MBA (IB)

MANAGEMENT SCIECNE

Compiled Reading material

Compiled by

Dr. H.K.Dangi

Video lecture available on

http://www.youtube.com/c/HAMENDRADANGI

Please post your query on – hkd2studets@gmail.com

GAME THEORY

GAME :- Game refers to the general situation of conflict and competition in

which participants are engaged in decision – making activities in anticipation of certain outcomes overtime.

Characteristics of Game :

1. Number of participants, (players) in any competitive situation is finite.

2. The number of alternatives, (moves) available to each of participant is finite.

3. Every participant has the knowledge about the alternatives available to his opponent.

4. Every combination of courses of action determine an outcome which result in a gain of each player.

5. All players are equally wise and behave rationally.

6. Each player attempts to maximize his gain or minimize loss.

7. Each player makes individual decision.

8. There exist set of rules according to which pay-off will be determined.

SOME DEFINITION

i) Strategy :- A rule or a set of rules on the basis of which a player select his alternative. The strategy may be of two kinds:

(a) Pure Strategy :- A pure strategy is a decision to select the same

course of action. (b) Mixed Strategy :- Mixed strategy is a decision to select more than

one of course of action with fixed probability. ii) Value of Game :- The value of game is expected gain of player A if boll

player use their optimal strategy. iii) Solution of Game :- The solution involves:

(a) Optimal strategy for A (b) Optimal strategy for B (c) The value of Game

TWO PERSON ZERO SUM GAME Two person zero sum game are games played by two persons, parties or group, with directly opposing interest one person’s gain is exactly equal to loss of another person and therefore sum told of the gains and losses equal to zero. PURE STRATEGY & SADDLE POINT Steps

(i) Enclose in a circle, minimum entry of each row. (ii) Enclose in a square, maximum entry of each column. (iii) If there is an entry which has been enclosed in a circle as well as in a

square, then it indicates the existence of saddle point in a game.

A: Tries to Maximize (Attacking Player) B: Tries to minimize (Defensive Player) B1 B2 B3

A1 15 14 16 Row m1

A2 12 13 17 14*

A3 14 13 10 12

10

Col. 15 14* 17

Maxima

Equilibrium (A1, B2)

1

7

5 3

1 0

-4 3

Find the solution of following game

B1 B2 B3

A1 -1 2 10

A2 1 3 11

A3 0 4 2

A4 0 5 2

Optimal strategy for A is A2

Optimal strategy for B is B1

Value of game = 1

Mixed Strategy :- Game without Saddle Point ODDOMENT METHOD Step 1 :- Subtract the smaller payoff in each row (each column) from larger payoff write these result against column(row). R1 1 7/8 R2 7 1/8 5 3

3/8 5/8

B A Step 2 :- Interchange each of these pair of subtracted number found is

Step I

B

1 0

-4 3

1 0

-4 3

1 0

-4 3

A 1 7

7 1

5 3

3 5

Step 3 :- Put each of inter changed numbers over the sum of pairs of numbers as

B

A 1 7/7+1 (7/8)

7 1/7+1 (1/8)

5 3

(3/8) 3/5+3 5/5+3 (5/8)

Eg. Two players A and B each put down a coin. If coin matches i.e., both are heads or both are tails. A gets rewarded otherwise B. However matching on heads gives a double premium. Obtain the best strategy for both player and value of game. Head Tail Head A Tail Step 1 3 2 3 2 Step 2

2 -1

-1 2

2 -1

-1 2

2 -1

-1 2

1 0

-4 3

3 2 2 3 3 2 2 3 Step 3 3 2/3+2 2 3/3+2 3 2 2/3+2 3/3+2 Optimal strategy for A (2/5, 3/5) Optimal strategy for B (2/5, 3/5) Value of game = 2 x 2/6 + (-1) x 3/5 = 1/5 Note 1 :- This arithmetic is very useful because it is less complex. But this technique can be applied only for solving 2 x 2 games. Note 2 :- This technique can be used only when sum of row differences and column difference obtained in step1 (called oddome) is same. DOMINANCE IN GAMES General rules 1. If each element of row (say ith row) of the payoff matrix is less than or equal

to corresponding element in another row (or average of the corresponding element of two rows) then ith strategy of row player is said to be dominated and row player will never use ith strategy, therefore ith strategy (ith row) can be delete from payoff matrix of the game.

2. If each element of a column (say ith column) of the payoff matrix is greater

than or equal to corresponding element in another column (average of corresponding element of two or more columns) then column player will never use jth strategy.

2 -1

-1 2

Example: B1 B2 B3 B4

A1 42 72 32 12

A2 40 30 25 10

A3 30 8 -10 0

A4 45 10 0 15

Solution

Comparing A1 and A2. A2 is dominate By A1 A1 42 72 32 12

A3 30 8 -10 0

A4 45 10 0 15

Comparing A3 and A4. A3 is dominate By A4 A1 42 72 32 12

A4 45 10 0 15

Again comparing B1 with B3and B2with B3, B1 and B2 are dominated by B2 the reduced matrix will be B3 B4 A1 20 15/35 A4 15 20/35 32 3 3/35 32/35 Optimal strategy for A (15/35, 0, 0, 20/35) Optimal strategy for B (0, 0, 3/35, 32/35)

