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of 66 - Quantitative Methods in Management

Jamnalal Bajaj Institute of Mgmt Studies

QUANTITATIVE METHODS IN MANAGEMENT

Software Operative

Science of data analysis Research

Optimisation techniques

There are three methods of quantitative management.

1. Conventional Method

2. Conventional Method + Software

3. Software – One of the software used is Microsoft Excel – Solver

Books.

For Exams – Quantitative techniques for management. By ND Vora

(Tata McGrawhill Publication)

For Assignments – Introduction to management science. By Anderson, Sweeney and

Williams. (Publisher -Thomson)

For Corporate Application -

(i) Practical management science. By Winston Broadie

(Publisher -Thomson)

(ii) Introduction to Management science. By Hillier & Hillier

(Publisher -McGrawhill)

How to load Microsoft Excel – Solver?

(a) Open Excel

(b) Click “Tools”. A drop down menu will open

(c) Check if “SOLVER” is there in the drop down menu?

(d) If No, Click “ADD-IN” button in the drop down menu.

(e) Check the box against SOLVER in the new window that had opened.

SOLVER will get installed.




History of Quantitative Management

Quantitative management is mix of statistics and operation research. The subject

was discovered during the WW-II due to paucity of resources for war due to

extraordinary scope and length of the war. People from different disciplines like

defence, scientists, production managers, etc got together to maximise the utility of the

resources available.

Industrial Engg is the application of the Operation Research in Engg.

Process of Applying QT.

1. Define the problem

2. Collect data

3. develop the mathematical model

4. Test the model by existing data

5. Get the solution (Optimisation of the model)

6. Implement the solution

a. At small (Micro) level to see if it is a better solution in the actual

working scenario.

b. Does not create any additional problems like people’s reaction.

7. Expand it.

Eg. Preparation of a Mathematical Model for a given problem




Product A B Availability

Selling Price Rs 10/- Rs 8/- NA

Raw Material Consumption 2Kg 3 Kg 6000 Kg

Man Hours 1 hr ½ hr 7500 Hrs

Lub 1/3 Hr ¼ Hr 5000 ltrs

Z (Turnover) = 10X1 + 8X2 (X1 and X2 being the number of pieces produced)

Purpose is to maximise Z with following restrictions imposed by the availability: -

2X1 + 3X2 <= 6000

X1 + ½ X2 <= 7500

1/3 X1 + ¼ X2 <= 5000

X1 >= 0, X2 >= 0

Model will contain 2 parts:-

1. The objective function ‘fn’ whose value is to be maximum or minimum.




2. Constraints on resources/other conditions on variables.

Following difficulties are encountered while preparing the model: -

1. Situation too complex to convert into a mathematical model.

2. Assumptions are not valid in the real situations.

Software Available: -

1. Crystal Ball – For simulation

2. Evolutionary Solver – Non Linear programming

3. @ Risk

4. MS Project – PERT and CPM charts preparation

5. Tree Plan – for Decision Tree

6. Prima Vera

STATISTICS

Book – Statistics for business and economics. By Anderson, Sweeney and Williams)

Definition – Science of collecting organising and analysing data for decision making.

Data Collection.




Census Survey. Data can be collected from all units in population which is

called Census Survey.

Sample Survey. From units in sample called Sample Survey. (Most preferred

method)

Reasons for adopting sample survey method -

(a) Resources limitations

(b) Destructive analysis. (Sample is damaged in the process)

Example

Objective – To find the annual income pattern in Mumbai.

Let n = 10000 house holds

Prepare frequency table and draw frequency curve.




FREQUENCY TABLE

Class Interval (CI) Mid Point of CI (Xi) Frequency (fi) Relative Frequency (Rfi)

f1 f1/fi

f2

f3

fk

fi

Points joined by straight lines are called frequency polygon

Points jopined by smooth curve is called frequency curve.

frequency polygon

fi

Xi

Relative frequency is useful when no of observations for two sets of data are different.

Mumbai (10000)

Chennai (8000)

fi

Xi




Class Interval (CI) Mid Point of CI (Xi) Frequency (fi) Relative Frequency (Rfi)

0-10 5 25 25/1000 = .025

10-20 15 30 30/1000 = .030

f3

fk

1000

Relative Frequency gives probability of an event.

Frequency ‘fi’ column shows how the frequency is distributed ever various class intervals.

Therefore, called Frequency Distribution.

Relative Frequency column ‘Rfi’ indicates how a total probability on ‘1’ is distributed over

various class intervals. Therefore, called probability distribution. Class Interval (CI) Mid Point of CI (Xi) Frequency (fi)

10-20 15 10

20-30 25 20

30-40 35 40

40-50 45 35

50-60 55 30

60-70 65 15

Question – What is the percentage of population lie between 32-61?




Answer can be found by integrating the function ‘fx dx’ for range 32 to 61.

However to draw a mathematical model for such a random function is literally impossible.

Therefore a graph is drawn and the curve is approximated to one of the standard curves.

And then use their known properties to arrive at the answer. There will definitely be loss of

accuracy but that is better than no solution.

Standard Frequency curves are given by theoretical probability distribution as follows

(a) Binomial

(b) Poisson

(c) Normal

Theoretical Probability Distribution

Variable. Any phenomenon that takes different quantitative values from one obhect ot

other object in sample /population is called a variable. For eg. Annual Income of house

holds, No of defectives in a batch production, percentage of marks obtained in exam, etc

Attribute. A phenomenon whose data can not be captured in quantitative values is called

attribute. eg Likes, perception, preferences etc.

Bach Size – 10000

X: No of defectives in batch production.

X can take values from 0,1,2,3,4,------------10000.

A variable that takes specific (discrete values is called “Discrete Variable”).

Y: percentage of marks obtained in exam.

Y can take values from ‘0 – Maximum Marks’, even in decimals.

A variable that can take any value in a given interval is called “Continuous Variable”.

Theoretical Probability Distribution

For Discrete Variable Continuous Variables

Binomial Normal

Poisson

There are even more theoretical distribution curves but most situations can be covered with these 3 curves. Balance curves will be covered in the next semester.

Binomial Distribution. This distribution can be used when




1. Trial batch has only two outcomes. Success or failure

2. An experiment consists of repeating the trial ‘n’ no of times

3. The probability of getting success in a trial is denoted by ‘q’ and remains

same for all trials in the exp.

4. The outcome of one trial doesn’t depend on outcome of other trials. ie Each

trial is independent.

Real Life situation – Interview of candidates for job selection.




QUANTITATIVE MANAGEMENT (09 Jul 04)

By Prof CY Nimkar

Properties of Binomial Distribution

If we denote number of successes by ‘r’ then probability ‘p’ of getting ‘r’ successes out

of ‘n’ trials is

nCr p

r (1-p)

n-r = ___n!____ x p

r (1-p)

n-r

(n-r)! x r!}

Example.

Five candidates were called for an interview. Two candidates are to be selected from

the group. Here

Number of trials (interviews) = n = 5

Probability denoted as ‘Prob{r = 2}’ = p .

Giving a diagrammatic representation of the situation,

Candidates Nos 1 2 3 4 5

Selected/Not Selected S S N N N

S N S N N (Possible combinations)

S N N S N

N S S N N

- - - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - -

Possible number of combinations can be calculated from the above COMBINATION

formula. Substituting the values in above formula we get

5C2 p

2 (1-p)

5-2 = ____5!____ x p

2(1-p)

5-2

(5-2)! x 2!}

Mean of Probability = np

Variance = np (1-p)

Mean is a measure of central tendency in data to cluster around a central value.

Variance is a measure of dispersion. ie tendency in data to move away from central value.




Example.

If x1, x2, x3, x4 - - - - - - - - - - - -xn are number of observations. Then

n

Mean = X = {x1+x2+x3 + - - - - - - - xn}/n = [∑ xi ]/n

i=1

Variance = {(x1 – X)2 +(x2 – X)

2 + (x3 – X)

2 + - - - - - - -+ (xn – X)

2}/ n

n

= ∑ (xi – X)2

i=1

Example:

A personnel manager has called 10 MBA candidates for interview. The probability

that a candidate will be short listed is 65%. Find

(a) Probability that he could short list 4 candidates.

(b) He will short list at the most 4 candidates

(c) Probability that he would short list at least 5 candidates.

(d) If he desires to short list 6 candidates, how many candidates should he call for

interview?

Solution:

Success = a candidate is short listed.

Probability {Success} = 0.65 = p

n = 10

(a) r = 4

Prob{r = 4} = 10

C4 (0.65)4 (0.35)

6

= {10!/(6! x 4!)}(0.65)4 x (0.35)

6

= {362880/(720) x (24)}(0.1785) (0.0018)

= 0.0689

(b) Calculate for Prob{r=0)+Prob{r=1}+Prob{r=2}+ Prob{r=3}+Prob{r=4}

(c) Short listing of at least 5 candidates

r = {0, 1, 2, 3, 4}, {5, 6, 7, 8, 9,

10}

Complementary Event Main Event




It may be noted that the maximum probability is 1 which is basically sum of

probability of all individual combinations possible.

