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- 1. TRANSPORTATION(SPECIALLY STRUCTURED LINEAR PROGRAMS) SUBMITTED BY: Transportation 1 Escober, Mark de la Cruz, Edghan Bryan Transportation 2 Kali, Norhasim Lagasca, Michelle Ronidel, Roderick
- 2. TRANSPORTATION AND ASSIGNMENT PROBLEM Transportation problem is concerned with selecting routes in a product – distribution networks among manufacturing plants to distribution warehouse or among regional warehouses to local distribution outlets. Assignment problem involves assigning employees to tasks, salesperson to towns, contract to bidders or jobs to plants.
- 3. TRANSPORTATION METHOD
- 4. The Transportation Problem:(demand equals supply) Let us consider the case of the Bulacan Gravel Company , which has received a contract to supply gravel for three road projects located in the towns of Marilao, Bocaue and Balagtas. Construction engineers have estimated the amount of gravel which be needed at three construction projects: Weekly Requirement Project Location Truckloads A Marilao 72 B Bocaue 102 C Balagtas 41 Total 215
- 5. The Transportation Problem:(demand equals supply) The Bulacan Gravel Company has three gravel plants located in the towns of San Rafael, San Ildefonso and San Miguel. The gravel required for the construction projects can be supplied by these three plants. Bulacan’s chief dispatcher has calculated the amounts of gravel which can be supplied by each plant: Amount Available/Week Plant Location Truckloads W San Rafael 56 X San Ildefonso 82 Y San Miguel 77 Total 215 The company has computed the delivery costs from each plant to each project site. Cost per Truckload From To project A To project B To project C Plant W P4 P8 P8 Plant X 16 24 16 Plant Y 8 16 24
- 6. Gravel plants, road construction projects and transportation costs for Bulacan GravelCompany P4 Plant W (San Rafael) Project A 56 loads available (Marilao) 72 loads required P8 P16 Project B (Bocaue) P24 Plant X ( San Ildefonso) 102 loads 82 loads available required P8 P8 P16 P16 Project C Plant Y (San Miguel) P24 (Balagtas) 77 loads available 41 loads required
- 7. The objective is to minimize the total transportation cost: 4WA + 8WB + 8WC + 16XA + 24XB + 16XC + 8YA + 16YB + 24YC There are three “origin constraints” which say that Bulacan cannot ship out more gravel than they have: WA + WB + WC < 56: Plant W XA + XB + XC < 82 : Plant X YA + YB + YC < 77: Plant Y There are three “destination constraints”, which say that each project must receive the gravel it requires: WA + XA + YA > 72: Project A WB + XB + YB > 102: Project B WC +XC + YC > 41:Project C
- 8. Using it all together, we get the final linear programming model:minimize 4WA + 8WB + 8WC + 16XA + 24XB + 16XC + 8YA + 16YB + 24YCsubject to WA + WB + WC < 56: Plant W XA + XB + XC < 82: Plant X YA + YB + YC < 77: Plant Y WA + XA + YA > 72: Project A > WB + XB + YB 102: Project B WC + XC + YC > 41: Project C All Variables > 0
- 9. STEP 1: Set up the transportation tableSTEP 2: Develop an initial solutionSTEP 3: Test the solution for improvementSTEP 4: Develop the improved solution
- 10. C Plant To Project Project Project From Capacit A B C y Plant 56 WA Plant X E 82 B Plant Y 77 Project 215 Requireme 72 102 41 215 nts D
- 11. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 56 X1 X2 X3 1 2 1 XA XB XC Plant X 6 4 6 82 X4 X5 X6 1 2 YA 8 YB YC Plant Y 6 4 77 X7 X8 X9 Project 72 102 41 215Requirem 215 ents
- 12. 1 WA 4 2 + - X1 3
- 13. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 56 56 1 2 1 XA XB XC Plant X 6 4 6 82 16 66 1 2 YA 8 YB YC Plant Y 6 4 77 36 41 Project 72 102 41 215Requirem 215 ents
- 14. From To Quantity,Plant Project truckloads/week W A 56 X A 16 X B 66 Y B 36 Y C 41 215 Used Squares = Total Rim Requirements – 1 5=6-1
- 15. Total Cost of the Initial Solution Source - destination Quantity Unit Total Combination Shipped x Cost = Cost WA 56 P 4 224.00 XA 16 16 256.00 XB 66 24 1,584.00 YB 36 16 576.00 YC 41 24 984.00 Total Transporation Cost P 3,624.00 Is this the best solution?
