CHAPTER 3 LINEAR PROGRAMMING: FORMULATION AND APPLICATIONS SOLUTION TO SOLVED PROBLEMS3.S1 Farm Management Dwight and Hattie have run the family farm for over thirty years. They are currently planning the mix of crops to plant on their 120-acre farm for the upcoming season. The table below gives the labor hours and fertilizer required per acre, as well as the total expected profit per acre for each of the potential crops under consideration. Dwight, Hattie, and their children can work at most 6,500 total hours during the upcoming season. They have 200 tons of fertilizer available. What mix of crops should be planted to maximize the family’s total profit. a. Formulate and solve a linear programming model for this problem in a spreadsheet. Crop Oats Wheat Corn Labor Required (hours per acre) 50 60 105 Fertilizer Required (tons per acre) 1.5 2 4 Expected Profit (per acre) $500 $600 $950 This is a resource-allocation problem. The activities are the planting of the three crops and the limited resources are land, labor, and fertilizer. We will therefore build a spreadsheet following the template for resource-allocation problems in Figure 3.4. Start by entering the data. The data for this problem are the labor required, fertilizer required, and expected profit for each crop (per acre). Following the template, the data in the spreadsheet would be entered as displayed below, where range names of ProfitPerAcre (C4:E4) and TotalAvailable (H7:H9) are assigned to the corresponding data cells. B 3 4 5 6 7 8 9 Profit (per acre) Resources Land (acres) Labor (hours) Fertilizer (tons) C Oats $500 D Wheat $600 E Corn $950 F G H Resources Used per Acre 1 1 1 50 60 105 1.5 2 4 Total Available 120 6500 200 1 The decisions to be made in this problem are how many acres of each crop to plant. Therefore, we add three changing cells in C12:E12 with range name AcresPlanted. The values in AcresPlanted (C12:E12) will eventually be determined by the Solver. For now, an arbitrary value of 1 is entered for each crop. B 3 4 Profit (per acre) 5 6 Resources 7 Land (acres) 8 Labor (hours) 9 Fertilizer (tons) 10 11 12 Acres Planted C Oats $500 D Wheat $600 E Corn $950 F G H Resources Used per Acre 1 1 1 50 60 105 1.5 2 4 1 1 1 Total Available 120 6500 200 The goal is to maximize the family’s total profit. Thus, the target cell should calculate the total profit. In this case, the total profit will be Total Profit = ($500)(acres of oats) + ($600)(acres of wheat) + ($950)(acres of corn) or Total Profit = SUMPRODUCT(ProfitPerAcre, AcresPlanted). This formula is entered into cell H12. With 1 acre of each crop planted, the total cost would be ($500)(1) + ($600)(1) + ($950)(1) = $2,050. B 3 4 Profit (per acre) 5 6 Resources 7 Land (acres) 8 Labor (hours) 9 Fertilizer (tons) 10 11 12 Acres Planted C Oats $500 D Wheat $600 E Corn $950 F G H Resources Used per Acre 1 1 1 50 60 105 1.5 2 4 Total Available 120 6500 200 Total Profit $2,050 1 1 1 11 12 H Total Profit =SUMPRODUCT(ProfitPerAcre,AcresPlanted) 2 this will be =SUMPRODUCT(C7:E7. we calculate the total resources used in TotalUsed (cells F7:F9). labor. this formula can be copied into cells F8:F9 to calculate the amount of labor and fertilizer used.The functional constraints in this problem involve the limited resources of land. The total resources used must be <= TotalAvailable (H7:H9). B 3 4 Profit (per acre) 5 6 Resources 7 Land (acres) 8 Labor (hours) 9 Fertilizer (tons) 10 11 12 Acres Planted C Oats $500 D Wheat $600 E Corn $950 F G H Resources Used per Acre 1 1 1 50 60 105 1. as indicated by the <= in G7:G9.AcresPlanted) 3 .050 1 1 1 5 6 7 8 9 F Total Used =SUMPRODUCT(C7:E7. For land. and fertilizer. Using a range name or an absolute reference for the acres planted.AcresPlanted) =SUMPRODUCT(C8:E8. Given the AcresPlanted (the changing cells