06IP/IM74 OPERATIONS RESEARCH. UNIT - 3: Transportation Problem

Transcription

1 06IP/IM74 OPERATIONS RESEARCH UNIT - 3: Transportation Problem Introduction: The objective of the transportation problem is to transport various quantities of a single homogenous commodity, which are initially stored at various origins to various destinations in such a way that the total transportation cost is minimum. Definitions: Basic Feasible solution: A feasible solution to a m-origin, n-destination problem is said to be basic if the number of positive allocations are equal to (m+n-1). Feasible Solution: A set of positive individual allocations which simultaneously removes deficiencies is called a feasible solution. Optimal Solution: A feasible solution (not basically basic) is said to be optimal if it minimises the total transportation cost. Mathematical Formulation of Transportation Problems Suppose there are m ware houses (w 1,w 2,w 3, _, _, w m ), where the commodity is stocked and n markets where it is needed. Let the supply available in wear houses be a 1, a 2, a 3, _,_,_ a m and The demands at the markets (m 1, m 2, m 3, _, _, m n ) be b 1, b 2, b 3, _, _, _ b n. The unit cost of shipping from ware house i to a market j is C ij (C 11,C 12,_, _ C n ), Let X 11, X 12,X 13,_, _, X mn be the distances from warehouse to the markets we want to find an optimum shipping schedule which minimises the total cost of transportation from all warehouses to all the markets 1

2 Ware houses Markets m 1 m 2 m m n Supplies W 1 C 11 X 11 C 12 X 12 C 13 X C 1n X 1n a 1 W 2 W w m C 21 X 21 C 22 X 22 C 23 X C 1n X 1n C 31 X 31 C 32 X 32 C 33 X C 1n X 1n - - C n1 X m1 C n2 X m2 C n3 X m Demand b 1 b 2 b b n i=1 Σ m a i = j=1 Σ n b j The total minimum transportation cost is Z = i=1 Σ m j=1σ n X ij * C ij i.e. Z = X 11 C 11 + X 12 C 12 + _ + X mn C mn Types of Transportation Problems C mn X mn 1. Minimisation Balanced Transportation Problems 2. Minimisation Unbalanced Transportation Problems 3. Maximisation Balanced Transportation Problems 4. Maximisation unbalanced Transportation Problems 5. All the above models with degeneracy. a 2 a a m 2

3 Methods of solving Transportation Problems 1. North- West Corner Rule method 2. Row-minima Method 3. Column minima method 4. Matrix Minima Method or least cost method 5. Vogel's Approximation method (VAM) Methods for checking Optimality 1. Modified Distribution Method, UV or MODI method PROBLEMS: 1. Solve the following transportation problem by North-West corner rule, Row Minima, Column Minima, Matrix Minima and VAM Method: Solution: Factories W1 W2 W3 W4 Supply F1 F2 F Demand This is a balanced transportation problem, since supply is equal to demand 3

4 1. North-West corner rule Method: Factories W1 W2 W3 W4 Supply F1 F2 F3 6(6) 4(8) (2) 2(14) (1) 2(4) Demand The Total feasible transportation cost = 6(6) + 4(8) + 9(2) + 2(14) + 6(1) + 2(4) = Rs. 128/- 2. Row Minima Method: Factories W1 W2 W3 W4 Supply F1 F2 F (14) 5 8(6) 9(5) 2(1) 7(4) 4 3(5) Demand The Total feasible transportation cost = 1(14)+8(6)+9(5)+2(1)+7(4)+3(5) = Rs.152/- 4

5 3. Column-Minima Method: Factories W1 W2 W3 W4 Supply F1 F2 6(1) 4(10) 1(3) (12) 7(4) F3 4(5) Demand The Total feasible transportation cost = 6(1)+4(10)+1(3)+2(12)+7(4)+4(5) = Rs. 121/- 4. Matrix-Minima Method or Least Cost method: Factories W1 W2 W3 W4 Supply F1 F2 F (14) 5 8(6) 9(9) 2(1) 7 4 3(1) 6 2(4) Demand The Total feasible transportation cost = 1(14)+8(6)+9(9)+2(1)+3(1)+2(4) = 156/- 5

