UNIVERSITY OF WATERLOODEPARTMENT OF MANAGEMENT SCIENCES MSCI603 - Principles of Operations Research Problem set 3 Problem 1 Cornco produces two products: PS and QT. The sales prices for each product and the maximum quantity of each that can be sold during each of the next three months are given in the table below. Month 1 Month 2 Month 3 Product Price ($) Demand Price ($) Demand Price ($) Demand PS 40 50 60 45 55 50 QT 35 43 40 50 44 40 Each product must be processed through two assemble lines: 1 and 2. The number of hours required by each product on each assembly line is given below: Hours Product Line 1 Line 2 PS 3 2 QT 2 2 The number of hours available on each assembly line during each month is given below: Month Line 1 2 3 1 1200 160 190 2 2140 150 110 Each unit of PS requires 4 pounds of raw material; each unit of QT requires 3 pounds. As many as 710 units of raw material can be purchased at $3 per pound. At the beginning of month 1, 10 units of PS and 5 units of QT are available. It costs $10 to hold a unit of either product in inventory for a month. Solve this LP in Excel Solver and use your output to answer the following questions. (Please include a print-out of your input and the sensitivity report.) Hint: Let Pi = units of PS produced in month i, PSi = units of PS sold in month i, IPi = inventory of product P at end of month i, Qi = units of QT produced in month i, QSi = units of QT sold in month i, IQi = units of QT in inventory at end of month i, RM = pounds of raw material purchased. a) Find the new optimal solution if it costs $11 to hold a unit of PS in inventory at the end of month 1. b) Find the companyβs new optimal solution if 210 hours on line 1 are available during month 1. c) Find the companyβs new profit level if 109 hours are available on line 2 during month 3. d) What is the most Cornco should be willing to pay for an extra hour of line 1 time during month 2? e) What is the most Cornco should be willing to pay for an extra pound of raw material? f) What is the most Cornco should be willing to pay for an extra hour of line 1 time during month 3? h) An increase of 6 is not within the allowable increase of 1.Q1 = 5 18) QS2 .75. so the new profit is 7705 β 7(1) = $7698. so the Shadow Price = 0 and allowable decrease is 1010.t. h) Find the new optimal solution if QT sells for $50 during month 3. d) The most Cornco will pay is equal to the shadow price of this constraint = $3. i) Suppose spending $20 on advertising would increase demand for QT in month 2 by 5 units. New Profit = 7705 .IP2 + IP3 . . Profit Down by ($1)* πΌπ1 = $25.IP1 + IP2 . The shadow price for the constraint is 7. so this 1 hour decrease is within the allowable limit. We are not willing to pay anything for any additional time (shadow price is 0).IQ2 + IQ3 .10(45) = $7255. 2) P1S β€ 50 3) P2S β€ 45 4) P3S β€ 50 5) Q1S β€ 43 6) Q2S β€ 50 7) Q3S β€ 40 8) 3 P1 + 2 Q1 β€ 1200 9) 3 P2 + 2 Q2 β€ 160 10) 3 P3 + 2 Q3 β€ 190 11) 2 P1 + 2 Q1 β€ 2140 12) 2 P2 + 2 Q2 β€ 150 13) 2 P3 + 2 Q3 β€ 110 14) PS1 + IP1 .Q3 = 0 20) .IQ1 + IQ2 .RM + 4 P1 + 3 Q1 + 3 Q2 + 4 P2 + 4 P3 + 3 Q3 β€ 0 21) RM β€ 710 All variables β₯ (See Excel file for Sensitivity and Answer Reports) a) Allowable increase for πΌπ1 is 4. meaning there is slack in this constraint.P1 = 10 15) PS2 . the shadow price $10. g) Find the new optimal solution if PS sells for $50 during month 2. This is what we would gain if given a "free pound of raw material. Should the advertising be done? Solution MAX 40 PS1 + 60 PS2 + 55 PS3 + 35 QS1 + 40 QS2 + 44QS3 . c) Allowable decrease is 10 hours.3 RM - 10(IP1+IP2+IP3+IQ1+IQ2+IQ3) s. Constraint has positive slack. so current basis is no longer optimal and question cannot be answered from current printout. so decision variables remain the same. Thus we would pay up to $10. so a $1 increase is within the allowable limit.Q2 = 0 19) QS3 . Thus new optimal solution remains the same. b) This is a non-binding constraint. g) A decrease of 10 is within the allowable decrease for this variable.P3 = 0 17) QS1 + IQ1 .25 = $7680. f) This is a non-binding constraint. New z-value = 7705 .33 e) From constraint 20.P2 = 0 16) PS2 . and sells them to customers either finished (F) or unfinished (U).67. c) The allowable decrease in the FC coefficient in the objective function is 5.66666667 1E+30 $C$4 FT 0 -6. so we can use sensitivity analysis.666666667 1E+30 $D$4 UC 0 -50 60 50 1E+30 $E$4 FC 1333.333333 0 110 1E+30 5 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R. .333333 0 6000 1E+30 666.67.Problem 2 A furniture