WQ HW1_2023_ZJUI (1)

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Finance

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Jan 9, 2024

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CEE 434 Fall 2023 - ZJUI Water Quality Assignment # 1 Due on 2023.12.23 Problem 1 (40 points) Use the MATLAB file provided for the laundromat problem to design regulatory policies ( ࠵? ࠵? , ࠵? ࠵? ) and calculate costs ( ࠵? ࠵? , ࠵? ࠵? and ࠵? ࠵? ) under both Least Cost and UPR programs for the following values of ࠵? = 2,5,7,8 . Assume that ࠵? ࠵?࠵? = 4.5 , ࠵? ࠵?࠵? = 3.5 , ࠵? ࠵? = 4.0 and ࠵? ࠵? = 7.5 . In one graph, plot ࠵? ࠵? and ࠵? ࠵? versus ࠵? for both programs, and in another graph, plot ࠵? ࠵? , ࠵? ࠵? and ࠵? ࠵? versus ࠵? for both programs. Discuss the differences between the two programs based on your results. Problem 2 (60) Now let’s assume that there are three laundries discharging their waste into the lake. The initial untreated discharges from these laundries are ࠵? ࠵? 1 , ࠵? ࠵? 2 and ࠵? ࠵? 3 . Part a) Formulate an optimization model for the least cost program.  Hint 1: Your model should be very similar to the least cost model discussed in lecture 12b, but there are now three dischargers.  Hint 2: Considering the steady state situation and full mixing in the lake, define: ࠵? = ࠵? ࠵? 1 + ࠵? ࠵? 2 + ࠵? ࠵? 3 − ࠵? . ࠵? ࠵? . Part b) Assume that the upper bound constraints for phosphorus removal are not binding. Derive the optimal solutions for the optimization model (phosphorus removals: ࠵? 1 , ࠵? 2 and ࠵? 3 ) using KKT approach. Part c) Assume that ࠵? ࠵? 1 = 55 , ࠵? ࠵? 2 = 5 , ࠵? ࠵? 3 = 18, ࠵? 1 = 1 , ࠵? 2 = 110 , and ࠵? 3 = 18 . Calculate the optimal phosphorus removals, marginal costs for each laundry, physical and financial costs for each laundry, and the total physical and financial costs for ࠵? = 25 and ࠵? = 55 . Part d) Redo Part c if we use a market-based program based on taxes instead of the least cost program. ⼀

23-12-20 下午 10:40 C:\Users\904...\untitled2.m 第 1 页，共 2 页 % Given parameters aw = 4; % empirical constant for Wembley as = 7.5; % empirical constant for Sno-White rtw = 4.5; % P generated by Wembley (before treatment) rts = 3.5; % P generated by Sno-White (before treatment) R_values = [2, 5, 7, 8]; % Values of R to loop over % Initialize results storage for Least Cost and UPR policies results_rw = zeros(2, length(R_values)); results_rs = zeros(2, length(R_values)); results_kw_LC = zeros(1, length(R_values)); results_ks_LC = zeros(1, length(R_values)); results_kt_LC = zeros(1, length(R_values)); results_kw_UPR = zeros(1, length(R_values)); results_ks_UPR = zeros(1, length(R_values)); results_kt_UPR = zeros(1, length(R_values)); % Cost objective function fun = @(x) aw*x(1)^2 + as*x(2)^2; % Optimization settings options = optimoptions( 'fmincon' , 'Display' , 'off' ); % Turn off display for cleaner output % Perform calculations for both policies for i = 1:length(R_values) R = R_values(i); % Least Cost Policy policy = 1; lb = [0, 0]; ub = [rtw, rts]; A = [-1, -1]; b = -R; Aeq = []; beq = []; x0 = [1, 1]; [x_LC, ~] = fmincon(fun, x0, A, b, Aeq, beq, lb, ub, [], options); results_rw(1, i) = x_LC(1); results_rs(1, i) = x_LC(2); results_kw_LC(i) = aw * x_LC(1)^2; results_ks_LC(i) = as * x_LC(2)^2; results_kt_LC(i) = results_kw_LC(i) + results_ks_LC(i); % UPR Policy policy = 2; Aeq = [rts, -rtw]; beq = 0; [x_UPR, ~] = fmincon(fun, x0, A, b, Aeq, beq, lb, ub, [], options); results_rw(2, i) = x_UPR(1); results_rs(2, i) = x_UPR(2); results_kw_UPR(i) = aw * x_UPR(1)^2; results_ks_UPR(i) = as * x_UPR(2)^2;

23-12-20 下午 10:40 C:\Users\904...\untitled2.m 第 2 页，共 2 页 results_kt_UPR(i) = results_kw_UPR(i) + results_ks_UPR(i); end % Plot rw and rs versus R for both programs figure; plot(R_values, results_rw(1,:), 'r-o' , R_values, results_rs(1,:), 'b-o' ); hold on ; plot(R_values, results_rw(2,:), 'r--*' , R_values, results_rs(2,:), 'b--*' ); xlabel( 'R' ); ylabel( 'rw and rs' ); title( 'rw and rs versus R for Least Cost (solid) and UPR (dashed)' ); legend( 'rw - Least Cost' , 'rs - Least Cost' , 'rw - UPR' , 'rs - UPR' ); grid on ; saveas(gcf, 'rw_rs_vs_R.png' ); % Save the figure as an image % Plotting costs versus R for both programs figure; plot(R_values, results_kw_LC, 'b--o' , R_values, results_ks_LC, 'r--o' , R_values, results_kt_LC, 'g--o' ); hold on ; plot(R_values, results_kw_UPR, 'b-*' , R_values, results_ks_UPR, 'r-*' , R_values, results_kt_UPR, 'g-*' ); xlabel( 'R' ); ylabel( 'Cost' ); title( 'Costs versus R for Least Cost and UPR Policies' ); legend( 'k_w LC' , 'k_s LC' , 'k_T LC' , 'k_w UPR' , 'k_s UPR' , 'k_T UPR' ); grid on ; saveas(gcf, 'costs_vs_R.png' ); % Save the figure as an image The Least Cost (LC) program prioritizes economic e ﬃ ciency by assigning more treatment to the less expensive Wembley facility, whereas the Uniform Percentage Reduction (UPR) program mandates equal percentage reductions for all, regardless of cost. As a result, LC achieves lower overall costs with uneven treatment loads, while UPR ensures equitable treatment across laundries but may incur higher costs.

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