32 12

0 15

V = 32 x 15 + 0 = 96 35 7 Solve the game graphically (2 x n) B1 B2 B3 B4

A1 2 1 0 -2

A2 1 0 3 2

A1 A2 B’s for A’s Payoff 3 3 B1 2(p) + 1 (1-p) = p +1 2 2 B2 1.p(+ 0 (1-p) = p 1 1 B3 0.p + 3 (1-p) = -3p + 3 0 0 B4 -2.p + 2 (1-p) = -4p + 2 -1 -1 -2 -2 A’s Maximin B2 B4 A1 3 2/5 A4 2 3/5 1 4 SA (2/5, 3/5) SB (0, 4/5, 0, 1/5) 4/5 4/5 V = 2/5

1 -2

0 2

EXERCISE NO 2

1 What do you understand by pure and mixes strategy?

2 Write critical note on Game theory?

3 Solve the game :

Strategy B1 B2 B3 B4

A1 8 10 9 14

A2 10 11 8 12

A3 13 12 14 13

4 Solve the Game

S T

P -3 7

Q 6 1

5 Solve the Game

X1 X2

Y1 1 4

Y2 5 3

6 Solve the Game

B1 B2

A1 1 -5

A2 4 2

7. Solve the game graphically

Player 2

Player 1 0 -2 2

5 4 -3

8. Find the value of game

Strategy B1 B2 B3

A1 -5 10 20

A2 5 -10 -10

A3 5 -20 -20

CASE

Munish Nanda

Present situation

1. Colonel I M Fannekhan an ex student of MBA(DM) course of GGSIP university

Delhi is now posted to a location on the Line of Control in the State of J&K. He has been

entrusted with the task to prevent infiltration in two sectors from across the border by

insurgents apart from the normal activities to maintain the sanctity of a highly active line

of control.

2. He finds he is able to carry out effectively the activities required on the line of

control with his Battalion less two companies. Two companies are available to prevent

infiltration in the given sectors A and B. One coy is deployed to prevent infiltration in

Sector B but Sector A poses problems due the given terrain configuration (a sketch of

Sector A is attached as Appendix).

3. Colonel IM Fannekhan decides to examine the problem confronting him applying

the decision making techniques learnt in the MBA course.

Statement of Problem

4 The security forces due to strength limitation can patrol only one route effectively

5. The security forces estimates that insurgent infiltrate in the groups of 100 over

route A and in group of 80 over route B An encounter results not only arrest on insurgent

on route but also in the arrest of several others due to information provided.

6. When security forces patrol route A, encounter leads to 70 arrests at an average

.Where as encounter on route B result in 60 arrests. Security forces intelligence set up

estimate that only 40% of insurgent on route A and 25 % on route B are intercepted when

security force patrol route A and B respectively

Determine optimal strategy

REPRESENTATIVE SKETCH OF SECTOR ‘A’

(MOUNTAINS NOT SHOWN)

PAKISTAN INDIA

LINE OF

CONTROL

ROUTE 1

INFILTRATION-100

ARRESTS-70

WATER BODY

ROUTE 2

INFILTRATION-80

ARRESTS-60

Assignment Problem

The assignment problem is special case of linear programming problem. It involves the

allocation of various productive resources having different efficiency to various task that

are to be completed.

Person Jobs

J1 J2 J3 Jn

P1 C11 C12 … C1n

P2 C21 C22 … C2n

P3

Pn Cn1 Cn2 … Cnn

Mathematical statement of problem

Let the decision variable Xij represent the value of assignment of the ith person to jth job.

The value of Xij is either 1 or 0, if the jth person is assigned to the ith job then value of Xij

is 1 otherwise 0.

n n

Minimize Z = Cij Xij

C=1 J=1

n n

Subjected to Xij = 1 = Xij

C=1 J=1

and Xij = 0 or 1 for i & j

Solution method for an assignment problem

(i) Enumerate method.

(ii) Simplex method.

(iii) Transportation method.

(iv) Hungarian method.

Hungarian Method: “D. Koing”

“If in an assignment problem we add or subtract, a constant to every element of a row (or

column) of the matrix, then an assignment plan which minimize the total cost matrix for

new matrix, also minimize the total cost for the original cost matrix”. Based on this

principle a computational technique known as Hungarian method was developed.

Minimization Case

Step 1: Develop the cost table for given problem if the number of rows not equal to number

of column or vice-versa dummy row or dummy column must be added. The assignment

cost of dummy row or column is always zero.

Step 2: Subtract the least entry of each row from all the entries of that row thus we will

have atleast one zero in each row of new table (first reduced cost table).

Subtract the least entry of reduce cost table from all the entries of that column.

Step 3: A minimal set of straight line are down horizontally & vertically to cover all the

zero of total opportunity cost matrix. If number of lines are exactly equal to number of

rows in matrix an optimal assignment can be obtained if not obtained then proceed as

follows.

Step 4: Select the smallest entry from all the entries which has not been covered by the

straight line. Subtract this smallest entry from all the entries of table which are not covered

by straight lines & add this smallest entry to all those entries which are intersection of two

lines.

All other elements covered by one straight line remain unchanged in the revised matrix.

Again apply optimality test to this revised total opportunity cost matrix by drawing straight

lines.

Step 5: Repeat step 3 & 4 until number of covering line become equal to row or column.

Step 6: (A) Starting from top row examine row successively until a row with exactly one

unmarked O is obtained. Make assignment to this single zero by making square ( )

around it and cross all the zero in the column of that zero.