In this case rather than calculating the probability with value r = 5 to 10,

calculate the probability of occurrence for r = 0 to 4 and subtract the

resulting answer from 1 to get the desired result.

This method is called complementary event method.

(d) r = 6

p = 0.65

np = r

n = 6/0.65

n = 9.2 ≈ 10

Never take a lesser value in approximation even if variation is just 0.0001.

9.00001 should be approximated to 10 and not 9.

Example:

Military radar and missile detection systems are designed to warn a country against

enemy attacks. A reliability question is whether a detection system will be able to identify

an attack. If a detection system has a 90% probability of detecting an attack. Answer the

following questions: -

(a) What is the probability that a single detection system will detect an attack?

(b) Of two systems are installed that operate independently, what is the probability

that at least one of them will detect the attack?

(c) If 3 systems are installed that are independent, what is the probability that at

least one of them would detect an attack?

(d) Do you recommend multiple detection system? Justify your answer.

Solution:

(a) Success: A system detects an attack.

n = 1, Therefore, Probability =

Prob{success} = 0.90 = p

(b) n = 2, Therefore, Probability =

= Prob{One system detects attack} + Prob{both detect attack}

= Prob{r=1} + Prob{r = 2}

= 2C1(0.90)

1 (0.10)

1 +

2C2 (0.90)

2 (0.10)

0

= 0.99

(c) n = 3, Therefore, Probability =




= Prob{One system detects attack} + Prob{two detect attack} + Prob{three

detect attack}

= 3C1(0.90)

1 (0.10)

2 +

3C2 (0.90)

2 (0.10)

1 +

3C3 (0.90)

3 (0.10)

0

= 0.999

Better way to attempt this situation is to calculate complementary value of the

situation, ie r = 0 which translated into logic would mean that no system is able

to detect the attack and subtract the result from ‘1’ to arrive at the result of at

least one of the three systems detecting the attack.

(d) I may be observed that with each addition of radar, the probability level is

improving only to next decimal. ie from 0.90 to 0.99 to 0.999 and so on.

Therefore, the decision would be based on possible losses arising out of failure

to detect the attack. In a nuclear attack situation, because the losses are colossal,

multiple detection system are well justified. Take the case of USA which is

investing trillions of dollars in Space based National Missile Defence System.

Example:

40% of the business travellers carry either a cell phone or a laptop. In a sample of

15 business travellers, find

(a) Probability that 3 have either a cell or a laptop.

(b) Probability that 12 travellers neither have a cell phone nor a laptop.

(c) Probability that at least 3 travellers have a cell phone or a laptop.

Solution:

n = 15, p = 0.40

Success = A traveller has either a cell phone or laptop.

Failure = A traveller neither has a cell phone nor a laptop.

Prob{success} = 0.40

Prob{failure} = 0.60

(a) It may be observed that answer for (a) and (b) are same because both questions

are complementary of each other.

Prob{r=3} = 15

C3(0.40)3 (0.60)

12

(b) Let Success = Neither cell nor laptop

Prob{success} = 0.60

Prob{r=12} = 15

C12(0.60)12

(0.40)3

Since nCr =

nC(n-r)., Therefore

15C3 =

15C12

(c) Take complimentary route, ie,

r = 0, 1, 2, 3

Probability = 1 – Prob{(r=0) + (r=1) + (r=2)}




Poisson Distribution

Examples of application area: -

1. A reservation counter opens from 8AM to 8 PM.

(a) No of people joining from 8 AM – 9 AM All values will

be

(b) No of people joining from 9AM – 10AM different

(c) No of people joining from 10AM – 11AM

(d) No of people joining from -----

(e) No of people joining from -----

(f) No of people joining from 7PM – 8 PM.

No of occurrences in one unit of time is discrete random variable.

Instead of time, take length as the unit. In a weaving mill, number of knots in particular

length of cloth will be variable.

Thus, Poisson Distribution is used when the result of trial is a random variable in

(a) One unit of time (as defined by the user, minute, hour, day, month, year,

decade, century, millennium, etc).

(b) One unit of length

P.S. No other unit of any kind other than above two is ever used in Poisson distn.

Properties of Poisson Distribution

Prob{‘r’ occurrences in one unit of time/length} = e-

r

r!

Where ‘e’ = Base of natural log = 2.718

‘’= Average occurrence per unit of time or length

Mean =

Variance =

Example:

In a reservation ‘Q’ on an average 20 customers join/hr. Find: -

(a) What is the probability that 10 customer will join a 1 hr?

(b) What is probability that 40 customer will join in 3 hr?




Solution:

r = 10, = 20

(a) {e-20

(20)10

}/10!}

= {(2.718)-20

(20)10

}/10!}

= 21106.22/3628800

= 0.0058

(b) r = 40, = 60 (20 x 3)

Prob = {(2.718)-60

(60)40

}/40!}

= 0.0014

Example:

Phone calls arrive @ 48/hr in a company, Find

(a) The probability of receiving 3 call in a 5 min interval of time.

(b) Suppose no calls are currently on hold. If a customer takes 5 minutes to

complete his call, how may calls do you expect to be waiting?

(c) If no calls are currently being processed, what is the probability that the

caller can take 3 min without being interrupted by incoming call?

Solution:

(a) One time unit = 5 min

Average call in 5 min = 48/12 since (48 calls in 60 min converted to ? calls

per 5 min)

= 4

Prob{r=3} = {e-4

43}/3!}

= 0.1952

(b) = 2.4, r = 0

Prob = 0.091

Home Work:

How many lines should be there on EPABX if a caller should not be kept on hold for over a

minute?

Determine No of calls/min




03 Oct 04

Quantitative Techniques

Mr Moradian

Simulation

Suppose Z is any function of (x, y)

Z = fn (x,y)

Z= 2x + 3y

Solving the equation,

If x = 0, y = 1 and Z = 3

x = 1, y = 3 and Z = 11

x = 4, y = 1 and Z = 07

Let probabilities of x and y be as follows: -

x Prob Cum Prob

Tag No y Prob Cum Prob

Tag No

0 0.1 0.1 00 - 9 1 0.15 0.15 00-14

1 0.2 0.3 10-29 2 0.3 0.45 15-44

2 0.25 0.55 30-54 3 0.4 0.85 45-84

3 0.3 0.85 55-84 4 0.15 1 85-99

4 0.1 0.95 85-94

5 0.05 1 95-99

1.00 1.00

Find Probability of Z.

Finding probability of Z is literally impossible since there is not a formula devised in maths

to find such probability. This is where Simulation comes to the rescue.

Simulation is employed when every thing else fails.

Trial No

x y

Z Zmin = 3 Zmax = 22

Random Derived Random Derived Z frequency Prob

1 54 2 85 4 16 3

2 10 1 51 3 11 4

3 29 1 62 3 11 5

4 64 3 35 2 12 6

5 89 4 16 2 14 7

6 20 1 18 2 8 8

7 84 3 54 3 15 9

8 68 3 98 4 18 10

9 22 1 92 4 14 11

10 82 3 62 3 15 12

13

14




15

16

17

18

19

20

21

22

10000 1.00

For a single channel ‘Q’ situation following statistics was gathered.

Time between arrivals Service time Distribution

Min Prob Cum Prob Tag # Min Prob Cum Prob Tag #

3 0.05 0.05 00-04 3 0.10 0.10 00-09

4 0.20 0.25 05-24 4 0.20 0.30 10-29

5 0.35 0.60 25-59 5 0.40 0.70 30-69

6 0.25 0.85 60-84 6 0.25 0.95 70-94

7 0.10 0.95 85-94 7 0.05 1.00 95-99

8 0.05 1.00 95-99

Determine ‘Q’ parameters Ws, Wq, Lρ, Ls, ρ.

Solution

Here the inter-arrival time is not negative exponential Nor is the service time negatrive

exponential. Therefore, there is no mathematical formula available to solve the problem.

Arr # Time betw’n

arrivals

Arr

time

Service

begin

at

Service time Service

ends at

Wq Ws Service

station

idle time Ran ∴ Ran ∴

Seed

1

49

5

8:05

8:05

90

6

8:11

-

6

5

2 15 4 8:09 8:11 60 5 8:16 2 7 0

3 18 4 8:13 8:16 81 6 8:22 3 9 0

4 13 4 8:17 8:22 15 4 8:26 5 9 0

5 97 8 8:25 8:26 22 4 8:30 1 5 0

6 93 7 8:32 8:32 40 5 8:37 0 5 2

7 05 4 8:36 8:37 44 5 8:42 1 6 0

8 72 6 8:42 8:42 73 6 8:48 0 6 0

9 99 8 8:50 8:50 13 4 8:54 0 4 2

10 93 6 8:56 8:56 51 5 9:01 0 5 2

11 53 5 9:01 9:01 23 4 9:05 0 4 0

12 88 7 9:08 9:08 16 4 9:12 0 4 3

13 79 6 9:14 9:14 81 6 9:20 0 6 2




Total 12

12

70

12

11

Start 8:09 Finish 9:20

In 71 min total operation time, idle time = 11 min

Therefore, utilisation time of service station = 60 min.

