Question.3138 - 1. Mineral Mining Company sends one truckload of iron and copper ore daily from the mine to the processing plant. The truck has a weight capacity of 10 tons and a volume capacity of 1200 cubic feet. Each pound of iron ore takes up 0.04 cubic feet of space and yields a net profit of $0.30 when processed. Each pound of copper ore uses 0.08 cubic feet of space and provides $0.50 of net profit. The following LP model is proposed, where I is the number of pounds of iron ore to load on the truck and C is number of pounds of copper ore to load on the truck: Maximize 0.3 I + 0.5 C Subject to I + C < 20000 0.04 I + 0.08 C < 1200 I , C > 0 a. Solve the problem graphically. Explain the optimal solution and objective function value in the context of the problem. (7 points) b. Graphically and algebraically determine the sensitivity range of each objective function coefficient at the optimal solution of the linear program in Problem 5. Explain the meaning of each range in the context of the problem. (6 points) For each constraint that holds with equality at the optimal solution of the linear program in Problem 1, perform the following: c. Graphically and algebraically determine the sensitivity range of the right- hand-side value. (4 points) d. Calculate the shadow price associated with each sensitivity range found in part (c). Explain the meaning of the shadow price in the context of the problem. (6 points) e. For the problem of Mineral Mining Company in Problem 1, management is considering leasing one of the following two new trucks to replace the existing one: Truck Type Weight Capacity (lb) Volume Capacity (ft)^3 Additional Cost ($/day) Tr 22/12 22000 1200 100 Tr 20/13 20000 1300 150 Should the company replace the old truck? And, if so, with which one of the new trucks should the company replace? Explain. (Use results from problem c and d) (5 points)
Answer Below:
a.) Objective function is the criterion for selecting the best values of the decision variables to maximize profits and minimize costs. Here the objective function is: 0.3 I + 0.5 C. Solving the problem graphically we get, ie I = 10000 pounds and C = 10000 pounds. 05000100001500020000250003000035000 0 5000 10000 15000 20000 25000 Every optimal solution is a boundary point. We can find an improving direction whenever we are at an interior point. Objective functional value = 8000 i.e maximum profit in dollars subject to the given constraints. b.) 05000100001500020000250003000035000 0 5000 10000 15000 20000 25000 Slope of the objective function = -3/5 Slope = -1 Slope = -1/2 05000100001500020000250003000035000 0 5000 10000 15000 20000 25000 Slope of the objective function = -3/5 Sensitivity Ranges for objective function coefficients Final Reduce d Objective Allowabl e Allowabl e Name Value Cost Coefficien t Increase Decrease Iron ore: 1000 0 0 0.3 0.2 0.05 Copper ore: 1000 0 0 0.5 0.1 0.2 The range of sensitivity for each coefficient provides the range of values over which the current solution will remain optimal. c.) RHS coefficients usually give some maximum limit for a resource or some minimum requirement that must be met. Changes to the RHS can happen when extra units of the resource become available or when some of the original resource becomes unavailable. Final Shadow Constraint Allowable Allowable Name Value Price R.H. Side Increase Decrease Constraint1: 1200 5 1200 400 400 Constraint2: 20000 0.1 20000 10000 5000 d.) Maximize Z = 0.3 I + 0.5 C Subject to I + C < 20000 0.04 I + 0.08 C < 1200 I , C > 0 Optimal solution is I =10000, C= 10000 With Z = 8000 Change the RHS value of the 1st constraint to 20000+k and resolve for the optimal point by binding the 1st and 2nd constraints: I + C =20000+k and 0.04 I + 0.08 C = 1200. The solution is I = 10000+2k, C = 10000-k, Z = 8000+0.1k. Z new - Z old = 0.1k, so the dual price or the shadow price = 0.1k Now Change the RHS value of the 2nd constraint to 1200+k and resolve for the optimal point by binding the 1st and 2nd constraints: I + C =20000 and 0.04 I + 0.08 C = 1200+k. The solution is I = 10000-25k, C = 10000+25k, Z = 8000+5k. Z new - Z old = 5k, so the dual price or the shadow price = 5k. Final Shadow Constraint Allowable Allowable Name Valu e Price R.H. Side Increase Decrease Constraint1: 1200 5 1200 400 400 Constraint2: 20000 0.1 20000 10000 5000 A shadow price always means the amount the objective function will change given a one unit increase in the RHS value of a constraint. If the objective function coefficients did not take the value of the resource into consideration, these are sunk costs. Shadow price = the value of an extra unit of the resource. If the objective function coefficients did take the value of the resource into consideration, these are included costs. Shadow price = a premium above the current price of the item that one would be willing to pay for an extra unit. e.) If we consider the two new trucks, both of them incorporates an additional cost. Even after the additional cost, the profit made is higher than the case discussed above. So the company should replace the old truck. We should replace the old truck with Tr 20/13 as the objective function value in this truck is $8,350 i.e maximum profit. By using the truck Tr 22/12 we get the objective function value as $8,100.More Articles From Computer
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