CIVE 646 Homework 2 Optimization, Fall 2011

AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 Homework # 2 - Solution CIVE 646 - Water Resources Systems: Planning and Management (Fall 2011-12) 4-24 An industrial firm makes two products, A and B. These products require water and other resources. Water is the scarce resource-they have plenty of other needed resources. The products they make are unique, and hence they can set the unit price of each product at any value they want to. However experience tells them that the higher the unit price for a product, the less amount of that product they will sell. The relationship between unit price and quantity that can be sold is given by the following two demand functions: P0(A) = 8 – A P0(B) = 6 – 1.5B (a) What are the amounts of A and B, and their unit prices, that maximize the total revenue that can be obtained? Formulate the problem and use optimization tool of Matlab to solve it. (b) Suppose the total amount of A and B could not exceed some amount Tmax. What are the amounts of A and B, and their unit prices, that maximize total revenue, if: i) Tmax = 10 ii) Tmax = 5 Formulate the problem and use optimization tool of Matlab to solve it. Water is needed to make each unit of A and B. The production functions relating the amount of water XA needed to make A, and the amount of water XB needed to make B are A = 0.5 XA, and B = 0.25 XB, respectively. (c) Find the amounts of A and B and their unit prices that maximize total revenue assuming the total amount of water available is 10 units. Formulate the problem and use optimization tool of Matlab to solve it. Solution: (a) Total Revenue Functions: TRA = (8 - A)A = 8A - A2 TRB = (6 - 1.5B)B = 6B – 1.5B2 Maximize TR = TRA + TRA Optimization tool of Matlab: function f = objfun(x) a = x(1); b = x(2); f = -8*a + a^2 - 6*b + 1.5*b^2; Quantity of product A Quantity of product B Po Unit price Po Unit Price 8 – A 6 – 1.5B AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 A = 4, P0(A) = 4 B = 2, P0(B) = 3 TR = 4*4 + 2*3 = 22 b) In this case: A + B  Tmax.  10 The solution for A and B would be same as (a) since A + B = 6  10 A + B  Tmax.  5 function f = objfun(x) a = x(1); b = x(2); f = -8*a + a^2 - 6*b + 1.5*b^2; Subject to A + B  5 AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 A = 3, P0(A) = 5 B = 2, P0(B) = 3 TR = 5*3 + 2*3 = 21 c) In this case: A = 0.5XA  XA = 2A B = 0.25XB  XB = 4B XA + XB  10  2A + 4B  10 function f = objfun(x) a = x(1); b = x(2); f = -8*a + a^2 - 6*b + 1.5*b^2; Subject to: 2A + 4B  10 AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 A = 3, P0(A) = 5 & XA = 6 B = 1, P0(B) = 4.5 & XB = 4 TR = 5*3 + 4.5*1 = 19.5 4.27 Assume that there are m industries or municipalities adjacent to a river, which discharge their wastes into the river. Denote the discharge sites by the subscript i and let Wi be the kg of waste discharged into the river each day at those sites i. To improve the quality downstream, wastewater treatment plants may be required at each site i. Let xi be the fraction of waste removed by treatment at each site i. Develop a model for estimating how much waste is removal is required at each site to maintain acceptable water quality in the river at a minimum total cost. Use the following additional notation: aij = decrease in quality at site j per unit of waste discharged at site i qj = quality at site j that would result if all controlled upstream discharges were eliminated (i.e., W1 = W2 = 0) Qj = minimum acceptable quality at site j Ci = cost per unit (fraction) of waste removed at site i AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 Solution: Minimize ( x ) C m i i i  1 Subject to qj - j i i m i ij Q x ) a w (     1 1 j  0  xi  1 i  4.28 Assume that there are two sites along a stream, i = 1, 2, at which waste (BOD) is discharged. Currently, without any wastewater treatment, the quality, q2 and q3, at each of sites 2 and 3 is less than the minimum desired, Q2 and Q3, respectively. For each unit of waste removed at site i upstream of site j, the quality improves by Aij. How much treatment is required at sites 1 and 2 that meets the standards at a minimum total cost? Following are the necessary data: Ci = cost per unit fraction of waste