in C12:E12).5 2 4 Total Used 3 215 8 <= <= <= Total Available 120 6500 200 Total Profit $2. AcresPlanted).AcresPlanted) =SUMPRODUCT(C9:E9. AcresPlanted) =SUMPRODUCT(C8:E8.The Solver information and solved spreadsheet are shown below. A 1 2 3 4 5 6 7 8 9 10 11 12 B C D E F G H Farm Management Profit (per acre) Resources Land (acres) Labor (hours) Fertilizer (tons) Oats $500 Wheat $600 Corn $950 Total Used 120 6. Formulate this same model algebraically.400 200 <= <= <= Total Available 120 6500 200 Total Profit $64. while not planting any corn. with a resulting total profit of $64. C = Acres of corn planted. oats should be planted on 80 acres and wheat on 40 acres.AcresPlanted) =SUMPRODUCT(C9:E9. and corn to plant. start by defining the decision variables.AcresPlanted) Thus. These variables are defined below: Let O = Acres of oats planted. b.000. W = Acres of wheat planted. the three decisions are how many acres of oats.000 Resources Used per Acre 1 1 1 50 60 105 1. 4 . To build an algebraic model for this problem. In this case.5 2 4 Acres Planted 80 40 0 Range Name AcresPlanted ProfitPerAcre TotalAvailable TotalProfit TotalUsed Cells C12:E12 C4:E4 H7:H9 H12 F7:F9 5 6 7 8 9 F Total Used =SUMPRODUCT(C7:E7. wheat.AcresPlanted) 11 12 H Total Profit =SUMPRODUCT(ProfitPerAcre. W ≥ 0. the goal is to achieve the highest possible total profit. 6500 hours of labor. and 200 tons of fertilizer. In this case. The total resources used need to be less than or equal to the amount available.Next determine the goal of the problem. C ≥ 0. 1. W = Acres of Wheat planted. each acre of wheat yields a profit of $600. 5 .5O + 2W + 4C ≤ 200 tons and O ≥ 0. The objective function is therefore Maximize Total Profit = $500O + $600W + $950C. the complete algebraic formulation is given below: O = Acres of Oats planted. subject to Land: Labor: Fertilizer: O + W + C ≤ 120 acres.5O + 2W + 4C ≤ 200 tons After adding nonnegativity constraints. 50O + 60W + 105C ≤ 6500 hours. while each acre of corn yields a profit of $950. Each acre of oats yields a profit of $500. There are three limited resources in this problem: 120 acres of land. 50O + 60W + 105C ≤ 6500 hours. The data for labor hours used and fertilizer used per acre planted can be used to calculate the total resources used as a function of the decision variables. Each acre of a given crop that is planted uses up one available acre. Maximize Total Profit = $500O + $600W + $950C. These constraints are therefore as follows: Land: Labor: Fertilizer: Let O + W + C ≤ 120 acres. 1. C = Acres of Corn planted. 25 Minimum Requirement 13 1.3. are shown below.0 0.8 1.3 0.19 32 0 93 6 . beans.) $2. along with the minimum nutritional requirements. Formulate and solve a linear programming model for this problem in a spreadsheet.0 0. Milk (gallons) 3. and MinimumRequirement (H9:H11) are assigned to the corresponding data cells. What diet should be fed to each prisoner? a.5 45.19 93.5 45 F G H Unit Cost Niacin Thiamin Vitamin C Nutritional Contents (mg) 3. The goal is to minimize cost.8 0. Following the template. Valencia) 0.9 1. subject to meeting the minimum nutritional requirements imposed by law.20 E Oranges $0.2 4.9 0.6. The cost and nutritional content of each food. and oranges. Start by entering the data.0 2. NutritionalContents (C9:E11). The data for this problem are the nutrient content of each food.0 Niacin (mg) Thiamin (mg) Vitamin C (mg) Cost ($) This is a cost-benefit-trade-off problem. She would like to offer some combination of milk.20 Oranges (large Calif. B 3 4 5 6 7 8 9 10 11 C Milk (gal.3 0.00 D Beans (cups) $0. the data in the spreadsheet would be entered as displayed below.12 32.0 1. The activities are the quantities of food to feed each prisoner and the required benefits are the minimum nutritional requirements. and the cost of