6 5. VAM- Vogel s approximation method: Step-I: Against each row and column of the matrix, denote the difference between the two least cost in that particular row and column. Step-II: Select the maximum value noted as per step-i, in this row or column select the cell which has the least cost Step-III: Allocate the maximum possible quantity Step-IV: After fulfilling the requirements of that particular row or column, Ignore that particular row or column and recalculate the difference by the two lowest cost for each of the remaining rows or columns, Again select the maximum of these differences and allocate the maximum possible quantity in the cell with the lowest cost in that particular / corresponding row or column. Step-V: Repeat the procedure till the initial allocation is completed 5. VAM- Vogel s approximation method: Factories W1 W2 W3 W4 Supply F1 F2 F3 6 (4) 4(10) 1 5 8(1) 9 2(15) 7 4(1) 3 6 2(4) Demand

7 The Total feasible transportation cost = 6(4) + 4(10) + 8(1) + 2(15) + 4 (1) + 2(4) = 114/- II- Check for degeneracy: If (m+n-1) is not equal to the number of allocated cells, then it is called degeneracy in transportation problems, Where m= number of rows, n= number of columns. This will occur if the source and destination is satisfied simultaneously. The degeneracy can be avoided by introducing a dummy allocation cell. To equate the number of allocated cells equal to (m+n-1) For the above problem (m+n-1) = (3+4-1) =6 = Number of allocations = 6 Hence there is no degeneracy. III- Checking optimality using MODI method: For allocated cells C ij (Ui +V j ) = 0 For unallocated cells Cij (U ij ) +Vj) 0 V1=0 V2=-2 V3 =-6 V4= -2 6(4) 4(10) 1 5 8(1) 9 2(15) 7 U1=6 U2=8 4(1) 3 6 2(4) U3= 4 The Total feasible transportation cost = 6(4)+4(10)+8(1)+2(15)+4(1)+2(4) = 114/- 7

8 Problem.2 There are 3 Parties who supply the following quantity of coal P1= 14t, P2=12t, P3= 5t. There are 3 consumers who require the coal as follows C1=6t, C2=10t, C3=15t. The cost matrix in Rs. Per ton is as follows. Find the schedule of transportation policy which minimises the cost: Solution: Factories W1 W2 W3 Supply F1 F2 6 8(5) 4(9) 4(6) 9 3(6) F3 1 2(5) 6 05 Demand

9 Therefore the total feasible transportation cost = 8(5)+ 4(9) + 4(6) +3 (6) + 2(5) = Rs. 128/- II. Check for Degeneracy: (m+n-1) = (3+3-1) = 5 = Number of allocations Hence there is no degeneracy III- Checking optimality using MODI method: For allocated cells C ij (U i +V j ) = 0 For unallocated cells C ij (U ij ) +V j ) 0 V1=5 V2=8 V3=4 6 8(5) 4(9) 4(6) 9 3(6) 1 2(5) 6 U1= 0 U2=-1 U3=-6 Since all the marginal costs for the unallocated cells are positive, it gives an optimal solution and the total minimum transportation cost = Rs. 128/- 9

10 Factories W1 W2 W3 W4 Supply F1 F2 F Demand Since the unit cost of production is Rs.4, 3 and 5 at the three factories, therefore Total cost = Production cost + cost Factories W1 W2 W3 W4 Supply F1 F (300) 12 7(100) 9(300) 12 8(100) F3 7(100) 11 9(100) Demand Therefore the total feasible solution = 7(300) + 7(100) +9(300) + 8(100) +7(100) + 9(100) = Rs. 7900/- 10

