x2 is chosen.

5. Lower bounds may be specified in an LPP. For example, over and above to the three usual constraints, it may be stipulated that x1 cannot be less than 25 or 40 or l1.i.e., x1 > l1. This can be handled quite easily by introducing a variable y1 such that x1 = l1 + y1. Substitute x1 = l1 + y1 wherever it occurs and solve



the LPP. Computations would be greatly reduced. Please see illustration on page no. 15.32 in this connection.

6. In all the simplex tables there is bound to be a unit matrix of size p × p where p is the no. of rows (excluding net evaluation row). The columns that constitute such a unit matrix need not be adjacent.

7. In view of the tediousness of computational aspects it is useful to make a check at each iteration. This can be done by deriving the net evaluation row in two ways. (i) just like any other row in the simplex tableau by deriving its fixed ratio (ii) by summing the product of the quantities column with the profit/cost column and subtracting this sum from the original profit contribution or cost co-efficient of variable. These should tally. Also, having obtained the optimal solution it is desirable to verify it if it obeys the given constraints.

8. The simplex method, the graphical and trial and error methods, the dual approach provide several ways of doing an LPP. The student may want to do each LPP in more than one way for the sake of verification of the answer and practice.

Multiple Optimal Solutions: Multiple solutions of a LP Problems are solution, each of which Maximize or

Minimize the Objective function. Under Simplex Method; the existence of multiple optimal solutions is indicated

by a situation under which a non-basic variable in the final simplex table showing optimal solution to a problem,

has a net zero contribution. In other words, if at least one of the non-basic variables in the (Ci=Zj) row of the final

simplex table has a zero value, it indicates that there is more than one optimal solution.

Unboundedness:- An unbounded solution of a linear programming problem is a situation where objective function is infinite. A Linear Programming problem is said to be unbounded solution if the solution can be made infinitely large without violating any of its constraints in the problem. Since there is no real applied problem which has infinite return, hence an unbounded solution always represents a problem that has been incorrectly formulated. Under Simplex Method; an unbounded solution is indicated when there is no positive value of Replacement ratio i.e. Replacement ratio value are either infinite or negative, in this case there is no outgoing variable. Infeasibility: An Linear Programming problem is said to be infeasible if there is no solution that satisfies all the constraints. It represents a state of inconsistency in the set of constraints. Under Simplex Method: The LP problems is said to have no feasible solution if at least one of the artificial variable remains in the final simplex table as basic variable with non-zero quantity. Degeneracy:- Degeneracy in a linear programming is said to occur when a basic feasible solution contains a smaller number of non-zero variables than the number of independent constraints when values of some basic variables are zero and the Replacement Ratio is same. In other words, under simplex method, degeneracy occurs, where there is tie for the minimum positive replacement ratio for selecting outgoing variable. In this case, the choice for selecting outgoing variable may be made arbitrarily.

Linear Programming


Limitations of Linear Programming: The limitations of linear programming problems are as follows:

(i) A primary requirement of linear programming is that the objective function and every constraint function must be linear.

(ii) It may not be possible to solve those problems using linear programming, in which non-linearity arises because of joint interactions between some of the activities regarding the total measure of effectiveness or total usage of some resource. Hence linear programming problem requires that the total measure of effectiveness and total resource usage resulting from the joint performance of the activities must equal the respective sums of these quantities resulting from each activity being performed individually. In some situations, it may not be true. For example, consider a situation where by a product is produced with the scrap material from the primary product. The material would still have to be procured if only one of the two products were produced. However, the total material requirements if both products are produced is less than the sum of requirements is each were produced individually. It may not be possible to handle such situation with linear programming problems.

(iii) In linear programming problem, fractional values are permitted for the decisions variables. However, many decision problems require that the solution for decision variable should be obtained in non-fractional values. Rounding-off the values obtained by linear programming techniques may not result into an optimal solution in such cases.

(iv) In linear programming problem, coefficients in the objective function and the constraint equations must be completely known and they should not change during the period of study i.e. they should be known constraints.

Applications of Linear Programming (i) Diet Problems: To determine the minimum requirement of nutrients subject to availability of foods

and their prices.

(ii) Manufacturing Problems: To find the number of items of each type that should be manufactured so as to maximize the profit subject to production restrictions imposed by limitation on the use of machinery and labour.

(iii) Transportation Problems: To find the least costly way to transporting shipments from the warehouses to customers.

(iv) Blending Problems: To determine the optimum amount of several constituents to use in producing a set of products while determining the optimum quantity of each product to produce.

(v) Assembling Problems: To have the best combinations of basic components to produce goods according to certain specifications.

(vi) Production Problems: To decide the production schedule to satisfy demand and minimize cost in face of fluctuating rates and storage expenses.

(vii) Job-Assigning Problems: To assign job to workers for maximum effectiveness and optimum results subject to restrictions of wages and other costs.

(viii) Trim-loss Problems: To determine the best way to obtain a variety of smaller rolls of paper from a standard width of roll that is kept in stock and, at the same time, minimize wastage.



Theoretical Questions TQ 1: Write Short Notes on the characteristic of the dual problem [N 12, 7e, 4M]

Ans: Characteristic of the dual problem:

(i) Any linear programming model (called primal model), there exists a companion model which is

called the dual model.

(ii) The number of constraints in the primal model equals the number of variables in the dual model.

(iii) The number of variables in the primal model equals the number of constraints in the dual model.

(iv) If the primal model has a maximization objective then the dual model will have a minimization

objective and vice-versa. Inequalities get reversed.

(v) The solution of the primal model yields the solution of the dual model. Also, an optimal simplex table

for the dual model yields the optimal solution to the primal model. Further, the objective functions

of the two optimal tables will have identical values.

(vi) Dual of the dual problem is the original primal itself.

(vii) Feasible solutions to a primal and dual problem are both optimal if the complementary slackness

conditions hold. If this relationship does not hold either the primal solution or the dual solution or

both are not optimal.

(viii) If the primal problem has no optimal solution due to infeasibility, then the dual problem will have

no optimal solution due to unboundedness.

(ix) If primal has no optimal solution due to unboundedness , than the dual will have no optimal

solution due to infeasibility.

TQ 2: Explain the limitations of linear programming.

[M 04, 1b, 4M], [N 00, 7(v), 5M] [M 98, 6b(ii), 5M] [M 96, 7b(ii), 5M]

Ans: The limitations of linear programming problems are as follows:

(i) A primary requirement of linear programming is that the objective function and every constraint function must be linear.

(ii) It may not be possible to solve those problems using linear programming, in which non-linearity arises because of joint interactions between some of the activities regarding the total measure of effectiveness or total usage of some resource. Hence linear programming problem requires that the total measure of effectiveness and total resource usage resulting from the joint performance of the activities must equal the respective sums of these quantities resulting from each activity being performed individually.

(iii) In linear programming problem, fractional values are permitted for the decisions variables. However, many decision problems require that the solution for decision variable should be obtained in non-fractional values. Rounding-off the values obtained by linear programming techniques may not result into an optimal solution in such cases.

(iv) In linear programming problem, coefficients in the objective function and the constraint equations must be completely known and they should not change during the period of study i.e. they should be known constraints.

Linear Programming


TQ 3: What are practical applications of linear programming? [M 07, 6b, 7M] [N 00, 7(v), 5M] [V2-7]

Ans: Linear programming can be used to find optional solutions under constraints. (a) In production:

i) Product mix under capacity constraints to minimise costs/maximise profits along with marginal costing.

ii) Inventory management to minimise holding cost, warehousing / transporting fromfactories to warehouses etc.

(b) Sensitivity Analysis: By providing a range of feasible solutions to decide on discounts on selling price, decisions to make or buy.

(c) Blending: Optional blending of raw materials under supply constraints. (d) Finance: Portfolio management, interest/receivables management. (e) Advertisement mix: In advertising campaign – analogous to production management and product

mix. (f) Assignment of personnel to jobs and resource allocation problems.

TQ 5: Enumerate the industrial applications of linear programming [M 03, 1b, 4M]

Ans: Industrial Applications of Linear Programming

(i) Diet Problems: To determine the minimum requirement of nutrients subject to availability of foods and their prices.

(ii) Manufacturing Problems: To find the number of items of each type that should be manufactured so as to maximize the profit subject to production restrictions imposed by limitation on the use of machinery and labour.

(iii) Transportation Problems: To find the least costly way to transporting shipments from the warehouses to customers.

(iv) Blending Problems: To determine the optimum amount of several constituents to use in producing a set of products while determining the optimum quantity of each product to produce.

(v) Assembling Problems: To have the best combinations of basic components to produce goods according to certain specifications.

(vi) Production Problems: To decide the production schedule to satisfy demand and minimize cost in face of fluctuating rates and storage expenses.

(vii) Job-Assigning Problems: To assign job to workers for maximum effectiveness and optimum results subject to restrictions of wages and other costs.

(viii) Trim-loss Problems: To determine the best way to obtain a variety of smaller rolls of paper from a standard width of roll that is kept in stock and, at the same time, minimize

TQ 6: Distinguish between a slack variable & an artificial slack variable in linear programming [M 03, 3b, 3M] Ans: Slack variable: In a LP problem, in order to convert every constraint of the type ‘less than equal to’ into

an equality constraint, so that solution of the problem can be arrived, we add a variable to each such

constraint. The variable so added in each constraint is known as slack variable. Slack Variables

represent idle or unused resources. The contribution per unit of a slack variable is always taken as zero

in objective function. A slack variable is always non negative.

