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PROBLEM 5 (A) SCHOOL DISTRICT ADDEENBASHIRONCOBBITHDAIMMAN STUDENT POPULATION NORTH 250 5,5,5,5,5,-,- SOUTH 340 6,6,6,-,-,-,- EAST 110 200 310 2,2,2,2,4,4,4,4.

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Presentation on theme: "PROBLEM 5 (A) SCHOOL DISTRICT ADDEENBASHIRONCOBBITHDAIMMAN STUDENT POPULATION NORTH 250 5,5,5,5,5,-,- SOUTH 340 6,6,6,-,-,-,- EAST 110 200 310 2,2,2,2,4,4,4,4."— Presentation transcript:

1 PROBLEM 5 (A) SCHOOL DISTRICT ADDEENBASHIRONCOBBITHDAIMMAN STUDENT POPULATION NORTH 250 5,5,5,5,5,-,- SOUTH 340 6,6,6,-,-,-,- EAST 110 200 310 2,2,2,2,4,4,4,4 WEST 210 14,14,-,-,-,-,- CENTRAL 40 60 190 290 5,5,5,5,1,1,8,- DUMMY 200 0,-,-,-,-,-,- CAPACITY 400 12,3,3,3,3,3,3,18 400 10,5,5,10,-,-,- 400 21,1,1,1,1,1,1 400 10,6,1,1,1,5,- 1600 1521 0 16 10 31 27 233517 20 24 10 00 23 35 22 12 18 26 29 15 0

2 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

3 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)….

4 PROBLEM 5 (A) SCHOOL DISTRICT ADDEENBASHIRONCOBBITHDAIMMAN STUDENT POPULATION NORTH 250 SOUTH 340 -40 +40 340 EAST 110 +40 200 -40 310 WEST 210 CENTRAL 40 -40 60 +40 190 290 DUMMY 200 CAPACITY 400 1600 1521 0 16 10 31 27 233517 20 24 10 00 23 35 22 12 18 26 29 15 0

5 PROBLEM 5 (A) SCHOOL DISTRICT ADDEENBASHIRONCOBBITHDAIMMAN STUDENT POPULATION NORTH 250 SOUTH 300 40 340 EAST 150 160 310 WEST 210 CENTRAL 100 190 290 DUMMY 200 CAPACITY 400 1600 1521 0 16 10 31 27 233517 20 24 10 00 23 35 22 12 18 26 29 15 0

6 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.

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8 CATEGORIZED INTO 2 TYPES: 1.MINIMIZATION PROBLEM 2.MAXIMIZATION PROBLEM

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

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

11 NOW WE WILL BALANCE THIS PROBLEM BY ADDING DUMMY COLUMN

12 NOW WE CAN SOLVE THE PROBLEM BY USING VAM

13 ONE OF THE WORKING HOW TO SOLVE PROBLEM BY USING VAM

14 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

15 1.Choose maximum allocation = 0 2.Let U2= 0 3.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.

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17 U1=1W11= -4 U2=0W13= -5 U3=3W14= -2 V1=2W24= -3 V2=3W31= 0 V3=0W33= -1 V4= -3Wij ≤ 0 (optimum allocation has been made) C ij =U i +V j W ij =U i +V j -C ij

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

19 SUGGESTED LINKS http://www.bms.co.in/how-to-solve- transportation-problems/ http://www.slideshare.net/beautifulneha/transp ortation-problem-in-operational-research https://www.youtube.com/watch?v=7_yPdc0m IHE https://www.youtube.com/watch?v=il78DKRw2 JU

20 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.


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