11 Factories W1 W2 W3 W4 Supply F1 F2 F (300) 12 7(100) 9(300) 12 8(100) 7(100) 11 9(100) Demand Therefore the total feasible transportation solution = 7(300) + 7(100) + 9(300) + 8(100) + 7(100) + 9(100) = Rs.7900/- II. Check for Degeneracy: (m+n-1) = (3+4-1) = 6 = Number of allocations Hence there is no degeneracy III- Checking optimality using MODI method: For allocated cells C ij (Ui +V j ) = 0 For unallocated cells Cij (U ij ) +Vj) 0 V1 V2 V3 V (300) 12 7(100) 9(300) 12 8(100) U1= -2 U2=0 7(100) 11 9(100) 10 U3=0 Since all the marginal costs for the unallocated cells are positive, it gives an optimal solution and the total minimum cost = Rs. 7900/- 11

12 Problem-4 A company has three plants supplying the same product to the five distribution centers. Due to peculiarities inherent in the set of cost of manufacturing, the cost/ unit will vary from plant to plant. Which is given below. There are restrictions in the monthly capacity of each plant, each distribution center has a specific sales requirement, capacity requirement and the cost of transportation is given below. Factories W1 W2 W3 W4 W5 Supply The cost of manufacturing a product at the different plants is Fixed cost is Rs 7x10 5, 4x 10 5 and 5x Whereas the variable cost per unit is Rs 13/-, 15/- and 14/- respectively. Determine the quantity to be dispatched from each plant to different distribution centers, satisfying the requirements at minimum cost. Solution: F1 F2 F Demand Factories W1 W2 W3 W4 W5 D Supply F F2 F Demand

13 Factories W1 W2 W3 W4 W5 D Supply F1 F2 F (55) 16(85) 19 17(60) (25) (100) (105) 17(10) Demand Therefore the total feasible transportation cost = = 16(55) + 16(85) + 17(60) + 20(25) + 0(100) +16(60) + 16(105) +17(10) = Rs6570/- II. Check for Degeneracy: (m+n-1) = (6+3-1) = 8 = Number of allocations Hence there is no degeneracy 13

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17 Therefore total feasible solution = 1(15) +4(10)+4(35)+2(5)+0+3(22)+2(20)+4(3)= Rs. 323/- Therefore total feasible solution = 1(15) +4(10)+4(35)+2(5)+0+3(22)+2(20)+4(3)= Rs. 323/- II. Check for degeneracy: (m+n-1) = (3+6-1) = 8 = Number of allocations Hence there is no degeneracy III. Check for optimality: V1=3 V2=1 V3=2 V4=4 V5=4 V6=2 4 1(15) 3 4(10) 4(35) 0(-2) (5) 3 0(30) 3(22) 5 2(20) 4(3) 4 0(-2) U1=0 U2=-2 U3=0 17

18 V1=3 V2=1 V3=2 V4=4 V5=4 V6=2 4 1(15) 3 4(10-x) 4(35) 0(x) (5+x) 3 0(30-x) 3(22) 5 2(20) 4(3) 4 0(-2) U1=0 U2=-2 U3=0 10-x=0 30-x=0 Min. value of x=10 V1=1 V2=1 V3=0 V4=2 V5=4 V6=0 4 1(15) 3 4 4(35) 0(10) (15) 3(-1) 0(20) 3(22) 5 2(20) 4(3) 4 0(-2) 20-X = 0: 3-x=0: therefore x=3 V1=3 V2=1 V3=2 V4=3 V5=4 V6=0 U1=0 U2=0 U3=2 4 1(15) 3 4 4(35) 0(10) (18) 3(-1) 0(17) 3(22) 5 2(20) 4 4 0(3) U1=0 U2=-1 U3=0 18