Constraint 3x + 2y ≤ 90 can be written as 3x + 2y + S1= 90, here S1 is a slack variable and is +ve.

Artificial variable: for finding the solution of the L. P. problem, in order to convert constraints of the

type ‘greater than equal to’ equality, we first subtract a surplus variable and then add a variable. This

variable is also added in the constraints of the type ‘equal to’ to start with the initial feasible solution.

The variable added in the constraints as explained above is known as artificial variable.

Artificial variable is a fictitious variable and cannot have any physical or economic meaning. It is

intentionally introduced to form an initial solution for further iterations. It has an infinitely large cost

coefficient. Artificial variables are always positive.

Constraints 3x + 4y ≥50 & 2x + 6y = 40 can be written as 3x + 4y - S1 + A1=50 & 2x + 6y + A2 = 40; Here

S1 is a surplus variable and A1 & A2 are artificial variables. S1, A1 & A2 are +ve.

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Transportation Problems


Year M-13 N-12 M-12 N-11 M-11 N-10 N-10* M-10 M-10* N-09 Q No. - 2 (b) 1b+3b 1d+4d 6 (a) 5 (a) 6 (b) 4 (b) 2 (b) Marks - 5# 5+3# 5+4# 8 10 4# 4# 4#

Year N-09* J-09 J-09* N-08 N-08* M-08 N-07 M-07 N-06 M-06 Q No. 4 (c) 4 (a) 4 (a) 6 (a) 1 (b) Marks 4# 4# 3# 10 6

Year N-05 M-05 N-04 M-04 N-03 M-03 N-02 M-02 N-01 M-01 Q No. 5(c) 2 (b) 3 (b) 3 (c) 3 (c) Marks 8 3# 10 10 10

Year N-00 M-00 N-99 M-99 N-98 M-98 N-97 M-97 N-96 M-96 Q No. 3 (b) 4 (b) 2 (c) 3 (b) 4 (b) 5 (c) 2 (b) 4 (b) 2 (b) 2 (b) Marks 10 10 10 10 10 10 10 10 10 10

* means questions from old syllabus # means theory questions

Learning Objectives:

Understanding the feature of Transportations Problems.

Formulation of Transportations Problems.

Initial Basic Feasible Solution by NWC, LCM and VAM Methods

Optimal Solution – MODI Method

Unbalanced Problems

Maximization Problems



Transportation: The transportation problem is concerned with the allocation of items between suppliers (called origins) and consumers (called destinations) so that the total cost of the allocation is minimized. The objective of the transportation problems is to determine the quantity to be shipped from each source to each destinations so as the transportation cost is lowest to maintain the supply and demand requirements. Applications:

(i) Minimize shipping costs from factories to warehouses (or from warehouses to retail outlets) (ii) Determine lowest cost location for new factory; warehouse; office or other facility. (iii) Find minimum cost production schedule that satisfies firm’s demand and production limitations

(called ‘Production Smoothing’) Conditions: To use the transportation algorithm; the following conditions must be satisfied.

(i) The cost per item of each combination of origin and destination must be satisfied. (ii) The supply of items at each origin must be known. (iii) The requirement of items at each destination must be known. (iv) The total supply must equal the total demand.

Stages in Transportation Problems: The transportation algorithm has four stages. Stages -1: Arrange the data in table format and set up the transportation table to find any feasible allocation. A feasible allocation is one in which all demand at the destinations is satisfied and all supply at the origin is allocated. Stages -2: Do preliminary check consist the following:

(i) Verify Objective = Minimization. In case of Maximization or profit matrix, convert the same in to opportunity loss matrix by subtracting each element from the highest element of the matrix.

(ii) Verify that the matrix is balanced i.e. Total Availability = Total Requirement. In case of unbalanced matrix, a dummy column or dummy row should be introduced with nil transportation cost.

Stages -3: Develop an initial basic feasible solution (IBFS) by using any of the following methods. (i) Least Cost method. (ii) North-West corner Method. (iii) Vogel’s Approximation Method.

Stages -4: Test the IBFS solution for optimality. Modified Distribution (MODI) is the most common method and widely used for test the optimality. Stages -5: In case the IBFS solution is not optimal, develop the improved solutions. Method 1:– North-West corner method: The practical steps in the North-West Corner cell are given below: Step 1: Make maximum possible allocation to the upper-left corner cell (known as North-West corner cell) in

the first row depending upon the availability to supply for that row and demand requirement for the column containing that cell.

Step 2: Move to the next cell of the first row depending upon remaining supply for that Row and the demand

requirement for the next column. Go on till the Row totally is exhausted. Step 3: Move to the next row and make allocation to the cell below the cell of the preceding row in which the

last allocation was made and follow Step-1 and Step-2. Step 4: Repeat the Step-1 to Step-3 till all requirements are exhausted i.e. entire demand and supply is

exhausted.

The Transportation Problems


Method 2:- Least Cost Method : The practical steps involved in the Least Cost are given below: Step 1: Make maximum possible allocation to the least cost cell depending upon the demand/supply for the

column/row containing that cell. In case of tie in the least cost cell, make allocation to the cell by which maximum demand or capacity is exhausted.

Step 2: Make allocation to the second least cost cell depending upon the remaining demand/supply for the

column/row containing that cell. Step 3: Repeat the Step-1 and Step-2 till all requirements are exhausted i.e. entire demand and supply is

exhausted. Method 3:- Vogel’s Approximation Method: This methods uses penalty costs. For each row and column the penalty cost is the difference between the cheapest available route and the next cheapest. Step 1: To calculate the penalty cost for each row and column, we look at the least cost cell and next least cell.

For each row and column subtract the least cost cell from the next least cell. This gives the penalty cost of not allocating into the cell with cheapest cost. Write the penalty cost of each row on the right-hand side of the concerned row and penalty cost of each column on the below of the concerned column.

Step 2: We choose the row or column with the largest penalty cost and allocate as much as possible to the least

cost cell in that row or column depending upon the quantity available. In this way, the highest penalty costs; are avoided as far as possible. In case of tie between the penalty cost, select the row or column having least cost cell however in case of tie even in the case of least cost cell; make allocation to that cell by which maximum requirement are exhausted.

Step 3: Shade the row/column whose availability or requirement is exhausted so that it shall not be considered

for further allocation. Step 4: Repeat the Step-1 to Step-3 till all requirements are exhausted i.e. entire demand and supply is

exhausted. Optimality Test: It can be applied to a transportation table if it satisfies the following conditions: a) It contains exactly m+n-1 allocations, where m & n represent the no. of rows & columns of the table. {Note: (i) In case degeneracy occurs (i.e. allocations are less than m+n-1), infinitely small allocation(s) i.e. α

is/are introduced in least cost & independent cell to non-degenerate the solution. If least cost cell(s) is/are not independent, then cell(s) with next lower cost(s) is/are chosen.

(ii) Degeneracy can occur at initial basic feasible solution or at a later stage. It may be removed once the purpose is over.}

b) These allocations are independent i.e. loop can not be performed by them. {Note: (i) Closed path or loop should have even no. of turns & is formed with an allocation on each corner,

which in turn is a join of horizontal & vertical (not diagonal) lines. A loop may not involve all the allocations. (ii) Generally, IBFS obtained by Northwest Corner rule or VAM are always in independent positions though they may be m+n-1 or less than m+n-1 in no.}



Modified Distribution Method (MODI) 1. For the current IBFS (Initial Basic Feasible Solution) with m+n-1 occupied cell, calculate index numbers

(dual variables) Ui (i= 1,2,3,…m) and Vj (j= 1,2,3,…n) for rows and columns, respectively. For calculating values of Ui andVj , the following relationship (formula) for occupied cells is used.

C ij = (Ui +Vj ) for all I,J.

2. Calculation Opportunity Cost for unoccupied cells by using the following formula: ∆ij = C ij - (Ui +Vj )

3. Examine unoccupied cells evaluation for all ∆ij. (a) If ∆ij >0; then the cost of transportation will increase. i.e. and optimal solution has been arrived at. (b) If ∆ij =0; then the cost of transportation will remain unchanged. But there exists are alternative

solution. (c) If ∆ij <0; then an improved solution can be obtained by introducing cell (i,j) in the basis and go to

next step.

4. Select an unoccupied cell with largest negative opportunity cost among all unoccupied cells. 5. Construct a closed path for the unoccupied cell determined in Step 7 and assign plus (+) and minus (-)

sign alternatively beginning with plus sign for the selected unoccupied cell in clock wise or the direction. eg:-

ᶲ 7-ᶲ ᶲmax i.e the maximum Qnty to be transferred from occupied cell to unoccupied cell =

10- ᶲ 5+ᶲ min of -ᶲ cells i.e. min of 7, 10 i.e. 7 units

6. Assign as many units as possible to the unoccupied cells satisfying rim conditions. The smallest allocation in a cell with negative sign on the closed path indicated the number of units that can be transported to the unoccupied cells. This quantity is added to all the occupied cells on the path marked with plus sign and subtract from those cells on the path marked with minus sign.