(B) When all the row have been examine by this way then starting from left examine

column successively until a column with unmarked zero is obtained make an assignment

by making square around it and cross all the zero is that row.

(C) If a row or column has two or more unmarked zero then choose the zero for assignment

arbitrarily.

Example: Find the assignment that minimize cost.

Machine Jobs

J1 J2 J3 J4 J5

M1 10 4 5 3 11

M2 13 11 9 12 10

M3 12 3 10 1 9

M4 9 1 11 4 8

M5 8 6 7 3 10

Solution

Step 1: The cost matrix is given

Step 2: Subtract least entry from each row

7 1 2 0 8

4 2 0 3 1

11 2 9 0 8

8 0 10 3 7

5 3 4 0 7

Subtract least entry from each column

3 1 2 0 7

0 2 0 3 0

7 2 9 0 7

4 0 10 3 6

1 3 4 0 6

Step 3: We draw minimum number of line to cover all zeros since three line cover all zero

an optimal assignment possible at this stage.

Step 4: Select the smallest entry not cover straight line. Subtract this entry from the

uncovered entries and add it all those entry which are at interests.

2 0 1 0 6

0 2 0 4 0

6 1 8 0 6

4 0 10 4 6

0 2 3 0 5

Repeat step No. 4 as number of lines are only 4

1 0 0 0 5

0 3 0 5 0

5 1 7 0 5

3 0 9 4 5

0 3 3 1 5

1 0 0 0 5

0 3 0 5 0

5 1 7 0 5

3 0 9 4 5

0 3 3 1 5

Machine 1 Job 3 Cost Rs. 5

Machine 2 Job 5 Cost Rs. 10

Machine 3 Job 4 Cost Rs. 1

Machine 4 Job 2 Cost Rs. 1

Machine 5 Job 1 Cost Rs. 8

Total: Rs. 25

Alternative Solution

Eg A department head has five subord and five jobs to be done. The subord differs in

efficiency and jobs differs in their intrinsic difficulty. The estimates of times each man

would take to perform each is given in effectiveness matrix. How should task be allocated

to min. total man hours.

Job I II III IV V

A 1 3 2 3 6

B 2 4 3 1 5

C 5 6 3 4 6

D 3 1 4 2 2

E 1 5 6 5 4

Step I: Subtracting least value in each row.

0 2 1 2 3

1 3 2 0 4

2 3 0 1 3

2 0 3 1 1

0 4 5 4 3

Step II: Subtracting least value in each column & covering by lines.

0 2 1 2 4

1 3 2 0 3

2 3 0 1 2

2 0 3 1 0

0 4 5 4 2

Step III: Add least entry at intersection, subtracting from uncovered column.

0 1 0 2 3

1 2 1 0 2

3 3 0 2 2

3 0 3 2 0

0 3 4 4 1

Step IV: Final step.

0 0 0 1 2

2 2 2 0 2

3 3 0 1 1

4 0 4 2 0

0 2 4 3 0

1. 2.

0 0 0 1 2 0 0 0 1 2

2 2 2 0 2 2 2 2 0 2

3 3 0 1 1 3 3 0 1 1

4 0 4 2 0 4 0 4 2 0

0 2 4 3 0 0 2 4 3 0

A I Time 1 hr A II Time 3 hrs

B IV 1 hr B IV 1 hr

C III 3 hrs C III 3 hrs

D II 1 hr D V 2 hrs

E V 4 hrs E I 1 hr

10 hrs 10 hrs

Maximizing Case

The transformation involves subtracting all the entries of the original pay off table from

the maximum pay off table from maximum entry of that table.

Unbalanced Assignment Problem

If the number of person is different from number of jobs the assignment problem is said to

be unbalanced. If the number of job is less than number of person some of the person cannot

be assigned any job. We introduce me or more dummy jobs of zero duration. On the other

hand if number of persons are less than number of jobs we add one or more dummy person

with zero duration.

eg: Four different Airplanes are to be assigned to handle three cargo. With a view to

maximize profit. The profit as follows (in thousands of rupees).

Airplane Cargo

I II III

W 8 11 12

X 9 10 10

Y 10 10 10

Z 12 8 9

Find the optimal assignment

Step 1: Convert it to minimize (By subtracting from highest value in profit matrix).

I II III

W 4 1 0

X 3 2 2

Y 2 2 2

Z 0 4 3

Step 2: Balancing the matrix by adding dummy column.

I II III Dummy

W 4 1 0 0

X 3 2 2 0

Y 2 2 2 0

Z 0 4 3 0

Step 3: Apply row & column reduction as mentioned in previous question.

4 0 0 0

3 1 2 0

2 1 2 0

0 3 3 0

5 0 0 1

3 0 1 0

2 0 1 0

0 2 2 0

Case I Case II

5 0 0 1 5 0 0 1

3 0 1 0 3 0 1 0

2 0 1 0 2 0 1 0

0 2 2 0 0 2 2 0

Profit

W III 12,000 W III 12,000

X II 10,000 X Dummy 0

Y Dummy 0 Y II 10,000

Z I 12,000 Z I 12,000

34,000 34,000

EXERCISE

1. Explain the Hungarian Method of Obtaining Optimal solution in Assignment

Problem?

2 Solve the following Assignment problem

A.

