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Linear Programming Models: Graphical and C H A P T E R 7 Computer Methods TEACHING SUGGESTIONS model). Here, the issue is the source of data. When accountants tell you a proﬁt contribution is $8.50 per unit, is that ﬁgure accurate Teaching Suggestion 7.1: Draw Constraints for a within 10% or within 10¢? The solution to an LP problem can Graphical LP Solution. change dramatically if the input parameters are not exact. Mention Explain constraints of the three types ( , , ) carefully the ﬁrst that sensitivity analysis also has other names, such as right-hand- time you present an example. Show how to ﬁnd the X1, X2 inter- side ranging, post-optimality analysis, and parametric programming. cepts so a straight line can be drawn. Then provide some practice in determining which way the constraints point. This can be done by picking a few X1, X2 coordinates at random and indicating ALTERNATIVE EXAMPLES which direction fulﬁlls the constraints. Alternative Example 7.1: Hal has enough clay to make 24 Teaching Suggestion 7.2: Feasible Region Is a Convex Polygon. small vases or 6 large vases. He only has enough of a special glaz- Explain Dantzing’s discovery that all feasible regions are convex ing compound to glaze 16 of the small vases or 8 of the large (bulge outward) polygons (many-sided ﬁgures) and that the opti- vases. Let X1 the number of small vases and X2 the number of mal solution must lie at one of the corner points. Draw both con- large vases. The smaller vases sell for $3 each, while the larger vex and concave ﬁgures to show the difference. vases would bring $9 each. Teaching Suggestion 7.3: Using the Iso-Proﬁt Line Method. (a) Formulate the problem. This method can be much more confusing than the corner point ap- (b) Solve graphically. proach, but it is faster once students feel comfortable drawing the SOLUTION: proﬁt line. Start your ﬁrst line at a proﬁt ﬁgure you know is lower (a) Formulation than optimal. Then draw a series of parallel lines, or run a ruler paral- OBJECTIVE FUNCTION: lel, until the furthest corner point is reached. See Figures 7.6 and 7.7. Maximize $3X1 $9X2 Teaching Suggestion 7.4: QA in Action Boxes in the LP Chapters. There are a wealth of motivating tales of real-world LP applica- Subject to: Clay constraint: 1X1 4X2 24 tions in Chapters 7–9. The airline industry in particular is a major Glaze constraint: 1X1 2X2 16 LP user. (b) Graphical solution Teaching Suggestion 7.5: Feasible Region for the Minimization Problem. 15 Students often question the open area to the right of the constraints in a minimization problem such as that in Figure 7.10. You need to explain that the area is not unbounded to the right in a mini- mization problem as it is in a maximization problem. X2 = Number of Large Vases Teaching Suggestion 7.6: Infeasibility. 10 This problem is especially common in large LP formulations since many people will be providing input constraints to the problem. (0, 8) This is a real-world problem that should be expected. Glaze Constraint Teaching Suggestion 7.7: Alternative Optimal Solutions. (0, 6) This issue is an important one that can be explained in a positive B way. Managers appreciate having choices of decisions that can be 5 (8, 4) Clay Constraint made with no penalty. Students can be made aware that alternative C optimal solutions will arise again in the transportation model, as- signment model, integer programming, and the chapter on net- Feasible Region work models. A (24, 0) Teaching Suggestion 7.8: Importance of Sensitivity Analysis. (0, 0) D (16, 0) 0 Sensitivity analysis should be stressed as one of the most important 0 5 10 15 20 25 LP issues. (Actually, the issue should arise for discussion with every X1 = Number of Small Vases 85 86 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS Point X1 X2 Income plied to minimization problems. Conceptually, isoproﬁt