7. Go to Step-2 and repeat procedure until all ∆ij >0 i.e. an optimal solution is reached. Calculate the associated total transportation cost.

The Transportation Problems


Theoretical Questions TQ 1: What do you mean by Degeneracy in transportation problem? How this can be solved?

[N 12, 2b, 5M] [M 10, 6b, 4M] [N 09*, 4c, 4M] Ans: In transportation problem, if the no of occupied cells is less than m+n+1, then such a situation is

called as degeneracy. Degeneracy can occur two ways: i) The initial basic solution can turn out to be a degenerate solution. Or ii) an improved solution can turn out to be a degenerate solution Degeneracy can be solved by introducing an infinitesimally small allocation e (epsilon) to least cost and independent empty cell , So that the total number of allocated cells becomes to m+n+1.

TQ2: In a transportation problem for cost minimization, there are 4 rows indicating quantities demanded and this totals up to 1,200 units. There are 4 columns giving quantities supplied. This totals up to 1,400 units. What is the condition for a solution to be degenerate?

Ans: The condition for degeneracy is that the number of allocations in a solution is less than m+n- 1.

The given problem is an unbalanced situation and hence a dummy row is to be added, since the Column quantity is greater than that of the Row quantity. The total number of Rows and Columns then = 9 i.e. (5+4). Therefore, m+n- 1 = 8, i.e. if the number of allocations is less than 8, then degeneracy would occur

TQ 3: Will the initial solution for a minimization problem obtained by Vogel’s Approximation Method and the

Least Cost Method be the same? Why? [M 11,4d, 4M] Ans: Vogel’s Approximation Method uses the differences between the minimum and the next minimum costs

for each row and column. This is the penalty or opportunity cost of not utilising the next best alterative. The highest penalty is given the 1st preference. This need not be the lowest cost. For example if a row has minimum cost as 3, and the next minimum as 2, penalty is 1; whereas if another row has minimum 4 and next minimum6, penalty is 2, and this row is given preference.

Least cost method given preference to the lowest cost cell, irrespective of the next cost. Vogel’s Approximation Method will to result in a more optimal solution than least cost. The Initial Basic Solution under VAM and LCM will be the same only when the maximum penalty and the minimum cost coincide.

TQ 4: In an unbalanced minimization transportation problem, with positive unit transport costs from 3

factories to 4 destinations, it is necessary to introduce a dummy destination to make it a balanced transportation problem. How will you find out if a given solution is optimal? [M10, 4b, 4M]

Ans: Test for optimality: Number of allocation = m+n-1 where there are m rows and n columns. The

allocations are independent i.e if no loop can be formed by them. Here, no of factories + no of destination + dummy -1 = m + n +1 -1 = 3+ 4+ 1-1 = 7. Hint: No, Dummy destination won’t be introduced if total units available at factories = total units required at destinations.



TQ 5: How do you know whether an alternative solution exists for a transportation problem? [N 09, 2b, 4M]

Ans: In the Δij matrix, Δij = Cij – (Ui + Vj) Where Cij is the cost matrix and (Ui + Vj)) is the cell evaluation matrix for allocated cell.

If the Δij matrix has one or more ‘Zero’ elements, means that, if that cell is brought into the solution, or if some quantity has been transferred to such cell, the total cost will not change. Thus, a ‘Zero’ element in the Δij matrix reveals the possibility of an alternative optimal solution.

TQ 6: State the methods in which initial feasible solution can be arrived at in a transportation problem

[N 08 *, 4a, 4M] [N-02, 2b, 3M] Ans: The methods by which initial feasible solution can be arrived at in a transportation model are as under:

(i) North West Corner Method. (ii) Least Cost Method (iii) Vogel’s Approximation Method (VAM)

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The Assignment Problems


Year M-13 N-12 M-12 N-11 M-11 N-10 N-10* M-10 M-10* N-09

Q No. - 4 (a) 7 (b) 2 (b) 3 (a) 7 (d) 3 (b) 4 (b)

Marks - 8 4 6 10 4 8 4#

Year N-09* J-09 J-09* N-08 N-08* M-08 N-07 M-07 N-06 M-06

Q No. 4 (b) 1 (b) 4 (a) 2 (a)

Marks 7 8 11 6

Year N-05 M-05 N-04 M-04 N-03 M-03 N-02 M-02 N-01 M-01

Q No. 6 (b) 2 (a) 2 (b) 2 (c) 2 (c) 2 (c)

Marks 7 10 10 10 10 10

Year N-00 M-00 N-99 M-99 N-98 M-98 N-97 M-97 N-96 M-96

Q No. 2 (b) 2 (c) 3 (c) 2 (b) 2 (b) 2 (b) 3 (b) 2 (b) 6 (b) 3 (b)

Marks 10 10 10 10 10 10 10 10 10 10




Understanding the feature of Assignment Problems.

Formulate an Assignment problem.

Hungarian Method

Unbalanced Assignment Problems

Profit Maximization Problems



The Assignment Problem:

The assignment problem, like the transportation problem, is a special case of the linear programming problem.

In general items, it is concerned with a one to one assignment. One person to one machine, one machine to one

job etc. The cost (or profit) of each person-machine, machine-job, or other assignment is known; the objective is

to minimize the total cost or to maximize the total profit of the resource.

Suppose there are 5 jobs to be performed and 5 persons are available for doing these jobs. Each persons can do

each job but with different efficiency. Now the problem is to find an assignment for allocation of jobs to the

available persons i.e. which jobs should be assigned to which person so that the total cost of performing all jobs

is minimum. thus in Assignment Problem, the objective is to assign a no. of person to the same number of

jobs in such a way that the total cost is minimum. The assignment is to be made on one to one basis i.e. each

person can associate with one and only one job.

The Assignment Algorithms (Hungarian Assignment Method)

The assignment problem can be solved by applying the following steps:

Step 1(Row Operation):

In the Cost Matrix, subtract the lowest (minimum) element of each row from every elements of the same row.

Step 2(Column Operation):

In the Cost Matrix obtained after Step 1, subtract the lowest (minimum) element of each column from every

elements of the same column. The resulting matrix is the starting matrix for the following procedure.

Step 3:

In the Cost Matrix obtained after Step 2, draw the minimum number of horizontal and vertical lines that cover all

the zeros.

(i) If number of lines (N) = Order of the matrix (n), optimal assignment can be made and follow Step 4.

(ii) If number of lines (N) ≠ Order of the matrix (n), then for optimal assignment, we have to follow Step

3A until we obtain the matrix with number of lines (N) = Order of the matrix (n) and then follow

Step 4

Step 3A:

Select the smallest element out of the remaining elements which do not lie on any line. Subtract this element

from all such (uncovered) elements and add it to the elements which are placed at the intersections of the

horizontal and vertical lines. Do not alter the elements through which only one line passes. Repeat this steps

until we get the number of lines (N) = Order of the matrix (n)

Step 4:

(i) Starting with first row, examine all rows of matrix in step 3 until a row containing exactly

one zero is found. Surround this zero by () , indication of an assignment there. Draw a

vertical line through the column containing this zero. This eliminates any confusion of

making any further assignments in that column. Process all the rows in this way.

The Assignment Problems


(ii) Apply the same treatment to columns also. Starting with the first column, examine all

columns until a column containing exactly one zero is found. Mark around this zero and

draw a horizontal line through the row containing this marked zero.

(iii) Repeat these steps, until one of the following situations arises:

a) No unmarked ( ) or uncovered (by a line) zero is left.

b) There may be more than one unmarked zero in one column or row. In this case, put

around one of the unmarked zero arbitrarily and pass 2 lines in the cells of the

remaining zeros in its row and column. Repeat the process until no unmarked zero is left

in the matrix.

Step 5:

After Step 4, we have a matrix which contains exactly one marked zero in each row and each column. The

assignment corresponding to these marked zeros will give the optimal assignment. After obtaining the optimal

assignment, the total actual cost can be calculated by adding the values of the corresponding cost in the assigned

cell from the optimal cost matrix.

Unbalanced Assignment Problems (No. of row ≠ No. of column)

An assignment problem is called an unbalanced assignment problem if number of sources is not equal to

number of destinations or the number of jobs is not equal to the number of persons. In such cases, dummy row

or dummy columns with zero costs are added in the matrix so as to form a square matrix.

Maximization Problems

When the objective of the assignment problem is to maximize the profits or Revenues, such types of assignment

problems is called Maximization Problems. To solve such problems, we first convert the maximization problem

to minimization problem and then apply the normal procedure of assignment algorithm.

Conversion of Maximization Problem to Minimization Problem can be done either of any following two ways:

(a) Identify the highest element in the given matrix from all the elements of the matrix and then subtract all

the elements of the matrix from the highest element. OR

(b) Put a negative sign before each of the elements in the given matrix and then subtract all the elements of

the matrix from the highest element.

Prohibited Assignment:

Prohibited Assignments are the cells where no assignment is to be made. Hence we put a very large value M or

∞ for the prohibited cells.



Theoretical Questions Q 1: [N 09, 4b, 4M]

In an assignment problem to assign jobs to men to minimize the time taken, suppose that one man does not

know how to do a particular job, how will you eliminate this allocation from the solution?

