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Management Science – MNG221

Management Science – MNG221. Lecture 1: Transportation, Transshipment, and Assignment. Transportation, Transshipment , and Assignment. T ransportation , transshipment, and assignment problems are special types of linear programming model formulations.

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Management Science – MNG221

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  1. Management Science – MNG221 Lecture 1: Transportation, Transshipment, and Assignment

  2. Transportation, Transshipment, and Assignment • Transportation, transshipment, and assignment problems are special types of linear programming model formulations. • They are part of a larger class of linear programming problems known as network flow problems. • There solution approaches are variations of the traditional simplex solution procedure.

  3. Transportation, Transshipment, and Assignment Transportation Model

  4. Transportation Model • Transportation models are formulated to solve problems where: • A product is transported from a number of sources to a number of destinations at the minimum possible cost. • Each source is able to supply a fixed number of units of the product, and each destination has a fixed demand for the product.

  5. Transportation Model • In a transportation problem, items are allocated from sources to destinations at a minimum cost. • The linear programming model for a transportation problem has constraints for supply at each source and demand at each destination.

  6. Transportation Model • A Transportation can be a: • Balanced transportation model in which supply equals demand, all constraints are equalities.Supply = Demand • Unbalanced transportation model in which supply exceeds demand or demand exceeds supply: Supply < Demand Supply > Demand Transportation problems are usually solved manually within a tableau format.

  7. Transportation, Transshipment, and Assignment The Transshipment Model

  8. Transshipment Model • The transshipment model is an extension of the transportation model in which intermediate transshipment points are added between the sources and destinations. • An example of a transshipment point is a distribution center or warehouse located between plants and stores.

  9. Transshipment Model • In a transshipment problem, items may be transported: • From sources through transshipment points on to destinations, • From one source to another, • From one transshipment point to another,

  10. Transshipment Model • From one destination to another, • Or directly from sources to destinations, • Or some combination of these alternatives. • The transshipment model includes intermediate points between sources and destinations.

  11. Transportation, Transshipment, and Assignment The Assignment Model

  12. Assignment Model • The Assignment Modelis a special form of a linear programming model that is similar to the transportation model. • The differences, however, is that the supply at each source and the demand at each destination are each limited to one unit.

  13. Assignment Model • The linear programming formulation of the assignment model is similar to the formulation of the transportation model, except all the supply values for each source equal one, and all the demand values at each destination equal one.

  14. Transportation, Transshipment, and Assignment Transportation: Worked Example

  15. Transportation: Worked Example • Wheat is harvested in the Midwest and stored in grain elevators in three different cities - Kansas City, Omaha, and Des Moines. • These grain elevators supply three flour mills, located in Chicago, St. Louis, and Cincinnati. • The cost of transporting one ton of wheat from each grain elevator (source) to each mill (destination) differs according to the distance and rail system. • The problem is to determine how many tons of wheat to transport from each grain elevator to each mill on a monthly basis in order to minimize the total cost of transportation.

  16. Transportation: Worked Example

  17. Transportation, Transshipment, and Assignment Transportation: Worked Example

  18. Transportation

  19. Transportation: Worked Example

  20. Transportation: Worked Example Minimize Z = $61A + $81B + $101C + $72A + $112B + $112C + $43A + $53B + $123C Subject to: Supply Constraints • x1A + x1B + x1C = 150 • x2A + x2B + x2C = 175 • x3A + x3B + x3C = 275 Demand Constraints • x1A + x2A + x3A = 200 • x1B + x2B + x3B = 100 • x1C + x2C + x3C = 300 Non-negativity • xij≥ 0 Xij- the number of tons of wheat transported from each grain elevator, i - (where i = 1, 2, 3), to each mill, j -(where j = A, B, C). Example Supply Constraint, x1A + x1B + x1C = 150 Kansas City to all three mills: Chicago (x1A), St. Louis (x1B), and Cincinnati (x1C). Constraints are an equation( = ) and not a inequality (≤), because all of wheat available is needed to meet demand of 600 tons.

  21. Transportation • Minimize Z = $61A + $81B + $101C + $72A + $112B + $112B + $43A + $53B + $123C • Subject to: • x1A + x1B + x1C = 150 x1A + x2A + x3A = 200 xij≥ 0 • x2A + x2B + x2C = 175 x1B + x2B + x3B = 100 • x3A + x3B + x3C = 275 x1C + x2C + x3C = 300 • Each cell in a transportation tableau is analogous to a decision variable that indicates the amount allocated from a source to a destination. • The supply and demand values along the outside of the rim of the tableau are called rim requirements. To From

  22. Transportation: Worked Example • 2 -Methods for solving a Transportation Model – • The Stepping-Stone Method • The Modified Distribution Method (also known as MODI) BUT • Must be given an Initial Solution • The Northwest Corner Method • The Minimum Cell Cost Method • Vogel’s Approximation Method

  23. Transportation Northwest Corner Method the largest possible allocation is made in the cell in the upper left hand corner of the tableau, followed by allocations to adjacent feasible cells. The initial solution is complete when all rim requirements are satisfied. To From 150 100 25 50 275 The cost of the initial cost is obtained by substituting the allocations in the objective function Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C = 6(150) + 8(0) + 10(0) + 7(50) + 11(100) + 11(25) + 4(0) + 5(0) + 12(275) = $5,925

  24. Transportation The Minimum Cost Method as much as possible is allocated to the cell with the minimum cost. The initial solution is complete when all rim requirements are satisfied. To From 25 125 175 200 75 The cost of the initial cost is obtained by substituting the allocations in the objective function Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C = 6(0) + 8(25) + 10(125) + 7(0) + 11(0) + 11(175) + 4(200) + 5(75) + 12(0) = $4,550

  25. Transportation • Vogel's Approximation Model is based on the concept of penalty cost or regret. • A Penalty Cost is the difference between the largest and next largest cell cost in a row (or column). • VAM allocates as much as possible to the minimum cost cell in the row or column with the largest penalty cost.