∴ ρ = 60/71

For calculating Lρ and Ls, take random observations, say

Observation at Lρ Ls,

8:10 1 2

8:20 1 2

8:30 0 1

8:40 0 1

8:50 0 1

9:00 0 1

9:10 0 1

Total 2 9

# of observations = 7

Average = 2/2 and 9/7

Table for multi(2) channel single phase

Arr

ival

#

Arr

ival

Tim

e Service Station A Service Station B

Ws Wq Lρ Ls

Ser

vic

e

beg

ins

at

Service

time

Ser

vic

e

ends

at

Ser

vic

e

beg

ins

at

Service

time

Ser

vic

e

ends

at

Ran ∴ Ran ∴

Question

A Rent a Car agency has gathered following statistics No of cars

demand daily

Prob Cum Prob

Tag No Length of rental in

days Prob

Cum Prob

Tag No

0 0.10 0.10 00 - 9 1 0.50 0.50 00-49

1 0.15 0.25 10-24 2 0.30 0.80 50-79

2 0.20 0.45 25-44 3 0.15 0.95 80-94




3 0.30 0.75 45-74 4 0.05 1.00 95-99

4 0.25 1.00 75-99

1.00 1.00

There is a net profit of Rs 800/- per day for each car rented. Each day a car is unutilised,

costs the company Rs 500/-. When there is a demand for a car and no car is available, the

company has estimated a goodwill cost of Rs 1000/-. How many car should the company

own?

Simulate for 4 cars.

Day

No

Car

s av

ail

at

beg

in o

f day

Demand

for cars

Rental Period Order

lost Car No 1 Car No 2 Car No 3 Car No 4

Ran ∴ Ran ∴ Ran ∴ Ran ∴ Ran ∴

1 4 65 3 58 2 11 1 00 1 UN 0

2 3 28 0 UN UN UN 0

3 4 47 3 26 1 17 3 86 3 UN 0

4 3 85 4 24 1 45 1 79 2 1

5 2 51 3 14 1 08 1 1

6 4 41 2 02 1 54 2 UN UN 0

7 3 48 3 64 2 11 1 82 3 0

8 2 11 1 65 2 UN 0

9 2 61 3 89 3 93 3 93 3 1

10 2 46 3 00 1 1

11 1 17 1 82 3 0

12 2 48 3 06 1 89 3 1

Total no of days = 12 days

Possible car days for rental = 48

Car days unutilised = 8

Car days utilised = 40

(a) Rental net profit = 40 X 800 = 32,000

(b) Goodwill cost = 5 x 1000 = 5000

(c) Non-utilisation cost = 8 x 500 = 4000

Net profit = 32000 – 5000 – 4000 = 23,000 in 12 days.

Operation Research by Wagner




QUANTITATIVE MANAGEMENT (16 Jul 04)

By Prof CY Nimkar

Probability Distribution (Contd …..)

Exponential Distribution. Poisson’s Distribution is concerned with number of arrivals/

occurrences in one unit of time/length, whereas Exponential Distribution is concerned with

time between tow consecutive occurrences/arrivals. (in this case time is the variable. In

Poisson’s distribution, occurrences/arrivals was the variable and time was a fixed unit).

For eg the time intervals between arrival of consecutive customers in a shop.

Theoretical Distribution for Continuous Variable.

There are two kind of variables, Discrete and Continuous. Discrete variables are

those which can take only integer values (means 1,2,3,4,---, 10,---50,--- 105,--- etc but no

decimal values like 1.2, ---, 2.5, -----, 4.9, ----, etc). Continuous variables are those, which

can take decimal values.

Normal Distribution. This method was invented by GAUSS and therefore it called

Gaussian Distribution also. However, since 85-90% of the real life problems can be solved

using this method, it came to be known as Normal Distribution.

Properties.

1. The shape of the probability graph (called probability density function

graph) is bell shaped.

0

10

20

30

40

50

60

70

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Bell Shape




Example:

Suppose X = Monthly rental paid by house holds in a town

Take a sample of 250 house holds from the town.

Get Raw data (value of X) for each sampled house hold.

Convert raw data into tabular form – Frequency Table

Let us assume that we have the following frequency table: -

CLASS

INTERVAL Xi fi Rfi

350-360 355 12 0.048

360-370 365 25 0.100

370-380 375 45 0.180

380-390 385 78 0.312

390-400 395 55 0.220

400-410 405 25 0.100

410-420 415 10 0.040

TOTAL 250

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

355 365 375 385 395 405 415

GRAPH OF RELATIVE FREQUENCY

If the relative frequency graph drawn above resembles a bell shape, we can infer that the

monthly rental paid can be approximated to Normal Distribution.

Symmetry is an essential feature of Normal Distribution. The bell shaped graph

characteristic is that the pattern of increase in Rfi from lowest Class Interval to middle

Class Interval and from highest Class Interval to Middle Class Interval is same (ie we have

symmetrical curve).

To be able to get a graph, the variable must be continuous. Otherwise you get only points

on the chart and not the lines since interval values between two measurement points are

non existent.

Functional form of Normal Distribution graph (also called probability density function) is




f(x) = ______1____ e -½ [x-]/

(√2)

f(x) is the height of the graph at x.

Parameter – The terms on the right hand side of the probability density function whose

values should be known to compute probability are called parameters of Distribution.

= Mean of Distribution

= Standard Deviation of the Distribution.

Lower

Higher

0.000

0.100

0.200

0.300

0.400

0.500

0.600

1 2 3 4 5 6 7

X1 - = 4, = 15 X2 - = 12, = 15

0

10

20

30

40

50

60

70

80

90

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15




Y1 - = 15, = 10 Y2 - = 15, = 15

= 10 = 15

For attaining the same level of accuracy in forecasting,

If variation is more – Sample should be large

If variation is small – Sample can be small. ie. The sample size is directly proportional to

the variation and is not dependent on the size of the population.

‘Z’ Transformation

If X is a continuous variable whose distribution can be approximated to normal

distribution with mean and standard deviation , then, if we define the new

variable Z as follows: -

Z = X-

Z will have normal distribution with mean ‘0’ and SD 1. For any value of

and of X mean of Z will always be ‘0’ and SD will always be ‘1’. Therefore,

Z is called Standard Normal Variable.

Eg. X = Monthly rental paid

Mean of X = = Rs 365/-

& SD = = Rs 50/-

Sample of 100 House Holds

X = 310, 410, ------

Z = 310 – 365 , 410-365 , - - - - - - - - - - - - - - - -

50 50




Probability Graph of ‘Z’

‘0’

Table of ‘Z’ values will be available in any statistics book.

Graph of ‘Z’

0.3508

‘0’ Z = 1.04

Z 0.01 0.02 0.03 0.04 0.09

0.1 0.000 0.004 0.0080 0.0120 0.0359

0.2 0.0398 ------ ------- -------

0.3 0.0793 ------ ------- -------

0.4

0.5 Area undr curve fro 0 - Z

0.6

0.7

0.8

0.9

1.0 0.3508

3.0




Eg.

It is observed that Annual Income of House Holds follows a Normal Distribution with

mean and SD = 72000/-. Find the Annual Income such that

1. The top 20% of population have AI higher than this value

2. Percentage of population having icome between 3-4 lakhs per annum.

X Graph

20%

3,25,000 ‘X’

X to be found out

Convert X into Z.

Z = X –325000

72000

Breadth of the curve on the right side of vertical line = 0.5 (Half of ‘1’)

Position of ‘X’ = 0.3

So, from the table find Area value very close to 0.3 = 0.2995

This value occurs in row of 0.8 and column of 0.04. Therefore corresponding value of Z is

0.84 (obtained by adding 0.8 and 0.04).

So Z = X-325000 = 0.84

72000

X = 385480.

For second part of the question, calculate 0 – 25 and 0 – 75 since 3.25 is the mean value

3,00,000 4,00,000




Z value corresponding to X = 300000 = 300000 – 325000 = 0.347

72000

X = 400000 = 400000 –325000 = 1.04

72000

So required %age is area between Z = -0.35 to Z = 1.04

Look for row 0.3 and column 0.05 = 0.1368

Next Look for row 1 and column 0.04 = 0.3508

Reqd value is 0.4876

ie 48.76%.