Important Notes

13


C

H

A

P

T

E

R

CPM & PERT


Year M-13 N-12 M-12 N-11 M-11 N-10 N-10* M-10 M-10* N-09

Q No. - 6c 5b+7a 1c+6a 2b+6b 4c 3b 5b 1b 4b+6a

Marks - 5# 8+4 5# +10 3#+6# 4# 7 4 5# 4# + 6

Year N-09* J-09 J-09* N-08 N-08* M-08 N-07 M-07 N-06 M-06

Q No. 1b 5b 4a 4a 3c 1b+4b 1a 3a 2b+5b

Marks 9 4# 12 10 11 9+4# 11 11 6+6

Year N-05 M-05 N-04 M-04 N-03 M-03 N-02 M-02 N-01 M-01

Q No. 5b 2c+6c 5b 6b - 6c 5c+4b 5b+7(iii) 5b+7(ii) 5b

Marks 7 4# +6 8 8 - 8 7+4# 10+5# 10+5# 10

Year N-00 M-00 N-99 M-99 N-98 M-98 N-97 M-97 N-96 M-96

Q No. 5c+7(iv) 3b+7(iv) 4b 6b+7(iii) 6b+7b(ii) 7b 4b+6b 5b(i)+3b 3b 5b+7b(ii)

Marks 10+5# 10+5# 10 10+5# 10+5# 10 10+10 5#+10# 10 10+5#




Understanding the importance of PERT & CPM.

Construction of Network diagram.

Determination of Critical Path and float associated.

Determining the Probability of Completing the project on or before the

schedule date.

Project Crashing.

Resource Smoothing & Resource Levelling

Updating of Network



Introduction: A project is an endeavor to create a unique method product or service. Projects are overall

Programme and are broken down into well define set of tasks, subtask and further if desired, all of which must

be completed in the specific time along with minimum cost. Hence, before starting any project, it is necessary to

prepare a plan for scheduling and controlling the various tasks (activities) involved in the given project.

Planning involves the formulation of objectives and goals that are subsequently translated into specific plans and projects. Scheduling is concerned about the implementation of activities necessary to achieve the laid down plans. Controlling: The function of control is to institute a mechanism that can trigger a warning signal if actual performance is deviating (in terms of time, cost and some other measures of effectiveness) from the plan. If such a deviation is unacceptable to the concerned manager, he will be required to take corrective action to bring performance in conformity with the plans. The PERT and CPM models are extremely useful for the purpose of planning, scheduling and controlling the progress and completion of large and complex projects or for carrying out the analysis of these three managerial functions. The working methodology of critical path analysis (CPA) which includes both CPM and PERT, consists of following five steps:

1. Analyse and break down the project in terms of specific activities and/ or events. 2. Determine the interdependence and sequence of specific activities and prepare a network. 3. Assign estimates of time, cost or both to all the activities of the network. 4. Identify the longest or critical path through the network. 5. Monitor, evaluate and control the progress of the project by re-planning, rescheduling and reassignment

of resources. The central task in the control aspect of these models is to identify the longest path through the network. The longest path is the critical path because it equals the minimum time required to complete the project. Advantages of Critical Path Analysis: There are a number of advantages in using critical path analysis.

1. It allows for a comprehensive view of the entire project. Because of the sequential and concurrent relationships, time scheduling becomes very effective. Identifying the critical activities keeps the executive alert and in a state of preparedness, with alternative plans ready in case these are needed. Breaking down the project into smaller components permits better and closer control.

2. Critical path analysis offers economical and effective system of control based on the principle of management by exception i.e. need for corrective action arises only in exceptional situations and in most of other cases, performance is in conformity with the plans.

3. It is a dynamic tool of management which calls for constant review, a reformulation of the network, and finding the current path of relevance and optimum resources allocation.

Project: A project can be defined as a set of large number of activities or jobs that are performed in a certain sequence determined logically or technologically and it has to be completed within (i) a specified time, (ii) a specified cost and (iii) meeting the performance standards. Network: A network is a graphical representation of a project, depicting the flow as well as the sequence of well-defined activities and events. Activity: An activity is any portion of a project which consumes time or resources and has a definable beginning and ending. Activities are graphically represented by arrows, usually with description and time estimates written along the arrows. The tail of the arrow portraying an activity represents the starting point of the activity and its head represents its completion.

CPM & PERT


Activity can be presented in alphabetically or in numbers. A 1 2

1 2 1 2

1 2

2 1 2 1

Activities may be forward, downward, upward however it can never be in curve or reverse direction or

descending order.

Events: The beginning and ending points of an activity or a group of activities are called events. Synonyms of an event are "node" and "connectors" An event is often represented graphically by a numbered circle All activities in a network must commence from some event. Such events are called the tail events because they are connected to the tail of an activity. Similarly, all activities in a network must have terminal points called the head event, because it is at the head of an activity Tail Event Head Event If an event represents the joint completion of more than one activity, it is called a merge event. If an event represents the joint initiation of more than one activity, it is called a burst event. Conventions adopted in drawing networks: There are two conventions normally adopted while drawing networks:

(a) Time flows from left to right. (b) Head events always have a number higher than that of the tail events.

Graphical representation of events and activities: Events are represented by numbers within circles. Activities are represented by arrows, the arrow-heads represent the completion of the activities. The length and orientation of the arrow are of no significance whatsoever (chosen only for the convenience of drawing).



Fundamental properties governing the representation of events and activities: The representation of events and activities is governed by one simple dependency rule which requires that an activity which depends upon another activity is shown to emerge from the head event of the activity upon which it depends and that only dependent activities are drawn in this way.

1. An event cannot occur until all activities leading to it are complete. 2. No activity can start until its tail event is reached.

The above two properties can be combined into a single one, namely that "no activity may start until all previous activities in the same chain are completed.

Logical sequencing and connection of activities: A project entails several activities. The arrows are arranged to show the plan of logical sequence in which the activities of the project are to be accomplished. The sequence is ascertained for each activity by answering the following three queries viz:

(i) Which activity or activities must be completed before the start of a particular activity? (ii) Which activity or activities should follow this? (iii) Which activities can be accomplished simultaneously?

The activity or activities which immediately come before another activity without any intervening activities are called predecessor activities to that activity. The activities which follow another activity without any intervening activities are called successor activities to that activity.

Errors in logical sequencing: Two types of errors in logic may arise while drawing a network, particularly when it is a complicated one. These are known as looping and dangling. Looping: Normally in a network, the arrow points from left to right. This convention is to be strictly adhered, as this would avoid illogical looping, as shown wrongly below : Dangling: The situation represented by the following diagram is also at fault, since the activity represented by the dangling arrow 9-11 is undertaken with no result. To overcome the problem arising due to dangling arrows, following rules may be adopted.

(i) All events, except the first and the last, must have at least one activity entering and one activity leaving them.

(ii) All activities must start and finish with an event. Duplicate activities: Consider the following figure 11: In the above figure, activities A and B may be called duplicate activities because they have same head event .One remedy for such a situation is the introduction of a dummy activity.

CPM & PERT


Dummy activity: It is a hypothetical activity which consumes no resource and time. It is represented by dotted lines and is inserted in the network to clarify activity pattern under the following situations:

(i) It is created to make activities with common starting and finishing events distinguishable. (ii) To identify and maintain the proper precedence relationship between activities that are not

connected by events. (iii) To bring all "loose ends" to a single initial and a single terminal event in each network using

dummies, if necessary. Case – I When two activities have the same tail (starting point) and same head events (end point) with different timing than dummy activity should be introduced to maintain the preceding/succeeding relationship. A (6 day) Wrong C B( 8 day) A (6 day) Correct 0 days B(8 day) C Case-II Whenever one activity repeats two times in the preceding column, one time separately and the other time with other activity, than in order to maintain the preceding/succeeding relationship, we have to introduce dummy activity. For ex. B-C and BD -E B C D E

Numbering of the events: The event numbers in a network should in some respect reflect their logical sequences. A rule devised by D.R. Fulkerson, involving the following steps may be followed to resolve the problem of numbering the events.

(i) An "initial"event is one which has arrow/arrows coming out of it and none of the arrow entering it. In a network there will be only one such event. Call it “1”.

(ii) Delete all arrows coming out from the event 1. This will give us at least one more "initial event". (iii) Number these events as “2, 3....” (iv) Delete all emerging arrows from these numbered events which will create new initial events. Then

follow step (iii). (v) Continue the above steps till last event is obtained which has no arrows coming out of it. Consider

the numbering of events in the following figure. Here we proceed from left to right. The event with least x-co-ordinate is assigned the smallest integer, say 1. Other events are assigned progressively higher integers with regard to x-co-ordinate. If two or more events (4 and 5 above) have the same x-co-ordinate, the one towards arrow should have higher number.



Basic steps involved in drawing a CPM/PERT network: Network is defined as a diagram representing the activities and events of a project, their sequence and interrelationships. The basic steps involved in drawing a network are:

(i) Breaking up of the entire project into smaller systems known as tasks. (ii) For each task ascertain the activities and events to be performed. (iii) For each activity determine the preceding and succeeding activities. (iv) For each activity determine or estimate the time and other resources needed. (v) Draw a network depicting the assembly of tasks into a project.

