Chapter 7 Transportation, Assignment and Transshipment Problems to accompany Operations Research: Applications & Al - PowerPoint PPT Presentation

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Chapter 7 Transportation, Assignment and Transshipment Problems to accompany Operations Research: Applications & Al

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  1. Chapter 7Transportation, Assignment and Transshipment Problems to accompanyOperations Research: Applications & Algorithms,4th edition, by Wayne L. Winston

  2. Description A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points. While trying to find the best way, generally a variable cost of shipping the product from one supply point to a demand point or a similar constraint should be taken into consideration.

  3. 7.1 Formulating Transportation Problems Example 1: Powerco has three electric power plants that supply the electric needs of four cities. The associated supply of each plant and demand of each city is given in the table 1. The cost of sending 1 million kwh of electricity from a plant to a city depends on the distance the electricity must travel.

  4. Transportation tableau A transportation problem is specified by the supply, the demand, and the shipping costs. So the relevant data can be summarized in a transportation tableau. The transportation tableau implicitly expresses the supply and demand constraints and the shipping cost between each demand and supply point.

  5. Table 1. Shipping costs, Supply, and Demand for Powerco Example Transportation Tableau

  6. Solution Decision Variable: Since we have to determine how much electricity is sent from each plant to each city; Xij = Amount of electricity produced at plant i and sent to city j X14 = Amount of electricity produced at plant 1 and sent to city 4

  7. 2. Objective function Since we want to minimize the total cost of shipping from plants to cities; Minimize Z = 8X11+6X12+10X13+9X14 +9X21+12X22+13X23+7X24 +14X31+9X32+16X33+5X34

  8. 3. Supply Constraints Since each supply point has a limited production capacity; X11+X12+X13+X14 <= 35 X21+X22+X23+X24 <= 50 X31+X32+X33+X34 <= 40

  9. 4. Demand Constraints Since each supply point has a limited production capacity; X11+X21+X31 >= 45 X12+X22+X32 >= 20 X13+X23+X33 >= 30 X14+X24+X34 >= 30

  10. 5. Sign Constraints Since a negative amount of electricity can not be shipped all Xij’s must be non negative; Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

  11. LP Formulation of Powerco’s Problem Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24 +14X31+9X32+16X33+5X34 S.T.: X11+X12+X13+X14 <= 35 (Supply Constraints) X21+X22+X23+X24 <= 50 X31+X32+X33+X34 <= 40 X11+X21+X31 >= 45 (Demand Constraints) X12+X22+X32 >= 20 X13+X23+X33 >= 30 X14+X24+X34 >= 30 Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

  12. General Description of a Transportation Problem A set of m supply points from which a good is shipped. Supply point i can supply at most si units. A set of n demand points to which the good is shipped. Demand point j must receive at least di units of the shipped good. Each unit produced at supply point i and shipped to demand point j incurs a variable cost of cij.

  13. Xij = number of units shipped from supply point i to demand point j

  14. Balanced Transportation Problem If Total supply equals to total demand, the problem is said to be a balanced transportation problem:

  15. Balancing a TP if total supply exceeds total demand If total supply exceeds total demand, we can balance the problem by adding dummy demand point. Since shipments to the dummy demand point are not real, they are assigned a cost of zero.

  16. Balancing a transportation problem if total supply is less than total demand If a transportation problem has a total supply that is strictly less than total demand the problem has no feasible solution. There is no doubt that in such a case one or more of the demand will be left unmet. Generally in such situations a penalty cost is often associated with unmet demand and as one can guess this time the total penalty cost is desired to be minimum

  17. 7.2 Finding Basic Feasible Solution for TP Unlike other Linear Programming problems, a balanced TP with m supply points and n demand points is easier to solve, although it has m + n equality constraints. The reason for that is, if a set of decision variables (xij’s) satisfy all but one constraint, the values for xij’s will satisfy that remaining constraint automatically.

  18. Methods to find the bfs for a balanced TP There are three basic methods: Northwest Corner Method Minimum Cost Method Vogel’s Method

  19. 1. Northwest Corner Method To find the bfs by the NWC method: Begin in the upper left (northwest) corner of the transportation tableau 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 this 2 values).

  20. According to the explanations in the previous slide we can set x11=3 (meaning demand of demand point 1 is satisfied by supply point 1).

  21. 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).

  22. 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).

  23. Finally, we will have the following bfs, which is: x11=3, x12=2, x22=3, x23=2, x24=1, x34=2

  24. 2. Minimum Cost Method The Northwest Corner Method dos not utilize shipping costs. It can yield an initial bfs easily but the total shipping cost may be very high. The minimum cost method uses shipping costs in order come up with a bfs that has a lower cost. To begin the minimum cost method, first we find the decision variable with the smallest shipping cost (Xij). Then assign Xijits largest possible value, which is the minimum of si and dj

  25. After that, as in the Northwest Corner Method we should cross out row i and column j and reduce the supply or demand of the noncrossed-out row or column by the value of Xij. Then we will choose the cell with the minimum cost of shipping from the cells that do not lie in a crossed-out row or column and we will repeat the procedure.