each food. the minimum daily requirement for each nutrient.00 Navy Beans (cups) 4.12 1. where range names of UnitCost (C5:E5). We will therefore build a spreadsheet following the template for cost-benefit-trade-off problems in Figure 3.25 Minimum Daily Requirement 13.S2 Diet Problem The kitchen manager for Sing Sing Prison is trying to decide what to feed its prisoners.2 1. we add three changing cells in C13:E13. Therefore.5 45 Total Cost $2.9 0.20)(cups of beans) + ($0.20 E Oranges $0.) $2. an arbitrary value of 1 is entered for each food type. with range name Quantity. For now.25 Minimum Requirement 13 1. the target cell should calculate this cost: Total Cost = ($2)(gallons of milk) + ($0. Quantity). The values in Quantity (C13:E13) will eventually be determined by the Solver.19 32 0 93 1 1 1 13 14 7 . The goal is to minimize the total cost per prisoner. This formula is entered into cell H14.2 4.12 1.The decisions to be made in this problem are how much of each food type should be fed to each prisoner.25)(number of oranges) or Total Cost = SUMPRODUCT(UnitCost.8 1.00 D Beans (cups) $0.Quantity) F G H Unit Cost Niacin Thiamin Vitamin C Quantity (per prisoner) Nutritional Contents (mg) 3. B 3 4 5 6 7 8 9 10 11 12 13 14 C Milk (gal.45 H Total Cost =SUMPRODUCT(UnitCost. Thus.3 0. The benefit achieved (total of each nutrient) must be >= the minimum needed (H9:H11).9 0.8 1.9 2.610 125 Minimum Requirement 13 1. Using an absolute reference for the acres planted. we calculate the total resources used in F9:F11. B 3 4 5 6 7 8 9 10 11 12 13 14 C Milk (gal.00 D Beans (cups) $0.25 Total Nutrients 8.20 E Oranges $0.Quantity) 10 =SUMPRODUCT(C10:E10.The functional constraints in this problem involve the minimum daily requirement of each nutrient.45 F G H Unit Cost Niacin Thiamin Vitamin C Quantity (per prisoner) Nutritional Contents (mg) 3.Quantity) 11 =SUMPRODUCT(C11:E11. as indicated by the >= in G9:G11.5 45 Total Cost $2. this formula can be copied into cells F10-F11 to calculate the thiamin and vitamin C. Given the amount of food fed each prisoner (the changing cells in C13:E13).) $2. this will be =SUMPRODUCT(C9:E9. $C$13:$E$13).Quantity) 8 .3 0. For niacin.19 32 0 93 1 1 1 >= >= >= F 7 Total 8 Nutrients 9 =SUMPRODUCT(C9:E9.2 4.12 1. Quantity) Thus.The Solver information and solved spreadsheet are shown below. the three decisions are how many gallons of milk.Quantity) 13 14 H Total Cost =SUMPRODUCT(UnitCost.64 Unit Cost Niacin Thiamin Vitamin C Quantity (per prisoner) Nutritional Contents (mg) 3.25 Total Nutrients 13 >= 3.574 0.64. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 B C D E F G H The Prison Diet Problem Milk (gal. To build an algebraic model for this problem.12 1. Formulate this same model algebraically. b.8 1.438 >= 45 >= Minimum Requirement 13 1.) $2.5 45 Total Cost $0.484 Range Name MinimumRequirement NutritionalContents Quantity TotalCost TotalNutrients UnitCost Cells H9:H11 C9:E11 C13:E13 H14 F9:F11 C5:E5 F 7 Total 8 Nutrients 9 =SUMPRODUCT(C9:E9.19 32 0 93 0 2.20 Oranges $0. In this case.Quantity) 11 =SUMPRODUCT(C11:E11.2 4.574 cups of beans and 0.9 0. and how many oranges to feed each prisoner.3 0. These variables are defined below: 9 . cups of beans.00 Beans (cups) $0.484 oranges for a total cost of $0. each prisoner should be fed a daily average of 2.Quantity) 10 =SUMPRODUCT(C10:E10. start by defining the decision variables. 25O. 32M + 93O ≥ 45mg. O = number of oranges fed to each prisoner. and vitamin C.3B + 0. After adding nonnegativity constraints.2M + 4. In this case. the goal is to meet the nutritional requirements at the lowest possible cost.12M + 1.00M + $0. Each gallon of milk costs $2.25O. subject to Niacin: Thiamin: Vitamin C: 3.9B + 0.25.20B + $0. The total nutrients need to be greater than or equal to the minimum requirement.19O ≥ 1. B = cups of beans fed to each prisoner.19O ≥ 1.9B + 0. each cup of beans costs $0. B ≥ 0. O = number of oranges fed to each prisoner.20.8O ≥ 13mg. Minimize Total Cost = $2.5mg. These constraints are therefore as follows: Niacin: Thiamin: Vitamin C: Let 3. 