CRITICAL PATH ANALYSIS: The purpose of the analysis is two-fold:

(i) To find the critical path, i.e. the sequence of activities with the longest duration; and (ii) To find the float associated with each non-critical activity.

Scheduling computations: After preparing the network diagram, we wish to know how long it will take to complete the project and also to identify the activities in the network that are to be placed under strict control. The basic scheduling computations involve a forward and then a backward pass through the network. The process of tracing the network from START to END is called forward pass, and from END to START is called backward pass.

(a) Forward pass computations: As stated above, the purpose of the forward pass is to compute the earliest start (EST) and finish time (EFT) for each activity. The EST time indicates the earliest time that a given activity can be scheduled. Earliest finish time (EFT) for an activity indicates the time by which the activity can be completed, at the earliest. A general rule can also be given here for determining the earliest event time as below:

Ej = Max (Ei + Dij) Where Ej is the earliest time for event j, Ei is the earliest time for event i, and Dij is the duration of the activity i-j.

Note:- In case two or more time estimates exist for a particular event, then the time estimate with maximum value is retained as the earliest event time and other values are discarded. (i) The earliest start time of an activity is given by the earliest allowable occurrence time of the tail

event of that activity. ESTij = Ei

(ii) The earliest finish time of an activity will be simply equal to the earliest start time of the activity plus the duration of that activity.

EFTij = ESTij + Dij

(b) Backward pass computations: The purpose of the backward pass is to compute the latest start and finish

times for each activity. These computations are precisely a "mirror image" of the forward pass computations.

In general, the latest allowable occurrence time of an event can be calculated by selecting an appropriate formula among the following two:

Li = Lj – Dij or Li = minimum (Lj – Dij)

The second formula is used for the event having two or more latest allowable occurrence time estimates. (i) The latest finish time of an activity is equal to the latest allowable occurrence time of the head

event of that activity. LFTij = Lj (ii) The latest start time of an activity is equal to its latest finish time minus its duration.

LSTij = LFTij – Dij.

CPM & PERT


The critical path determination: A network will consist of a number of paths. A path is a continuous series of activities through the network that leads from the initial event (or node) of the network to its terminal event. Critical path: A path in a project network is called critical if it is the longest path. The activities lying on the critical path are called the critical activities. Floats: We know that for every critical activity in a network, the earliest start and latest start time are the same. This is so since the critical activities cannot be scheduled later than their earliest schedule time without delaying the total project duration, they do not have any flexibility in scheduling. However, non-critical activities do have some flexibility i.e. these activities can be delayed for some time without affecting the project duration. This flexibility is termed as slack in case of an event and as float in case of an activity. Slack time for an event The slack time or slack of an event in a network is the difference between the latest event time and the earliest event time. Mathematically it may be calculated using the formula Li –Ei. where Li is the latest allowable occurrence time and Ei is the earliest allowable occurrence time of an event i. Total float of an activity: The total activity float is equal to the difference between the earliest and latest allowable start or finish times for the activity in question. Thus, for an activity (i-j), the total float is given by TFij = LST – EST or LFT – EFT ree Float: It is defined as that portion of the total float within which an activity can be manipulated without affecting the float of the succeeding activities. It can determine by subtracting the head event slack from the total float of an activity. i.e Total Float - Slack of Head Event Independent float: It is defined as that portion of the total float within which an activity can be delayed for start without affecting float of the preceding activities. It is computed by subtracting the tail event slack from the free float of an activity. Free Float - Slack of Tail Event The independent float is always either equal to or less than the free float of an activity. If a negative value is obtained, the independent float is taken to be zero. PTR:

1. Ej = Max (Ei + Dij)

2. Li = Lj – Dij or Li = minimum (Lj – Dij) 3. E of Tail Event (Ei) = EST of the activity (ESTij)

4. L of Head Event (Lj) = LFT of the activity (LFTij)

5. EFTij = ESTij + Dij 6. LSTij = LFTij – Dij. 7. Total Float = LFT-EFT or LST –EST

8. Slack of Event = E-L

9. Free Float = Total Float - Slack of Head Event

10. Independent Float = Free Float – Slack of Tail Event



PERT: CPM is incapable of handling uncertainty in timing which is a rule rather than an expectation for innovational projects such as introducing a new product or oil exploration project. PERT (Program Evaluation and Review Technique) is more relevant for handing such projects which have a great deal of uncertainty associated with the activity durations. To take these uncertainty into account, three kinds of times estimates are generally obtained. These are:

(a) The Optimistic Times Estimate: This is the estimate of the shortest possible time in which an activity can be completed under ideal conditions. For this estimate, no provision for delays or setbacks are made. We shall denote this estimate by to.

(b) The Pessimistic Time Estimate: This is the maximum possible time which an activity could take to accomplish the job. If everything went wrong and abnormal situations prevailed, this would be the time estimate. It is denoted by tp.

(c) The Most Likely Time Estimate: This is a time estimate of an activity which lies between the optimistic and the pessimistic time estimates. It assumes that things go in a normal way with few setbacks. It is represents by tm.

Average time (te) =

Beta distribution is a assumed for these “guess estimates” and PERT analysts have found that beta-distribution curve happens to give fairly satisfactory results for most of the activities. For a distribution of this type, the standard deviation is approximately one sixth of the range, i.e.

Standard Deviation ( ) = (

) = √

= √

Variance ( ) =

Steps:

1. Draw the Network

2. Calculated te for each activity.

3. Calculate project duration on the basis of te

4. Calculate variance for each activity which is = Variance ( ) =

5. Calculate Project variance i.e. the sum of variances of critical activities which lies on critical path.

6. Calculate Standard Deviation of the Project = √

7. To Calculate the Probability (Area) (i) First we should convert the normal values to the standardized value with the help of relation

P (X ≤ … days) = Z ≤ ( ̅

)

(ii) Then find out the corresponding values of the Z from the table which is to be termed as required area.

PROJECT CRASHING In the discussion on PERT, we saw how the probability of completion of a project can be computed for a specified duration. There are usually compelling reasons to complete the project earlier than the originally estimated duration of the critical path computed on the basis of normal activity times, by employing extra resources. In the present section we will deal with those situations which will speak of the effect of increase or decrease in the total duration for the completion of a project and are closely associated with cost considerations. In such

CPM & PERT


cases when the time duration is reduced, the project cost increases, but in some exceptional cases project cost is reduced as well. Activity Cost: It is defined as the cost of performing and completing a particular activity or task. Normal Time(Nt): This is the minimum (scheduled) time required to complete an activity at normal cost. Normal Cost(Nc): This is the lowest possible direct cost required to complete an activity. Crash Time(Ct): This is the minimum time required to complete an activity. Crash Cost (Cc): This is the direct cost that is anticipated in completing an activity within the crash time. Activity cost slope: The cost slope indicates the additional cost incurred per unit of time saved in reducing the duration of an activity.

Activity Cost Slope =

=

Assumptions:

(a) Direct Cost: The total cost which can be directly traceable with a particular activity. if the duration of

that activity is to reduced, the direct Cost will increased.

(b) Indirect Cost: The cost, which cannot be directly identify with a particular activity. If the duration of

any activity reduced, the total project duration will reduced and accordingly the Indirect cost will

reduce.

(c) We should always reduce the duration of critical activities, which lies on critical path, so that the total

project duration will reduce. Or we can say that there is no logic to reduce non-critical activities.

(d) Critical Paths always to remain critical. It may be possible that non-critical path become critical after

crashing.

Steps:

1. Draw the network

2. Identify the critical path along with other possible paths.

3. Calculate the Total Cost of the Project i.e. Direct Cost of each activity + Indirect Cost of Project Duration.

4. Identify the Activity Cost Slope for each activity and possible reduction of days.

5. Reduce the duration of critical activity, which lie on critical path. So that the project duration reduces

and equals to the duration of next critical path. The above process is to be achieved by selecting the

activity which has least cost slope.

6. When two path have the same project duration then reduce both the path simultaneously by preparing

Statement of comparative cost.

7. When all the critical activities which lies on the critical path has been crashed than further crashing

would not be possible.

Resource Smoothing: It is a network technique used for smoothening peak resource requirement during different periods of the

project network. Under this technique the total project duration is maintained at the minimum level. The

resources required for completing different activities of a project are smoothened by utilizing floats available on

non-critical activities. These non-critical activities having floats are rescheduled or shifted so that a uniform

demand on resources is achieved. In other words, the constraint in the case of resource smoothing operation



would be on the project duration time. Resource smoothing is a useful technique for business managers to

estimate the total resource requirements for various project activities.

In resources smoothing, the time-scaled diagram of various activities and their floats (if any), along with

resource requirements are used. The periods of maximum demand for resources are identified and non critical

activities during these periods are staggered by rescheduling them according to their floats for balancing the

resource requirements.

Resource Levelling: It is also a network technique used for reducing the requirement of particular resource due to its paucity. It utilizes the large floats available on non-critical activities of the project and thus cuts down the demand on the resources.