Jobs/ Machine D1 D2 D3

O1 20 27 30

O2 10 18 16

O3 14 16 12

B.

I II III IV

A 8 26 17 11

B 13 28 4 26

C 38 19 18 15

D 19 26 24 10

C.

A1 A2 A3 A4

C1 15 29 35 20

C2 21 27 33 17

C3 17 25 37 15

C4 14 31 39 21

Q.No 3 Solve the following Maximization problem by Assignment Problem

Q.No 4 Solve the following assignment problem:

A B C D E

1 30 37 40 28 40

2 40 24 27 21 36

3 40 32 33 30 35

4 25 38 40 36 36

5 29 62 41 34 39

1 2 3 4 5

1 5 8 12 4 9

2 5 11 12 11 7

Q No 5 Solve the following assignment problem:

Job/ machine W X Y Z

A 18 24 28 32

B 8 13 17 18

C 10 15 19 32

Q.No 6 Four salesman are to be assigned four district Estimate of the sales return in

hundred of rupees for each are as under

Salesman/

District

A B C D

1 320 350 400 280

2 400 250 300 220

3 420 270 340 300

4 250 390 410 350

Q.No 7 Solve the assignment problem for minimization

1 2 3 4

A 20 25 22 28

B 15 18 23 17

C 19 17 21 24

D 25 23 24 24

Q.No 8: Define/ differentiate the following term

a. Balanced assignment problem

b. Unbalanced assignment problem

c. Dummy job/facility

d. An infeasible assignment

e. Assignment and Transportation model

3 9 14 18 7 6

4 14 15 11 12 8

5 8 9 8 11 7

CASE P.K Sinha

Background:

The Office of Divisional Commissioner is involved in various works where there is

always a need to arrive at a decision which is accurate time effective and keeps the

money expenditure under the control, Here we have recognized a major event conducted

by this department, i.e. to conduct Kawar Camp and Urs transit camp in which we call for

tenders and thus award work to each firm based on the rates quoted.

Present procedure:

The procedure followed at present tales only into account the rates quoted by the firm and

the work is awarded invariably only to that has quoted the lowest rate. Since, in this way

if the same firm has quoted rates, which is lowest in al sectors of work, he is awarded the

work for all the work.

Merit: This minimizes the expenditure of the Government

Demerit: The limitation of the firm in implementing the work simultaneously is not taken

in consideration.

Proposed Procedure:

Thus to overcome this difficulty and to keep the cost at its minimal we can award one

work to each firm and decision regarding this could be made by applying the method of

Quantitative techniques to arrive at the most economical decision.

Assignment problem: The method that could be used here fruitfully to arrive at a

decision is detailed as under.

The rate of firms A, B, C & D and the kind of work to be taken by each of the firm is also

given in the table

Name of firm /

Works

A B C D

Putting up tent

age items

1200 1125 1139 950

Putting flood

lights

650 435 989 666

Providing foods

per day

250 225 129 239

Generators supply

per hr.

59 69 80 95

Suggest an optimal assignment

CASE Deepak Hastir and AK Sharma

Back Ground: MCD is a civic body; it has to perform a large number of activities in

public interest by deploying a large number of resources wherever required. There e are

certain important functions, which are routinely carried out by deploying special kind of

vehicles, which are limited in numbers therefore, the deployment of these vehicles

requires an effective decision of assigning these to the tasks, MCD carries out demolition

of unauthorized structures, removal of garbage and collection of property tax MCD

carries out demolition of unauthorized structures, removal of garbage and collection of

property tax to name a few. MCD has a limited number Bulldozers or JCB Machines

which are required to demolish unauthorized structures and to clear the land after leveling

of the same and lifting and loading of huge quantity of debris so produced after

demolition of the sites.

Now, since the MCD HAS hired 4 Bulldozers JCB s to clear 4 different sites where

unauthorized occupation exists. Since the Bulldozers are to be paid hourly rate of hire

charges, the efficiently and optimum utilization is to be required to save revenue, through

best deployment of Machines as per their specification

An estimate by the experts has been prepared showing the number of Hours required by

each JCB for clearance of 4different sites as given under;

JCB ID Site 1 Site 2 Site3 Site4

1A 120 100 80 90

2A 80 90 110 70

3A 110 140 120 100

4A 90 90 80 90

Suggest optimal assignment

TRANSPORTATION MODEL Transportation technique/model is special case of linear programming. It is called transportation method because most of the problems solved by this method deal with the transportation of a product manufactured at different plant to a number of different destinations. The objected is to satisfy the demand at different destination given the supply constraints at the minimum total transportation cost. The transportation model in original form described by F.L. Hitchcock in 1941. The LPP formulation was initially stated by George B. Dantiz in 1951. General form of Transportation Model Let O1, O2, . . . Om be m plants where a homogeneous product in the amounts a1, a2, . . . am respectively is available and let d1, d2, . . . dn be n destination each of which requires the amount b1, b2, . . . bn respectively. Let us further assumes that transportation cost per unit from origin o i to destination dj is Cij and directly proportional to amount shipped. Again suppose Xjj be the number of units of product shipped. From Origin oi to destination dj. The problem is then to distribute the product from m origin to n destination such that total transportation cost is minimum. Mathematically

General transportation table Origin Destination Available D1 D2 D3 . . . Dn

O1 a1 O2

.

.

.

a2

.

.

