HW7_2023_Solution

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University of Illinois, Urbana Champaign *

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434

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Industrial Engineering

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Jan 9, 2024

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1 Assignment # 7, CEE 434 Fall 2023 Solution Problem 1 (80 points): Groundwater Management Part a) ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? = -[5࠵? ! " + ࠵? ! " ] + 310࠵? ! # + 2࠵? ! # 78 + 9:5࠵? $ " + 5 3 ࠵? $ " < + :10࠵? $ # + 5 9 ࠵? $ # < + :15࠵? $ % + 5 3 ࠵? $ % <> Subject to: ?3࠵? ! " + (5 ∙ ࠵? ! # + ࠵? ! # BC + 3࠵? $ " + D3࠵? $ # + ࠵? $ # B + (12࠵? $ % + ࠵? $ % )7 = ࠵? & (࠵?࠵? ≥ ࠵? & ) ࠵? ! " ≤ 5࠵? ! " ࠵? ! # ≤ 11࠵? ! # ࠵? $ " ≤ 3࠵? $ " ࠵? $ # ≤ 9࠵? $ # ࠵? $ % ≤ 8࠵? $ % ࠵? ! " + ࠵? ! # ≤ 1 ࠵? $ " + ࠵? $ # + ࠵? $ % ≤ 1 Decision variables: ࠵? ! " , ࠵? ! # , ࠵? $ " , ࠵? $ # , ࠵? $ % (the pumping rate located in each piece of the cost function), and ࠵? ! " , ࠵? ! # , ࠵? $ " , ࠵? $ # , ࠵? $ % (indicating which piece of the linearized cost function should be used) ࠵? is the demand. ?3࠵? ! " + (5 ∙ ࠵? ! # + ࠵? ! # BC provides the amount of water pumped from well A. 3࠵? $ " + D3࠵? $ # + ࠵? $ # B + (12࠵? $ % + ࠵? $ % )7 provides the amount of water pumped from well B. Part b) Use Excel solver (LP-Simplex) to solve the optimization model for different values of ࠵? & = 6,12,16,20,25,30,35 . For each case, report your optimal decision variables and the optimal value of objective function. Plot two graphs: optimal cost versus demand, and well withdrawal ( ࠵? ! and ࠵? $ ) versus demand.
2 demand cost ࠵? ! " ࠵? ! # ࠵? $ " ࠵? $ # ࠵? $ % ࠵? ! " ࠵? ! # ࠵? $ " ࠵? $ # ࠵? $ % ࠵? ! ࠵? $ 6 11.67 0 0 0 3 0 0 0 0 1 0 0 6 12 15 0 0 0 0 0 0 0 0 0 1 0 12 16 21.67 0 0 0 0 4 0 0 0 0 1 0 16 20 28.33 0 0 0 0 8 0 0 0 0 1 0 20 25 38.33 0 0 0 0 8 0 1 0 0 1 5 20 30 48.33 0 5 0 0 8 0 1 0 0 1 10 20 35 58.33 0 10 0 0 8 0 1 0 0 1 15 20 All pumping rates in the above table have a unit of gal/min. 0 5 10 15 20 25 0 5 10 15 20 25 30 35 40 demad (gal/min) water demand (gal/min) well withdraw verse water demand well A well B 0 10 20 30 40 50 60 70 0 5 10 15 20 25 30 35 40 optimal cost water demand (gal/min) optimal cost verse water demand
3 Excel Solver Setup for Part b) Part c) Now let’s assume that additional information regarding the impact of pumping on groundwater table in two critical points (P1 and P2) are available in the form of response matrix ࠵? = [࠵? ’,) ] . Note that ࠵? ’,) shows the drawdown in points j ( ࠵? ∈ {࠵?1, ࠵?2} ) per unit flow rate pumped from well i ( ࠵? ∈ {࠵?, ࠵?} ). Also, the maximum possible drawdown at these two points are given as ࠵?࠵? *+ and ࠵?࠵? *, . Add the new constraint(s) to your model in part a. The response matrix and the maximum drawdown MD are given below. With the new constraint(s), what is the largest water demand ( ࠵? & ) that can be met, while minimize the cost? ( Hint: Update the Excel model from Part b with the new constraints, and do trial-and-error with ࠵? ࠵? = ࠵?, ࠵?࠵?, ࠵?࠵?, ࠵?࠵?, ࠵?࠵?, ࠵?࠵?, ࠵?࠵?࠵? ࠵?࠵? . Up to a certain value of ࠵? ࠵? , the model will be infeasible. This is the largest ࠵? ࠵? you can obtain. )
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4 The new constraints are ࠵? !,*+ 3࠵? ! " + (5 ∙ ࠵? ! # + ࠵? ! # )7 + ࠵? $,*, 3࠵? $ " + D3࠵? $ # + ࠵? $ # B + (12࠵? $ % + ࠵? $ % )7 ≤ MD .+ ࠵? !,*, 3࠵? ! " + (5 ∙ ࠵? ! # + ࠵? ! # )7 + ࠵? $,*, 3࠵? $ " + D3࠵? $ # + ࠵? $ # B + (12࠵? $ % + ࠵? $ % )7 ≤ MD ., The largest water demand is 25 gal/min by trial-and-error. In this case, the amount of water pumped from well A is 16 gal/min, and the amount of water pumped from well B is 9 gal/min. The cost is 45.33 and the drawdown in two critical points are 0.6 and 0.478 respectively. That is, the drawdown constraint at P1 is binding. Besides, well A has lower drawdown response coefficients for the two critical points than well B ( ࠵? !,*+ < ࠵? $,*+ , ࠵? !,*, < ࠵? $,*, ). And when the water demand is 25 gal/min, the pumping rate from well A reaches its upper limit, which also indicates that no larger water demand can be satisfied without increasing the maximum drawdown. [ Note : Larger ࠵? & is not a feasible solution since some binary variables will become non-binary. To determine whether a solution is feasible, check all the constraints and binary decision variables.] Excel Solver Setup with optimal solution for Part c)