In resource levelling, the maximum demand of a resource should not exceed the available limit at any point of

time. To achieve this, non-critical activities are rescheduled by utilizing their floats. It may lead to enlarging the

completion time of the project. The constraint here is on the limit of the resource availability

Updating the Network: The progress of various activities in a project network is measured periodically. Normally, either most of the

activities are ahead or behind the schedule. It is therefore, necessary to update or redraw the network

periodically to know the exact position of completion of each activity of the project. Even the logic may also

change i.e. some of the existing activities may have to be dropped and new activities may be added up. In brief

the network should be amended accordingly in the light of new developments.

DISTINCTION BETWEEN PERT AND CPM

The PERT and CPM models are similar in terms of their basic structure, rationale and mode of analysis.

However, there are certain distinctions between PERT and CPM networks which are enumerated below: .

(i) CPM is activity oriented and results of various calculations are considered in terms of activities of

the project. However, PERT is event oriented.

(ii) CPM is a deterministic model i.e. It does not take into account the uncertainties involved in the

estimation of time. However PERT, is a probabilistic model. It uses three estimates of the activity

time; optimistic, pessimistic and most likely; with a view to take into account time uncertainty.

(iii) CPM places dual emphasis on time and cost and evaluates the trade-off between project cost and

project time. By deploying, additional resources, it allows the critical path project manager to

manipulate project duration within certain limits so that project duration can be shortened at an

optimal cost. On the other hand, PERT is primarily concerned with time. It helps the manager to

schedule and coordinate various activities so that the project can be completed on scheduled time.

(iv) CPM is commonly used for those projects which are repetitive in nature and where one has prior

experience of handling similar projects. PERT is generally used for those projects where time

required to complete various activities are not known as prior. Thus, PERT is widely used for

planning and scheduling research and development projects.

CPM & PERT


Limitations of the assumption of PERT and CPM

(i) Beta distribution may not always be applicable.

(ii) The formulae for expected duration and standard deviation are simplification. In certain cases,

errors due to these have been found up to 33 %

(iii) The above errors may get compounded or may cancel each other

(iv) Activities are assumed to be independent. But the limitations on the resources may not justify the

assumption.

(v) It may not always be possible to sort out completely identifiable activities and to state where they

begin and where they end

(vi) If there exist alternatives in outcome, they need to be incorporated by way of a decision tree

analysis.

(vii) Time estimates have a subjective element and to this extent, techniques could be weak. Contractors

can manipulate and underestimate timein cost plus contract bids. In incentive contracts,

overestimation is likely.

(viii) Cost-time tradeoffs / cost curve slopes are subjective and even experts may be widely off the mark

even after honest deliberations



Theoretical Questions TQ1: Write difference between – PERT and CPM

[N 12, 6c, 5M] [M 11, 2b, 3M] [N 00, 7(iv), 5M] [N 98, 7b, 5M] [RTP N 12]

Ans: Mentioned on Page No. 13.10

TQ2: State any 5 limitations of the assumptions of PERT and CPM [N11, 1c, 5M]


TQ3: Point out the errors in the network given below, going by the usual conventions while drawing a

network to use CPM [M 11, 6b, 6M]

Ans: Flows

2 – 3: There are 2 activities which are duplicate. In case they are two different activities, one may pass

through a dummy

2 – 5 is a dangling activity; No complete path exists. This can be joined to (9) with a dummy

4 – 6 & 6 – 4: looping exists; This is not proper sequencing

TQ4: List the 5 steps involved in the methodology of critical path analysis. [N 10, 4c, 4M]

Ans: The working methodology of PERT which includes both CPM and PERT, consists of following five steps:

1. Analyze and break down the project in terms of specific activities and/or events.

2. Determine the interdependence and sequence of specific activities and prepare a net-work.

3. Assign estimates of time, cost or both to all the activities of the network.

4. Identify the longest or critical path through the network.

5. Monitor, evaluate and control the progress of the project by re-planning, rescheduling and

reassignment of resources

TQ5: [M 10*, 1b, 5M]

A Project Manager has to manage various projects. For each project given below, you are required to advise

him whether to use PERT or CPM and briefly state the reason:

i. Project K is yet to begin. The manager has recently successfully handled similar projects. He is able to break down the project into smaller modules and knows when he may comfortably finish each module.

ii. Project L has been sanctioned some fixed amount. Though the manager is familiar about what time it will take, he expects pressure towards the end to finish the project slightly earlier, by deploying additional resources of the company.

iii. Project M is new to the manager. He has never handled such a project. He can break up the project into smaller modules, but even then, he is not sure of their exact times.

iv. Project N has a limitation on the skilled workforce available. But the manager knows from earlier experience, the slack on each event in the project. He is confident of handling the bottleneck of labour.

v. Project O is a research project, bound to produce immense benefit to the company in future. Ans:

Project To use Reason K CPM No uncertainty regarding timing of activities L CPM Known timing; optional crashing required M PERT Unknown activity timing; Probabilistic model is necessary N CPM Known activity timing; Limited availability of skilled labour, calls for

resource smoothening O PERT Research Project; uncertainty about of timing of activities

CPM & PERT


TQ6: Explain the following in the context of a network: (i) Critical path (ii) Dummy activity

[M 09, 5b, 4M]

Ans: Critical Path:

Critical Path is a chain of activities that begin with the starting event and ends with ending event of a

particular project. It is that path that runs through a network with the maximum length of time or it

indicates the maximum possible time required for completion of a project. Critical path indicates the

minimum time that will be required to complete a project. It is determined after identifying critical

events. Critical path goes through critical events.

Dummy Activities:

Dummy Activity is that activity which does not consume time or resources. It is used when two or more

activities have same initial and terminal events. As a result of using dummy activities, other activities

can be identified by unique end events. These are usually shown by arrows with dashed lines.

TQ7: What do you mean by a dummy activity? Why it is used in networking? [M 08, 4b, 4M]

Ans: Dummy activity is a hypothetical activity which consumes no resource or time. It is represented by

dotted lines and is inserted in the network to clarify an activity pattern under the following situations.

(i) To make activities with common starting and finishing events distinguishable.

(ii) To identify and maintain the proper precedence relationship between activities that are not

connected by events.

(iii) To bring all “loose ends” to a single initial and single terminal event.

e.g.

Dummy (2) – (3) is used to convey that (3) – (4) can start only after events numbered (2) and (3) are

over:

TQ8: Write short notes on Resource Smoothing and Resource Leveling. [

[M 05, 2c, 4M] [N 02, 4b, 4M] [M02, 7(iii), 5M] [M00,7(iv), 5M] [M 99, 7(iii), 5M] [M 97, 5b, 5M]


1

2

3 4



Important Notes

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Simulation


Year M-13 N-12 M-12 N-11 M-11 N-10 N-10* M-10 M-10* N-09 Q No. - 3 (b) 7 (c) 7 (a) 7 (a) 4 (e) 6 (b) 5 (a) 5 (c) Marks - 7 4 4# 5 4# 4# 7 6

Year N-09* J-09 J-09* N-08 N-08* M-08 N-07 M-07 N-06 M-06 Q No. 3 (a) - 3 (b) 6 (b) 4 (c) 3 (b) Marks 9 - 5# 9 4# 5#

Year N-05 M-05 N-04 M-04 N-03 M-03 N-02 M-02 N-01 M-01 Q No. 6 (b) 6 (c) 6c+4a 6 (b) 6 (b) 5 (a) 6 (b) 7(iii) 6 (b) Marks 8 7 7+4# 4# 7 4# 10 5# 10

Year N-00 M-00 N-99 M-99 N-98 M-98 N-97 M-97 N-96 M-96 Q No. 6 (b) 5 (c) 6 (b) 4 (b) 5 (b) 4 (b) 7 (b) 5(b)(ii) 5(b) (i) 6 (b) Marks 10 10 10 10 10 10 10 5# 5# 10




Understanding the feature of Simulations.

Advantage & Disadvantage of Simulation.

Monte Carlo Simulation Technique



Simulation

Simulation is a quantitative procedure which describes a process by developing a model of that process and then

conducting a series of organized trial and error experiments to predict the behavior of the process over time.

Steps in the Simulation Process

i. Define the problem or system you intend to simulate.

ii. Formulate the model you intend to use.

iii. Test the model; compare its behavior with the behavior of the actual problem environment.

iv. Identify and collect the data needed to test the model.

v. Run the simulation.

vi. Analyze the results of the simulation and, if desired, change the solution you are evaluating.

vii. Rerun the simulation to test the new solution.

viii. Validate the simulation, that is, increase the chances that any inferences you draw about the real

situation from running the simulation will be valid.

Monte Carlo Simulation

The Monte Carlo method employs random numbers and is used to solve problems that depend upon probability.

The Monte Carlo method is a simulation technique in which statistical distribution functions are created by

using a series of random numbe₹ This method is used to solve problems where physical experimentation is

impracticable and the creation of a mathematical formula is impossible or where solution of the model is not

possible by analytical method.

In other words, it is method of Simulation by the sampling technique. First of all, the probability distribution of

the variable under consideration is determined; then a set of random numbers is used to generate a set of values

that have the same distributional characteristics as the actual experience it is devised to simulate. Monte-Carlo

simulation yields a solution which should be very close to the optimal, but not necessarily the exact solution.