- 16. Choose the unused square to be evaluated Beginning with the selected unused square, traced a closed path (moving horizontally and vertically only) Assign plus (+) and minus (-) signs alternately at each corner square of the closed path Determine the net change in costs as a result of the changes made in tracing the path Repeat the above steps until an improvement index has been determined for each unused squares
- 17. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 56 56 _ + 1 2 1 XA XB XC Plant X 6 4 6 82 16 + 66 _ 1 2 YA 8 YB YC Plant Y 6 4 77 36 41 Project 72 102 41 215Requirem 215 ents
- 18. Addition to cost: From plant W to project B P 8 From plant X to project A 16 P24Reduction to cost: From plant W to project A P 4 From plant X to project B 24 28 -P4 Improvement index for square WB = WB – WA + XA – XB = P8 - P4 + P16 - P24 WB = -P4
- 19. Improvement index for WC = WC – WA + XA – XB + YB – YC = P8 – 4P + P16 -P24 + P16 - P 24 WC = - P 12
- 20. Improvement index for XC= XC – XB + YB – YC = P16 - P24 + P16 - P24 XC = -P16
- 21. Improvement index for YA = YA - XA + XB – YB = P8 - P16 + P24 - P16 YA = P 0
- 22. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 56 56 -4 -12 1 2 1 XA XB XC Plant X 6 4 6 82 16 66 -16 1 2 YA 8 YB YC Plant Y 6 4 77 0 36 41 Project 72 102 41 215Requirem 215 ents
- 23. 24 16XB XC `6 _6 + 16 24YB YC3 4 + _6 1
- 24. XB XC66 - 41 0 + 41 ` 2 4 5 1 YB YC36 + 41 41 - 41 7 7 0
- 25. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 56 56 1 2 1 XA XB XC Plant X 6 4 6 82 16 25 41 1 2 YA 8 YB YC Plant Y 6 n 4 77 77 Project 72 102 41 215Requirem 215 ents
- 26. Shipping Quantity Unit TotalAssignments Shipped x Cost = Cost WA 56 P4 P224 XA 16 16 256 XB 25 24 600 XC 41 16 656 YB 77 16 1,232 P 2,968
- 27. Unused Closed Computation ofSquares path Improvement Index WB + WB - WA + XA - XB + 8 - 4 + 16 - 24 = - 4 WC + WC - WA + XA - XC + 8 - 4 + 16 - 16 = + 4 YA + YA - XA + XB - YB + 8 - 16 + 24 - 16 = 0 YC + YC - XC + XB - YB + 24 - 16 + 24 -16 = + 16
- 28. WA WB56 - 25 0 + 25 ` 3 2 1 5 XA XB16 + 25 = 25 - 25 4 1 0
- 29. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 56 31 25 1 2 1 XA XB XC Plant X 6 4 6 82 41 41 1 2 YA 8 YB YC Plant Y 6 n 4 77 77 Project 72 102 41 215Requirem 215 ents
- 30. Shipping Quantity Unit TotalAssignments Shipped x Cost = Cost WA 31 P4 P124 WB 25 8 200 XA 41 16 656 XC 41 16 656 YB 77 16 1,232 P 2,868