and iso- cost are the same. A 0 0 $0 The major differences between minimization and maximiza- B 0 6 54 tion problems deal with the shape of the feasible region and the di- C 8 4 60* D 16 0 48 rection of optimality. In minimization problems, the region must be bounded on the lower left, and the best isocost line is the one *Optimum income of $60 will occur by making and sell- closest to the zero origin. The region may be unbounded on the top ing 8 small vases and 4 large vases. and right and yet be correctly formulated. A maximization prob- Draw an isoproﬁt line on the graph on page 85 from (20, 0) to lem must be bounded on the top and to the right. The isoproﬁt line (0, 6X\c) as the $60 isoproﬁt line. yielding maximum proﬁt is the one farthest from the zero origin. Alternative Example 7.2: A fabric ﬁrm has received an order 7-2. The requirements for an LP problem are listed in Section for cloth speciﬁed to contain at least 45 pounds of cotton and 25 7.2. It is also assumed that conditions of certainty exist; that is, co- pounds of silk. The cloth can be woven out on any suitable mix of efﬁcients in the objective function and constraints are known with two yarns, A and B. Material A costs $3 per pound, and B costs $2 certainty and do not change during the period being studied. An- per pound. They contain the following proportions of cotton and other basic assumption that mathematically sophisticated students silk (by weight): should be made aware of is proportionality in the objective func- tion and constraints. For example, if one product uses 5 hours of a Yarn Cotton (%) Silk (%) machine resource, then making 10 of that product uses 50 hours of machine time. A 30 50 LP also assumes additivity. This means that the total of all ac- B 60 10 tivities equals the sum of each individual activity. For example, if the objective function is to maximize P 6X1 4X2, and if X1 What quantities (pounds) of A and B yarns should be used to mini- X2 1, the proﬁt contributions of 6 and 4 must add up to produce mize the cost of this order? a sum of 10. Objective function: min. C 3A 2B 7-3. Each LP problem that has been formulated correctly does Constraints: 0.30A 0.60B 45 lb (cotton) have an inﬁnite number of solutions. Only one of the points in the 0.50A 0.10B 25 lb (silk) feasible region usually yields the optimal solution, but all of the Simultaneous solution of the two constraint equations reveals that points yield a feasible solution. If we consider the region to be A 39 lb, B 55 lb. continuous and accept noninteger solutions as valid, there will be The minimum cost is C $3A $2B 3(39) (2)(55) an inﬁnite number of feasible combinations of X1 and X2. $227. 7-4. If a maximization problem has many constraints, then it can be very time consuming to use the corner point method to solve it. 300 Such an approach would involve using simultaneous equations to solve for each of the feasible region’s intersection points. The iso- proﬁt line is much more effective if the problem has numerous con- 250 straints. 7-5. A problem can have alternative optimal solutions if the isoproﬁt or isocost line runs parallel to one of the problem’s con- 200 Pounds of Yarn B straint lines (refer to Section 7-8 in the chapter). 7-6. This question involves the student using a little originality 150 to develop his or her own LP constraints that ﬁt the three condi- tions of (1) unboundedness, (2) infeasibility, and (3) redundancy. These conditions are discussed in Section 7.8, but each student’s 100 graphical displays should be different. min C 7-7. The manager’s statement indeed had merit if the manager 50 understood the deterministic nature of linear programming input data. LP assumes that data pertaining to demand, supply, materi- als, costs, and resources are known with certainty and are constant during the time period being analyzed. If this production manager 50 100 150 200 250 