1.2M + 4.Let M = gallons of milk fed to each prisoner. the complete algebraic formulation is given below: M = gallons of milk fed to each prisoner. The nutritional requirements include minimum requirements for niacin. The objective function is therefore Minimize Total Cost = $2. B = cups of beans fed to each prisoner. and M ≥ 0. and each orange costs $0. The data for the nutritional contents of each type of food can be used to calculate the total level of each nutrient achieved as a function of the decision variables. 1.3B + 0.5mg.12M + 1. 10 .00M + $0. 32M + 93O ≥ 45mg. Next determine the goal of the problem.8O ≥ 13mg.20B + $0. thiamin. O ≥ 0. We will build a spreadsheet following the template for cost-benefit-trade-off problems in Figure 3. For example. Following the template. For each length of rod. or 5 12” rods (with no waste). Start by entering the data. However. Decora needs 25 12” rods. 18”. or 3 12” rods and an 18” rod (with 6” scrap). MinimumRequirement (P7:P11). the data in the spreadsheet would be entered as displayed. the minimum number of each rod length required and. and 12 60” rods. 52 18” rods. and Waste (C12:M12) are assigned to the corresponding data cells. the required benefit is obtaining the minimum number needed of those rods. etc. What is the fewest number of 60” rods that can be purchased to meet their production needs? Formulate and solve an integer programming model in a spreadsheet. less than 12”). or 3 12” rods and a 24” rod. For example. This is a cost-benefit-trade-off problem. The data for this problem are the total length of the uncut rods. the number of each length rod created.e. including decorative towel rods and shower curtain rods.3. Decora purchases 60” rods from an outside supplier and then cuts the rods as needed for their products. Each of the accessories includes a rod made out of stainless steel. Each possible way of cutting a 60” rod will represent an activity. Each 60” rod can be used to make a number of smaller rods. B C D 3 Length of Uncut Rods 4 5 Length of 6 Cut Rod 1 2 7 12 5 3 8 18 1 9 24 10 40 11 60 12 Waste 0 6 E F G 60 4 2 2 H I J K L M N O P 3 3 1 0 Pattern 5 6 7 1 1 1 1 1 2 1 6 0 8 8 3 9 2 1 0 10 11 1 1 1 0 Minimum Requirement 25 52 45 30 12 0 6 2 12 B Waste C =UncutRodLength-SUMPRODUCT($B$7:$B$11. a 60” rod could be used to make a 40” and an 18” rod (with 2” of waste).C7:C11) 11 .. a 60” rod can be cut into 5 12” rods. for each possible pattern of cutting a 60” rod. many different lengths are needed: 12”.S3 Cutting Stock Problem Decora Accessories manufactures a variety of bathroom accessories. and 60”. 45 24” rods. There are 11 possible patterns of cutting a 60” rod that either leave no scrap or an unusable piece of scrap (i. where range names of UncutRodLength (F3). For the next production period. 30 40” rods.6. 40”. 24”. Therefore. B 3 4 5 6 7 8 9 10 11 12 13 14 C D E F G Length of Uncut Rods 60 H I J K L M N O P Length of Cut Rod 12 18 24 40 60 Waste 1 5 2 3 1 3 3 1 4 2 2 Pattern 5 6 7 1 1 1 1 1 2 1 6 1 0 1 8 1 8 3 9 2 1 10 11 1 1 0 6 1 0 1 0 1 6 1 0 1 2 1 1 0 1 Minimum Needed 25 52 45 30 12 Pattern Quantity 1 The goal is to minimize the total number of 60” (uncut) rods needed. we add 11 changing cells in C14:M14. B 3 4 5 6 7 8 9 10 11 12 13 14 C D E F G Length of Uncut Rods 60 H I J K L M N O P Length of Cut Rod 12 18 24 40 60 Waste 1 5 2 3 1 3 3 1 4 2 2 Pattern 5 6 7 1 1 1 1 1 2 1 6 1 0 1 8 1 8 3 9 2 1 10 11 1 1 0 6 1 0 1 0 1 6 1 0 1 2 1 1 0 1 Minimum Needed 25 52 45 30 12 Total Uncut Rods Needed 11 Pattern Quantity 1 13 14 P Total