- 31. Unused Closed Computation ofSquares path Improvement Index WC +WC - WA + XA - XC +8 – 4 + 16 – 16 = + 4 XB +XB - WB + WA - XA +24 – 8 + 4 – 16 = + 4 YA +YA - WA + WB – YB +8 – 4 + 8 – 16 = - 4 YC - YB + WB - WA + YC + XA - XC + 24 – 16 + 8 – 4 + 16 – 16 = + 12
- 32. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 56 56 1 2 1 XA XB XC Plant X 6 4 6 82 41 41 1 2 YA 8 YB YC Plant Y 6 n 4 77 31 46 Project 72 102 41 215Requirem 215 ents
- 33. Unused Closed Computation ofSquares path Improvement Index WA +WA – YA + YB - WB +4 – 8 + 16 – 8 = + 4 WC +WC – XC + XA – YA + YB - WB +8 – 16 + 16 – 8 + 16 – 8 = + 8 XB +XB – YB + YA - XA +24 – 16 + 8 – 16 = 0 YC +YC – XC + XA - YA +24 – 16 + 16 – 8 = +16
- 34. Shipping Quantity Unit TotalAssignments Shipped x Cost = Cost WB 56 P8 P448 XA 41 16 656 XC 41 16 656 YA 31 8 248 YB 46 16 736 Total Transportation Cost P 2,744
- 35. R i = value assigned to row iK = value assigned to row j jC i j = cost in square ij ( the square at the intersection of row i and column j
- 36. Kj K1Ri K2 K3 Plant To Project Project Project Capacit From A B C y WA 4 WB 8 WC 8R1 Plant W 5 56 6 1 2 1 XA XB 4 XC 6R2 Plant X 1 6 6 82 6 6 1 2 YA 8 YB 6 YCR3 Plant Y 3 4 4 77 6 1 Project 215 Requirem 72 102 41 215 ents
- 37. R1 + K1 To solve the five equations, then we proceed as follows. If R1 = 0, then =4R2 + K1 R1+ K1= 4 R2 + K2 = 24 = 16R2 +K = 24 0 + K1 = 4 12 + K2 = 24 2R3 +K = 16 K1 = 4 K2 = 12 + 2 R3 + K3 = 24R3 K3 = 24 4 + K3 = 24 K3 = 20 R2 + K1 = 16 R3 + K2 = 16 R2 + 4 = 16 R3 + 12 = 16 R2 = 12 R3 = 4
- 38. Kj K1 = 4 K2 = 12 K3 = 20Ri To Plant Project Project Project Capacit From A B C y WA 4 WB 8 WC 8R1 = 0 56 Plant W 5 6 1 2 1R2 =12 XA 6 XB 4 XC 6 Plant X 1 6 82 6 6 1 2 R3 = 4 YA 8 YB 6 YC 4 Plant Y 3 4 77 6 1 Project 215 Requirem 72 102 41 215 ents
- 39. Unused C ij – Ri – Kj ImprovementSquare Index 12 C12 - R1 – K2 -4 8 – 0 – 12 13 C13 - R1 - K3 -12 8 – 0 - 20 23 C23 – R2 - K3 -16 16 – 12 - 20 31 C31 – R3 – K1 0 8–4-4
- 40. Traced a close path for the cell having the largest negative improvement index Placed plus and minus signs at alternative corners of the path beginning with a plus sign at the unused square. The smallest stone in a negative position on the close path indicates the quantity that can be assigned to the unused square. Finally, the improvement indices for the new solution are calculated.