Significance of 6

99.98%

- 3.0 ‘0’ 3.0

Smaller value of means more accurate procedure.

All the values of variable must lie from a point which is 3 to the left of the mean to 3to

the right of the mean. Hence the concept of 6The quality of the process is based on the

value of . Smaller the value of , more precise the process is.




23 JUL 04

QUANTITAVE METHODS OF MANAGEMENT Mr RS Sardesai

Normal Probability Distribution

Properties – Contd ……

Mean = Median = Mode

Mean = Arithmetic Average of all observations – (x + y+ z + ……)/n

Median = Positional average of all observation. It is half population divide. 50% of

the observations are on left side of the median and remaining 50% observations are on

other side. 50 – 50.

50% observations 50% observations

0 Median 100%

2, 4, 8, 10, 12, 12, 15, 16, 17, 20, 27, 28, 28, 29, 37

Median

If all the observations are arranged in ascending or descending order, then the central value

is called the median value. For e.g. In case of 9 observations arranged in ascending order,

fifth value will be median value because there are 4 observations on either side of the 5th

the median value. In case of even number of observations, the average of two central values

is taken as the median value.

Mode is the value at which maximum frequency occurs.

2. ‘Z’ Transform.

3. If X is a continuous variable with normal distribution having mean and Std

Deviation , then - z to + z will contain (1-)% of values where area

between –Z to Z is (1-)% where is the balance area.

This interval will contain 95% of the Z values.

X N(,)

(1-)% = 95%

/2 /2

95%

0 -(1.96) +(1.96)




Q. From a survey it is observed that the monthly consumption of detergent per person

follows a Normal Distribution with mean 2.6 Kg and Standard Deviation 0.25Kg. Find

out the lower and upper values of consumption that contains 95% of population around

mean.

Sol. Distribution of per capita of detergent

95%

from Z transform table

2.65

2.65-(1.96)(0.25) 2.65+(1.96)(0.25)

[2.16 3.14]

X N(,)

99.74% of values

( -3) (+3)

Z N(0,1)

-Z = -3 0 Z= +3




Decision Analysis

Bayes Theorem.

Vendor Defective Good Total

A 25 90 115

B 35 110 145

Total 60 200 260

Seeing from the above diagram, what is the probability that the item would be defective?

Sol. There are total of 260 items and there are in all 60 defective item. If no other details

are available in the question, then

Prob(item defective) = 60/240. This situation is called “A Priori” where no

additional info is available or used.

Q. What is the probability that the defective item was supplied by ‘A’?

Sol. Out of every 60 defective items, 25 defective items are being supplied by ‘A’. So

Prob(defective item supplied by ‘A’) = 25/60

In “Posterior Prob”, we have additional information and use it to arrive at more precise

and accurate analysis.

Bayes Theorem is used when the layers of Posterior Prob keep increasing. Then presenting

the data in tab form becomes difficult. In such case we use Bayes theorem to calculate

posterior prob.

Bayesian Decision Models – will be done subsequently.

Bayes Theorem Definition – If A1, A2, A3,-,-,-,-,-,AK are mutually exclusive and

exhaustive events and B is any event then probability of occurring Ai knowing out come of

event B is P(Ai/B)

= P(Ai/B) P(B/Ai)

k

P(Ai/B) P(B/Ai)

i =1

Taking last example,

A1 = Supplied by Vendor A

A2 = Supplied by Vendor B From Table

B = Item is found defective

P(A1/B)= 25/60




Compute the probability using Bayes Theorem

P(A1/B) = P(A1) P(B/A1)

P(A1) P(B/A1) + P(A2) P(B/A2)

= (115/260) x (25/115)

(115/260) x (25/115) + (145/260) x (35/145)

= (25/260)

(25/260) + (35/260)

= 25/60

Q. A company has a recruitment system for recruiting salesmen. From the past data it

has fond that probability that selected salesman will be successful is 80%. Company

receives a proposal sent by HR consultant recommending a new test that will

increase the probability of selecting successful salesmen. For validating this claim,

the test was administered to the existing sales person of the company. Based upon

their performance following data has been compiled: -

Test Result Performance of

Salesmen Pass Fail

Successful 210 30 240

Unsuccessful 90 60 150

300 90 390

Should company purchase the test?

Sol: -

Let

S= Selected candidate will be successful salesmen

U= Selected candidates will be unsuccessful salesmen

P(S) = 0.80 - A Priori Prob

Q = Candidate qualifies the test

F = Candidate fails the test

P(S/Q) = P(S) x P(Q/S)

P(S) x P(Q/S) + P(U) x P(Q/U)

= 0.80 x (210/240)

0.80 x (210/240) + (0.2) x (90/150)

= (0.80 x 7/8) = 0.85

(0.80 x 7/8) + (9/15)




Next Lecture – Repeat Bayes Theorem, Decision Tree.




30 Jul 04

Bayes Probability Prof Nimkar

Tabular form in computing Bayes Probability

P(Ai/B) = P(Ai) P(B/Ai)

k

P(Ai) P(B/Ai)

i =1

Table

Apriori Conditional Joint Probability Posterior Probability

Probability Probability P(Ai/B)

P(B/Ai)

A1 P(A1) P(B/A1) P(A1) P(B/A1) P(A1)

P(B/A1)

k

P(Ai) P(B/Ai)

i =1

A2 P(A2) P(B/A2) P(A2) P(B/A2)

A3 P(A3) P(B/A3) P(A3) P(B/A3)

A4 P(A4) P(B/A4) P(A4) P(B/A4)

--

--

--

--

An P(An) P(B/An) P(An) P(B/An)

n

Prob(B) = P(Ai) P(B/Ai)

i =1

Q. The Indian Economists feel that the probability that economy will

grow @ > 7% is 40%, between 6-7% is 45% and <6% is 15%. The Govt has approached

the World Bank for providing a soft loan. From the previous experience it is found that

when the economy has grown by > 7%, the World Bank has given loan 40% of the times.

Similarly, when the economy has grown between 6-7% and less then 6% World Bank loan




was available 30% and 25 % of times. The Govt has received a commitment from World

Bank that they will provide the loan. What is the probability that the economy will grow

between 6 & 7%?

Sol. Let A1 = Economy grows by > 7%.

A2 = Economy grows between 6-7%.

A3 = below 6%.

B = Probability that World Bank gives the loan

Event Apriori Conditional Joint Posterior

Probability Probability Probability Probability

A1 0.40 0.40 0.16 0.16/0.3325 = 48.12%

A2 0.45 0.30 0.135 0.135/0.3325 = 40.60%

A3 0.15 0.25 0.0375 0.0375/0.3375 = 11.28%

0.3325

DECISION MODELS

Decision making Decision making Decision making under certainty under Risk under uncertainty

Decision Making under Certainty – If values of all the variables are known or can be

controlled, then the decision taken under such conditions is called Decision making under

certainty. In other words, All decision variables are under control so that their values are

known with certainty.

Decision Making under Risk - If some of the variables are not under control but values

can be calculated by probability distribution method, then the decision taken such

conditions is called decision making under Risk. In other words, Few variables which are

not under be control, can be approximated using probability distribution.

Decision Making under Uncertainty - If no probability distribution is possible of

unknown variables, then decision taken under such conditions is called Decision Making

under Uncertainty. In other words, there is no possibility of even approximating some of

the decision variables.




Example. A company is contemplating to expand its manufacturing base. For this

purpose, it has three options: -

(a) Expand internal capacity.

(b) Acquire competitor’s venture

(c) Float a joint venture

Above decisions are based on the market conditions.

The expected profit based on market conditions are as follows: -

Mkt

Condition

Decision Poor Mkt Avg Mkt Good Mkt

Internal Expansion 10 15 14

Acquision 2 7 20

Joint Venture -20 -4 10

What should company do?

Sol: -

Decision Making under Certainty - Suppose, company comes to know with certainty the

market condition to be avaerage, then it will evaluate the payoff of each alternative

available to company, apply decision criteria to pay off and select the alternative that

satisfies decision criteria.

Decision under Risk – Suppose company can not predict with certainty condition of the

market but marketing deptt predicts that probability that the market will be poor, average,

good will be 40, 50, and 10% respectively.

In such a case we calculate expected pay off of each alternative available to the company.

Therefore expected pay of expanding internally will be 0.4x10 + 0.5x15 + 0.1x14 = 12.9

Expected pay off of acquiring competitors company will be 0.4x2 + 0.5x7 + 0.1x20 = 6.3

Expected pay off of Joint Venture will be 0.4x-20 + 0.5x-4 + 0.1x10 = -9

Since highest expected pay off which is 12.9 is for internal expansion, company should

decide to expand internally.

Suppose an agency offers to do the market research and provide perfect condition of the

market in future at a fees of Rs 10 Lakh, should the agency be hired for the research.