operates in a very unstable environment (for example, prices and Pounds of Yarn A availability of raw materials change daily, or even hourly), the model’s results may be too sensitive and volatile to be trusted. The SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS application of sensitivity analysis might be trusted. The applica- 7-1. Both minimization and maximization LP problems employ tion of sensitivity analysis might be useful to determine whether the basic approach of developing a feasible solution region by LP would still be a good approximating tool in decision making. graphing each of the constraint lines. They can also both be solved 7-8. The objective function is not linear because it contains the by applying the corner point method. The isoproﬁt line method is product of X1 and X2, making it a second-degree term. The ﬁrst, used for maximization problems, whereas the isocost line is ap- second, fourth, and sixth constraints are okay as is. The third and CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 87 ﬁfth constraints are nonlinear because they contain terms to the Let: X1 number of air conditioners to be produced second degree and one-half degree, respectively. X2 number of fans to be produced 7-9. For a discussion of the role and importance of sensitivity Maximize proﬁt 25X1 15X2 analysis in linear programming, refer to Section 7.9. It is needed subject to 3X1 2X2 240 (wiring) especially when values of the technological coefﬁcients and con- tribution rates are estimated—a common situation. When all 2X1 1X2 140 (drilling) model values are deterministic, that is, known with certainty, sen- X1, X2 0 sitivity analysis from the perspective of evaluating parameter ac- Proﬁt at point a (X1 0, X2 0) $0 curacy may not be needed. This may be the case in a portfolio se- lection model in which we select from among a series of bonds Proﬁt at point b (X1 0, X2 120) whose returns and cash-in values are set for long periods. 25(0) (15)(120) $1,800 7-10. Sensitivity analysis is important in all quantitative model- Profit at point c (X1 40, X2 60) ing techniques. Especially common is the analysis of inventory 25(40) (15)(60) $1,900 model results in which we test the model’s sensitivity to changes Profit at point d (X1 70, X2 0) in demand, lead time, cost, and so on. 25(70) (15)(0) $1,750 7-11. The computer is valuable in (1) solving LP problems quickly and accurately; (2) solving large problems that might take The optimal solution is to produce 40 air conditioners and 60 fans days or months by hand; (3) performing extensive sensitivity during each production period. Proﬁt will be $1,900. analysis automatically; and (4) allowing a manager to try several 7-15. ideas, models, or data sets. 140 7-12. The student is to create his or her own data and LP formula- tion. (a) The meaning of the right-hand-side numbers (resources) is to be explained. (b) The meaning of the constraint coefﬁcient (in 120 terms of how many units of each resource that each product re- quires) is also to be explained. (c) The problem is to be solved graphically. (d) A simple sensitivity analysis is to be conducted by 100 changing the contribution rate (Cj value) of the X1 variable. For ex- (26.67, 80) ample, if C1 was $10 as the problem was originally formulated, the b student should resolve with a $15 value and compare solutions. 80 e 7-13. A change in a technological coefﬁcient changes the feasi- X2 ble solution region. An increase means that each unit produced re- c Optimal Solution quires more of a scarce resource (and may lower the optimal 60 proﬁt). A decrease means that because of a technological advance- ment or other reason, less of a resource is needed to produce 1 40 unit. Changes in resource availability also change the feasible re- gion shape and can increase or decrease proﬁt. Feasible 7-14. 