- 41. Kj K1 K2 K3Ri To Plant Project Project Project Capacit From A B C y WA 4 WB 8 WC 8R1 56 Plant W 5 6 1 2 1R2 XA 6 XB 4 XC 6 Plant X 1 2 4 82 6 5 1 1 2 R3 YA 8 YB 6 YC 4 Plant Y 7 77 7 Project 215 Requirem 72 102 41 215 ents
- 42. Stone square 11: Stone square 22: R1+ K1= 4 R2 + K2 = 24 0 + K1 = 4 12 + K2 = 24 Stone Square 32: K1 = 4 K2 = 12 R3 + K2 = 16 R3 + 12 = 16Stone square 21: Stone square 23: R3 = 4 R2+ K1= 16 R2 + K3 = 16 R2+ 4 = 16 12 + K3 = 16 R2 = 12 K3 = 4
- 43. Unused C ij – Ri – Kj ImprovementSquare Index 12 C12 - R1 – K2 -4 8 – 0 - 12 13 C13 - R1 – K3 +4 8–0-4 31 C31 – R3 – K1 0 8–4-4 33 C33 – R3 – K3 +16 24 – 4 - 4
- 44. Kj K1 = 4 K2 = 8 K3 = 4Ri To Plant Project Project Project Capacit From A B C y WA 4 WB 8 WC 8R1 = 0 56 Plant W 3 2 1 5 1 2 1R2 =12 XA 6 XB 4 XC 6 Plant X 4 4 82 1 1 1 2 R3 = 8 YA 8 YB 6 YC 4 Plant Y 7 77 7 Project 215 Requirem 72 102 41 215 ents
- 45. Unused C ij – Ri – Kj ImprovementSquare Index 13 C13 - R1 – K3 +4 8–0-4 22 C22 – R2 – K2 +4 24 – 12 - 8 31 C31 – R3 – K1 -4 8–8-4 33 C33 – R3 – K3 +12 24 – 8 - 4
- 46. Kj K1 K2 K3Ri To Plant Project Project Project Capacit From A B C y WA 4 WB 8 WC 8R1 56 Plant W 5 6 1 2 1R2 XA 6 XB 4 XC 6 Plant X 4 4 82 1 1 1 2 R3 YA 8 YB 6 YC 4 Plant Y 3 4 77 1 6 Project 215 Requirem 72 102 41 215 ents
- 47. Stone square 12: Stone square 23: R1+ K2= 8 Stone Square 32: R2 + K3 = 16 0 + K2 = 8 R3 + K2 = 16 16 + K3 = 16 K2 = 8 8 + K2 = 16 K3 = 0 K2 = 8Stone square 21: Stone square 31: R2+ K1= 16 R3 + K1 = 8 R2 + 0 = 16 R3 + 0 = 8 R2 = 16 R3 = 8
- 48. Unused C Ri – Kj ImprovementIs this the Optimal Solution? Square ij – Index 11 C11 - R1 – K1 +4 4–0-0 13 C13 – R1 – K3 +8 8–0-0 22 C22 – R2 – K2 0 24 – 16 - 8 33 C33 – R3 – K3 +16 24 – 8 - 0
- 49. 1. For each solution, compute the R and K values for the table2. Calculate the improvement indices for all unused squares3. Select the unused square with the most negative index4. Trace the closed path for the unused square having the most negative index.5. Develop an improved solution6. Repeat steps 1 to 5 until an optimal solution has been found.
- 50. Demand Less than SupplyConsidering the original Bulacan Gravel Company problem, suppose that plant W has a capacity of 76 truckloads for week rather than 56. The company would be able to supply 235 truckloads per week. However, the project requirements remain the same .
- 51. Plant To Project Project Project Dummy CapacitFrom A B C D y WA 4 WB 8 8 0Plant W WC 76 7 4 2 1 2 1 XA 6 XB 4 XC 6 0Plant X 8 82 2 1 2 YA 8 YB 6 YC 4 0Plant Y 1 4 2 77 6 1 0Project 235Requirem 72 102 41 20 235 ents Total Cost: 72 x P4 = P 288 4 x P8 = 32 82 x P24= 1,968 16 x P16 = 256 41 x P24 = 984 20x P0 = 0 P 3,528
- 52. Plant To Project Project Project Dummy CapacitFrom A B C D y WA 4 WB 8 8 0Plant W WC 76 7 6 1 2 1 XA 6 XB 4 XC 6 0Plant X 2 4 2 82 1 1 0 1 2 YA 8 YB 6 YC 4 0Plant Y 7 77 5 2Project 235Requirem 72 102 41 20 235 ents Total Cost: 76 x P8 = P 608 21 x P24= 504 41 x P16= 656 20 x P0 = 0 72 x P8 = 576 5x P16 = 80 P 2,424
- 53. Demand Greater than Supply Assume that project A will require 10 additional truckloads per week and that project C estimates additional requirements of 20 truckloads. The total project requirements now would be equal to 245 truckloads, as opposed to the 215 available from the plants. Total Cost: 56 x P4 = P224 26 x P16= 416 56 x P24= 1,344 46 x P16= 736 31 x P24= 744 30 x P 0= 0 P 3,464
- 54. Total Cost:56 x P 8 = P 44821 x P16 = 33661 x P16 = 97661 x P 8 = 48816 x P 16 = 25630 x P 0 = 0 P 2,504
- 55. There maybe an excessive number of stone squares in a solution; the number of stone squares is greater than the number of rim requirements minus 1. There may be an insufficient number of stone squares in a solution.