. Om am

Required b1 b2 bi . . . bn

Balanced Transportation Problem

In transport model supply equals demand then it balanced transportation model & if it is not the case then we balance it by introducing dummy origin or destination depending upon the given condition. IMPORTANT DEFINITION

i) Feasible solution :- A set of non negative individual allocation is called a feasible solution.

ii) Basic feasible solution :- A feasible solution to m-origin, n destination problem is said to be basic if the number of occupied cells are m + n – 1.

iii) Optimal solution :- A basic feasible s01n which minimize total transportation cost is called optimal.

METHODs OF INITIAL ASSIGNMENT

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

Solve the following problem by (i) NWCM (ii) LCM (iii) VAM

Origin Destination

d1 d2 d3 d4 Capacity

O1 13 11 15 40 2

O2 17 14 12 13 6

O3 18 18 15 12 7

demand 3 3 4 5 15

(i) NWCM N

Working W E

S We first fill cell (o1 d1). The quantity of supply & demand, whichever is less is allocated in the case of first row and first column 2 is chased & allocated to this cell. This allocation is feasible because O1 has 2 units available & requirement of destination d1 is of 3 units. This completely exhaust the capacity of origin O1 but col. d1 still needs 1 (3-2) unit to satisfy its requirement and so on we move to next column & row following top-left criteria.

The limitation with this method is it doesn’t consider the transportation cost. In this case total transportation cost will be : 2 x 13 + 1 x 17 + 3 x 14 + 2 x 12 + 2 x 15 + 5 x 12 = Rs. 199. ii) Least Cost Method (LCM) The allocation according to least cost method give an improved starting solution in comparison to the initial 001n obtained by NWCM, because this method takes in to consideration the lowest cost and thus reduces computation. Step 1 (a) Select the cell with min. transp. Cost among all the row or column of the transportation table.

(b) If this min. cost is not unique then select the cells corresponding to lower numbered row. In case they appear in same row, select the cell with lower numbered column.

Step 2 Allocate as many units as possible to this cell determined in step 1 & eliminate the row / col. whom rim requirement is fulfilled Step 3 Repeat step 1 & 2 for reduced table until entire supply is exhausted to satisfy demand. LCM :- In the given problem Total cost 2 x 11 + 1 x 17 + 1 x 14 + 4 x 12 + 2 x 18 + 5 x 12 = 197 Vogel’s Approximation Method (VAM)

The vogels method or penalty method is preferred over the two method discussed earlier. Step 1 compute penalty for each row and each column, which is incurred as a result of not assigning a non zero value to the cell with the lowest cost i.e., find out difference betn two least cost entry. Step 2 Identify row or column among all the rows & column with maximum penalty. In this identified row or col. select the cell with least cost & allocate maximum possible units to this cell. Delete the row or col. in which supply is exhausted. The corresponding supply & demand quantities are adjusted accordingly. If largest penalty corresponding to two or more rows are equal, we select the top most row if the largest penalty corresponding to two or more colm are equal we select the column in extreme lest. Step 3 Repeat step 1 and 2 for reduced table until the entire supply at various plants are exhausted to satisfy the demand at different destinations. 2 x 13 + 3 x 14 + 3 x 12 + 1 x 18 + 1 x 15 + 5 x 12 = 197 Optimality test Once initial solution has been found the next step is to test that solution for optimality. Stepping – Stone Method

In this method net change in the cost that may occur by introducing any one of the empty cell into the solution is calculated. The important rule to keep in mind is that every increase or decrease in supply at one occupied cell must associate with a decrease/increase at another occupied cell. This indicates that if empty cell (O1d2) is occupied then transp. Cost will decrease by Re. 1 TRANSHIPMENT MODEL In transportation problem it is assumed that item can be shipped only from am origin to a destination. There could be a situation where it might be economical to transport item in several stage. This problem of allowing an item to pass through one or more intermediary points before reaching the final destination is called trans-shipment problem. Maximization Transportation Problem A maximization transportation problem can be converted into usual minimization problem by subtracting all the contribution from highest contribution involved in the problem.

EXERCISE (1) Write short note on “Transportation Model”. (2) Explain in brief with example a) NWCM (b) LCM (c) VAM (3) Differentiate between transportation & trans-shipment. (4) Solve the following transportation problem

Origin Destination

S1 S2 S3 Supply

D1 15 10 16 50

D2 11 10 11 40

D3 13 14 12 47

demand 60 35 42 137

(5) Solve the following transportation problem

Origin Destination

S1 S2 S3 Supply

D1 6 4 1 14

D2 8 9 2 12

D3 4 3 6 5

demand 6 10 15 31

Case A.K Sinha

1. Disaster created by terrorists/ insurgents/ militants in North Eastern State of

Manipur offers a plethora of extreme problematic situations that require

decision making in complex situations where a number of equally feasible and

equally attractive alternatives present themselves to the Director General of

Police (DGP). The DGP would invariably be in a decision situation for

effectively managing the policing of 4 No. of hill Districts Namely, Senapati

(S), Tamenglong( T), Churachandpur (C) and Ukhrul(U). The community

compositions, no. of different militant outfits, ability of Superintendents of

Police(SP) to handle problems in different situations , operational skills, public

support, knowledge of ground situations, etc creates complex situation before

the DGP to post most suitable SPs namely , Kailun(K), Achin (A) , Biren( B),

Tomba(T), Chaoba(C) and Lalropuia(L) in different districts.