Steps in Monte Carlo Simulation

The steps involved in carrying out Monte Carlo Simulation are:

i. Select the objectives or measure of effectiveness of the problem.

ii. Identify the variables which influence the measure of effectiveness significantly.

iii. Determine the proper cumulative probability distribution of each variable selected under step (ii). Plot

these, with the probability on the vertical axis and the values of variables on horizontal axis.

iv. Get a set of random numbe₹

v. Consider each random number as a decimal value of the cumulative probability distribution. With the

decimal, enter the cumulative distribution plot from the vertical axis. Project this point horizontally,

Simulation


until it intersects cumulative probability distribution curve. Then project the point of intersection down

into the vertical axis.

vi. Record the value (or values if several variables are being simulated) generated in step (v) into the

formula derived from the chosen measure of effectiveness. Solve and record the value. This value is the

measure of effectiveness for that simulated value.

vii. Repeat steps (v) and (vi) until sample is large enough for the satisfaction of the decision maker.

Applications of Simulation

Simulation has its utility and applications in the following areas-

(i) Analysis of mathematical model of real-life systems like Inventory Control, Production Scheduling;

Network Analysis etc.

(ii) Planning Military strategy, traffic control, management games and role playing, medical diagnosis,

hospital emergency facilities, gambling and analysis.

(iii) Learning about operating features of a new airplane by simulating flight conditions in a wind tunnel.

(iv) Study of electronic or hydraulic analog models of production processes of economic systems, etc.

Advantage of Simulation

(i) Simulation models are comparatively flexible and can be modified to adjust the variation in the

environments of real situations.

(ii) Simulations methods are easier to apply than pure analytical methods.

(iii) Simulation has the advantage of being relatively free from complicated mathematics and thus can be

easily understood by the operating staff and also by non-technical managers.

(iv) It enables us to assess the possible risk involved in a new policy before actually implementing it.

(v) The simulation of complicated systems helps us to locate which variables have the important

influences on system performance.

Limitation of Simulation

(i) Simulation does not produce optimum results, the results of simulation are only reliable

approximations subject to statistical errors.

(ii) A good simulation model may be very expensive and it is a time consuming and long exercise.

(iii) Probability Distribution based on the past data may not reflect in future.

(iv) Choice of random number is subjective.

(v) All situations cannot be evaluated using simulation. Only situation involving uncertainty can be

measured.



Theoretical Questions

TQ 1: What is Simulation? What are the steps involved in the simulation process?

[N 11, 7a, 4M] [N06, 3b, 5M] [N01, 7(iii), 5M] [M97, 5b, 5M], [V2-4]

Ans: Simulation is a quantitative procedure which describes a process by developing a model of that process

and then conducting a series of organized trial and error experiments to predict the behaviour of the

process over time.

Steps in Simulation Process

1. Define the problem or system you intend to simulate.

2. Formulate the model you intend to use.

3. Test the model; compare its behavior with the behavior of the actual problem environment.

4. Identify and collect the date needed to test the model.

5. Run the simulation.

6. Analyze the results of the simulation and, if desired, change the solution you are evaluating.

7. Rerun the simulation to test the new solution.

8. Validate the simulation, that is, increase the chancesthat any inferences you draw about the real

situation from running the simulation will be valid.

TQ 2: What are the steps involved in carrying out Monte Carlo Simulation Model? [N 10, 4e, 4M]

Ans: Steps involved in Monte Carlo simulation are:

(i) To select the measure of effectiveness of the problem, that is, what element is used to measure

success in improving the system model ed. This is the element one wants to maximize or minimize.

(ii) Identifying the variables which influence the measure of effectiveness significantly.

(iii) Determining the proper cumulative probability distribution.

(iv) To get a set of random numbers.

(v) Consideration of each random number as a decimal value of the cumulative probability distribution.

With the decimal, enter the cumulative distribution plot from the vertical axis, Project this point

horizontally, until it intersects cumulative probability distribution curve.

(vi) Recording the value generated in step(v) into the formula derived from the chose measures of

effectiveness. Solve and record the value.

(vii) Repeating steps (V) and (VI) until sample is large enough for the satisfaction of the decision maker.

TQ 3: How can be simulation be applied in practical situation? [N 10*, 6b, 4M]

Ans: Simulation may be applied in the following areas:

(i) Inventory control – Raw material spares Demand simulated and re-order point determined.

(ii) Inventory of finished goods:– How much to produce, by simulating external demand.

(iii) Waiting time for servicing by simulating arrival and service times used by banks, doctors, etc.

(iv) Checking defectives and determining percentage to formulate standards.

TQ 4: State major reason for using a simulation technique to solve a problem and also describe basic steps in a

general simulation process. [J 09*, 3b, 5M]

Ans: Reasons:

(i) It is not possible to develop a mathematical model and solutions with-out some basic

assumptions.

(ii) It may be too costly to actually observe a system.

(iii) Sufficient time may not be available to allow the system to operate for a very long time.

(iv) Actual operation and observation of a real system may be too disruptive.

Steps:

(i) Define the problem or system which we want to simulate.

Simulation


(ii) Formulate an appropriate model of the given problem.

(iii) Ensure that model represents the real situation/ test the model, compare its behaviour with the

behaviour of actual problem environment.

(iv) Identify and collect the data needed to list the model.

(v) Run the simulation

(vi) Analysis the results of the simulation and if desired, change the solution.

(vii) Return and validate the simulation.

TQ 5: How would you use Monte Carlo Simulation method in inventory control?

[M08, 4c, 4M] [N03, 6b, 4M] [N02, 5a, 4M]

Ans: The Monte Carlo Simulation: It is the earliest mathematical Model of real situations in inventory

control:

Steps involved in carrying out Monte Carlo simulation are:

a) Define the problem and select the measure of effectiveness of the problem that might be inventory

shortages per period.

b) Identify the variables which influence the measure of effectiveness significantly for example,

number of units in inventory.

c) Determine the proper cumulative probability distribution of each variable selected with the

probability on vertical axis and the values of variables on horizontal axis.

d) Get a set of random numbers.

e) Consider each random number as a decimal value of the cumulative probability distribution with the

decimal enter the cumulative distribution plot from the vertical axis. Project this point horizontally,

until it intersects cumulative probability distribution curve. Then project the point of intersection

down into the vertical axis.

f) Then record the value generated into the formula derived from the chosen measure of effectiveness.

Solve and record the value. This value is the measure of effectiveness for that simulated value.

Repeat above steps until sample is large enough for the satisfaction of the decision maker

Uses in Inventory Control:

a) Determination of ROQ and ROL

b) Computation of Stock out Cost and Impact on Profit.

c) Analysis of value of storage facilities for avoiding stock-outs and impact on profit, and

d) Analysis of demand distribution during lead time, etc.

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Learning Curve


Year M-13 N-12 M-12 N-11 M-11 N-10 N-10* M-10 M-10* N-09

Q No. - 5 (b) - 7 (b) - 7 (b) 1 (c) 4 (a) - 6 (c)

Marks - 4# - 4# - 4# 5 11 - 6

Year N-09* J-09 J-09* N-08 N-08* M-08 N-07 M-07 N-06 M-06

Q No. 1(c) - 1(c) 1 (b) 2 (c) 2(b) 5 (c) 3 (c) 6 (c)

Marks 4# - 4 3 7 9# 4# 2# 4#

Year N-05 M-05 N-04 M-04 N-03 M-03 N-02 M-02 N-01 M-01

Q No. - 3 (b) - 3 (b) - 4 (a) 5 (b) - - -

Marks - 8 - 4# - 4# 8 - - -

Year N-00 M-00 N-99 M-99 N-98 M-98 N-97 M-97 N-96 M-96

Q No. - - - - - - - - - -

Marks - - - - - - - - - -




Understanding the meaning of Learning Curve.

Understanding the meaning of Learning Curve Effect.

Feature and Limitations of Learning Curve Theory.



Learning Curve Theory

The Theory of learning curve was first introduced by T.P. Wright of Curtiss—Wright, Buffalo, U.S.A. engaged in

production of airframes. As the production quantity of a given item doubled, the cost of that item decrease at a

fixed rate. This phenomenon is the basic premise on which the theory of learning curve has been formulated.

The key words, ‘doubled’ and ‘rate’ are important as the quantity produced doubles, the absolute amount of cost

increase will be successively smaller but the rate of decrease will remain fixed. This is the essence of the

learning curve theory.

Distinctive Features of Learning Curve Theory in manufacturing environment:

(i) Better tooling methods are developed and used;

(ii) More productive equipments are designed and used to make the product;

(iii) Designed bugs are detected and corrected;

(iv) Engineering changes decrease over the time. Designed engineering are prompted to achieve better

design for reducing material and labour cost.

(v) (v) Earlier teething problems are overcome. As the product involves monthly, management is

prompted to strive for better planning and better management.

(vi) Rejections and rework tend to diminish over time.

There happen a significant influence of all these features on labour as the quantity produced increases and the

cost per unit drops. The reasons for this are that each unit will entail: (a) less labour; (b) less material; (c) more

units produced from the same equipments; (d) cost of fewer delays and less loss time. Every Time Study

engineer or industrial engineer experienced in work measurement has seen this phenomenon happening.