  26. An example for Minimum Cost MethodStep 1: Select the cell with minimum cost.

  27. Step 2: Cross-out column 2

  28. Step 3: Find the new cell with minimum shipping cost and cross-out row 2

  29. Step 4: Find the new cell with minimum shipping cost and cross-out row 1

  30. Step 5: Find the new cell with minimum shipping cost and cross-out column 1

  31. Step 6: Find the new cell with minimum shipping cost and cross-out column 3

  32. Step 7: Finally assign 6 to last cell. The bfs is found as: X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6

  33. 3. Vogel’s 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.

  34. An example for Vogel’s MethodStep 1: Compute the penalties.

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

  36. Step 3: Identify the largest penalty and assign the highest possible value to the variable.

  37. Step 4: Identify the largest penalty and assign the highest possible value to the variable.

  38. Step 5: Finally the bfs is found as X11=0, X12=5, X13=5, and X21=15

  39. 7.3 The Transportation Simplex Method In this section we will explain how the simplex algorithm is used to solve a transportation problem.

  40. How to Pivot a Transportation Problem Based on the transportation tableau, the following steps should be performed. Step 1. Determine (by a criterion to be developed shortly, for example northwest corner method) the variable that should enter the basis. Step 2. Find the loop (it can be shown that there is only one loop) involving the entering variable and some of the basic variables. Step 3. Counting the cells in the loop, label them as even cells or odd cells.

  41. Step 4. Find the odd cells whose variable assumes the smallest value. Call this value θ. The variable corresponding to this odd cell will leave the basis. To perform the pivot, decrease the value of each odd cell by θ and increase the value of each even cell by θ. The variables that are not in the loop remain unchanged. The pivot is now complete. If θ=0, the entering variable will equal 0, and an odd variable that has a current value of 0 will leave the basis. In this case a degenerate bfs existed before and will result after the pivot. If more than one odd cell in the loop equals θ, you may arbitrarily choose one of these odd cells to leave the basis; again a degenerate bfs will result

  42. Illustration of pivoting procedure on the Powerco example. We want to find the bfs that would result if X14 were entered into the basis. 5 0 4 3 2 1 Nothwest Corner bfs and loop for Powerco

  43. New bfs after X14 is pivoted into basis. Since There is no loop involving the cells (1,1), (1,4), (2,1), (2,2), (3,3) and (3, 4) the new solution is a bfs. After the pivot the new bfs is X11=15, X14=20, X21=30, X22=20, X33=30 and X34=10.

  44. Two important points! In the pivoting procedure: Since each row has as many +20s as –20s, the new solution will satisfy each supply and demand constraint. By choosing the smallest odd variable (X23) to leave the basis, we ensured that all variables will remain nonnegative.

  45. Pricing out nonbasic variables To complete the transportation simplex, now we will discuss how to row 0 for any bfs. For a bfs in which the set of basic variables is BV, the coefficient of the variable Xij (call it čij) in the tableau’s row ) is given by čij = cBV B-1aij – cij Where cij is the objective function coefficient for Xij and aij is the column for xij in the original LP.

  46. Since the example is a minimization problem, the current bfs will be optimal if all the čij‘s are nonpositive; otherwise, we enter into the basis with the most positive čij. After determining cBVB-1 we can easily determine čij. Since the first constraint has been dropped, cBVB-1 will have m+n-1 elements. cBVB-1= [u2 u3…um v1 v2…vn] Where u2 u3…um are elements of cBVB-1 corresponding to the m-1 supply constraints, and v1 v2…vn are elements of cBVB-1 corresponding to the n demand constraints.

  47. To determine cBVB-1 we use the fact that in any tableau, each basic variable Xij must have čij=0. Thus for each of the m+n-1 variables in BV,cBV B-1aij – cij=0For the Northwest corner bfs of Powerco problem, BV={X11, X21, X22, X23, X33, X34}. Applying the equation above we obtain:č11= [u2 u3 v1 v2 v3 v4] -8 = v1-8=0

  48. č21= [u2 u3 v1 v2 v3 v4] -9 = u2+v1-9=0 č22= [u2 u3 v1 v2 v3 v4] -12 = u2+v2-12=0 č23= [u2 u3 v1 v2 v3 v4] -13 = u2+v3-13=0

  49. č33= [u2 u3 v1 v2 v3 v4] -16 = u3+v3-16=0 č34= [u2 u3 v1 v2 v3 v4] -5 = u3+v4-5=0 For each basic variable Xij (except those having i=1), we see that the equation we used above reduces to ui+vj=cij. If we define u1=0, we must solve the following system of m+n equations.

  50. u1=0 u1+v1=8 u2+v1=9 u2+v2=12 u2+v3=13 u3+v3=16 u3+v4=5 By solving the system above we obtain: v1=8, u2=1, v2=11,v3=12, u3=4, v4=1