2. The process of decision making of posting of SPs in 4 districts in response to

the problem situation of districts requires skilful analysis of the objective, the

relevant constraints, identification of various viable alternatives and evaluation

of alternatives and selection of best course of action. The process of analysis

could take either of the two forms- Qualitative or Quantitative.

3. Skills in qualitative Analysis may be inherent in the DGP and he has achieved

it through knowledge and experience. Skills in quantitative Analysis requires

study of Operation Research and Managerial Statistical techniques. These

techniques are extremely helpful in arriving at a decision as well as useful in

validation and reinforcement of the decision arrived by the DGP through

qualitative analysis based on judgment and experience.

4. Aim of this case study is to exemplify and illustrate the use of Quantitative

Techniques in making Decision of best SPs in the 4 districts to tackle disaster

situation created by terrorist activities.

5. On the basis of internal quality assessment of the SPs, top 4 SPs have got

quality points out of 20 points for suitability for the benefit of different district

situations. These points can be tabulated as given below:

Districts

SP

s

T C U S

A 16 10 14 11

K 14 11 15 15

T 15 15 13 12

L 13 12 14 15

Solve the problem by transportation method

CASE 7 Vijay Sagar

In every summer season the water scarcity problem arising. There are

some colonie]\where DJB not providing the regular piped water supply network in

unauthorized colonies not regularized by the town planner and residents are depends on the

tanker supply. In summer season the water born diseases increased like cholera, gastritis due

to scarcity of water. It is the prime responsibility of DJB to provide the potable water in most

deficit area to avoid any kind of outbreaks of any water borne disease

The most deficit area under the division of North-East II where water regular water supply

does not exist are as under:-

Sonia vihar no of tankers required daily 6nos.

Mandoli village no of tankers required daily 10nos.

Sabhapur village no of tankers required daily 8nos

Karawal nagar village no of tankers required daily 6nos

This division has set up the four nos of water emergencies having the following strength

of water tanker are as under

Yamuna vihar 8nos.

East of Loni road 7nos

Wazirabad road 7nos

Tahirpur 8nos

This division has to manage in such a way that implication of transportation cost be minimum

for supplying of potable water through tanker in most deficit area so DJB have to incurred

the minimum his expenditure to avoid the loss.

D1 D2 D3 D4

O1 8

O2 7

O3 7

O4 8

6 10 8 6

Solve the transportation problem by: North-west corner method, Least cost method,

Vogel,s Approximination method

100

11

150 200 250

200 250 150 100

350

250

400 450

150

300

350 300

Network Techniques – PERT & CPM

PERT

Programme Evaluation and Review Technique (PERT) developed in late 1950’s for

planning, scheduling and controlling the polari blast missile projects. PERT incorporate

uncertain i.e. approach in PERT is probabilistic. Fundamental to PERT is the concept of

“Event”.

CPM

Closely akin to PERT, but developed independently is the technique of CPM (Critical Path

Method). The CPM was used and developed for construction projects. This technique deals

with time-cost trade-offs.

Both, the technique of PERT and technique of CPM share the motion of critical path and

are based on the network analysis that determine the most critical activities to be controlled

so as to meet completion dates.

Though the use of these is based on individual characteristic but mainly the PERT is useful

for non repetitive and complex projects in which time estimates are uncertain where as

CPM is best utilized for repetitive and non complex projects.

DEFINITIONS

Activity: An activity is an act or work which requires time & resources for its completion.

An activity is depicted by single arrow line on the project network. The activity arrow is

not sealed. The head of the arrow shows the sequence or flow of activity one activity is

represented by one arrow line only.

Event: Event is an indicator which indicate the end of an activity and beginning of another.

An event does not consume any time and resources. An event is represented by a circle

[O]. All activities must begin and end with event nodes.

Event

Tail event (1) Head event

Predecessor Activity: Activity that must be completed immediately prior to start or another

activity are called predecessor activity.

Successor Activity: The activity which can start immediately after the completion of an

activity is called successor activity.

Concurrent Activities: Activities which can start simultaneously if resources permits are

called concurrent activities.

Figure

1 – 2 – 5 – 6 – critical path (longest path).

FLOAT v/s SLACK

Event Slack: Slack of an event is the difference between latest time & earliest time.

Event slack = TLL – TE

L

Total float

FLOAT Free float

Independent

Total Float: The total float of an activity is the time by which it can be delayed without

extending the project duration.

Guidelines for construction of network

1. Each activity is represented by one and only one arrow in the network.

2. No two activities are identified by the same tail & head events.

3. Two events are numbered in such a way that event of higher number can happen

only after completion of lower number event.

4. Cycling must be avoided.

5. Arrow should always move from left to right.

6. Arrow should not cross each other.

Projects duration & critical path

eg Activity Predecessor Time estimate

A — 14

B — 13

C A 12

D A 17

E B 16

F C 14

G D, E 12

Find project duration & critical path.

Figure

(i) Optimistic time: It is the time required to complete an activity under favourable

condition (denoted by to).

(ii) Most likely time: It is the time required to complete an activity under normal

condition (denoted by tm).

(iii) Pessimistic time: It is the longest time required for completion of an activity under

adverse condition (tp).

Expected time = te =

2

Variance = 2 =

& Stand deviation =

Free float: It is that part of total float within which early start of an activity can be

manipulated without delaying the start of successor of activity.

