qtm project lakshay pawar

7/27/2019 QTM Project Lakshay Pawar

1/26

QUANTITATIVE TECHNIQUESOF MANAGEMENT

LAKSHAY PAWAR 50284

BBS 2-C


2/26

ACKNOWLEDGEMENT

Firstly, we would like to thank our supervisor of this project, Mr.Abhishek Tandon for his expert guidance and support. He continuously

motivated us to work for this project and make it a success. His

willingness to inspire us contributed tremendously in its completion. He

had been very kind and patient while suggesting us the outlines of this

project and correcting our doubts.

Also, we would like to thank the Delhi University and Shaheed Sukhdev

College of Business Studies for providing us with a platform and

environment to work and nurture our talents. We would like to extend

heartfelt thanks to our Principal, Ms. Poonam Verma for her invaluable

support.

We would also like to express our gratitude towards our parents for

their kind co-operation and encouragement that helped us in

completion of this project.

Our thanks and appreciations also go to our batch mates in developing

the project and people who have willingly helped us out with their

abilities.


3/26

TABLE OF CONTENTS:

S. NO PARTICULARS PAGE NO.1 TRANSPORTATION PROBLEM (INTRO) 4

2 NORTH WEST CORNER METHOD 7

3 APPLYING MODI METHOD 10

4 VOGEL APPROXIMATION METHOD 17

5 APPLYING MODI METHOD 19

6 DEGENERACY 24

7 MULTIPLE OPTIMAL SOLUTIONS 25

8 SOME MORE SPECIAL CASES 26


4/26

TRANSPORTATION PROBLEM

One of the most important and successful applications of quantitative analysis to

solving business problems has been in the physical distribution of products,

commonly referred to as TRANSPORTATION PROBLEM.

Basically, the purpose is to minimize the cost of shipping goods from one locationto another so that the needs of each arrival area are met and every shipping

location operates within its capacity.

SPECIMEN OF A TRANSPORTATION TABLE:

From To

City 1 City 2 City 3 City 4 Supply (Million

kwh)

Plant 1 $8 $6 $10 $9 35

Plant 2 $9 $12 $13 $7 50

Plant 3 $14 $9 $16 $5 40

Demand

(Million kwh)

45 20 30 30


5/26

FORMULATION OF A LPP:

Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24+14X31+9X32+16X33+5X34

S.T.: X11+X12+X13+X14 = 30

X14+X24+X34 >= 30

Xij >= 0 (i= 1, 2, 3; j= 1, 2, 3, 4)


6/26

TRANSPORTATION PROBLEM is a special branch of LPP problem.

These problems are solved by transportation method.

1. Obtain the initial solution North West Corner Method (NWC method) Vogel Approximation Method (VAM method)

2. Ascertain whether the solution is optimal or not Modify Distribution Method (MODI method)

3. If solution is optimal we stop, otherwise we revise the solution.


7/26

3 2

6

2

X 5 2 3

1.North West Corner Method (NWC method)-Begin in the upper left (northwest) corner of the transportation table and

set x11 as large as possible (here the limitations for setting x11 to a larger

number, will be the demand of demand point 1 and the supply of supply

point 1. Your x11 value can not be greater than minimum of these 2 values).

According to the explanation above we can set x11=3 (meaning demand of

demand point 1 is satisfied by supply point 1).


8/26

After we check the east and south cells, we saw that we can go east (meaning

supply point 1 still has capacity to fulfill some demand).

After applying the same procedure, we saw that we can go south this time

(meaning demand point 2 needs more supply by supply point 2).

3 2 X

6

2

X 3 2 3

3 2 X

3 3

2

X X 2 3


9/26

Continuing the same procedure we will get:-

x11=3, x12=2, x22=3, x23=2, x24=1, x34=2

3 2 X

3 2 1

2

X X X 3

3 2 X

3 2 1 X

2

X X X 2

3 2 X

3 2 1 X

2 X

X X X X


10/26

Applying MODI METHOD for optimal solution:-


11/26

By following the similar procedure, we assign the values as:-

Cost comes out to be


12/26

Similarly, we found out all the rest of the values:-


13/26

Now we find out the opportunity cost of all uncovered cells as:-

We get the optimal solution when opportunity cost of all the uncovered cells

comes out to be negative.

But here, since all the opportunity costs are positive we will go to the next stage

i.e. making loops.