20 Region 140 a d 0 b 0 20 40 60 80 100 120 120 X1 Drilling Constraint Maximize profit 25X1 15X2 100 subject to 3X1 2X2 240 Number of Fans, X2 2X1 1X2 140 80 Optimal Solution X1 20 (X1 = 40, X2 = 60) X2 80 c X1, X2 0 60 Profit at point a (X1 20, X2 0) 25(20) (15)(0) $500 40 Wiring Profit at point b (X1 20, X2 80) Feasible Region Constraint 25(20) (15)(80) $1,700 20 a d 0 0 20 40 60 80 100 120 Number of Air Conditioners, X1 88 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS Profit at point c (X1 40, X2 60) 300 25(40) (15)(60) $1,900 Profit at point d (X1 70, X2 0) 250 25(70) (15)(0) $1,750 Profit at point e (X1 26.67, X2 80) 25(26.67) 15(80) $1,867 200 Hence, even though the shape of the feasible region changed from Problem 7-14, the optimal solution remains the same. 7-16. X2 150 300 a 100 Optimal Solution, 250 $2862.50 Profit 50 a Feasible Region b 200 0 c 0 50 100 150 200 250 300 350 400 X2 150 b X1 7-18. 100 60 X1 + X2 = 60 Feasible 50 Region 50 Feasible c Region 0 40 0 50 100 150 200 250 300 350 X1 X2 30 b X2 = 20 X1 number of model A tubs produced X2 number of model B tubs produced 20 a Optimal Maximize proﬁt 90X1 70X2 Solution subject to 125X1 100X2 25,000 (steel) X1 = 30 10 20X1 30X2 6,000 (zinc) X1, X2 0 Proﬁt at point a (X1 0, X2 200) $14,000 0 0 10 20 30 40 50 60 Proﬁt at point b (X1 85.71, X2 142.86) $17,714.10 X1 a Proﬁt at point c (X1 200, X2 = 0) $18,000 X1 number of undergraduate courses optimal solution X2 number of graduate courses 7-17. X1 number of benches produced Minimize cost $2,500X1 $3,000X2 X2 number of tables produced Maximize proﬁt $9X1 $20X2 subject to X1 30 X2 20 subject to 4X1 6X2 1,200 hours X1 X2 60 10X1 35X2 3,500 pounds Total cost at point a (X1 40, X2 20) X1, X2 0 2,500(40) (3,000)(20) Profit at point a (X1 0, X2 100) $2,000 $160,000 Profit at point b (X1 262.5, X2 25) $2,862.50 Profit at point c (X1 300, X2 0) $2,700 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 89 Total cost at point b (X1 30, X2 30) 60 2,500(30) (3,000)(30) $165,000 Point a is optimal. 7-19. a 40 40 X2 b Feasible Region Optimal Solution 30 20 Optimal Solution a Feasible Region is Heavily Shaded Line X2 20 0 c b 0 20 40 60 X1 10 7-21. 250,000 X1 + X2 = 100,000 0 0 10 20 30 40 X1 200,000 X1 = 125,000 X1 number of Alpha 4 computers X2 number of Beta 5 computers 150,000 Maximize proﬁt $1,200X1 $1,800X2 X2 X2 = 100,000 subject to 20X1 25X2 800 hours a 100,000 (total hours 5 workers Feasible 160 hours each) Region is this Line X1 10 50,000 X2 15 Corner points: a(X1 10, X2 24), profit $55,200 b(X1 1 21 , X2 15), profit $52,500 b 4 0 0 50,000 100,000 150,000 200,000 250,000 Point a is optimal. X1 7-20. Let: X1 number of pounds of compost in each bag X1 $ invested in Treasury notes X2 number of pounds of sewage waste in each bag X2 $ invested in bonds Minimize cost 5X1 4X2 (in cents) Maximize ROI 0.08X1 0.09X2 subject to X1 X2 60 (pounds per bag) X1 $125,000 X1 30 (pounds compost per bag) X2 $100,000 X2 40 (pounds sewage per bag) X1 X2 $250,000 Corner point a: X1, X2 0 (X1 30, X2 40) ⇒ cost 5(30) (4)(40) $3.10 a Point a (X1 150,000, X2 100,000), ROI $21,000 Corner point b: optimal solution (X1 30, X2 30) ⇒ cost 5(30) (4)(30) $2.70 Point b (X1 250,000, X2 0), ROI $20,000 Corner point c: (X1 60, X2 0) ⇒ cost 5(60) (4)(0) $3.00 90 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 7-22. 7-23. Point a lies at intersection of constraints (see ﬁgure below): 3X1 2X2 120 X1 3X2 90 50 Multiply the second equation by 3 and add it to the ﬁrst (the 5X1 + 3X2 ≤ 150 method of simultaneous equations): 3X1 2X2 120 40 Isoprofit Line Indicates 3X1 9X2 270 that Optimal Solution 7X2 150 ⇒ X2 21.43 and X1 25.71 Lies at Point a Cost $1X1 $2X2 $1(25.71) ($2)(21.43) 30 3 3 $68.57 X2 (X1 = 18 4 , X2 = 18 4 , Profit = $150) 7-24. X1 $ invested in Louisiana Gas and Power 20 X1 – 2X2 ≤ 10 X2 $ invested in Trimex Insulation Co. a Minimize total investment X1 X2 Feasible subject to $0.36X1 $0.24X2 $720 Region 3X1 + 5X2 ≤ 150 10 $1.67X1 $1.50X2 $5,000 0.04X1 0.08X2 $200 Investment at a is $3,333. 