treatment at site i (both C1 and C2 are unknown but for the same amount of treatment, whatever that amount, C1 > C2) Ri = decision variables, unknown waste removal fractions at sites i = 1, 2 A12 = 1/20 W1 = 100 Q2 = 6 A13 = 1/40 W2 = 75 Q3 = 4 A23 = 1/30 q2 = 3 q3 = 1 Solution: Site 1 Site 2 Site 3 Park W1 W2 Stream W1 W1(1-x1) Industry 1 WWTP 1 Stream Wi Wi(1-xi) Industry i WWTP i W3 W3(1-x3) Industry 3 WWTP 3 Qj = minimum acceptable quality at site j Site j AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 Minimize C1R1 + C2R2 Subject to q2 + A12W1R1  Q2 q3 + A13W1R1 + A23W2R2  Q3 AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 4.40 Consider a crop production problem involving three types of crops. How many hectares of each crop should be planted to maximize total income? Resources: Max Limits Resource requirements Crops: Corn Wheat Oats Water 1000/week 3.0 1.0 1.5 units/week/ha Labor 300/week 0.8 0.2 0.3 person hrs/week/ha Land 625 hectares Yield $/ha 400 200 250 Formulate a linear programming model?. Solution: Maximize 400*Corn + 200*Wheat + 250*Oats Subject to: Water constraint: 3*Corn + Wheat + 105*Oats  1000 Labor constraint: 0.8*Corn + 0.2*Wheat + 0.3*Oats  300 Land constraint: Corn + Wheat + Oats  625 4.43 In Indonesia there exists a wet season followed by a dry season each year. In one area of Indonesia all farmers within an irrigation district plant and grow rice during the wet season. This crop brings the farmer the largest income per hectare; thus they would all prefer to continue growing rice during the dry season. However, there is insufficient water during the dry season to irrigate all 5000 hectares of available irrigable land for rice production. Assume an available irrigation water supply of 32  106 m3 at the beginning of each dry season, and a minimum requirement of 7000 m3/ha for rice and 1800 m3/ha for the second crop. (a) What proportion of the 5000 hectares should the irrigation district manager allocate for rice during the dry season each year, provided that all available hectares must be given sufficient water for rice or the second crop? (b) Suppose that crop production functions are available for the two crops, indicating the increase in yield per hectare per m3 of additional water, up to 10,000 m3/ha for the second crop. Develop a model in which the water allocation per hectare, as well as the hectares allocated to each crop, is to be determined, assuming a specified price or return per unit of yield of each crop. Under what conditions would the solution of this model be the same as in part (a)? Solution: a) Let XR be the hectares of rice to be grown during the dry season and 5000 - XR be the hectares of the second crop. The total amount of water required is: 7000* XR + 1800(5000 - XR) = 32 x 106 Hence XR = 4435 ha of rice. b) Let the known parameters: PR, PS equal the price per unit yield of rice and the second crop Define as the unknown variables: AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 XR, XS = the hectares of rice and second crop WR, WS = the additional water allocations per ha. Assuming the objective is to maximize total income, the model can be written: Maximize PR (7000 + WR) XR + PS(1800 + WS) XS Subject to Water availability: (7000 + WR) XR + (1800 + WS) XS  32 x 106 Land availability: XR + XS  5000 Bounds WS  10, 000 m3/ha 4.45 In Algeria there are two distinct cropping intensities, depending upon the availability of water. Consider a single crop that can be grown under intensive rotation or extensive rotation on a total of A hectares. Assume that the annual water requirements for the intensive rotation policy are 16000 m3 per hectare, and for the extensive rotation policy they 4000 m3 per hectare. The annual net production returns are 4000 and 2000 dinars, respectively. If the total water available is 320,000 m3, show that as the available land area A increases, the rotation policy that maximizes total net income changes from one that is totally intensive to one that is increasingly extensive. Solving graphically: AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12 Optimization tool of Matlab: For A = 50: XI = 10, XE = 40 AUB-FEA-CEE CIVE646- Water Resources Systems: Planning and Management Fall 2011-12