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Transportation simplex method. B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 Balanced?

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Presentation on theme: "Transportation simplex method. B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 Balanced?"— Presentation transcript:

1 Transportation simplex method

2 B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 Balanced?

3 Methods for finding basic feasible solution We need to find a feasible solution, which satisfy all the constraints (supply and demand). - Northwest corner - Minimum cost - Vogel’s method Solution:

4 Northwest corner Northwest cornerB1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 20 42 10 6 634 |10 |34 8 |42 8

5 B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 20 12 30 16 1240 |30 2 |14 1. row: Penalty: 4-2 = 2 2. row: Penalty: 3-2 = 1 1 3. row: Penalty: 5-2 = 3 30 3 4 2 |2 40 Vogel’s method

6 Minimum cost B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 16 |2|30 |14 28 2 2 14 20 40

7 Solution SolutionB1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 20 42 10 634 8 (1;1)(1;2)(2;2)(2;3)(3;3)(3;4) IB=IB=

8 (1;1) (1;2)(2;2)(2;3)(3;3)(3;4) IB=IB= 8 = U1 + V1 2 = U1 + V2 4 = U2 + V2 3 = U2 + V3 5 = U3 + V3 9 = U3 + V4 6 equation 7 variable  Fix U1=0 U1=0V1=8 V2=2U2=2 V3=1U3=4 V4=5

9 U1=0 V1=8 U2=2 V2=2 U3=4 V3=1 V4=5 B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 20 42 10 634 8 d 13 =4-(U1+V3)=3 3 -10 -3-5 2

10 B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 120 20 42 10 634 8 3 -10 -3-5 2 Θ 34+ Θ 8- Θ 6- Θ 20- Θ 10+ Θ

11 B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 Θ 8- Θ 34+ Θ 6- Θ 10+ Θ 20- Θ 120 42 Θ =6

12 B1B2B3B4 R130 8247 R240 7432 R350 2559 20164242 6 2 40 1614 120 42 Θ =6

13


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