0 Investment at b is $3,179. k optimal solution 0 10 20 30 40 50 X1 Investment at c is $5,000. Short-term growth is $927.27. Intermediate-term growth is $5,000. Note that this problem has one constraint with a negative sign. This may cause the beginning student some confusion in plotting Dividends are $200. the line. See graph on page 91. Figure for Problem 7-23. 80 8X1 + 2X2 ≥ 160 X2 ≤ 70 60 Feasible Region X2 40 Iso cos t Li ne =$ 100 =1 X 1 +2 20 a X 2 X1 + 3X2 ≥ 90 3X1 + 2X2 ≥ 120 0 0 20 40 60 80 100 X1 Isoprofit Line Indicates That Optimal Solution Lies At Point a CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 91 Figure for Problem 7-24. 4,000 a(X1 = 0, X2 = 3,333) 3,000 Feasible Region X2 2,000 b(X1 = 1,359, X2 = 1,818.18) 1,000 c (X1 = 5,000, X2 = 0) 0 0 1,000 2,000 3,000 4,000 5,000 X1 7-25. Point c (X1 25, X2 20), exposure 875,000 400,000 3,000X1 + 1,250X2 ≤ 100,000 80 1,275,000 X1 ≥ 5 X1 ≤ 25 Point d (X1 25, X2 10), exposure 875,000 b Optimal Exposure 200,000 60 Rating 1,075,000 7-26. Let: X1 number of barrels of pruned olives X2 number of barrels of regular olives X2 40 Maximize proﬁt $20X1 $30X2 subject to 5X1 2X2 250 (labor hours) Feasible Region 1X1 2X2 150 (acres) X2 ≥ 10 20 c X1 40 (barrels) X1, X2 0 a d a. Corner point a (X1 0, X2 0), proﬁt 0 0 0 5 10 15 20 25 30 35 Corner point b (X1 0, X2 75), proﬁt $2,250 X1 Corner point c (X1 25, X2 62Z\x), proﬁt $2,375 k optimal proﬁt Let: X1 number of TV spots Corner point d (X1 40, X2 25), proﬁt $1,550 X2 number of newspaper ads Corner point e (X1 40, X2 0), proﬁt $800 Maximize exposures 35,000X1 20,000X2 b. Produce 25 barrels of pruned olives and 62Z\x barrels subject to 3,000X1 1,250X2 $100,000 of regular olives. X1 5 c. Devote 25 acres to pruning process and 125 acres to X1 25 regular process. X2 10 Point a (X1 5, X2 10), exposure 375,000 (optimal) Point b (X1 5, X2 68), exposure 175,000 1,360,000 1,535,000 92 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 125 X1 2X2 2 line—this is also on the same slope as the isoproﬁt line X1 2X2 and hence there will be more than one optimal solu- tion. 100 As a matter of fact, every point along the heavy line will pro- vide an “alternate optimum.” b Formulation 3: 75 X2 c 5 Optimal Solution 50 Unbounded Region 4 Feasible d 25 Region 3 a e 0 X2 0 25 50 75 100 125 150 X1 2 7-27. 1 Formulation 1: 8 0 0 1 2 3 4 5 6 X1 6 Formulation 4: Infeasible Solution 8 Region X2 4 6 2 X2 4 0 0 2 4 6 8 10 12 Feasible Region X1 2 1 Formulation 2: 0 0 1 2 3 4 6 8 10 12 2 X1 Formulation 4 appears to be proper as is. Note that the constraint 4X1 6X2 48 is redundant. X2 1 Line For X1 + 2X2 Feasible Region 0 0 1 2 3 X1 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 93 7-28. Using the isoproﬁt line or corner point method, we see that 8 point b (where X1 37.5 and X2 75) is optimal if the proﬁt $3X1 $2X2. If the proﬁt changes to $4.50 per unit of X1, the optimal solution shifts to point c. If the objective function be- comes P $3X1 $3X2, the corner point b remains optimal. 6 150 Profit Line for 3X1 + 3X2 Profit = 4X1 + 6X2 = $21.71 X2 4 Optimal solution at a Profit Line for 4.50X1 + 2X2 100 X1 = 26/7, X2 = 15/7 2 1X1 + 3X2 = 8 X2 b 50 0 Profit Line for 0 2 4 6 8 3X1 + 2X2 X1 7-30. a. 100 c 0 0 50 100 150 X1 7-29. The optimal solution of $26 proﬁt lies at the point X1 2, 75 X2 3. Isoprofit Line for 1X1 + 1X2 = $66.67 8 a X2 50 (X 1 = 331/3, X2 = 331/3 ) 6 b 25 X2 4 c 0 Profit = 4X1 + 6X2 = $26 0 25 50 75 100 X1 2 0 0 2 4 6 8 X1 If the ﬁrst constraint is altered to 1X1 3X2 8, the feasible re- gion and optimal solution shift considerably, as shown in the next column. 94 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS b. 7-32. 