19 35-x=0 and 17-x = 0, therefore x= 17 V1=3 V2=1 V3=2 V4=3 V5=4 V6=0 4 1(15) 3 4 4(18) 0(27) (18) 3(17) 0 3(22) 5 2(20) 4 4 0(3) 27-x=0 : 22-X= 0: 17-X=0: therefore X=17 U1=0 U2=-1 U3=0 V1=3 V2=1 V3=2 V4=3 V5=4 V6=0 4 1(15) 3 4 4(35) 0(10) 2(17) 3 2 2(18) 3 0 3(5) 5 2(20) 4 4 0(20) U1=0 U2=-1 U3=0 19

20 Factories A B C Supply M1 M2 M Demand

21 Factories A B C Supply M1 M2 M Demand Factories A B C Supply M (70) 118(90) 160 M2 4(90) 48 45(30) 120 M3 1 0(140) Demand

22 V1=4 V2=42 V3= (70+x) 18(90-x) 4(90) 48 45(30) 1 0(140-x) 0(x) U1=73 U2=0 U3= X=0 and 140-X=0, therefore X=90 V1=4 V2=45 V3= (160) 18 U1=70 4(90) 48 45(30) U2=0 1 0(50) 0(90) U3=-45 22

23 Factories A B C Supply M1 M2 M Demand

24 Factories A B C Supply M1 M2 M Demand Factories A B C Supply M1 M2 M3 116(90) (70) 48 45(140) 1 0(120) Demand

25 V1=0 V2=-1 V3= (90) (70) 48 45(140) 1(θ) 0(120) 0 U1=116 U2=46 U3=1 Factories A B C Supply M1 M2 M3 116(90) (70) 48 45(140) 1(θ) 0(120) Demand

26 Factories A B C D Supply I II III Demand Factories A B C D Supply I II III D Demand A B C D Supply 4(150) 0(300) (50) 2 6(130) 7(120) (150) 0 0 0(20)

27 V1= 5 V2=1 V3= 6 V4=7 4(150) 0(300) 6 5(-1) 5(50) 2 6(130) 7(120) (150) 0 0 0(20) 0 V1= 5 V2=1 V3= 6 V4=7 4(150-X) 0(300) 6 5(X) 5(50+X) 2 6(130) 7(120-X) (150) 0 0 0(20) 0 U1=-1 U2=0 U3=-1 U4=-6 U1=-1 U2=0 U3=-1 U4= X= 0, 120-X= 0, therefore X=120 27

28 V1= 5 V2=1 V3= 6 V4=7 4(150-X) 0(300) 6 5(X) 5(50+X) 2 6(130) 7(120-X) (150) 0 0 0(20) 0 U1=-1 U2=0 U3=-1 U4=-6 V1= 4 V2=0 V3=5 V4=5 4(30) 0(300) 6 5(120) 5(170) 2 6(130) (150) 0 0 0(20) 0 U1=0 U2=1 U3=1 U4=-5 28

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31 III. CHECK FOR OPTIMALITY 31

32 Factories D E F G D1 Supply A B C OT.A OT.B 11(200) 13(50) (θ) 16 18(175) 14 10(175) (275) 10(125) (50) (75) Factories D E F G Supply A B C OT.A OT.B Demand

33 Factories D E F G D1 Supply A B C 11(200) 13(50) (175) 14 10(175) (275) 10(125) OT.A OT.B (50) (75) Demand

34 V1=16 V2=18 V3=13 V4=10 V5=0 11(200) 13(50+θ) (175-θ) 14 10(175) 0(θ) (275) 10(125) (-1) (50) (75) U1=-5 U2=5 U3=5 U4=0 U5=0 34

35 11(200) 13(50+θ) (175-θ-x) 14 10(175) 0(θ+x) (275) 10(125) (x) (50-x) (75) 35

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39 Reference Books: 1. Taha H A, Operation Research - An Introduction, Prentice Hall of India, 7 th edition, Ravindran, Phillips and Solberg, Operations Research : Principles and Practice, John Wiely & Sons, 2 nd Edition 3. D.S.Hira, Operation Research, S.Chand & Company Ltd., New Delhi,