- 56. DEGENERACY IN ESTABLISHING AN INITIAL SOLUTION To Plant Project A Project B Project C From Capacity WA 4 WB 8 WC 8 Plant W 55 35 20 1 2 1 XA XB XC Plant X 6 4 6 25 16 25 1 2 YA 8 YB YC Plant Y 6 4 35 35 Project 35 45 35 115Requirem 115 ents
- 57. To Plant Project A Project B Project CFrom Capacity WA 4 WB 8 WC 8Plant W 55 35 20 1 2 1 XA XB XC Plant X 6 4 6 25 16 25 0 1 2 YA 8 YB YC Plant Y 6 4 35 35 Project 35 45 35 115Requirem 115 ents
- 58. To Plant A B CFrom Capacity 9 8 5 W 25 6 8 4 X 35 7 6 9 Y 40 Project 30 25 45 100Requirem 100 ents TRANSPORTATION TABLE FOR SUPER FERRY SHIPPING
- 59. Shipping Quantity x Unit = TotalAssignments Shipped Cost Cost WC 25 5 P125 WX 15 6 90 WC 20 4 80 YA 15 7 105 YB 25 6 150 Total Transportation Cost P550
- 60. Shipping Quantity x Unit = Total Assignments Shipped Cost Cost WA 25 9 P225 XA 5 6 30 XB 25 8 200 XC 5 4 20 YC 40 9 360 Total Transportation Cost P8350
- 61. Shipping Quantity X Unit = Total Assignments Shipped Cost Cost WA 15 9 P135 WC 10 5 50 XC 35 4 140 YA 15 7 105 YB 25 6 150 Total Transportation Cost P5800
- 62. Opportunity Costs for the first allocation Row W Row X First Allocation using the VAM (Vogel Approximation Method)0
- 63. Opportunity costs for the Second allocation Second Allocation Using the VAM0
- 64. Opportunity costs for the Third allocation Second Allocation Using the VAM0
- 65. •THE ASSIGNMENT PROBLEM
- 66. •THE ASSIGNMENT PROBLEM The Meycauayan Machine Shop does custom metalworking for a number of local plants. Meycauayan currently has three jobs to be done (let us symbolize them A, B and C). Meycauayan also has three machines on which to do the work (X, Y, and Z). Anyone of the jobs can be processed completely on any of the machines. Furthermore, the cost of processing any job on any machine is known. The assignment of jobs to machine must be on one-to-one basis; that is, each job must be assigned exclusively to one and only one machine. T jobs to the objective is to assign the jobs to the machines so as to minimize the cost. Machine Total Job X Y Z A P25 P31 P35 91 B 15 20 24 59 C 22 19 17 58 Total 62 70 76 208
- 67. STEP 1: Determine the opportunity- cost table Column X Column Y Column Z 25-15=10 31-19 = 12 35 – 17 = 18 Step 1; part a 15-15 =0 20 – 19 = 1 24 – 17 = 7 22-15 = 7 19 – 19 = 0 17 – 17 = 0 Job X Y Z Computations A 10 12 18 B 0 1 7 C 7 0 0
- 68. Step 1; part b Computations: X Y Z Row A 10 – 10 = 0 12 – 10 = 2 18 – 10 = 8 Row B 0–0=0 1–0=1 7–0=7 Row C 7–0=7 0–0=0 0–0=0 STEP 2: Determine whether an optimal assignment can be made Line 1 Line 20
- 69. STEP 3: Revise the total opportunity-cost table Computations: 0 2–1=1 8–1=7 a) Subtract lowest number 0 1–1=0 7 – 1 = 6 from all uncovered number 7+1= 8 0 0 Line 1 b) Add same smallest number to Line 2 numbers lying at the intersection of two lines Job X Y Z revised opportunity cost table A 0 1 7 B 0 0 6 Line 2 C 8 0 0 Line 3 Line 10
- 70. Total Cost: Assignment Cost A to X P25 B to Y 20 C to Z 17 P620
- 71. •MAXIMIZATION PROBLEMS