150 12 10 Isoprofit Line for 3X1 + 1X2 = $150 100 8 6X1 + 4X2 = 36 X2 X2 6 (X 1 = 5, X2 = 11/2 ; Profit = $29 ) 50 4 2 1X1 + 2X2 = 8 Optimal Solution is Now Here 0 0 0 50 100 150 0 2 4 6 8 10 12 X1 X1 Using the corner point method, we determine that the optimal so- c. If X1’s proﬁt coefﬁcient was overestimated, but should lution mix under the new constraint yields a $29 proﬁt, or an in- only have been $1.25, it is easy to see graphically that crease of $3 over the $26 proﬁt calculated. the solution at point b remains optimal. 7-33. Let: X1 number of coconuts carried 7-31. X2 number of skins carried 100 Maximize proﬁt 60X1 300X2 (in rupees) subject to 5X1 15X2 300 pounds 1 8 X1 1X2 15 cubic feet 75 X 1, X 2 0 At point a: (X1 0, X2 15), P 4,500 rupees At point b: (X1 24, X2 12), P 1,440 3,600 5,040 rupees X2 50 (X 1 = 426/7, X2 = 142/7; Profit = $571/7 ) At point c: (X1 60, X2 0), P 3,600 rupees The three princes should carry 24 coconuts and 12 lions’ skins. This will produce a wealth of 5,040 rupees. 25 20 Optimal Solution Remains at b Point b 0 a 0 25 50 75 100 15 Number of Lion Skins, X2 X1 Optimal Solution b The optimal solution is at point b, but proﬁt has decreased from 10 $662 to $571, and the solution has changed considerably. 3 7 Feasible 5 Region c 0 0 30 60 90 120 Number of Coconuts, X1 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 95 7-34. a. Let: X1 number of pounds of stock X purchased per 0.2X1 0.4X2 0.1X3 0.2X4 26,000 (hours of assembly cow each month time available) X2 number of pounds of stock Y purchased per 0.5X1 0.1X2 0.5X3 0.5X4 1,200 (hours of inspection cow each month time) X3 number of pounds of stock Z purchased per X1 150 (units of XJ201) cow each month X2 100 (units of XM897) Four pounds of ingredient Z per cow can be transformed to: X3 300 (units of TR29) 4 pounds (16 oz/lb) 64 oz per cow X4 400 (units of BR788) 5 pounds 80 oz 7-36. Maximize Z [220 (0.45)(220) 44 20]X1 1 pound 16 oz [175 (0.40)(175) 30 20]X2 8 pounds 128 oz 57X1 55X2 3X1 2X2 4X3 64 (ingredient A requirement) Constraints: 2X1 3X2 1X3 80 (ingredient B requirement) X1 X2 390 production limit 1X1 0X2 2X3 16 (ingredient C requirement) 2.5X1 2.4X2 960 labor hours 6X1 8X2 4X3 128 (ingredient D requirement) Corner points: X3 5 (stock Z limitation) X1 384, X2 0, proﬁt $21,888 Minimize cost $2X1 $4X2 $2.50X3 X1 0, X2 390, proﬁt $21,450 b. Cost $80 X1 240, X2 150, proﬁt $21,930 X1 40 lbs. of X Students should point out that those three options are so close in X2 0 lbs. of Y profit that production desires and sensitivity of the RHS and cost X3 0 lbs. of Z coefficient are important issues. This is a good lead-in to the dis- cussion of sensitivity analysis. As a matter of reference, the 7-35. Let: X1 number units of XJ201 produced right-hand side ranging for the first constraint is a production X2 number units of XM897 produced limit from 384 to 400 units. For the second constraint, the hours X3 number units of TR29 produced may range only from 936 to 975 without affecting the solution. The objective function coefﬁcients, similarly, are very sensi- X4 number units of BR788 produced tive. The $57 for X1 may increase by 29 cents or decrease by $2. Maximize proﬁt 9X1 12X2 15X3 11X4 The $55 for X2 may increase by $2 or decrease by 28 cents. subject to 7-37. a. Let: X1 number of MCA regular modems made and 0.5X1 1.5X2 1.5X3 0.1X4 15,000 (hours of wiring sold in November time available) X2 number of MCA intelligent modems made 0.3X1 0.1X2 0.2X3 0.3X4 17,000 (hours of drilling and sold in November time available) Data needed for variable costs and contribution margin (refer to the table on the bottom of this page): Table for Problem 7-37(a) MCA REGULAR MODEM MCA INTELLIGENT MODEM Total Per Unit Total Per Unit Net sales $424,000 $47.11 $613,000 $58.94 Variable costsa Direct labor 60,000 6.67 76,800 7.38 Indirect labor 9,000 1.00 11,520 1.11 Materials 90,000 10.00 128,000 12.31 General expenses 30,000 3.33 35,000 3.37 Sales commissions $231,000 $23.44 $360,000 $25.76 Total variable costs $220,000 $24.44 $311,320 $29.93 Contribution margin $204,000 $22.67 $301,680 $29.01 a Depreciation, ﬁxed general expense, and advertising are excluded from the calculations. 