Transportation Problems - Answer Key [PDF]

Citation preview

Case study-1: Transportation Problem Case study-1: Problem-1: Powerco has three electric power plants that supply the needs of four cities. Each power plant can supply the following numbers of kilowatt-hours (kwh) of electricity: plant 1—35 million; plant 2—50 million; plant 3—40 million (see Table 1). The peak power demands in these cities, which occur at the same time (2 P.M.), are as follows (in kwh): city 1—45 million; city 2—20 million; city 3—30 million; city 4—30 million. The costs of sending 1 million kwh of electricity from plant to city depend on the distance the electricity must travel. Formulate an LP to minimize the cost of meeting each city’s peak power demand.



Solution: Xij: i = Plant; j = City Decision Variables: X11: Number of KWh transferred from Plant-1 to City-1 X12: Number of KWh transferred from Plant-1 to City-2 X13: Number of KWh transferred from Plant-1 to City-3 X14: Number of KWh transferred from Plant-1 to City-4 X21: Number of KWh transferred from Plant-2 to City-1 X22: Number of KWh transferred from Plant-2 to City-2 X23: Number of KWh transferred from Plant-2 to City-3 X24: Number of KWh transferred from Plant-2 to City-4 X31: Number of KWh transferred from Plant-3 to City-1 X32: Number of KWh transferred from Plant-3 to City-2 X33: Number of KWh transferred from Plant-3 to City-3 X34: Number of KWh transferred from Plant-3 to City-4 Objective function: Min. Z = 8X11+ 6X12+ 10X13+ 9X14+9X21+12X22+13X23+7X24+14X31+9X32 +16X33+5X34 Constrains: Supply constraints Plant-1: X11+X12 +X13 +X14 = 30 City-4: X14+ X24+ X34 >= 30 X11,X12,X13,X14,X21,X22,X23,X24,X31,X32,X33,X34 >=0 Solver answer: Z = 1020; X12=10; X13=25; X21=45; X23=5; X32=10; X34=30



Case study-1: Problem-2: Powerco has three electric power plants that supply the needs of four cities. Each power plant can supply the following numbers of kilowatt-hours (kwh) of electricity is given below Table. The peak power demands in these cities, which occur at the same time (2 P.M.), are shown in below Table. The costs of sending 1 million kwh of electricity from plant to city depend on the distance the electricity must travel. Formulate an LP to minimize the cost of meeting each city’s peak power demand. Table: Shipping Costs, Supply, and Demand for Powerco From City-1 8 9 14 40



Plant-1 Plant-2 Plant-3 Demand Solution: Supply =125;



City-2 6 12 9 20



To City-3 10 13 16 30



Supply City-4 9 7 5 30



35 50 40



Demand = 120



Supply ≠ Demand Add one Dummy City  City-5 (Dummy) If total supply exceeds total demand, we can balance a transportation problem by creating a dummy demand point that has a demand equal to the amount of excess supply. Because shipments to the dummy demand point are not real shipments, they are assigned a cost of zero. Xij: i = Plant; j = City Decision Variables: X11: Number of KWh transferred from Plant-1 to City-1 X12: Number of KWh transferred from Plant-1 to City-2 X13: Number of KWh transferred from Plant-1 to City-3 X14: Number of KWh transferred from Plant-1 to City-4 X15: Number of KWh transferred from Plant-1 to City-5 X21: Number of KWh transferred from Plant-2 to City-1 X22: Number of KWh transferred from Plant-2 to City-2 X23: Number of KWh transferred from Plant-2 to City-3 X24: Number of KWh transferred from Plant-2 to City-4 X25: Number of KWh transferred from Plant-2 to City-5 X31: Number of KWh transferred from Plant-3 to City-1 X32: Number of KWh transferred from Plant-3 to City-2 X33: Number of KWh transferred from Plant-3 to City-3 X34: Number of KWh transferred from Plant-3 to City-4 X35: Number of KWh transferred from Plant-3 to City-5



Objective function:



Min. Z = 8X11+ 6X12+ 10X13+ 9X14+9X21+12X22+13X23+7X24+14X31+9X32 +16X33+5X34 + 0X15+0X25+0X35 Constrains: Supply constraints Plant-1: X11+X12 +X13 +X14 +X15 = 30 City-4: X14+ X24+ X34 >= 30 City-5: X15+ X25+ X35 >= 5 X11,X12,X13,X14, X15,X21,X22,X23,X24, X25.X31,X32,X33,X34,X35 >=0 Solver answer: Z = 975; X13=20; X12=15; X21=40; X23=10; X32=5; X34=30; X35=5 Because X35=5  5 million Kwh of plant-3 capacity will be unused.