Expected Pay off with perfect information = 0.4x10 + 0.5x15 + 0.1x 20 = 13.5 (Highest

earning in each of the market conditions taken into consideration)




Expected pay off with perfect information – Expected pay off without perfect information

= Expected value of perfect information (EVPI) = 13.5 – 12.9 = 0.6 Lakh

3. Decision making under uncertainty – Following criteria are available

(a) Maximax

(b) Maximin

(c) Criteria of Equally Likely (Laplace Criterian)

(d) Minimax (On regeret matrix)

(e) Hurvicz Alpha

Maximax – From each row representing our decision, select maximum and from these

maximums, select maximums, ie Max pay off in matrix. This approach can be adopted by

optimistic decision maker. (ie Who is confident that all factors will be in his favour).

Maximin – Select min from each row and select max out of minimum. This approach can

be adopted by pessimistic decision maker.

Criterian of Equally Likely (Laplace Criterion) - It is assumed that probability of each

happening is assumed to be equal. Thus the problem changes its condition to Decision

making under Risk from Uncertainty. This criteria suggests to assign equal probability to

states of nature (ie Outcome of uncontrollable variables)

Expected Pay off of Internal Expansion = 0.33 (10+15+14) = 12.87

Expected Pay off of acquiring competitor’s company = 0.33 (29) = 9.57

Expected Pay off of Joint Venture = 0.33 (-14) = -4.62

Hurvicz Alpha –

The criteria is based on two values viz Max and Minimum pay off of an alternative.

= Probability of achieving highest pay off

(1-) – Probability of achieving lowest pay off

Assuming = (0.7) (Based on gut feeling. No scientific method available for this

assumption)

Expected pay off of internal expansion = 0.7 x 15 + 0.3 x 10 = 13.5

Expected pay off of Joint Venture = 0.7 x 10 + 0.3 x -20 = 1.0

Expected pay off of Acquiring Competitors company = 0.7 x 20 + 0.3 x 2 = 14.6

Minimax – If the objective is of maximisation of pay off then this criterion is to be used

for Regret Matrix.

Regret is loss of the decision maker for not taking correct decision had he known

the states of nature. It is calculated by finding maximum pay off for a state of nature and

deducting all elements fro, this maximum.

10 15 14 0 0 6 6

2 7 20 8 8 0 8




-20 -4 10 30 19 10 30

Next Lecture – Decision Tree Analysis




06 Aug 04

Decision Models Prof Nimkar

Eg. The imperical distribution of sale compiled by a seller is given below: -

No of cases sold per day Probability

Probability of selling

all the case purchased

4 0.05 1.0

5 0.15 0.95

6 0.15 0.80

7 0.20 0.65

8 0.25 0.45

9 0.10 0.20

10 0.10 0.10

Selling Price per case = Rs 60/-

Purchase price per case = Rs 40/-

What quantity of cases should be ordered to maximisse profits?

Sol: -

Calculate the expected pay of for each case (order qty)

Prepare a expected pay off matrix

.05 .15 .15 .20 .25 .10 .10

Sale Qty

Order Qty

4

(240)

5

(300)

6

(360)

7

(420)

8

(480)

9

(540)

10

(600)

Exp

pay

off

4 (160) 80 80 80 80 80 80 80 80

5 (200) 40 100 100 100 100 100 100 97

6 (240) 0 60 120 120 120 120 120 105

7 (280) -40 20 80 140 140 140 140 104

8 (320) -80 -20 40 100 160 160 160 91

9 (360) -120 -60 0 60 120 180 180 63

10 (400) -160 -100 -40 20 80 140 200 29

Highest Expected Pay Off is for order quantity = 6

While above method is quite easy to understand and workout, it becomes painfully

cumbersome when the data is large. For large data another method is available.

Method 2.

Calculate probability of sale viz a viz not selling additional unit.

If probability of sale is > probability of not selling, then increase the order quantity.




Let p = Probability of selling 1 additional unit

Then (1-p) = Probability of not selling the additional unit

Therefore p (1-p)

p will be (1-p) as long as probability of selling one additional unit Probability of not

selling the additional unit

i.e. as long as Prob (Demand Supply) he will keep going on adding the additional unit to

order quantity.

Expected profit by adding one unit in order Expected loss by adding one unit in order.

p x marginal profit (1-p) x marginal loss

p x MP ML – p[ML] Where MP = Marginal Loss and ML = Marginal Profit

p[MP – ML] = ML

p = ML

MP + ML

In the above example ML = cost of one unit = Rs 60

And MP = (Selling Price – Cost Price) of one unit = (60 – 40) - Rs 20

p = 40 = 0.67

20 + 40

We see from the table that for 6 cases the value of p is 0.80 and for 7 cases it is 0.65 which

is less than 0.67.

06 cases should be ordered for maximising the profit

Question – For the above data, calculate the expected value with perfect information.

No of cases ordered Expected pay off with perfect data 4 80

5 100

6 120

7 140

8 160

9 180

10 200

= 0.05 x 80 + 0.15 x 100 + 0.15 x 120 + .20 x 140 + .25 x 160 + .10 x180 + .10 x 200

= 4 + 15 +18 + 28 + 40 + 18 + 20 = 143

Expected value of Perfect Information (EVP) = 143 – 105 = 38

The above question was based on discrete values. However, when the data is large, then the

data, though discrete, can be assumed to be continuous and calculated using Normal

Distribution or ‘Z’ transform.




Question: - The daily demand for news paper is ovserved to follow the normal distribution

with mean of 50000 and variation of 2500 units. How many copies of news paper should be

printed per day to maximise the profits if marginal profit is Rs 1.50 and marginal loss is Rs

0.80?

Solution: -

Selling Price = Marginal Loss + Marginal Profit = 0.80 + 1.50 = 2.30

p = ML = 0.80

MP + ML 2.30

p = 0.35

p 0.35

50000 S

Demand

We want p 0.35

Prob[Demand Supply] 0.35

Convert this Normal Distribution to ‘Z’ transform

Z = (S – 50000)

2500

0.38 = (S – 50000) (0.38 has been taken from the Z transform table for

2500 area value of 0.35)

S = 50950

Decision Tree

Eg A company wants to decide whether to construct a large plant or a small plant or

not to enter the business. Available data is as follows:-

Favourable Mkt Unfavourable Mkt

Construction of 200000 -180000

large plant

Construction of 100000 -20000

small plant




No Plant 0 0

If market survey id done the cost is Rs 10000.

Prob(Survey Results are favourable) = 0.45

Expert feel that

(i) All favourable survey results for new products were correct 70% of times.

(ii) Unfavourable survey results for new products were correct 80% of times.

Draw the decision tree and suggest the best action.

- Decision Node (Where is decision making is in our hands)

- Event Node (Result/Out come is unknown)




13 Aug 04

QUANTITAVE METHODS OF MANAGEMENT Mr RS Sardesai

Expected Pay Off of Expected Monetary Value (EMV)

EMV for a decision tree should always be calculated from right to left.

With reference to last problem (last lecture)

Prob[Mkt Fav/Survey Results are fav] =

= P(Mkt Fav) P(Survey Fav/Mkt Fav)

P(Mkt Fav) P(Survey Fav/Mkt Fav) + P(Mkt Unfav) P(Survey Fav/Mkt Unfav)

= (0.50)x(0.70)

(0.50)x(0.70) + (0.50)x(0.20)

= 0.78

Prob[Mkt Unfav/Survey Results are fav] =

= P(Mkt Unfav) P(Survey Fav/Mkt Unfav)

P(Mkt Fav) P(Survey Fav/Mkt Fav) + P(Mkt Unfav) P(Survey Fav/Mkt Unfav)

= (0.50)x(0.20)

(0.50)x(0.70) + (0.50)x(0.20)

= 0.22

Note: Instead of calculating this second probability, this could have been deduced from

earlier calculation by complement method also, ie 1- Probability = 1 – 0.78 = 0.22

Pay off of Node 2 (EMV at 2)

= 0.78x190000 + 0.22x(-190000)

= 1,06,400

EMV at Node 3

= 0.78x90000 + 0.22 x 30000

= 63,600

EMV at event node is computed by summing the products of probabilities and

corresponding monetary values

EMV of decision node is computed by selecting maximum or mininmum EMV of

subsequent branches.




Thus company should go for conducting survey.

If survey results are positive company should go for large plant.

If survey result are negative company should go for small plant.

Q. An investor has two investment options A & B. He can invest in A or B but not in

both simultaneously. The initial investment in A and B is Rs 10000 each. If he invests in A,

but A turns out to be failure, then he can not proceed. Similar is situation for investment in

B. If A is successful, either he can stop or invest in B. Similarly, if B is successful, he can

stop or invest in A. The probability that A is successful is 65% and B will be successful is

45%. He gets a net return of Rs 24000 from A and Rs 20000 from B. If both investments

are independent, what should investor do?