  26. Transportation • Vogel's Approximation Model Steps • Determine the penalty cost for each row and column by subtracting the lowest cell cost in the row or column from the next lowest cell cost in the same row or column. • Column A: L C -4, Next LC -6. As such 6-4 = 2 • Select the row or column with the highest penalty cost (breaking ties arbitrarily or choosing the lowest-cost cell).

  27. Transportation • Vogel's Approximation Model Steps • Allocate as much as possible to the feasible cell with the lowest transportation cost in the row or column with the highest penalty cost. • Repeat steps 1, 2, and 3 until all rim requirements have been met.

  28. Transportation To From 2 4 175 1 2 1 3

  29. Transportation To From 2 175 1 8 100 2 1 3 After each VAM cell allocation, all row and column penalty costs are recomputed

  30. Transportation To From 2 150 175 8 25 100 150 2 1 After each VAM cell allocation, all row and column penalty costs are recomputed

  31. Transportation To From 150 175 25 100 150 The cost of the initial cost is obtained by substituting the allocations in the objective function Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C = 6(0) + 8(0) + 10(150) + 7(175) + 11(0) + 11(0) + 4(25) + 5(100) + 12(150) = $5,125

  32. Transportation • After obtaining the initial solution, the problem may be solved using: • Stepping-Stone Method • Modified Distribution Method (MODI) Stepping Stone Method – determines if there is a cell with no allocation that would reduce cost if used. Modified Distribution Method (MODI)– a modified version of the stepping-stone method. However, the individual cell cost changes are determined mathematically.

  33. Transportation • Stepping Stone Method • Step 1: Determine the stepping-stone path and cost changes for each empty cell in the tableau. • Step 2: Allocate as much as possible to the empty cell with the greatest net decrease in cost. • Step 3: Repeat steps 1 and 2 until an empty cell have positive cost changes that indicates an optimal solution.

  34. Transportation Step 1: Determine the stepping-stone path and cost changes for each empty cell in the tableau. But First Identify Empty Cells : 1A 2A 2B 3C To From X 25 125 X 175 X 200 75 X

  35. Transportation Evaluation of Cell 1A In evaluating the empty cells the constraint of the problems cannot be violated, and feasibility must be maintained. Review of the cost increase/reduction of the process. $6 - $8 + $5 - $4 = -$1 To From +1 -1 25 125 175 -1 +1 200 75 1A 1B 3B 3A

  36. Transportation Evaluation of Cell 2A Review of the cost increase/reduction of the process. To From -1 +1 25 125 -1 +1 175 -1 +1 200 75 2A 2C 1C 3B 3A 1B +$7 - $11 + $10 - $8 +$5 - $4 = -$1

  37. Transportation Evaluation of Cell 1A 2A 2B 3C Review of the cost increase/reduction of the process. +$11 - $11 + $10 - $8 = +$2 To From -1 +1 25 125 +1 -1 175 200 75 2B 2C 1C 1B

  38. Transportation Evaluation of Cell 1A 2A 2B 3C Review of the cost increase/reduction of the process. +$12 - $10 + $8 - $5 = +$5 To From +1 -1 25 125 175 -1 +1 200 75 2B 2C 1C 1B

  39. Transportation Summary 1A – reduction $1 2A – reduction $1 2B – increase $2 3C – increase $5 To From 25 125 175 200 75

  40. Transportation Evaluation of Cell 1A 2A 2B 3C Identify the minimum in the stepping stone path. To From +1 -1 25 -25= 0 125 25 175 -1 +1 200 -25 = 175 75 + 25 = 100 The cost of the initial cost is obtained by substituting the allocations in the objective function Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C = 6(25) + 8(0) + 10(125) + 7(0) + 11(0) + 11(175) + 4(175) + 5(100) + 12(0) = $4,525

  41. Transportation, Transshipment, and Assignment The Assignment Model

  42. Assignment Model An assignment problemis a special form of transportation problem where all supply and demand values equal one. Steps of the assignment solution method are: • Perform row reductions by subtracting the minimum value in each row from all row values.

  43. Assignment Model Steps of the assignment solution method (Continued) • Perform column reductions by subtracting the minimum value in each column from all column value. • In the completed Opportunity Cost Table, cross out all zeros, using the minimum number of horizontal or vertical lines.

  44. Assignment Model Steps of the assignment solution method (Continued) • If fewer than m lines are required (where m = the number of rows or column), subtract the minimum uncrossed value from all the other uncrossed values, and add this same minimum uncrossed value where two lines intersect. Leave all other values unchanged.

  45. Assignment Model Steps of the assignment solution method (Continued) • If m lines are required, the tableau contains the optimal solution and m unique assignments are made. If fewer than m lines are required, repeat step 4.

  46. Transportation, Transshipment, and Assignment The Assignment Model- A Worked Example

  47. Assignment Model - Example • The Atlantic Coast Conference has four basketball games on a particular night. • The conference office wants to assign four teams of officials to the four games in a way that will minimize the total distance traveled by the officials. • The distances in miles for each team of officials to each game location are shown below.

  48. Assignment Model Step 1: Perform row reductions by subtracting the minimum value in each row from all row values.

  49. Assignment Model Step 1: Perform column reductions by subtracting the minimum value in each column from all column value.

  50. Assignment Model Step 3: In the completed opportunity cost table, cross out all zeros, using the minimum number of horizontal or vertical lines.

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