96 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS Hours needed to produce each modem: Solution: Produce: 5,000 hours 1,100 units of W75C—backorder 300 units MCA regular 0.555 hour/modem 9,000 modems 250 units of W33C—backorder 0 units 10,400 hours 0 units of W5X—backorder 1,510 units MCA intelligent 1.0 hour/modem 10,400 modems 600 units of W7X—backorder 516 units Maximize proﬁt $22.67X1 $29.01X2 Maximized proﬁt will be $59,900. By addressing quality problems subject to 0.555X1 1.0X2 15,400 (direct labor hours) listed earlier, we could increase our capacity by up to 3% reducing X2 8,000 (intelligent modems) our backorder level. b. 2. Bringing in temporary workers in the Drawing Department would not help. Drawing is not a binding constraint. However, if X2 these former employees could do rework, we could reduce our re- work inventory and ﬁll some of our backorders thereby increasing 15,400 proﬁts. We have about a third of a month’s output in rework inven- tory. Expediting the rework process would also free up valuable cash. P = $534,339 3. The plant layout is not optimum. When we install the new equip- ment, an opportunity for improving the layout could arise. Exchang- 8,000 ing the locations for packaging and extrusion would create a better b ﬂow of our main product. Also, as we improve our quality and re- Optimal duce our rework inventory, we could capture some of the space now P = $629,000 used for rework storage and processing and put it to productive use. Our machine utilization of 63% is quite low. Most manufac- c turers strive for at least an 85% machine utilization. If we could 27,750 X1 determine the cause(s) of this poor utilization, we might ﬁnd a key c. The optimal solution suggests making all MCA to a dramatic increase in capacity. regular modems. Students should discuss the implications of shipping no MCA intelligent modems. INTERNET CASE STUDY: 7-38. Minimize cost 12X1 9X2 11X3 4X4 Agri-Chem Corporation subject to X1 X2 X3 X4 50 This case demonstrates an interesting use of linear programming in a production setting. X1 X2 X3 X4 7.5 Let X1 ammonia X1 X2 X3 X4 22.5 X2 ammonium phosphate X1 X2 X3 X4 15.0 X3 ammonium nitrate Solution: X4 urea X1 7.5 pounds of C-30 X5 hydroﬂuoric acid X2 15 pounds of C-92 X6 chlorine X3 0 pounds of D-21 X7 caustic soda X4 27.5 pounds of E-11 X8 vinyl chloride monomer Cost $3.35. Objective function: SOLUTION TO MEXICANA WIRE WORKS CASE Maximize Proﬁt 80X1 120X2 140X3 140X4 90X5 70X6 60X7 90X8 1. Maximize P 34 W75C 30 W33C 60 W5X 25 W7X Subject to the following constraints: subject to: X1 1,200 X5 560 1 W75C 1,400 X2 540 X6 1,200 1 W33C 250 X3 490 X7 1,280 1 W5XC 1,510 X4 160 X8 840 1 W7XC 1,116 Current natural gas usage 85,680 cu. ft. 103/day 1 W75C 2 W33C 0 W5X 1 W7X 4,000 20 percent curtailment 68,554 cu. ft. 103/day 1 W75C 1 W33C 4 W5X 1 W7X 4,200 1 W75C 3 W33C 0 W5X 0 W7X 2,000 1 W75C 0 W33C 3 W5X 2 W7X 2,300 1 W75C 150 1 W7X 600 CHAPTER 7 LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS 97 Hence, the ninth constraint is: After 8 simplex iterations, optimal solution is reached. The 8X1 10X2 12X3 12X4 7X5 18X6 20X7 14X7 following is the production schedule: 68,544 X1 1200 X5 560 The following is the production schedule (tons/day); X2 540 X6 720 X1 1,200 X5 560 X3 490 X7 0 X2 540 X6 1,200 X4 160 X8 840 X3 490 X7 425 Objective function value: $428,200 X4 160 X8 840 The caustic soda production is eliminated completely and the chlorine production is reduced from 1,200 to 720 tons/day. Objective function value $487,300 Because of the natural gas curtailment, the caustic soda pro- duction is reduced from 1280 tons/day to 425 tons/day. For a 40 percent natural gas curtailment, the ninth constraint is: 8X1 10X2 12X3 12X4 7X5 18X6 20X7 14X8 51,408