1




EMV at 1 = 0.45 x 20000 + 0.55x (-10000)

EMV at 2 = 0.55 (24000 + 3500) + 0.35 (-10000)

EMV at 3 = 0.55 (24000) –0.35 (-10000)

Internal Assessment

04 Cases to be given later. – 20 Marks

Decision Tree - 01 Case

Linear Programming – 02 Cases

Transportation Model – 01 Case

Internal Test - 30 Marks

Final Test 50 Marks

Syllabus –Topics covered up to 13 Aug 04

Test will comprise of 04 questions

Time – 02 Hours

P(Ai/B) P(B/Ai)

k

P(Ai/B) P(B/Ai)

i =1

Taking last example,

A1 = Supplied by Vendor A

A2 = Supplied by Vendor B From Table

B = Item is found defective

P(A1/B)= 25/60

Compute the probability using Bayes Theorem

P(A1/B) = P(A1) P(B/A1)

P(A1) P(B/A1) + P(A2) P(B/A2)

= (115/260) x (25/115)

(115/260) x (25/115) + (145/260) x (35/145)

= (25/260)

(25/260) + (35/260)

= 25/60

R. A company has a recruitment system for recruiting salesmen. From the past data it

has fond that probability that selected salesman will be successful is 80%. Company

receives a proposal sent by HR consultant recommending a new test that will




increase the probability of selecting successful salesmen. For validating this claim,

the test was administered to the existing sales person of the company. Based upon

their performance following data has been compiled: -

Test Result Performance of

Salesmen Pass Fail

Successful 210 30 240

Unsuccessful 90 60 150

300 90 390

Should company purchase the test?

Sol: -

Let

S= Selected candidate will be successful salesmen

U= Selected candidates will be unsuccessful salesmen

P(S) = 0.80 - A Priori Prob

Q = Candidate qualifies the test

F = Candidate fails the test

P(S/Q) = P(S) x P(Q/S)

P(S) x P(Q/S) + P(U) x P(Q/U)

= 0.80 x (210/240)

0.80 x (210/240) + (0.2) x (90/150)

= (0.80 x 7/8) = 0.85

(0.80 x 7/8) + (9/15)

Next Lecture – Repeat Bayes Theorem, Decision Tree.




27 Aug 04

Quantitative Methods Prof Nimkar

Statistics. Statistics is used for analysis of survey/research data results.

Operation Research. It is a tool for data analysis for optimising objectives.

Data Analysis

Research Purpose – Statistics For optimisation of results (Part of the Research Methodology Sem II Quantitative Techniques in Sem V

1. Linear Programming. Simplex method developed by Dantzig, used by Solver of

MS Excel.

Eg. A company is selling two products A & B. SP are Rs 40 and Rs 35 respectively.

Consumption of resources per unit of A & B and availability of resources are given below:

Item A B Availability

Raw Material (Kg) 2 4 60

Labour 3 3 96

What quantity of A & B should be produced and sold so that total sales are maximum.

Linear programming is used to find a solution that would achieve our objective under

certain constraints.

Step 1. To formulate the objective function and constraints.

Let X1 = Quantity of A to be sold

X2 = Quantity of B to be sold

Objective function = Total Sales (Z) = 40 X1 + 35 X2

Our objective is to maximise Z with respect to following constraints:

2X1 + 4 X2 ≤ 60 (Raw Material constraint)

3X1 + 3 X2 ≤ 96 (Labour Constraint)

X1 ≥ 0, X2 ≥ 0 (Non Negativity)

Equation or inequality of degree one is called Linear programming.




Step 2. To convert inequalities of constraints into equalities and update objective function,

add an imaginary variable in each inequality,

2X1 + 4 X2 + S1 = 60 Where S1 ≥ 0

Similarly

3X1 + 3X2 + S2 = 96 Where S2 ≥ 0

Interpretation of S1 and S2:

1. S1 and S2 are called surplus/slack variables.

2. S1 is the quantity of an artificial/imaginary product such that one unit of S1 will

consume 1 Kg of Raw Material and no labour hr.

3. Similarly S2 is the quantity of another artificial/imaginary product such that one

unit of S2 will require 01 unit of labour hr and no material.

Now update the Objective function

Z = 40 X1 + 35 X2 + 0S1 + 0S2

Because coefficients in this equation are indicating Selling price of the product and Selling

price of S1 and S2 are ‘0’ being imaginary products.

Step 3. To write first simplex table

1. Coefficients of variable in constraints equation are consumption or resources which

depends on technology used to produce product. Therefore, these coefficients are called

technological coefficients. These are denoted by ‘aij’ where suffix i = row and j =

column

2. RHS of constraint equations are denoted by bi. In our example b1 = 60 and b2 = 96.

3. Coefficients of variables in our objective function are denoted by ‘Cj’. In our

objective example C1 = 40 and C2 = 35.

Basic Var Coeff X1 X2 S1 S2 bi bi/aij




03 Sep 04

Quantitative Methods

Prof CY Nimkar

Example – A company is selling 3 products A, B and C. Selling price are Rs 10, Rs 6 and

Rs 4 respectively. Consumption and availability of resources are as follows:

Consumption/unit Availability

A B C

Chemical 1 (Kg) 10 4 5 600

Chemical 2 (Kg) 2 2 6 300

Labour (Hrs) 1 1 1 100

Determine quantities of A, B and C to be sold that will maximise total sales.

Solution – Objective Function is Z = 10 X1 + 6 X2+ 4 X3

Where X1 = Quantity of A to be sold

X2 = Quantity of B to be sold

X3= Quantity of C to be sold

Our objective is to maximise the value of Z subject to following constraints:

10 X1 + 4 X2+ 5 X3≤ 600

2 X1 + 2 X2+ 6 X3≤ 300

X1 + X2+ X3≤ 100

Step 2. Convert inequalities into equalities by inserting imaginary products

10 X1 + 4 X2+ 5 X3+ S1 = 600

2 X1 + 2 X2+ 6 X3+ S2= 300

X1 + X2+ X3 + S3 = 100

S1, S2 and S3 are all ≥ 0

So Z = 10 X1 + 6 X2 + 4 X3 + S1 + S2+ S3

Step 3 Basic Variable Coeff in

objective fn

X1 X2 X3 S1 S2 S3 bi bi/aij

S1 0 10 4 5 0 0 0 600 60

S2 0 2 2 6 0 0 0 300 150

S3 0 1 1 1 0 0 0 100 100




Cj 10 6 4 0 0 0 0

Zj 0 0 0 0 0 0 0

Cj – Zj 10 6 4 0 0 0 0

Incoming variable = (Cj – Zj). Pick the highest value of variables among (Cj – Zj)

Column containing incoming variable is called pivotal column

bi/aij = Technological Coeff / pivotal column

bi/aij indicates the maximum no of items that can be manufactured using that resource.

Lowest positive variable in bi/aij is called outgoing variable.

The row containing outgoing is called pivotal row.

The intersection point element is called pivotal row

The intersection point element is called pivotal element.

Basic Variable Coeff in

objective fn


X1 10 1 2/5 ½ 1/10 0 0 60 150

S2 0 0 6/5 5 -1/5 1 0 180 150

S3 0 0 3/5 ½ -1/10 0 1 40 67

Cj 10 6 4 0 0 0 0

Zj 10 4 5 1 0 0 600

Cj – Zj 0 2 -1 -1 0 0 -600

In new table replace out going variable with incoming variable row

The construct pivotal row of last table in new table by dividing with pivotal element.

Second row – (Element at corresponding pos

n from old table) –

(Element at ⊥ posn

on pivotal row) x (Element at ⊥ posn on pivotal column)

Pivotal element

= 2 – 4 x 2

10

= 6/5

= 6 – 5 x 2 = 5

10

= 0 – 2 = -1/5

10

= 300 – 2 x 600 = 180

10

= 2 – 2 x 10 = 0

10

= 1 – 4 x 1 = 3/5




10

= 1 – 5 x 1 = ½

10

= 0 – 0 x 1 = 0

10

Chemical 1 Chemical 2 Labour

2/5 X1 4 4/5 2/5

6/5 X2 6/5 3/5

4 2 1

Check whether the table is table of optimal solution by applying test for optimality. The

test for optimality is the value of (Cj – Zj) should be ‘0’ or ‘-‘ ve.

If the table is not the optimal solution, select the highest +ve value of (Cj – Zj).

Variable corresponding to this value will be incoming varible. Repeat the further step till

we get optimal solution.