Learning curve applications are fast growing with time in manufacturing environment. A company should never

blindly adopt another company's learning curve. The product approach for a company should be to develop

knowledge of its own learning preference in its plant.

Learning Curve Effect

Learning is the process during which a person acquires the skill to do a job. When a job is taken up for the first

time, the worker is new to the job and he takes considerable time to complete it because his performance is not

at its best. But, when the same job is repeated he is able to improve his performance because of the skill he has

acquired by doing a similar job earlier. This phenomenon is known as learning effect. It applies only to labour

oriented operations.

Learning Curve


Learning Curve

Learning curve is a geometrical progression, which reveals that there is steadily decreasing cost for the

accomplishment of a given repetitive operation, as the identical operation is increasingly repeated. The amount

of decrease will be less and less with each successive unit produced. The slope of the decision curve is expressed

as a percentage. The other names given to learning curve are Experience curve, Improvement curve and

Progress curve.

The Learning Curve Ratio

In the initial stage of a new product or a new process, the learning effect pattern is so regular that the rate of

decline established at the outset can be used to predict labour cost well in advance. The effect of experience on

cost is summaries in the learning ratio or improvement ratio:

Average labour cost of first 2N units

Average labour cost of first N units

Learning Curve Equation

Mathematicians have been able to express relationship in equations. The basic equation is Y = aXb

Where,

X is the cumulative number of units or lots produced

Y is the cumulative average unit cost of those units X or lots.

a is the average cost of the first unit or lots is the improvement exponent or the learning coefficient or

the index of learning which is calculated as follows:

b = log of learning ratio / log 2

Uses of Learning curve:

1) To analyse CVP Relationship during familiarization phase: Learning curve helps to analyse cost-

volume-profit relationship during familiarization phase of product or process and thus it is very useful

for cost estimates. Learning curve is of immense value as a tool for forecasting.

2) Budgeting and Profit Planning: Budget executive should select those costs which reflect learning effect

and then he should be able to incorporate this effect in process of developing budgets or in the exercises

relating to project planning.

3) Pricing: The use of cost data adjusted for learning effect helps in development of advantageous pricing

policy.

4) Design makers: It helps design engineers in making decisions based upon expected (predictable from

past experience) rates of improvement.

5) Negotiations: It is very helpful to Government in negotiations about the contracts. Government receives

full advantage of the decreasing unit cost in establishing the contract price.

6) Setting Standards: The learning curve is extremely useful in setting standards in learning phase.



Limitations of Learning Curve:

1. All activities of a firm are not subject to learning effect i.e. activities being performed by already

experienced workmen, thoroughly familiar with all activities, activities where workers are shifted even

before completion of one unit of work etc.

2. The following types of activities are subject to learning effect:

(a) Those that have not been performed in this present operational mode.

(b) Those which are being performed by new workmen, new employees or others not familiar with the

particular activity. In contrast, activities being performed by experienced workmen, who are

thoroughly familiar with those activities will not be subject to learning effect.

(c) Those involving utilization of material not used by firm so far.

3. It is correct that learning effect does take place and average time taken is likely to reduce. But in practice

it is highly unlikely that there will be a regular consistent rate of decrease, as exemplified earlier.

Therefore any cost predictions based on covernational learning curves should be viewed with caution.

4. Considerable difficulty arises in obtaining valid data, that will form basis for computation of learning

effect.

5. Even slight change in circumstances quickly renders the learning curve obsolete. While the regularity of

conventional learning curves can be questioned, it would be wrong to ignore learning effect altogether

in predicting future costs for decision purposes.

Remember:

Learning Curve Effect applies only to direct labour costs and those variable overheads, which are direct function of labour hours of input. It does not apply to material costs, non-variable costs or items which vary with output.

Incremental hours cannot be directly determined from the learning curve graph or formula, as the results are expressed in terms of cumulative average hours.

Learning Curve


Previous Examinations :- Theoretical Questions

TQ 1: What are the distinctive features of learning curve theory in manufacturing environment? Explain the

learning curve ratio. [N12, 5 b, 4M] [N10, 7b, 4M] [N07, 2b, 9M] [M03, 4a, 4M]

Ans: The Learning Curve Ratio

In the initial stage of a new product or a new process, the learning effect pattern is so regular that the

rate of decline established at the outset can be used to predict labour cost well in advance. The effect of

experience on cost is summaries in the learning ratio or improvement ratio:



As the production quantity of a given item is doubled, the cost of the item decreases at a fixed rate. This

phenomenon is the basic premise on which the theory of learning curve has been formulated. As the

quantity produced doubles, the absolute amount of cost increase will be successively smaller but the

rate of decrease will remain fixed. It occurs because of the following distinctive features of

manufacturing environment.

(i) Better tooling methods are developed and used

(ii) Design bugs are detected and corrected

(iii) More productive equipment is designed and used.

(iv) Engineering changes decrease over time.

(v) Earlier teething problems are overcome.

(vi) Rejections and rework tend to diminish overtime.

For example, if the average labour cost for the first 500 units is Rs. 25 and the average labour cost for the

first 1,000 units is Rs. 20, th e learning curve ratio is (Rs. 20/25) or 80%. Since the average cost per unit

of 1,000 units is Rs. 20, the average cost per unit of first 2,000 units is likely to be 80% of Rs. 20 or Rs.

16.

TQ 2: [SA-2] [N09*, 1c, 4M]

The following information is provided by a firm. The factory manager wants to use appropriate average learning

rate on activities, so that he may forecast costs and prices for certain levels of activity.

(i) A set of very experienced people feed data into the computer for processing inventory records in the

factory. The manager wishes to apply 80% learning rate on data entry and calculation of inventory.

(ii) A new type of machinery is to be installed in the factory. This is patented process and the output

may take a year for full fledged production. The factory manager wants to use a learning rate on the

workers at the new machine.

(iii) An operation uses contract labour. The contractor shifts people among various jobs once in two

days. The labour force performs one task in 3 days. The manager wants to apply an average learning

rate for these worker

You are required to advise to the manager with reasons on the applicability of the learning curve theory on the

above information.



Ans: The learning curve does not apply to very experienced people for the same job, since time taken can

never tend to become zero or reduce very considerably after a certain range of output. This is the

limitation of the learning curve.

(i) Data entry is a manual job so learning rate theory may be applied. Calculation of inventory is a

computerized job. Learning rate appl ies only to manual labour

(ii) Learning rate should not be applied to a new process which the firm has never tried before.

(iii) The workers are shifted even before completion of one unit of work. Hence learning rate will not

apply.

TQ 3: What are the limitations of the learning curve theory? [N11, 7b, 4M]

Ans: Limitations of Learning Curve Theory

1. All activities of a firm are not subject to learning effect. (Activities that have not been performed in the

present operational mode, those performed by new or unfamiliar employees are subjected to learning

effect, while those performed by familiar or experienced workmen will not be subjected to learning

effect)

2. It is correct that learning effect does take place and average time taken is likely to reduce. But in practice

it is highly unlikely that there will be a regular consistent rate of decrease. Therefore any cost prediction

based on conventional learning curves should be viewed with caution.

3. Considerable difficulty arises in obtaining valid data that will form basis for computation of learning

effect.

4. Even slight change in circumstances quickly renders the learning curve obsolete. While the regularity of

conventional learning curvescan be questioned, it would be wrong to ignore learning effect altogether in

predicting costs for decision purposes.

TQ 4: Discuss the application of the learning curve. [M07, 5c, 4M]

Ans: Application of Learning curve:

1. Learning curve helps to analyze cost-volume profit relationships during familiarization phase of

product or process to arrive at cost estimates.

2. It helps in budgeting and profit planning.

3. It helps in pricing and consequent decision making –e.g. acceptance of an order, negotiations in

establishing contract prices etc. with the advantage of the knowledge of decreasing unit cost.

4. It helps in setting standards in the learning phase.

Learning Curve


TQ 5: Briefly explain the learning curve ratio [N06, 3c, 2M]

Ans: In the initial stage of a new product or a new process, the learning effect pattern is so regular that the

rate of decline established at the outset can be used to predict labour cost well in advance. The effect of

experience on cost is summaries in the learning ratio or improvement ratio:



TQ 6: Explain the concept ‘Learning Curve’. How can it be applied for Cost Management?

[RTP M 09] [M06, 6c, 4M]

Ans: The first time when a new operation is performed, both the workers and the operating procedures are

untried. As the operation is repeated and the workers become more familiar with work, labour

efficiency increases and the labour cost per unit declines. This process continues for some time and a

regular rate of decline in cost per unit can be established. This rate can be used to predict future labour

costs. The learning process starts from the point when the first unit comes out of the production line. In

other words µ Learning curve¶ is a function that measures how labour hours per unit decline as units of

production increase because workers are learning and becoming better at their jobs.

Cost Management Application:

1. Learning curve is useful in analysing cost volume profit relationship. The company can set low

price of its product to generate high demand. As the production increases, cost per unit drops.

2. It helps in budgeting and profit planning.

3. It enables the company in price fixation. In particular, the company can fix a lower price for

repeat orders.

4. It helps the design engineers to take suitable decisions based on expected rates of improvement.

5. It helps in price negotiations.

6. It is useful in setting standards and in performance evaluation.

Important Notes

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