company makes tables (T) and chairs (C). so the company doesnβt need to change its product mix. b) How much rise in the price of finished tables is needed for the company to start producing them? c) If the company reduces the selling price of finished chair by $5.000 board feet of wood and 6.000 ft2.H.666666667 40000 5000 40000 $F$6 Labor LHS 5333. will it need to change its product mix to optimize its profit. Shadow price of wood = 3.66666667 70 76. and the selling price of each product are shown in the next table: Product UT FT UC FC Wood (ft2) 40 40 30 30 Labor (hr) 2 5 2 4 Price ($) 70 140 60 110 If 40. so the price of finished tables must increase by this amount before the company starts producing them. Use the sensitivity analysis output report to answer the next parts. so increasing the quantity of wood by 3000 will increase the profit by $11. a) How much improvement in the profit can be achieved by: i. Solution Below is a copy of the sensitivity analysis report generated by Excel Variable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$4 UT 0 -76. The amount of wood and labor needed. independently. ii. Allowable increase of wood quantity without changing the basis is 5000.000.666666667 140 6. Side Increase Decrease $F$5 Wood LHS 40000 3. ii. so increasing the quantity of labor hours will not affect the profit. The shadow price of labor hour is zero (the constraint is non-binding). increasing the quantity of wood available by 3. b) The reduced cost of finished tables (FT) is 6.6666667 a) i. increasing the number of labor hours available by 500 hours.000 hours of skilled labor are available. For type 1 cheese: 1 c1 2 β 2 β₯ β 110 β₯ β 2 so the price of type 1 cheese can range between 390 and 445 $/unit without changing the optimal solution.10) The maximum amount that should be paid for an additional hour per week to worker 1 is 45. Type W1 (hr) W2 (hr) Raw material cost ($) Selling Price ($) 1 1 2 250 400 2 2 2 200 420 a.5) and a profit of 1850. The optimal solution is the intersecting point of the two constraints. The companyβs profit decreases by 10*(50β48) = 20. The time for each worker and raw materials costs required to produce a unit of each type of cheese and its selling price are shown in the table below. what would the companyβs profit be? Verify your answer graphically. For type 2 cheese: 1 65 1 β 2 β₯ β c2 β₯ β 1 so the price of type 2 cheese can range between 375 and 440 $/unit without changing the optimal solution. x2 >=0 b. The current solution will remain optimal as long as the slope of the iso-profit line remains between the slopes of the binding constraints.Problem 3 A dairy producer makes two types of cheese. d. If W2 worker is willing to work only 48 hours. f. and the profit is 2280. b. The first (W1) is willing to work for up to 40 hours per week and is paid $25 per hour. The maximum amount that should be paid for an additional hour per week to worker 2 is 10. If W1 worker is willing to work only 30 hours per week. f. The company has two specialized workers. .x2) = (8. The second (W2) is willing to work for up to 50 hours per week and is paid $30 per hour. Solution a. From the graphical solution obtained in (b) determine the range of prices of types 1 and 2 cheese at which the current basis remain optimal.16).15) and a profit of 2300. would the current basis remain optimal? Would the optimal solution change? e. The only scarce resource that is needed to produce cheese is skilled labor. c. x1+2x2 <= 40 2x1+2x2 <= 50 x1>=0.x2) = (20. d. Yes. which is (10. The graphical solution is (x1. c. Solve the LP model graphically.t. e. The shadow price is (y1. Formulate an LP model for this problem to maximize the profit of the dairy producer. Determine the maximum amount that should be paid to each worker for an additional hour of work every week.y2) = (45. The new optimal solution will be (x1. x1 = the number of type 1 cheese produced x2 = the number of type 2 cheese produced max (400β250)x1+(420β200)x2 β25(x1+2x2) β 30(2x1+2x2) s. 5 By plugging these values in the objective function. We need to find the range of optimality. What is the profit range of belts that will keep the optimal solution unchanged? f. respectively. . What is the profit range of wallets that will keep the optimal solution unchanged? e. By how much the workshop can increase or decrease the quantity of leather available without changing its optimal product mix? g. W + 2B <= 500 (Leather constraint) a. the binding constraints must remain binding.Problem 4 A workshop makes two types of