Basic Variable Coeff in

objective fn


X1 10

S2 0

X3 6 0 1 5/6 -1/6 0 5/3 66.7

Cj

Zj

Cj – Zj

Optimal solution table is

Basic

Variable

Coeff in

objective fn


X2 6 0 1 5/6 5/3 -1/6 0 200/3 100/3

X1 10 1 0 1/6 -2/3 1/6 0 100/3 100

S3 0 0 0 4 -2 0 1 100 0

Cj 10 6 4 0 0 0 0

Zj 10 6 20/3 0 10/3 2/3 2200/3

Cj – Zj 0 0 -8/3 0 -10/3 -2/3 -2200/3

Interpretation of the optimal solution table:

1. Solution is X2= 200/3, X1 = 100/3, X3= 0

Z = 10 x (100/3) + 6 x (200/3) + 4 x (0)




= 2200/3 bi

2. Cj – Zj Value of slack variable. If Cj – Zj value of any slack variable is ‘0’, it means

that we have surplus of that resource.

3. If value is negative means those resources are consumed fully.

4. | Cj – Zj | value is called opportunity cost/shadow price/net worth of the resource. It

means, if resource availability is increased by 1 unit, value of Z will increase by a

quantity equal to opportunity cost of the resource.

Transportation Model

This model give the transportation route for supplying material from ‘n’ supply

points to ‘n’ destinations such that : -

(a) The constraints on no of units that can be supplied freom each supply point

are met.

(b) The demand at each destination is met.

(c) The total cost of transportation is lowest.

Eg. Following is the data on cost of supplying one unit to various ldestinnations from

supply points, requirements at each destination, quantitiers that would be supplied from

each supply oint. Find out the transportation route of lowest transportation cost.

Destination

Supply Point D1 D2 D3 Supply

S1 3 2 4 20

S2 4 3 6 35

S3 1 4 5 40

Demand 35 40 20 95

Solution.

Step 1. To check whether total demands = total supply. When this is true, the problem is

called Balanced Transportation Problem. If so, go to the nest step.

Step 2. To arrive at the first feasible solution (It may be correct but not the best solution) by

least penalty cost (LPC) method. Feasible solution is that solution which satisfies the

constraints on supply on supply and demand but not necessarily the most economical

solution.

D1 D2 D3 Supply LPC

S1 3 2 4 20

S2 4 3 6 35

S3 1 4 5 40

Demand 35 40 20




LPC




24 Sep 04

Quantitative Techniques Mr Nimkar

Assignment Model

The model deals with optimum assignment

J1 J2 J3 J4 J5

W1 10 6 7 5 9

W2

W3

W4

W5

Wi = Worker

Ji = Job

It is a model that gives assignment to worker of jobs in such a way that

a. Each worker is assigned only one job Constraints

b. Each job gets only one worker

c. Total cost/time of assignment is least - Objective

Eg. The following matrix is time in hours required by 4 workers to finish 4 jobs. Find

assignment of each worker to jobs such that each worker is assigned only one job., each job

gets only one worker and total time of assignment is least.

J1 J2 J3 J4

W1 5 3 4 3

W2 3 2 1 4

W3 3 3 1 2

W4 4 1 3 4

Solution Step 1. To check whether problem is balanced

Are No of rows = no of columns ?

If problem is balanced, go to step 2.

Step 2. To find lowest element form row and deduct from all elements of that row. Such a

matrix is called Row Minima matrix.

J1 J2 J3 J4

W1 2 0 1 0

W2 2 1 0 3

W3 2 2 0 1

W4 3 0 2 3




Step 3. To consider row minima matrix find minimum element of each column and deduct.

Such a matrix is called “Column Minima matrix.

J1 J2 J3 J4 Tot time

W1 0 0 1 0 3

W2 0 1 0 3 3

W3 0 2 0 1 1

W4 1 0 2 3 1

All the Zeros denote optimum allocation. There will be at least one 0 in each row

and one 0 in each column.

In column minima matrix there must be at least one zero in each row and one zero

in each column.

To find the optimal solution we must follow the following steps: -

d. Priority must be given to the row and columns that contain only one zero. Give

allocation to such zeros first. For this purpose we first start considering rows.

Identify the row that contains only one zero. Give allocation to this row zero. Once

allocation is given, any zeros in the column where allocation is given are cancelled.,

Once the operation in the rows is over, we perform similar operation on columns.

For this purpose identify columns with only one zero. Give allocation to this zero.

After giving allocation, any zeros in the row where allocation is given are

cancelled.

e. Once allocation to priority rows and columns are over, rest of the allocation is

given by judgement.

Eg.

J1 J2 J3 J4

W1 7 8 5 7

W2 6 7 3 5

W3 12 11 10 12

W4 3 7 2 5

Row Minima

J1 J2 J3 J4

W1 2 3 0 2

W2 3 4 0 2

W3 2 1 0 2

W4 1 4 0 3

Column Minima




J1 J2 J3 J4 Tot Time

W1 1 2 0 0 5

W2 2 3 0 0 5

W3 1 0 0 0 11

W4 0 3 0 1 3

Eg.

J1 J2 J3 J4 J5

W1 7 5 12 6 11

W2 4 12 5 3 9

W3 12 7 11 6 10

W4 14 15 15 10 12

W5 7 3 4 9 15

Row Minima

J1 J2 J3 J4 J5

W1 2 0 7 1 6

W2 1 9 2 0 6

W3 5 0 4 2 3

W4 4 5 5 0 2

W5 4 0 1 6 9

Column Minima

J1 J2 J3 J4 J5

W1 1 0 6 1 4

W2 0 9 1 0 4

W3 4 0 3 2 1

W4 3 5 4 0 0

W5 3 0 0 6 7

Eg

J1 J2 J3 J4 J5

W1 12 6 11 10 14

W2 17 5 6 9 13

W3 14 3 4 12 12

W4 11 9 15 9 10

W5 13 7 13 16 5

J1 J2 J3 J4 J5

W1 6 0 5 4 8

W2 12 0 1 4 8




W3 11 0 1 9 9

W4 2 0 6 0 4

W5 8 2 8 11 0

J1 J2 J3 J4 J5

W1 4 0 12 4 8

W2 10 0 0 4 8

W3 9 0 0 9 9

W4 0 0 5 0 4

W5 6 2 7 11 0

Since J4 job is not allotted to any worker and W3 is not allotted any job, this not the

optimal solution. So we go to next step.

Step. To draw minimum no of straight lines to cover all zeros. For this purpose follow the

following steps: -

i. Tick the rows or rows where allocation is not given.

ii. Consider this row and tick the columns in which zeros are appearing.

iii. Consider the columns and tick the row(s) where allocation is given.

iv. Repeat steps (ii) and (iii) till the process stops, cancel un-ticked rows and

ticked column.

Step. Find the lowest element form uncovered element (Elements not cancelled by

straight lines). (Here it is 4.

Deduct this element from all uncovered elements and add it to those elements that lie on

intersection of straight lines. All other elements will remain same.

J1 J2 J3 J4 J5

W1 0 0 4 0 4

W2 6 0 0 0 4

W3 5 0 0 5 5

W4 0 4 9 0 1

W5 6 6 11 11 0

There are more than one optimal solutions for this problem.

Unbalanced Assignment Problem.

When no of rows is not equal to no of columns, the problem is unbalanced. In such a case

we first balance the problem by adding dummy row/column.




Eg J1 J2 J3 J4

W1 7 8 5 7

W2 6 7 3 5

W3 12 11 10 12

Wd 0 0 0 0

J1 J2 J3 Jd

W1 7 8 5 0

W2 6 7 3 0

W3 12 11 10 0

W4 3 7 2 0

In the final solution, there will be some worker who will be assigned dummy job, so that

worker will actually be unassigned. Similarly, there will be unassigned job

Eg.

J1 J2 J3 J4

W1 5 7 8 4

W2 6 9 5 4

W3 8 12 10 5

Wd 0 0 0 0

Solution

Row Minima

J1 J2 J3 J4

W1 1 3 4 0

W2 2 5 1 0

W3 3 7 5 0

Wd 0 0 0 0

J1 J2 J3 J4

W1 0 2 3 0

W2 1 4 0 0

W3 2 6 4 0

Wd 0 0 0 1




10 Sep 04

Quantitative Techniques In Management

Prof CY Nimkar

Eg. Following is the cost of sending one unit of a product from soucing point to various

destinations. The demand at destinations and supply quantity from sourcing points are also

given. Find out the optimal transportation route.

D1 D2 D3 D4 D5 Supply

S1 12 4 9 5 9 55

S2 8 1 6 6 7 45

S3 1 12 4 7 7 30

S4 10 15 6 9 1 50

Demand 40 20 50 30 40 180

Solution.

Step 1. Check if the problem is balanced. Is demand = Supply?

Yes? Go to next step.