Uncut Rods Needed =SUM(PatternQuantity) 12 . This is just the sum of the changing cells: Total Uncut Rods needed = SUM(PatternQuantity). with range name PatternQuantity. an arbitrary value of 1 is entered for each pattern. For now.The decisions to be made in this problem are how many rods should be cut into each of the possible patterns. The values in PatternQuantity (C14:M14) will eventually be determined by the Solver. This formula is entered into cell P14. PatternQuantity) =SUMPRODUCT(C8:M8. this formula can be copied into cells N8:N11 to calculate the total number of other lengths created. For each rod length. the benefit achieved (the total number of that rod length cut) must be >= MinimumRequirement (P7:P11).The functional constraints in this problem require that the number of cut rods of each length must be at least the minimum number needed. Using a range name or absolute reference for the quantity of each pattern.PatternQuantity) 13 . we calculate the total number of each rod length produced in TotalProduced (N7:N11). this will be =SUMPRODUCT(C7:M7.PatternQuantity) =SUMPRODUCT(C10:M10. PatternQuantity). Given PatternQuantity (the changing cells in C14:M14). For 12” rods. 3 4 5 6 7 8 9 10 11 12 13 14 B C D E F G Length of Uncut Rods 60 Length of Cut Rod 12 18 24 40 60 Waste H I J K L M N O P 1 5 2 3 1 3 3 1 4 2 2 Pattern 5 6 7 1 1 1 1 1 2 1 6 1 0 1 8 1 8 3 9 2 1 10 11 1 1 0 6 1 0 1 0 1 6 1 0 1 2 1 1 0 1 Total Produced 16 10 5 2 1 >= >= >= >= >= Minimum Needed 25 52 45 30 12 Pattern Quantity 1 Total Uncut Rods Needed 11 5 6 7 8 9 10 11 N Total Produced =SUMPRODUCT(C7:M7.PatternQuantity) =SUMPRODUCT(C11:M11.PatternQuantity) =SUMPRODUCT(C9:M9. as indicated by the >= in O7:O11. The Solver information and solved spreadsheet are shown below.PatternQuantity) =SUMPRODUCT(C10:M10. 14 .PatternQuantity) 13 14 P Total Uncut Rods Needed =SUM(PatternQuantity) 12 B Waste C =UncutRodLength-SUMPRODUCT($B$7:$B$11.C7:C11) Thus. we constrain the changing cells to be integer. TotalRodsNeeded (P14) indicates that a total of 72 uncut rods are needed. PatternQuantity (C14:M14) show the number of rods that should be cut into the respective patterns. Therefore.Also. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 B C D E F G H I J K L M N O P Cutting Stock Problem Length of Uncut Rods 60 Length of Cut Rod 12 18 24 40 60 Waste Pattern 5 6 7 1 1 1 1 1 2 1 6 0 0 17 8 0 Total Minimum 10 11 Produced Needed 27 >= 25 2 1 52 >= 52 1 45 >= 45 1 30 >= 30 1 12 >= 12 0 2 0 Total Uncut Rods Needed 11 30 12 72 9 1 5 2 3 1 3 3 1 4 2 2 8 3 0 6 0 0 0 0 0 6 0 Pattern Quantity 2 Range Name MinimumNeeded PatternQuantity TotalProduced TotalRodsNeeded UncutRodLength Waste Cells P7:P11 C14:M14 N7:N11 P14 F3 C12:M12 5 6 7 8 9 10 11 N Total Produced =SUMPRODUCT(C7:M7. an integer number of each pattern must be cut.PatternQuantity) =SUMPRODUCT(C11:M11.PatternQuantity) =SUMPRODUCT(C8:M8.PatternQuantity) =SUMPRODUCT(C9:M9. Note that space has been reserved in the middle of the spreadsheet (C11:E12) for the changing cells. and how should they be distributed to each of the three wholesaler warehouses so as to minimize the combined cost of production and shipping? Formulate and solve a linear programming model in a spreadsheet. The data for this problem are the unit shipping cost. How many AEDs should be produced in each plant.S4 Producing and Distributing AEDs at Heart Start Heart Start produces automated external defibrillators (AEDs) in each of two different plants (A and B). MonthlyCapacity (H11:H12). and the monthly demand at each warehouse. The unit production costs and monthly production capacity of the two plants are indicated in the table below.5 of the text. the unit production cost and monthly production capacity at each plant. Start by entering the data. and MonthlyDemand (C15:E15) are assigned to the corresponding data cells. The shipping cost from each plant to the warehouse of each wholesaler along with the monthly demand from each wholesaler are also indicated in the table.3. The AEDs are sold through three wholesalers. where range names of UnitProductionCost (H5:H6). The data are shown below. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B C D E F G H Unit Production Cost $600 $625 Monthly Capacity 100 120 AED Production and Distribution Shipping Cost (per unit) Warehouse 1 Plant A $22 Plant B $16 Warehouse 2 $14 $20 Warehouse 3 $30 $24 Monthly Demand 80 60 70 15 . Unit Shipping Cost Warehouse Warehouse Warehouse 1 2 3 $22 $14 $30 $16 $20 $24 80 60 70 Unit Production Cost $600 $625 Monthly Production Capacity 100 120 Plant A Plant B Monthly Demand This is a transportation problem as described in Section 3. The total production costs and the total shipping costs are calculated as follows.UnitsShipped) Overall Cost =TotalProductionCost+TotalShippingCost 16 .The decisions to be made in this problem are how many units to ship from each plant to each warehouse. UnitsShipped). Total Production Cost = SUMPRODUCT(UnitProductionCost. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B C D E F G H Unit Production Cost $600 $625 Monthly Capacity 100 120 AED Production and Distribution Shipping Cost (per unit) Warehouse 1 Plant A $22 Plant B $16 Units Shipped Warehouse 2 $14 $20 Warehouse 3 $30 $24 Warehouse 1 Plant A 0 Plant B 0 Warehouse 2 0 0 Warehouse 3 0 0 Monthly Demand 80 60 70 The goal is to minimize the total cost. Total Shipping Cost = SUMPRODUCT(ShippingCost. For now. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 B C D E F G H Unit Production Cost $600 $625 Monthly Capacity 100 120 AED Production and Distribution Shipping Cost (per unit) Warehouse 1 Plant A $22 Plant B $16 Units Shipped Warehouse 2 $14 $20 Warehouse 3 $30 $24 Warehouse 1 Plant A 0 Plant B 0 Warehouse 2 0 0 Warehouse 3 0 0 Monthly Demand 80 60 70 Total Production Cost Total Shipping Cost Overall Cost $0 $0 $0 16 17 18 G H Total Production Cost =SUMPRODUCT(UnitProductionCost. These formulas are entered into H16 and H17. UnitsShipped). The values in UnitsShipped (C11:E12) will eventually be determined by the Solver.TotalShippedOut) Total Shipping Cost =SUMPRODUCT(ShippingCost. Therefore. with the overall total cost calculated by summing these two costs in H18. as shown in the spreadsheet below. zeroes are entered. we add a table of changing cells with range name UnitsShipped (C11:E12). including production costs and shipping costs. and each warehouse needs to receive its monthly demand. while the units shipped to each warehouse are caculated by summing each column of changing cells. The units shipped out of each plant are calculated by summing each row of changing cells.The functional constraints in this problem are that each plant can’t ship out more than its monthly capacity. These constraints are shown in the spreadsheet below. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 B C D E F G H Unit Production Cost $600 $625 Total Shipped Out 0 0 Monthly Capacity 100 120 AED Production and Distribution Shipping Cost (per unit) Warehouse 1 Plant A $22 Plant B $16 Units Shipped Plant A Plant B Total to Warehouse Monthly Demand Warehouse 2 $14 $20 Warehouse 3 $30 $24 Warehouse 1 0 0 0 = 80 Warehouse 2 0 0 0 = 60 Warehouse 3 0 0 0 = 70 <= <= Total Production Cost Total Shipping Cost Overall Cost $0 $0 $0 F 8 Total 9 Shipped 10 Out 11 =SUM(C11:E11) 12 =SUM(C12:E12) 13 B C D E Total to Warehouse =SUM(C11:C12) =SUM(D11:D12) =SUM(E11:E12) 17 . UnitsShipped) Overall Cost =TotalProductionCost+TotalShippingCost Thus. from Plant B they should ship 40 units to Warehouse 1 and 70 units to Warehouse 3.040 $132.790.The Solver information and solved spreadsheet are shown below. 