hand-made genuine leather products: wallets and belts. So. So this is the range of leather availability at which both products are in the optimal product mix. W/2 + B/3 <= 200 (Labor constraint) b.t. Each day there are 500 square feet of leather and 200 labor-hours available. the slope of the iso-profit line must remain within the slopes of the binding constraints. The same method as in part (b) d. What is the maximum amount the workshop shall be willing to pay for an extra hour of labor? c. Formulate an LP for this problem and solve it using the graphical method. The RHS of the leather constraint at these points has values of 400 and 1200. The workshop earns $40 profit from each wallet sold and $50 from each belt sold. The workshop shall not pay more than the dual price of the second constraint. the first constraint can shift parallel to itself between points (400. B >= 0 By solving using the graphical method we find that: W* = 350. B* = 75. b. Z_new β Z_old = change in RHS * dual price = 10 * 45 (from part b) = $450 increase in the profit. g. Z* = 17750 b. (-3/2) <= (-c/50) <= (-1/2) 25 <= c <= 75 So the profit range of wallets can range between 25 and 75 without changing the optimal solution e. To maintain the same product mix.600). Each belt requires 2 square feet of leather and 20 minutes of labor time.0) and (0. Max 40 W + 50 B s. What is the maximum amount the workshop shall be willing to pay for an extra square foot of leather? d. The same method as in part (d) f. For the optimal solution to remain unchanged. To find it we solve the two equations: W + 2B = 500 W/2 + B/3 = 201 The solution is: W = 353. Let W be the number of wallets and B be the number of belts produced daily. B = 73. we get Z_new = 17795 Dual price = Z_new β Z_old = $45 c. W. a. What is the impact of increasing the labor time by 10 hours daily on the workshop profit? Solution a. Each wallet requires 1 square foot of leather and 30 minutes of labor time. b) What is the range of wheat and corn prices that will keep the current basis optimal? c) What is the range of land and labor availability that will keep the current basis optimal? Solution: a) Decision Variables π₯π€ ππ’ππππ ππ πππππ πππππ‘ππ π€ππ‘β π€βπππ‘ π₯π ππ’ππππ ππ πππππ πππππ‘ππ π€ππ‘β ππππ Constraints 5π₯π€ β€ 140 Demand limit for wheat 4π₯π β€ 120 Demand limit for corn π₯π€ + π₯π β€ 45 Land availability 6π₯π€ + 10π₯π β€ 350 Labour availability Objective Function max 5 β 30π₯π€ + 4 β 50π₯π = 150π₯π€ + 200π₯π The graphical solution is shown in the figure below. 50 45 40 35 C2 b 30 25 C1 20 Optimal solution 15 a C4 10 5 C3 5 10 15 20 25 30 35 40 45 50 55 60 Xw Iso-profit line β π₯π€ = 25 ππππ π₯πβ = 20 ππππ π§ β = $7. He can sell at most 140 bushels of wheat and 120 bushels of corn. Each planted acre yields either 5 bushels of wheat or 4 bushels of corn.Problem 5 Farmer Leary grows wheat and corn in his 45-acre farm. so the price of wheat can range between 24 and 40 5 200 1 $/bushel without changing the optimal solution. and ten hours are needed to harvest an acre of corn. and corn sells for $50 per bushel. As many as 350 hours of labor are available.750 b) The current solution will remain optimal as long as the slope of the iso-profit line remains between the slopes of the binding constraints> For wheat: 3 π 1 β β₯β π€ β₯β or 120 β€ ππ€ β€ 200 . and solve it graphically. For corn: . Six hours of labor are needed to harvest an acre of wheat. a) Formulate the problem as an LP model. Wheat sells for $30 per bushel. 33. The RHS of C3 at this point is 46. This point is (8. The RHS of C4 at this point is 390 hours. A new assessment reached a conclusion that gold is no longer a suitable alternative for this investor.5 5 ππ 1 $/bushel without changing the optimal solution.5 and 62. c) Land availability: C3 can shift up parallel to itself without changing the basis until it hits the intersection of C1 and C4. Solve it using Excel. What is the minimum expected return that justifies allocating funds to this product? Solution Let G. If possible without resolving the problem. If the investment policy requires at least 5% of the investment to be held as gold.30). certificates of deposit. bonds and stocks. The investor lowers its acceptable profit threshold to 4. B and S be the percentages of funds invested in gold. The return of stocks is known to be volatile and may fall short of expectations. The RHS of C4 at this point is 338 hours.33 acre.1%. How would that affect the allocation? i. Problem 6 An investor has 4 investment alternatives for the next year: Gold (G). respectively.17). so the price of corn can range between 37. C4 can shift down parallel to itself without changing the basis until it hits the intersection of C1 and C3.18. Labor availability: C4 can shift up parallel to itself without changing the basis until it hits the intersection of C2 and C3. Certificates of deposit (C). expected total return (i. The RHS of C3 at this point is 38. average risk score does not exceed 2 3. How would this affect the allocation and the profit? g. profit) is at least 5% 2.2). If the investorβs risk tolerance becomes higher such that a risk score of 3 is now acceptable. This point is (15.e.2 acre. What is the minimum expected return of stocks such that the current allocation is not changed? e. The minimum percentage of funds to be invested in fixed income instruments in increased to 75%. how would this affect the allocation and the profit? b. use the sensitivity report to answer the following questions: a. If the expected return of the certificates of deposit falls by 0. C3 can shift down parallel to itself without changing the basis until it hits the intersection of C2 and C4. What is the minimum expected return of bonds that justifies buying them? c. This point is (28. how would that affect the allocation and the profit? f.30). How would this affect the allocation and the profit? h.. C. A new investment product called βleveraged ETFβ is introduced. which has a risk score of 8.5%. The problem can be formulated as: Max z= 2G+4C+6B+10S . Bonds (B) and Stocks (S). at most half of the funds are invested in fixed income instruments (C and B) Formulate an LP to determine the optimal investment plan. The expected return and the risk score for these investments are shown below: Investment type G C B S Expected return 2% 4% 6% 10% Risk score 2 1 3 6 Available funds are to be allocated between the different alternatives such that: 1. how would that affect the profit? d. 3 150 1 β β₯β β₯β or 150 β€ ππ β€ 250 . This point is (28. Solving the equations: C+6S=3 and C+S=1 yields: C=0. Side Increase Decrease $F$3 LHS 5.4 6 0.2 0 10 1E+30 1 Final Shadow Constraint Allowable Allowable Cell Name Value Price R.3 which is greater than 0.4%. Thus.2. The reduced cost of B is -0.25. The expected return constraint is non-binding and the allowable decrease in the RHS is infinite. g. but the profit will decrease by 0.2+1. The allowable increase in the RHS of constraint 3 without changing the basis is 0. z=5.2% Below is the Excel sensitivity report Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$2 G 0 -3. The decrease in C coefficient is within the allowable limit (<0.6 and S=0. The shadow price of the risk score constraint is +1. a 0.05*3.5 (minimum fixed income investment ratio) G+ C + B+ S = 1 (all funds are invested) G. So.3 1E+30 $F$6 LHS 1 2.8 1 1 0. h.071428571 a.166666667 $F$5 LHS 0. Adding a unit of the new product reduces the RHS of constraint 2 by 8 units and constraint 4 by 1 unit. Thus. . This is the minimum expected profit that justifies adding the new product to the portfolio.2 units to 5. B=0.2 2 3. no effect on the allocation or profit will result from this change in parameters.2 0 5 0.4. C=0.8. Thus.66667).12% b.5 0. C. it has no effect on the allocation or profit.2 2 1.2=6. so the optimal solution (allocation) will not change.08 to 5.4)=6. S=0. B.6)+10(0. This change is equivalent to adding the constraint G=0.5>=1. A unit of G forced into the optimal solution will lower the objective value by the reduced cost of G. Since the current optimal solution does not violate the new constraint.8 0 4 6 0.4.16% in profit to 5. i. meaning it reduces the profit by 8(1.8=0. add bonds to the portfolio) c.2 1E+30 $F$4 LHS 2 1.8)=12.s.4% for B to enter the basis (assume a non-zero value.4%. if the RHS of this constraint increases by one unit the objective function improves by 1.1*0.04% d. the expected return of bonds must increase by 0.666666667 $D$2 B 0 -0. thus nothing will change.8 0 0.4 to 6. so z=4(0.4% (the same result found using the shadow price) f.05 units will cause a decline of 0.2 and the allowable increase is 1.2)+1(2. This is a non-binding constraint and will remain non-binding. Allowable decrease in the coefficient of S is 1. S >=0 The optimal solution is: G=0.t 2G+4C+6B+10S>=5 (expected return) 2G+ C+3B+ 6S<=2 (risk score) C+ B >=0. The basis (and hence the binding constraints) remain unchanged.2 1E+30 $C$2 C 0.4 1E+30 $E$2 S 0. the minimum expected return of stocks so the current allocation remains optimal is 10-1=9% e.2= 0. Thus.H.5 0.