- 72. TRANSPORTATION FOR MAXIMIZATION PROBLEMS Juan dela Cruz is Marketing Vice President with the Malolos Company. Malolos is planning to expand its sales of computer software into new towns – Angat, Norzagaray, and Baliuag. It has been determined that 20, 15 , and 30 salespersons will be required to service each of the three areas, respectively. The company recently hired 65 new salespersons to cover these areas, and based on their experience and prior sales performance, the new hires have been classified as Type A, B, or C salesperson. Depending upon the regions to which each of the three types are assigned, Malolos estimates the following annual revenues per salesperson: Annual Revenues Type Number of Salespersons Angat Norzagaray Baliuag available A 24 P100,000 120,000 130,000 B 27 90,000 106,000 126,000 C 14 84,000 98,000 120,000 Dave wishes to determine how many salespersons of each type to assign to the three regions so that total annual revenues will be maximized.0
- 73. evaluation of unused squares0
- 74. evaluation of unused squares0
- 75. Salesper Territor Numbe Annual son Type y r Revenue Assign ed A Angat 9 P900, 000 A Norzaga 15 1,800,000 ray B Angat 11 990,000 B Baliuag 16 2,016,000 C Baliuag 14 1,680,000 P7,386,0000
- 76. STEP 1: Select the highest and second-highest revenue alternatives from among those not already allocated. The difference between these two revenues will be the opportunity cost for the row or column. STEP 2: Scan these opportunity-cost figures and identify the row or columns with the largest opportunity cost. STEP 3: Allocate as many units as possible to this row or column in the square with the greatest revenue.0
- 77. ASSIGNMENT FOR METHOD OF MAXIMIZATION PROBLEMS Josefa Santos manages the Aves Car Rental Agency. This year, she plans to purchase five new automobiles to replace five older vehicles. The older vehicles are to be sold at auction. Josefa has solicited bid from five individuals, each of whom wishes to purchase only one vehicle but has agreed to make a sealed bid on each of the five. The bids are as follows: Automobile Buyer Ford Honda Kia Mitsubishi Toyota Amalia P 3,000 P 2,500 P 3,300 P 2,600 P 3,100 Berto 3,500 3,000 2,800 2,800 3,300 Carlos 2,800 2,900 3,900 2,300 3,600 Dolores 3,300 3,100 3,400 2,900 3,500 Edgardo 2,800 3,500 3,600 2,900 3,0000
- 78. •REGRET VALUES FOR AUTOMOBILE Automobile Buyer Ford Honda Kia Mitsubishi Toyota Amalia 500 1000 600 300 500 Berto 0 500 1100 100 300 Carlos 700 600 0 600 0 Dolores 200 400 500 0 100 Edgardo 700 0 300 0 600 •OPTIMAL ASSIGNMENT OF BID AWARDS Automobile Buyer Ford Honda Kia Mitsubishi Toyota Amalia 200 700 200 0 100 Berto 0 500 1000 100 200 Carlos 800 700 0 700 0 Dolores 200 400 400 0 0 Edgardo 700 0 200 0 5000
- 79. •OPTIMAL SOLUTION TO Aves CAR RENTAL AGENCY Buyer Bid Accepted Bid Price Amalia Mitsubishi P 2,600 Berto Ford 3,500 Carlos Kia 3,900 Dolores Toyota 3,500 Edgardo Honda 3,500 P 17,000 Reference: Quantitative Approaches to Management, 8th edition; Richard I. Levin, et. al0
- 80. Thank You

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