Free float = total float – head event sla

Independent float: This is the time by which an activity can be delayed without affecting

preceding activity.

PERT

1. Activity time are statistically independent & usually associated with distribution.

2. There are enough activities involved in the network that sum of activity times based

on their means & variance will be normal distributed.

3. In PERT for each activity three time estimates can be obtained.

Example The table given below gives different time estimates

Activity Time Estimate (week)

to tm tp

tp – to

6

to + 4 tm + tp

6

tp – to

6

1 – 2 3 5 13

1 – 3 1 2 15

2 – 4 6 7 8

3 – 4 2 5 14

2 – 6 2 4 12

4 – 5 4 6 8

4 – 6 5 9 13

5 – 7 1 2 3

6 – 7 1 4 7

(a) Draw the project diagram.

(b) Calculate expected time & variance for each activity.

(c) Find variance of critical path.

(d) Find the probably that project will be completed in 23, 29 weeks.

(e) If project due date is 27 weeks time probability of not meeting due date.

Figure

(B)

Activity Time estimate te Variance

1 – 2 3 5 13 6 (10/6)2

1 – 3 1 2 15 4 (14/6)2

2 – 4 6 7 8 7 (2/6)2

3 – 4 2 5 14 6 (12/6)2 = 4

2 – 6 2 4 12 5 (10/6)2

4 – 5 4 6 8 6 (4/6)2

4 – 6 5 9 13 9 (8/6)2

5 – 7 1 2 3 2 (2/6)2

6 – 7 1 4 7 4 (6/6)2 = 1

(C) Variance of critical path (1 – 2 – 4 – 6 – 7)

(10/6)2 + (10/6)2 + (10/6)2 + (10/6)2 = 5.67

(D)

(E) Prob. of not meeting due date

= 1 – 0.6628 = 0.3372 or 33%

CRASHING

Normal Cost: Its minimum direct cost required to complete an activity.

Crash cost: Minimum cost required to complete an activity in crash time.

Crash time: It is minimum time required to complete an activity.

Normal time: Its time within which time activity can be completed.

Figure

NETWORK TECHNIAQUES

Q. 1 Draw the network for the following activities and find critical path and total duration

of projects

Activity Duration (Days)

1-2 20

1-3 25

2-3 10

2-4 12

3-4 5

4-5 10

Q.No 2 Draw the network and find critical path

Activity Duration (Days)

1-2 2

1-3 4

1-4 3

2-5 1

3-5 6

4-6 5

5-6 7

Q.3 A project has the following activities and characteristics

Activity Optimistic Most likely Pessimistic

1-2 2 5 8

1-3 4 10 16

1-4 1 7 13

2-5 5 8 11

3-5 2 8 14

4-6 6 9 12

5-6 4 7 10

Required

a) Find expected duration of each activity

b) Draw the project network and duration of the project

c) Find SD of activities on critical path

Q.No 4 A small project consist of six activities with the following information

Activity Normal time Normal Cost

1-2 9 200

1-3 8 300

1-4 15 250

2-4 5 350

3-4 10 200

4-5 2 100

The following information is also available

Normal duration Normal Cost

20 1400

17 1445

16 1485

15 1530

13 1660

12 1735

Indirect cost is Rs 50 per day

a) Draw the network , find out normal project duration and associated cost

b) Find out minimum project duration with associated cost

c) Find out optimum duration with cost

Q.No 5 A project has the following activities

Activity Immediate predecessor Duration

A - 3

B - 2

C - 2

D A 4

E B 4

F B 7

G C 4

H D 2

I E 5

J F, G 6

K H, I 3

Compute total, free and independent float for each activity

CASE A.S Shekhawat

A devastating and mostly unknown disaster hit the Andaman & Nicobar Island 28th Dec.

2004 in the form Tsunami. While the country was still trying to figure out as to what ahs

hit part of its population the CISF, sensing the disaster alerted their Disaster

MANAGEMENT Teams located in HYDERABAD. As soon as the directions from the

Control Room of Ministry of Home Affairs received in the CISF HQ rs for opeationalizing

the Search &Rescue Teams and airlift them to Andaman & NICOBAR Island, the teams

were asked to move immediately

However, the task was not as simple as it appeared. The operationalization of the Search

&Rescuer Teams involved a number of activities to make the teams fully equipped and

operational. A number of other stores, equipments and relief material were to be brought

from different parts like clothing and tentage from Bangalore, additional equipments and

diesel generator sets from different stations in Hyderabad, food packets from Chennai etc.

Similarly, medical relief team located in Trivandrum was also to be brought to Chennai to

merry up with the main teams. The airlifting of men and material from Chennai to Port

Blair was also to be done in batches.

The various activities involved in mobilization of resources are described

below:

Activities Details of work Duration

1-2 Mobilization of SAR Team from Hyderabad to

Chennai.

4hours

1-3 Mobilization of Medical Relief Team from

Kerala to Chennai.

5 hours

1-4 Mobilization of Relief Materials (Clothing

&Tentage) from Bangalore to Chennai.

4hours

2-5 Inoculation of SAR Team 1 hours

2-6 Mobilization of Additional Search&Resuce

Equipments &DG sets s from hyderabad to

Chennai.