Following are the conditions for making a loop:-

Loop will start from the cell having maximum opportunity cost with apositive sign

Always in a clockwise direction Cover even number of cells, of which only 1 cell will be uncovered cell No single row or column will have similar signs


14/26


15/26

Again we follow the same procedure:


16/26

By following the same procedure again and again we will

ultimately reach the optimal solution wherein all the

opportunity costs become negative:

The least cost comes out to be: $6,250


17/26

2.Vogel Approximation Method (VAM Method):-Begin with computing each row and column a penalty. The penalty will be

equal to the difference between the two smallest shipping costs in the row

or column. Identify the row or column with the largest penalty. Find the

first basic variable which has the smallest shipping cost in that row or

column. Then assign the highest possible value to that variable, and cross-

out the row or column as in the previous methods. Compute new penalties

and use the same procedure.

Step 1: Compute the penalties.

Step 2: Identify the largest penalty and assign the highest possible value to the

variable.

Supply Row Penalty

6 7 8

15 80 78

Demand

Column Penalty 15-6=9 80-7=73 78-8=70

7-6=1

78-15=63

15 5 5

10

15

Supply Row Penalty

6 7 8

5

15 80 78

Demand

Column Penalty 15-6=9 _ 78-8=70

8-6=2

78-15=63

15 X 5

5

15


18/26

By following same procedure we get tables as:

Supply Row Penalty

6 7 8

5 5

15 80 78

Demand

Column Penalty 15-6=9 _ _

_

_

15 X X

0

15

Supply Row Penalty

6 7 8

0 5 5

15 80 78

Demand

Column Penalty _ _ _

_

_

15 X X

X

15

Supply Row Penalty

6 7 8

0 5 5

15 80 78

15

Demand

Column Penalty _ _ _

_

_

X X X

X

X


19/26

Applying MODI METHOD for optimal solution:-

Since SS>DD, the problem is an UNBALANCED PROBLEM.

So we first balance the problem by introducing a dummy column as

follows:


20/26

Highest of all calculated penalty costs is for S3. Therefore, allocation is to be made

in row of source S3. The route (or cell), which one must select, should be the

lowest cost of this row. Route S3D5. Hence, first allocation is as follows.

Now, with the first allocation, destination D5 is consumed. We exclude this

column and work on the remaining matrix for calculating the penalty cost. We get

the following matrix.

Now for this, source S1 has highest penalty cost. For this row, the least cost route

is S1D1. Hence, next assignment is due in this route:


21/26

After second allocation, since destination D1 is consumed, we leave this column

and proceed for calculation of next penalty cost.


22/26

D2 and S1 is cancelled now, we move further:


23/26

Column D4 is consumed. In the only left column D3, the allocations of 100 units

and 150 units are done in route S2D3 and S4D3 respectively. Thus, we get the

following allocations in the Vogels approximation method.

The initial cost for this allocation is (13 100 + 16 150 + 16 100 + 15 100 + 17

150 + 0 100) or equal to Rs. 9350.


24/26

CHECK FOR DEGENERACY:

(m + n 1) = 7

Number of filled cell = 6, which is one less than (m + n - 1). Hence, go to step 4 for

removing the degeneracy.

We allocate in the least-cost un-filled cell. This cell is route S1D5 or S2D5. Let us

select route S1D5. Thus, we get following matrix after removing degeneracy.


25/26

OPPORTUNITY COST:

Since all opportunity costs are negative or zero, the initial assignment

of Vogels solution is optimal with total cost of Rs. 9350.

We had 8 uncovered cells, out of which 7 had a negative value and one

with a zero. This is a case ofMULTIPLE OPTIMAL SOLUTIONS.We make next table, start the loop making process from zero and get

that other solution.

Unassigned route Opportunity cost (ui + vj Cij)

S1D3

S1D4

S2D1

S2D2S2D5

S3D1

S3D2

S3D4

0 + 17 19 =2

0 + 16 17 =1

1 + 13 17 =5

1 + 16 19 =41 + 0 0 =1

0 + 13 15 =2

0 + 16 17 =1

0 + 16 16 = 0


26/26

SOME MORE SPECIAL CASES:

1.PROHIBITED ROUTE:-It is a special case of transportation problem wherein there is a restriction in the

supply of a commodity from a given source to a destination. In such a case a large

cost of M is entered for the restriction. This discourages the solution from using

such cells.

Rest of the question is solved in the same manner as discussed above.

2.MAXIMISATION PROBLEM:-If we are given a maximization type of problem, then we first need to convert it in

a minimization problem.

To convert it, we subtract all the values of the table from the maximum value of

the table.

Then solve the question accordingly and in the same manner as discussed earlier.

qtm project lakshay pawar

Documents