18 .TotalShippedOut) Total Shipping Cost =SUMPRODUCT(ShippingCost. from Plant A they should ship 40 to Warehouse 1 and 60 units to Warehouse 2.790 Range Name MonthlyCapacity MonthlyDemand OverallCost UnitShippingCost TotalProductionCost TotalShippedOut TotalShippingCost TotalToWarehouse UnitProductionCost UnitsShipped F 8 Total 9 Shipped 10 Out 11 =SUM(C11:E11) 12 =SUM(C12:E12) Cells H11:H12 C15:E15 H18 C5:E6 H16 F11:F12 H17 C13:E13 H5:H6 C11:E12 13 B C D E Total to Warehouse =SUM(C11:C12) =SUM(D11:D12) =SUM(E11:E12) 16 17 18 G H Total Production Cost =SUMPRODUCT(UnitProductionCost.750 $4. giving an overall total cost of $132. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 B C D E F G H Unit Production Cost $600 $625 Total Shipped Out 100 110 Monthly Capacity 100 120 AED Production and Distribution Shipping Cost (per unit) Warehouse 1 Plant A $22 Plant B $16 Units Shipped Plant A Plant B Total to Warehouse Monthly Demand Warehouse 2 $14 $20 Warehouse 3 $30 $24 Warehouse 1 40 40 80 = 80 Warehouse 2 60 0 60 = 60 Warehouse 3 0 <= 70 <= 70 = 70 Total Production Cost Total Shipping Cost Overall Cost $128. Finance. Start by entering the data. Each class has a limit of five students. Each class is limited to 5 students. where the corresponding data cells are assigned a range name of Points (C5:F14). Each student is assigned to take two classes. The data for this problem are the bid points for each student for each class. B 3 4 5 6 7 8 9 10 11 12 13 14 Points George Fred Ann Eric Susan Liz Ed David Tony Jennifer C Management Science 60 20 45 50 30 50 70 25 35 60 D Finance 10 20 45 20 30 50 20 25 15 10 E Operations Management 10 40 5 5 30 0 10 35 35 10 F Marketing 20 20 5 25 10 0 0 15 15 20 19 .S5 Bidding for Classes In the MBA program at a prestigious university in the Pacific Northwest. There are four electives available: Management Science. The assignees are the students and the tasks are the classes.3. students bid for electives in the second year of their program. This is a special kind of assignment problem. The bids submitted for each of the 10 students are shown in the table below. Student George Fred Ann Eric Susan Liz Ed David Tony Jennifer Management Science 60 20 45 50 30 50 70 25 35 60 Finance 10 20 45 20 30 50 20 25 15 10 Operations Management 10 40 5 5 30 0 10 35 35 10 Marketing 20 20 5 25 10 0 0 15 15 20 a. Each student has 100 points to bid (total) and must take two electives. Operations Management. Formulate and solve a spreadsheet model to determine an assignment of students to classes so as to maximize the total bid points of the assignments. and Marketing. a table of changing cells is created for each student and class combination in C18:F27. Therefore. The values in Assignment (C18:F27) will eventually be determined by the Solver. For now. and given a range name of Assignment. B 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Points George Fred Ann Eric Susan Liz Ed David Tony Jennifer C Management Science 60 20 45 50 30 50 70 25 35 60 Management Science 1 0 1 0 1 0 1 0 1 0 D Finance 10 20 45 20 30 50 20 25 15 10 E Operations Management 10 40 5 5 30 0 10 35 35 10 Operations Management 1 0 1 0 1 0 1 0 1 0 F Marketing 20 20 5 25 10 0 0 15 15 20 Assignment George Fred Ann Eric Susan Liz Ed David Tony Jennifer Finance 0 1 0 1 0 1 0 1 0 1 Marketing 0 1 0 1 0 1 0 1 0 1 20 . arbitrary values of 0 and 1 are entered.The decisions to be made in this problem are whether or not to assign each student to each class. Assignment) 21 . Assignment).The goal is to maximize the total bid points of the assignments. B 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 Points George Fred Ann Eric Susan Liz Ed David Tony Jennifer C Management Science 60 20 45 50 30 50 70 25 35 60 Management Science 1 0 1 0 1 0 1 0 1 0 D Finance 10 20 