Step 2. Find first feasible solution by LPC method

D1 D2 D3 D4 D5 Supply LPC

S1 12 4 9 25

5 30

9 55 25 1 4 3

S2 8 10

1 20

6 15

6 7 45 35 5 0 2

S3 1 30

12 4 7 7 30 3

S4 10 15 6 10

9 140

50 10 5 3 4

Demand 40 10

20 50 30 40 180

7 3 2 1 6

2 3 0 1 6

4 3 3




26 Sep 04

Queuing Theory Mr Moradian

Following four topics will be covered by Mr Moradian on successive Sundays. However,

he will not be available from mid Oct to early Nov and therefore last lecture will be on 9

Nov 04.

1. Queuing Theory – Waiting Line Model

2. Simulation

3. Decision Analysis Min 02 Questions in exam from

4. Decision Tree these four topics

Queuing Theory

Service Station

Unit being serviced

Units in

System Units in ‘Q’

Arrivals

Check Off list while attempting any problem (real or exam)

1. Calling Population – Finite or infinite

2. Arrivals in system – Single

Batch (Like a group in a hotel)

- Constant

- Varying

3.




01 Oct 04

Quantitative Methods in Management

Mr CY Nimkar

Use of Solver to slove LP, Transportation and Assignment problems.

The Clark’s County Sherif’s department schedules police officers for 8 – hour shifts. The

beginning times for the shifts are 8 AM, noon, 4 PM, 8 PM, midnight and 4 AM. An officer

beginning a shift at one of these times works for next 8 hours. During normal weekday

operations, the number of officers needed varies depending on time of day. The department

staffing guidelines require following minimum number of officers on duty:

Time of day Minimum officers on duty

8 AM – Noon 5

Noon- 4 PM 6

4PM – 8PM 10

8PM-midnight 7

Midnight-4AM 4

4AM-8AM 6

Determine number of police officers that should be scheduled to begin 8-hour shifts at each

of the six times (8AM, noon, 4PM, 8PM, midnight, 4AM) to minimize total number of

officers required.

Solution.

Let X1 = No of officers starting duty at 8:00 AM

X2 = No of officers starting duty at 12 Noon

X3 = No of officers starting duty at 4:00 PM

X4 = No of officers starting duty at 8:00 PM

X5 = No of officers starting duty at midnight

X6 = No of officers starting duty at 4:00 AM

Objective function = Z = X1 + X2 + X3 + X4 + X5 + X6

Subject to X1 + X2 + X3 + X4 + X5 + X6 >= 0

8--12 12--4 4--8 8--12 12--4 4--8

X1 X1

X2 X2

X3 X3




X4 X4

X5 X5

X6 X6

5 6 10 7 4 6

Component of work sheet

1. Matrix of aij (Coefficient of constraints)

2. Value of bi (RHS of constraints equations)

3. Cell address where optimum values of X1 – X6 will be stored.

4. Cell address where formula for Z is stored

5. Formula for LHS of constraint equation

Transportation Problem

D1 D2 D3 D4 D5 SUPPLY

S1 20 36 41 19 22 406

S2 18 15 21 19 27 450

S3 16 18 10 25 22 550

S4 19 22 25 27 21 300

DEMAND 100 200 300 150 100

1. Reproduce the matrix as it is.

Step 1. Prepare the matrix containing Cij, demand and supply.

Step 2. To check whether problem is balanced. Check if Demand = Supply. Here demand =

850 and supply = 1700.

Step 3. Select the cells where solution will be stored. (Copy the matrix table).

Step 4. Give the formulas for the constraints.

Step 5. Formula for Z = total cost of transport. Use “Sumproduct” function in formula in

Excel.

Assignment Problem.

J1 J2 J3 J4 J5

W1 10 9 14 12 24 1

W2 16 17 25 18 27 1

W3 15 22 20 22 30 1

W4 18 11 10 19 12 1

1 1 1 1 1

Treat it like a transportation problem where in each supply and demand point has only 1

unit to demand and supply.




Book – Introduction to Management Science by Anderson Sweeney and Williams.




08 Oct 04

Quantitative Techniques Mr Nimkar

Topics for Examination

1. Maximisation (under “≤” type constraints)

2. Transportation Models (Balanced/Unbalanced)

3. Assignments (Minimisation – Balanced/Unbalanced)

4. Decision Tree analysis(EMV, EVPI, EVSI)

5. Bayes Theorem

6. Sampling Probability Distributions

7. Inventory

8. Simulation – Monte Carlo Techniques

9. Queuing Theory

Inventory Models. – Objective is to reduce total inventory carrying cost.

Components of total inventory cost

(a) Holding cost

(b) Ordering cost

(c) Stock Out cost

There are two approaches of solving inventory management problems: -

(a) Mathematical Modelling approach

(b) Simulation – This model is used when Mathematical Model is not feasible

due to complexity of the problem.

For solving the Mathematical model, we use Minima of function (Total Cost)

Various approaches available to categorise the items.

(a) ABC Approach (Based on Annual consumption on monetary basis)

(b) VED Analysis (Based on Criticality of the item for production process)

(c) HML Analysis (Based on Cost of the item)

(d) S-OS Analysis (When item is required)

(e) FSN Analysis (Based on consumption rate of the item)

ABC Approach.

Step 1. To find the unitwise annual consumption in Rs.

Item Code Item Description Annual Consumption




Step 2. To prepare listing of items in descending order of annual Rs consumption

Item Code Item Description Annual Consumption Cumulative %

58 58/1000

42 (58+42)/1000

--

--

--

--

1000

100%

90%

80%

Cum %

A C

B

Items

VED Approach.

V- Vital (Production will stop in case of absence of this item)

E- Essential (Production though will not stop, capacity will get affected) D- Desirable (Production will not get affected immediately. But it may have

effect on production process in the long run)

HML Approach

H – High unit cost

M – Medium unit cost

L – Low unit cost

S-OS

S- Seasonal

OS – Off seasonal

FSN Approach

F – Fast moving

S – Slow moving

N – Non moving

Inventory Systems




- Fixed Order System (‘Q’ System) Order quantity is fixed and time of placement

of order is variable.

- Fixed Time System – Interval of orders is fixed. Order Quantity varies each

time depending on the consumption of items during the period.

Fixed Order System models.

(i) Economic Order Qty (EOQ) without price breaks.

Let A = Cost of placing one order

D = Consumption in units in 1 year

H = Cost of stocking 1 unit per annum

Q = Qty to be ordered

Total cost = total ordering cost + total holding cost

= (D) A + Qh

Q 2

Since we want to find ‘Q’ that will minimise total cost,

∴We differentiate that total cost function wrt Q and equate first order

derivative to zero and then check the sign of second order derivative.

d (total cost) = DA(-1)(Q)-2

+ h/2 (1) = 0

dQ

= -DA + h/2 = 0

∴ DA = h

Q2 2

2DA = Q2

h

Q = √2DA

√h

Second derivative

d2 (total cost) = (-DA) (-2Q

-3)

dQ2

= 2DA > 0

Q3

2. EOQ with price breaks

e.g. The purchased price of an item is given below: -

Qty to be ordered in 1 order Price limit

Q < 300 10.00

300 ≤ Q < 600 9.90

600 ≤ Q < 1000 9.70

1000 ≤ Q <1500 9.50

Q ≥ 1500 9.40

Cost of placing one order = Rs 2800/-

Demand per annum = 25000 units




Cost of stocking one unit per annum = Rs 550/-

Find EOQ.

Solution.

Total cost = Cost of ordering + cost of holding + cost of purchase

Q Cost Total Cost

Ordering Holding Purchase

280 250000 77000 250000 577000

300 233333 82500 247000 563333

600 116667 165000 242500 524167

1000 70000 275000 237500 582500

1500 46667 412500 235000 694167

Periodic Review System. In this system, stock is reviewed at regular intervals.

Let D = Demand in units per annum

A = Cost of placing one order

H = Cost of holding one unit per annum

T = Time interval expressed as % of year between two successive orders.

Solution

Total cost = total ordiering cost = total holding cost

Here we assume that annual demand is uniformly distributed over year

∴ Demand during time T = DT

No of orders in one year =D/DT = 1/T ---------- (1)

Annual ordering cost = (1/T) A = A/T

Average inventory = (DT/2)h ---------- (2)

Total cost = A/T + DTh/2

Differentiating wrt T we get

D(Total Cost) = - A/T2

+ Dh/2 = 0

dT

A/T2 = Dh/2

T2 = 2A/Dh

T = √ 2A

√ Dh

Practice Problem

A TV manufacturing company requereds 20000 TV tubes per annum. Cost of stocking one

tube per annum is Rs 240 and cost of placing order is Rs 1500. If the consucptionis uniform,

find optimum time to review the inventory periodically.

Solution

T = √2X1500

√2000 X 240

= 30

20X240




= 0.025 X 365 (FOR CONVERTING TO PER ANNUM)

= 9.125

≈ 9 or 10 days as chosen by you. Both answers would be considered correct.

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