1 Winston Chapter 3.4, Page 73, Number 1 (Linear Programming). Problem Statement: There are three factories on the Momiss River (1, 2, and 3). Each emits two types of pollutants (1 and 2) into the river. If the waste from each factory is processed, the pollution in the river can be reduced. It costs $15 to process a ton of factory 1 waste, and each ton processed will reduce the amount of pollutant 1 by 0.10 ton and the amount of pollutant 2 by 0.45 ton. It costs $10 to process a ton of factory 2 waste, and each ton processed will reduce the amount of pollutant 1 by 0.20 ton and the amount of pollutant 2 by 0.25 ton. It costs $20 to process a ton of factory 3 waste, and each ton processed will reduce the amount of pollutant 1 by 0.40 ton and the amount of pollutant 2 by 0.30 ton. The state wants to reduce the amount of pollutant 1 in the river by at least 30 tons and the amount of pollutant 2 in the river by at least 40 tons. Formulate an LP that will minimize the cost of reducing pollution by the desired amounts. Do you think that the LP assumptions (Proportionality, Additivity, Divisibility, and Certainty) are reasonable for this problem? A. Summarize the problem in a table format. B. Formulate the problem. Explain your decision variables and details of your formulation. C. Create the first simplex tableau for this problem. Do not solve manually. D. Use Quant software to solve this problem. Print the input data as free format. Print the solution as summary reports for the variables and the constraints. E. Write a report of your solution for a hypothetical manager. Use a language understandable to most people. Do not use mathematical abbreviations. A. Problem Summarized in Table Format: Factory (Tons) F1 F2 F3 Cost (per ton) $15 $10 $10 Pollution 1, P1 Pollution 2, P2 Limitations Reduction (tons) Reduction (tons) 0.10 0.45 None 0.20 0.25 None 0.40 0.30 None P1 to be reduced P2 to be reduced by by at least 30 tons at least 40 tons B. Problem Formulation: 1. Decision Variables: F1 = Waste of Factory 1 to be Processed, Tons, Positive F2 = Waste of Factory 2 to be Processed, Tons, Positive F3 = Waste of Factory 3 to be Processed, Tons, Positive P1 = 0.10F1 + 0.20F2 + 0.40F3, Reduction of Pollution 1, Positive P2 = 0.45F1 + 0.25F2 + 0.30F3, Reduction of Pollution 2, Positive 2 The Fx decision variables are the unknowns that are needed to complete a formulation to help minimize the cost of reducing pollution by the desired amounts. P1 and P2 are used only for a better understanding of this problem. 2. Objective Function: Since costs are being minimized, the objective function is created in the form of the sum of all costs. The objective function to minimize the cost of reducing pollution by the desired amounts is: o.f.: Min Z = 15F1 + 10F2 + 20F3. 3. Constraints: There are five constraints for this problem. Pollution 1 must be reduced by at least 30 tons and Pollution 3 must be reduced by at least 40 tons. Additionally, all factories must process a non-negative amount of pollution. 0.10F1 + 0.20F2 + 0.40F3 30 0.45F1 + 0.25F2 + 0.30F3 40 F1, F2, F3 0 s.t.: C. Initial Simplex Tableau: Row 0 1 2 Z 1 0 0 F1 F2 F3 15 10 20 0.10 0.20 0.40 0.45 0.25 0.30 Art1 0 1 0 Art2 0 0 1 D. Quant Input and Output: Quant Input: Free Format Model for 0073N01 Min 15F1+ 10F2+ 20F3 Subject to (1) .1F1+ .2F2+ .4F3 >= 30 (2) .45F1+ .25F2+ .3F3 >= 40 Ex1 0 -1 0 Ex2 0 0 -1 RHS 0 30 40 3 Quant Output: |------------------------------------------------------------------------------| | Summarized Report for 0073N01 Page : 1 | |------------------------------------------------------------------------------| | | | |Opportunity| Objective | Minimum | Maximum | |Number | Variable | Solution | Cost |Coefficient|Obj. Coeff.|Obj. Coeff.| |-------+----------+-----------+-----------+-----------+-----------+-----------| | 1 | F1 | +7.6923099| 0 | +15.000000| +5.0000000| +18.000000| | 2 | F2 | +146.15384| 0 | +10.000000| +8.3333340| +12.666666| | 3 | F3 | 0 | +6.1538458| +20.000000| +13.846154| + Infinity| |------------------------------------------------------------------------------| | Minimized OBJ = 1576.923 Iteration = 4 Elapsed CPU second = 0 | |------------------------------------------------------------------------------| |------------------------------------------------------------------------------| | Summarized Report for 0073N01 Page : 2 | |------------------------------------------------------------------------------| | | | | Shadow | Slack or | Minimum | Maximum | |Constr.| Status | RHS | Price | Surplus | RHS | RHS | |-------+---------+------------+-----------+-----------+-----------+-----------| | 1 | Tight | >+30.000000| +11.538460| 0 | +8.8888893| +32.000000| | 2 | Tight | >+40.000000| +30.769232| 0 | +37.500000| +135.00000| |------------------------------------------------------------------------------| | Minimized OBJ = 1576.923 Iteration = 4 Elapsed CPU second = 0 | |------------------------------------------------------------------------------| E. Report for a Manager: The minimum amount of funding to be spent in this scenario is $1,576.92 as seen by Quant’s minimized objective value. This minimized cost can be obtained when the optimum amount of each factory’s pollution is processed. Quant shows that 7.69 tons of Factory 1 pollution should be processed, 146 tons of Factory 2 pollution should be processed, and zero tons of Factory 3 pollution should be processed. Factory 3 pollution is undesirable to process because the overall results are too costly. Additionally, the objectives of treating the pollution with minimum cost can be successfully obtained without treating Factory 3’s pollution. Since it costs $15 to process one ton of Factory 1 pollution, it will cost $115.35 to process all 7.69 tons. Additionally, since it costs $10 to process one ton of Factory 2 pollution, it will cost $1461.50 to process all 146 tons. No pollution from Factory 3 should be processed. The total cost for processing the pollution is approximately $1,576.85, the lowest possible cost of treating the pollution while abiding to the constraints.