Case study-1: Problem-3: Powerco has three electric power plants that supply the needs of four cities. Each power plant can supply the following numbers of kilowatt-hours (kwh) of electricity is given below Table. The peak power demands in these cities, which occur at the same time (2 P.M.), are shown in below Table. The costs of sending 1 million kwh of electricity from plant to city depend on the distance the electricity must travel. Formulate an LP to minimize the cost of meeting each city’s peak power demand. Table: Shipping Costs, Supply, and Demand for Powerco From City-1 8 9 14 45



Plant-1 Plant-2 Plant-3 Demand Solution: Supply =115;



City-2 6 12 9 20



To City-3 10 13 16 30



Supply City-4 9 7 5 30



35 40 40



Demand = 125



Demand ≠ Supply Add one Dummy Plant  Plant-4 (Dummy) If total demand exceeds total supply, we can balance a transportation problem by creating a dummy plant -4 that has a supply equal to the amount of excess demand. Because shipments from the dummy plant-4 are not real shipments, they are assigned a cost of zero. Xij: i = Plant; j = City Decision Variables: X11: Number of KWh transferred from Plant-1 to City-1 X12: Number of KWh transferred from Plant-1 to City-2 X13: Number of KWh transferred from Plant-1 to City-3 X14: Number of KWh transferred from Plant-1 to City-4 X21: Number of KWh transferred from Plant-2 to City-1 X22: Number of KWh transferred from Plant-2 to City-2 X23: Number of KWh transferred from Plant-2 to City-3 X24: Number of KWh transferred from Plant-2 to City-4 X31: Number of KWh transferred from Plant-3 to City-1 X32: Number of KWh transferred from Plant-3 to City-2 X33: Number of KWh transferred from Plant-3 to City-3 X34: Number of KWh transferred from Plant-3 to City-4 X41: Number of KWh transferred from Dummy Plant-4 to City-1 X42: Number of KWh transferred from Dummy Plant-4 to City-2 X43: Number of KWh transferred from Dummy Plant-4 to City-3 X44: Number of KWh transferred from Dummy Plant-4 to City-4



Objective function: Min. Z = 8X11+ 6X12+ 10X13+ 9X14+9X21+12X22+13X23+7X24+14X31+9X32 +16X33+5X34 + 0X41+0X42+0X43+0X44 Constrains: Supply constraints Plant-1: X11+X12 +X13 +X14 = 30 X11,X12,X13,X14,X21,X22,X23,X24,X31,X32,X33,X34,X41,X42,X43,X44 >=0 Solver answer: Z = 900; X11=5; X12=10; X13=20; X14=0; X21= 40; X22=0; X23=0 ; X24=0 ; X31 =0 ; X32= 10; X33 =0 ; X34 = 30; X41 = 0; X42= 0; X43= 10; X44=0; X43=10  10 million Kwh transferred from dummy plant-4 to City-3. So City-3 power shortage is 10 million Kwh.



Case study-2: Transportation Problem (Handling Shortages) Two reservoirs are available to supply the water needs of three cities. Each reservoir can supply up to 50 million gallons of water per day. Each city would like to receive 40 million gallons per day. For each million gallons per day of unmet demand, there is a penalty. At city 1, the penalty is $20; at city 2, the penalty is $22; and at city 3, the penalty is $23. The cost of transporting 1 million gallons of water from each reservoir to each city is shown in below Table. Formulate a balanced transportation problem that can be used to minimize the sum of shortage and transport costs.



Solution: From Resorvoir-1 Resorvoir-2 Demand(million gallons) Supply =100;



City-1 7 9 40



To City-2 8 7 40



City-3 10 8 40



Supply (million gallons) 50 50



Demand = 120



Demand ≠ Supply Add one Dummy Reservoir  Resorvoir-3 (Dummy) If total demand exceeds total supply, we can balance a transportation problem by creating a dummy Reservoir that has a supply equal to the amount of excess demand. Because transfer from the dummy reservoir is not real transfer, they are assigned a cost of zero. Decision Variables:



Xij : i = Reservoir; j = City



X11: Number of million gallons transferred from Reservoir-1 to City-1 X12: Number of million gallons transferred from Reservoir-1 to City-2 X13: Number of million gallons transferred from Reservoir-1 to City-3 X21: Number of million gallons transferred from Reservoir-2 to City-1 X22: Number of million gallons transferred from Reservoir-2 to City-2 X23: Number of million gallons transferred from Reservoir-2 to City-3 X31: Number of million gallons transferred from dummy reservoir-3 to City-1 X32: Number of million gallons transferred from dummy reservoir-3 to City-2 X33: Number of million gallons transferred from dummy reservoir-3 to City-3



