P R O B L E M 5 ( A )

SCHOOL

DISTRICT

ADDEEN BASHIRON COBBITH DAIMMAN STUDENTPOPULATION

NORTH 250

2505,5,5,5,5,-,-

SOUTH 340340

6,6,6,-,-,-,-

EAST 110 200 310

2,2,2,2,4,4,4,4

WEST 210210

14,14,-,-,-,-,-

CENTRAL 40 60 190290

5,5,5,5,1,1,8,-

DUMMY 200200

0,-,-,-,-,-,-

CAPACITY 40012,3,3,3,3,3,3,18

40010,5,5,10,-,-,-

40021,1,1,1,1,1,1

40010,6,1,1,1,5,-

1600

15

21

0

16

10

31

27

23

35

17

20

24

10

0 0

23

35

22

12

18

26

29

15

0

Initial total traveling time:

250(12)+110(18)+40(15)+340(15)+60(10)+200(22)

+200(0)+210(10)+190(16)

=20 820 mins

Evaluate using stepping-stone method:

1) N to B: +23-12+15-10=16

2) N to C: +35-12+18-22=19

3) N to D: +17-12+15-16=4

4) S to A: +26-15+10-15=6

5) S to C: +21-15+10-15+18-22=-3

6) S to D: +27-15+10-16=6

7) ….

8) ….

9) ….

10)….

P R O B L E M 5 ( A )

SCHOOL

DISTRICT

ADDEEN BASHIRON COBBITH DAIMMAN STUDENTPOPULATION

NORTH 250

250

SOUTH 340-40 +40 340

EAST 110+40 200-40 310

WEST 210 210

CENTRAL 40-40 60+40 190 290

DUMMY 200 200

CAPACITY 400 400 400 400 1600

15

21

0

16

10

31

27

23

35

17

20

24

10

0 0

23

35

22

12

18

26

29

15

0

P R O B L E M 5 ( A )

SCHOOL

DISTRICT

ADDEEN BASHIRON COBBITH DAIMMAN STUDENTPOPULATION

NORTH 250

250

SOUTH 300 40 340

EAST 150 160 310

WEST 210 210

CENTRAL 100 190 290

DUMMY 200 200

CAPACITY 400 400 400 400 1600

15

21

0

16

10

31

27

23

35

17

20

24

10

0 0

23

35

22

12

18

26

29

15

0

Total traveling time for new solution:

250(12)+150(18)+300(15)+100(10)+40(21)+160(22)

+200(0)+210(10)+190(16)

=20 700 mins

Evaluate the new solution using stepping-stone method:

-All indices are zero and more than zero, the solution is an optimal solution at 20700 mins.

PROBLEM 5 (B)

TRANSPORTATION PROBLEMC AT E G O R I Z E D I N TO 2 T Y P E S :

1. M I N I M I Z AT I O N P R O B L E M

2. M A X I M I Z AT I O N P R O B L E M

THIS IS AN UNBALANCED MAXIMIZATION PROBLEM FOR FINDING OUT MAXIMUM INCREASE IN PRODUCT OUTPUT

TO SOLVE THIS FIRST WE WILL CONVERT THIS IN MINIMIZATION PROBLEM BY SUBTRACTING THE LARGEST ELEMENT FROM ALL ELEMENTS

NOW WE WILL BALANCE THIS PROBLEM BY ADDING DUMMY COLUMN

NOW WE CAN SOLVE THE PROBLEM BY USING VAM

ONE OF THE WORKING HOW TO SOLVE PROBLEM BY USING VAM

NOTE : FOR CALCULATING FINAL SCHEDULE, WE MUST MULTIPLY QUANTITY BY THE ORIGINAL ELEMENT VALUE

Initial Basic Feasible Solution = (45x10) + (60x8) + (25x9) + (6x5) + (35x12) + (65x0)= 1605

MODI

1.Choose maximum allocation = 02.Let U2= 03.Find the value of U1, U3, V1, V2, V3, V4, using

the formula Cij=Ui+Vj , only for those squares that are currently used or occupied.

4.Compute the improvement index for each unused square by the formula Wij =Ui+Vj-Cij

5. In minimization, if all Wij ≤ 0, optimum allocation has been made.

MODI

MODI

U1=1 W11= -4

U2=0 W13= -5

U3=3 W14= -2

V1=2 W24= -3

V2=3 W31= 0

V3=0 W33= -1

V4= -3 Wij ≤ 0 (optimum allocation has been made)

Cij=Ui+Vj Wij =Ui+Vj-Cij

MODI

Optimal Solution = (45x10) + (60x8) + (25x9) + (6x5) + (35x12) + (65x0)= 1605

SUGGESTED LINKS

http://www.bms.co.in/how-to-solve-transportation-problems/

http://www.slideshare.net/beautifulneha/transportation-problem-in-operational-research

https://www.youtube.com/watch?v=7_yPdc0mIHE

https://www.youtube.com/watch?v=il78DKRw2JU

C)1.Unbalance transportation model

Unbalance transportation model is the total supply does not equal to the total demand. The dummy supply exactly not really be able to supply but to make adjustment at the end. The costs nothing to transport from dummy because it does not really exist. So we put ‘cost coefficient ‘ is zero to each dummy location.

2.Sentitivity analysis

Sensitivity analysis is a way to predict the outcome of a decision if a situation turns out to be different compared to the key predictions. A sensitivity analysis shows how sensitive to businesses to change.

3.Modified distribution(MODI)

The modified distribution method is an improvement over the stepping stone method for testing and finding optimal solutions. This method allows us to compute indices for each unused square without drawing all the closed paths.