45 20 30 50 20 25 15 10 E Operations Management 10 40 5 5 30 0 10 35 35 10 Operations Management 1 0 1 0 1 0 1 0 1 0 F Marketing 20 20 5 25 10 0 0 15 15 20 G H I Assignment George Fred Ann Eric Susan Liz Ed David Tony Jennifer Finance 0 1 0 1 0 1 0 1 0 1 Marketing 0 1 0 1 0 1 0 1 0 1 Total Points 535 29 H I Total Points =SUMPRODUCT(Points. This formula is entered into cell I29. Total Bid Points = SUMPRODUCT(Points. for Management Science it is =SUM(C18:C27). This formula is copied into C28:F28. The number of students assigned to a class is the sum of the column of changing cells for each class. and each student can take each class at most once. for George it is =SUM(C18:F18). each class is limited to five students.The functional constraints in this problem are that each student must be assigned to two classes. This formula is copied into G18:G27. For example. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 C Management Points Science George 60 Fred 20 Ann 45 Eric 50 Susan 30 Liz 50 Ed 70 David 25 Tony 35 Jennifer 60 Management Science 1 0 1 0 1 0 1 0 1 0 5 <= Capacity 5 B D Finance 10 20 45 20 30 50 20 25 15 10 E F Operations Management Marketing 10 20 40 20 5 5 5 25 30 10 0 0 10 0 35 15 35 15 10 20 Operations Management Marketing 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 5 5 <= <= 5 5 G H I Assignment George Fred Ann Eric Susan Liz Ed David Tony Jennifer Finance 0 1 0 1 0 1 0 1 0 1 5 <= 5 Total Classes 2 2 2 2 2 2 2 2 2 2 = = = = = = = = = = Classes to Take 2 2 2 2 2 2 2 2 2 2 535 Total Points K 16 17 18 19 20 21 22 23 24 25 26 27 Total Classes =SUM(C18:F18) =SUM(C19:F19) =SUM(C20:F20) =SUM(C21:F21) =SUM(C22:F22) =SUM(C23:F23) =SUM(C24:F24) =SUM(C25:F25) =SUM(C26:F26) =SUM(C27:F27) 28 B C D E F Total in Class =SUM(C18:C27) =SUM(D18:D27) =SUM(E18:E27) =SUM(F18:F27) 22 . For example. The number of classes a student is assigned to is just the sum of the row of changing cells for each student. C19:F19) 20 =SUMPRODUCT(C7:F7.C18:F18) 19 =SUMPRODUCT(C6:F6.C20:F20) 21 : 22 : 28 B C D E F Total in Class =SUM(C18:C27) =SUM(D18:D27) =SUM(E18:E27) =SUM(F18:F27) 23 . A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 B C D E F G H I J K Bidding for Classes Points George Fred Ann Eric Susan Liz Ed David Tony Jennifer Management Science 60 20 45 50 30 50 70 25 35 60 Management Science 1 0 1 0 0 1 1 0 0 1 5 <= 5 Finance 10 20 45 20 30 50 20 25 15 10 Operations Management 10 40 5 5 30 0 10 35 35 10 Operations Management 0 1 0 0 1 0 1 1 1 0 5 <= 5 Marketing 20 20 5 25 10 0 0 15 15 20 Total Classes 2 2 2 2 2 2 2 2 2 2 Classes to Take 2 2 2 2 2 2 2 2 2 2 705 Student Points 80 60 90 45 60 100 80 60 50 80 Assignment George Fred Ann Eric Susan Liz Ed David Tony Jennifer Total in Class Capacity Finance 0 0 1 1 1 1 0 1 0 0 5 <= 5 Marketing 1 1 0 1 0 0 0 0 1 1 5 <= 5 = = = = = = = = = = Total Points Range Name Assignment Capacity ClassesToTake Points StudentPoints TotalClasses TotalInClass TotalPoints Cells C18:F27 C30:F30 I18:I27 C5:F14 K18:K27 G18:G27 C28:F28 I29 G 16 Total 17 Classes 18 =SUM(C18:F18) 19 =SUM(C19:F19) 20 =SUM(C20:F20) 21 : 22 : K 16 Student 17 Points 18 =SUMPRODUCT(C5:F5.The Solver information and solved spreadsheet are shown below. c. achieving a total of 705 points. while Ann got in with only 45 points.Assignment) Thus. Which alternative objectives might lead to a fairer assignment? Perhaps maximizing the minimum total number of bid points achieved by each student. For example. Eric did not get into Management Science despite bidding 50 points. Also. b. Does the resulting solution seem like a fair assignment? No. Eric got into classes worth only 45 total bid points to him while Liz got classes worth 100 bid points to her. the 1’s in Assignment (C18:F27) show the assignments that should be made.29 H I Total Points =SUMPRODUCT(Points. 24 .