2 hours

3-6 Briefing and Task Allocation of Medical Relief

Teams

1 hours

4-7 Packing of Food Packets, Potable Water and

Clothing

2 hours

5-8 Dispatch of SAR Team to Kar-Nicobar Island 4 hours

6-8 Dispatch of Medical Relief Team and

Additional SAR Equipment from Chennai to

Kar-Nicobar Island.

5 hours

7-8 Dispatch of Relief Material to Kar-Nicobar

Island

4 hours

Draw the project network and identify critical path?

CASE -9 L.K Jain

Disaster manager is required to handle highly complex situation. The state of nature/

events is often uncertain. Various decision alternatives, are available to him, which seems

equally attractive and associated with various degrees of risk attached to them. The

process of decision making in such situation requires a skillful analysis of the objective,

the relevant constraints, identification & evaluation of various alternatives and selection

of best course of action. This analysis will be qualitative as well as quantitative. Skill in

qualitative analysis are inherent in managers by virtue of their experience A certain industrialist decides to put up a textile mill. He discusses the project with

consultants, friends, industrialists, traders, consumers etc. to understand the nerve of the

market, demand & supply position, market locations etc. The sequence of events for

implementing the project would be as under:

Activity Brief description duration preceding

activity

Starting event: Letter of Intent ----- -----

1—2 Acquisition of land 4 months -----

2—3 Construction of factory building 5 months 1—2

1—4 Finalization of machinery suppliers 3 months -----

4—5 Ordering of equipments 2 months 1—4

5—6 Delivery period of machines 6 months 4—5

3—6 Dummy activity (completed building

to accommodate delivered machines.) ----- 2—3

6—7 Erection of machines 2 months 5—6

7—8 Trial run & commercial production 2 months 6—7

Draw the project network and compute project duration

CASE 10 M.S. Khan

Delhi Jal Board, constituted under Delhi Jal Board Act 1998, is responsible for production

and distribution of drinking water in Delhi. The Board is also responsible for collection,

treatment and disposal of waste water/sewage in the capital. Delhi Jal Board has provided

approximately 16.0 lacs water connections up to 1.4.2006. Delhi Jal Board is committed to

provide efficient and prompt services to the citizens of Delhi and to be courteous in

personal behavior and professional conduct.

Delhi Jal Board has decided to computerize and refit of its cash collection centres in order

to minimize waiting time and maximize customer satisfaction. One of such cash collection

centre has been selected at Kanhya Nagar, Tri Nagar where some of the existing office

equipments will be disposed off but remaining will be returned to Head quarter on

completion of the renovation work. Tenders are invited from a number of selected

contractors. The contractors will be responsible for all the activities in connection with the

renovation work excepting the prior removal of the old equipment and its subsequent

replacement.

The major elements of the project have been identified as follows along with their durations

and immediately preceding elements.

Activity Description Duration

(weeks)

Immediate

Predecessors

A Design new customer care centre 14 -

B Obtain tenders from the contractors 4 A

C Select the contractor 2 B

D Arrange details with selected contractor 1 C

E Decide which equipment is to be used 2 A

F Arrange storage of equipment 3 E

G Arrange disposal of other equipments 2 E

H Order new equipment 4 E

I Take delivery of new equipment 3 H,L

J Renovations take place 12 K

K Remove old equipment for storage or

disposal

4 D,F,G

L Cleaning after the contractor has finished 2 J

M Return old equipment for storage 2 H,L

(a) Draw the network diagram showing the inter-relations between the various activities of the

project.

(b) Calculate the minimum time that the renovation can take from the design stage.

(c) Find the effect on the overall duration of the project if the estimates or tenders can be

obtained in two weeks from the contractors by reducing their numbers.

(d) Calculate the ‘independent float’ that is associated with the non-critical activities in the

network diagram.

CASE S.K SINGH

In Delhi Jal Board, there are two main Engineering Departments- Civil and

Electrical & Mechanical (E&M). In E&M, major activities involve installation of

motors and pumps (from 5 HP to say 800 HP) during construction of water and

Sewage Treatment Plants. While designing the capacity of pumps and motors

various factors such as flow, head, losses etc. are of paramount importance, there

always remains some uncertainty while considering other important parameters

related with the design. In general, a PERT network of a Water or Sewage

Treatment Plant takes into account these uncertainties and accordingly three time

estimates for each activity are considered. The three time estimates are indicated along the activity arrows for the project shown in

Figure below:

Calculate (a) the expected or average time tE and the variance for each activity, (b) the

earliest expected time, and (c) the latest allowable occurrence time for each event. Make

the entries in a tabular form. Also, enter the last two values against the respective event

circles.

10

50

100

80

60

30

2

0

40

7

0

90

11

0

120

2-4-6

4-8-12

8-10-12

4-6-9

3-5-9 14-16-18

0-0-0

6-8-10 0-0-0

7-9-11

6-8-12

1-2-3

2-3-5

5-9-12

3-4-5

6-8-10

\Reference: source of compilation

1. Hillier, F.S & Hillier, M.S Introduction to Management Science, Tata

McGraw Hill

2. Khandelwal R.S ,& Gupta B.L Quantitative Techniques ,Shriyans Publication

Limited

3. Sharma J.K , Business Statistics ,Second edition ,Pearson education

4. Taha Hamdy Operation Research, eight edition, Pearson education.

5. Vohra N.D Quantitative Techniques in Management Tata McGraw

Hill.