Objective function: Min.Z = Total cost Min.Z = (Transfer cost + Penalty Cost) Min.Z = 7X11 +8X12 +10X13 +9X21 +7X22 +8X23 +0X31 +0X32+0X33+20X31 +22X32+23X33 Constraints: Supply constraints Reservoir -1: X11+X12 +X13 = 40 X11,X12,X13,X21,X22,X23,X31,X32,X33 >=0 Solver answer: Z = 1170 (including penalty); X11=20; X12=30; X13=0 ; X21=0; X22=10; X23= 40; X31=20; X32=0; X33= 0



Case study-3: Transportation Problem Foster Generators involves the transportation of a product from three plants to four distribution centers. Foster Generators operates plants in Cleveland, Ohio; Bedford, Indiana; and York, Pennsylvania. Production capacities over the next three-month planning period for one particular type of generator are as follows:



The firm distributes its generators through four regional distribution centers located in Boston, Chicago, St. Louis, and Lexington; the three-month forecast of demand for the distribution centers is as follows:



Management would like to determine how much of its production should be shipped from each plant to each distribution center.



Solution:



Case study-4: Transportation Problem A company is to subcontract work on four assemblies. The five subcontractors have agreed to submit a bid price on each assembly type and a limit on the total number of assemblies (if any combination) for which they are willing to contract. These bids, the contract times, and time requirements for assemblies are given in the following matrix.



Formulate the LP model to minimize the total assembly cost. Solution: Let x11 be Assembly 1 to Subcontractor A, x12 be assembly to Subcontractor B and so on.



It is a minimization problem since the objective is to minimize the cost of subcontract price of the assembly. Hence, objective function is given by —



Minimum Z =10x11 + 11x12 + 12x13 + 13x14 + 14x15 + 11x21 + 12x22 + 11x23 + 10x24 + 9x25 + 12x31 + 13x32 + 8x33 + 9x34 + 10x35+ 13x41 + 8x42 + 9x43 + 10x44 + 11x45 Subject to the following constraints: x11 + x21 + x31 + x41 ≤ 250 x12 + x22 + x32 + x42 ≤ 280 x13 + x23 + x33 + x43 ≤ 330 x14 + x24 + x34 + x44 ≤ 360 x15 + x25 + x35 + x45 ≤ 380 x11 + x12 + x13 + x14 + x15 ≤ 500 x21 + x22 + x23 + x24 + x25 ≤ 300 x31 + x32 + x33 + x34 + x35 ≤ 300 x41 + x42 + x43 + x44 + x45 ≤ 400 xij ≥ 0 for all i and j



(No. of Contract limits) (No. of Contract limits) (No. of Contract limits) (No. of Contract limits) (No. of Contract limits) (No. of Assemblies required) (No. of Assemblies required) (No. of Assemblies required) (No. of Assemblies required) (Non-Negativity constraint)



Case study-5: Transportation Problem North-East Aircraft Company, which operates out of a central terminal, has 8 aircraft of Type I, 15 aircraft of Type II, and 12 aircraft of Type III available for today’s flights. The tonnage material capacities are 4.5 ton/aircraft for Type I; 7 ton/aircraft for Type II and 4 ton/aircraft for Type III. The company dispatches it’s through aircraft to cities A and B. Tonnage requirements are 75 at city A and 55 at city B; excess tonnage capacity supplied to a city has no value. A plane can fly once only during the day (From Central Terminus to city-A or city-B). The cost of sending per ton material from the terminal to each city using type-I/II/III aircrafts are given by the following table: City-A City-B



Type-I 230$ 580$



Type-II 500$ 100$



Type-III 140$ 380$



Formulate the LP model to minimise the air-transportation cost.



Solution: Central Terminus Aircraft Type-I



City-A



Aircraft Type-II



City-B Aircraft Type-III



Decision Variables: Xij: i=Type of aircraft; j= City X11: No. of tons transferred from central terminus to City-A using Aircraft Type-I X12: No. of tons transferred from central terminus to City-B using Aircraft Type-I X21: No. of tons transferred from central terminus to City-A using Aircraft Type-II X22: No. of tons transferred from central terminus to City-B using Aircraft Type-II X31: No. of tons transferred from central terminus to City-A using Aircraft Type-III X32: No. of tons transferred from central terminus to City-B using Aircraft Type-III Objective: Min.Z = 230X11+500X21+140X31+580X12+100X22+380X32



Constraints Demand: City-A: X11+X21+X31 =75 City-B: X12+X22+X32 =55 Capacity: Aircraft Type-I: X11 + X12