Chapter One: Management Science

Transcription

1 Chapter One: Management Siene PROBLEM SUMMARY 1. Total ost, reenue, proit, and break-een 2. Total ost, reenue, proit, and break-een 3. Total ost, reenue, proit, and break-een 4. Break-een olume 5. Graphial analysis (1 2) 6. Graphial analysis (1 4) 7. Break-een sales olume 8. Break-een olume as a perentage o apaity (1 2) 9. Break-een olume as a perentage o apaity (1 3) 10. Break-een olume as a perentage o apaity (1 4) 11. Eet o prie hange (1 2) 12. Eet o prie hange (1 4) 13. Eet o ariable ost hange (1 12) 14. Eet o ixed ost hange (1 13) 15. Break-een analysis 16. Eet o ixed ost hange (1 7) 17. Eet o ariable ost hange (1 7) 18. Break-een analysis 19. Break-een analysis 20. Break-een analysis 21. Break-een analysis; olume and prie analysis 22. Break-een analysis; proit analysis 23. Break-een analysis 24. Break-een analysis; proit analysis 25. Break-een analysis; prie and olume analysis 26. Break-een analysis; proit analysis 27. Break-een analysis; proit analysis 28. Break-een analysis; proit analysis 29. Linear programming 30. Linear programming 31. Linear programming 32. Linear programming 33. Foreasting/statistis 34. Linear programming 35. Waiting lines 36. Shortest route PROBLEM SOLUTIONS 1. a) 300, $8,000, $65 per table, p $180; TC + $8, (300)(65) $27, 500; TR p (300)(180) $54, 000; Z $54, , 500 $26, 500 per month 8,000 b) tables per month a) 12, 000, $60, 000, $9, p $25; TC + 60, (12, 000)(9) $168, 000; TR p (12, 000)($25) $300, 000; Z $300, , 000 $132, 000 per year 60, 000 b) 3,750 tires per year a) 18, 000, $21, 000, $.45, p $1.30; TC + $21,000 + (18,000)(.45) $29,100; TR p (18, 000)(1.30) $23, 400; Z $23,400 29,100 $5,700 (loss) 21, 000 b) 24, yd per month $25, 000, p $.40, $.15, 25, ,000 lb per month

2 Reenue, ost, and proit ($1,000s) (lbs.) Break-een point Total Reenue Total Cost 1,000 2,000 3,000 4,000 5,000 6,000 7,000 8,000 Volume, ( ) Break-een point Volume, ($1,000s) $25, 000 1,250 dolls , % k 8,000 Total Reenue Total Cost Break-een olume as perentage o apaity Break-een olume as perentage o apaity 24, % k 25, 000 Break-een olume as perentage o 100, 000 apaity % k 120, , 000 2, tires 31 9 per year; it redues the break-een olume rom 3,750 tires to2,727.3 tires per year. 25, , lb per month; it redues the break-een olume rom 100,000 lb per month to 55, lb , , lb per month; it inreases the break-een olume rom 55, lb per month to 65, lb per month. 39, , lb per month; it inreases the break-een olume rom 65, lb per month to 102, lb per month. 15. Initial proit: Z p (9,000)(.75) 4, 000 (9, 000)(.21) 6, 750 4, 000 1,890 $860 per month; inrease in prie: Z p (5,700)(.95) 4,000 (5,700)(.21) 5,415 4,000 1,197 $218 per month; the dairy should not raise its prie. 35, ,750 p The inrease in ixed ost rom $25,000 to $35,000 will inrease the break-een point rom 1,250 to 1,750 or 500 dolls; thus, he should not spend the extra $10,000 or adertising. 17. Original break-een point (rom problem 7) 1,250 New break-een point: 17,000 1, Redues BE point by dolls. $27, a) 5, pizzas ,192.3 b) days 20 ) Reenue or the irst 30 days 30(p ) 30[(8.95)(20) (20)(3.75)] $3,120 $27,000 3,120 $23,880, portion o ixed ost not reouped ater 30 days. $23,880 New 5,685.7 pizzas Total break-een olume , ,285.7 pizzas 1-2

3 5,685.7 Total time to break-een days 19. a) Cost o Regular plan $55 + (.33)(260 minutes) $ Cost o Exeutie plan $100 + (.25)(60 minutes) $115 Selet Exeutie plan. b) 55 + (x 1,000)(.33) (x 1,200)(.25) x.25x 200 x minutes per month or hrs. 7, a) 14, 000 p.35 p $0.89 to break een b) I the team did not perorm as well as expeted the rowds ould be smaller; bad weather ould redue rowds and/or aet what ans eat at the game; the prie she harges ould aet demand. ) This will be a subjetie answer, but $1.25 seems to be a reasonable prie. Z p Z (14,000)(1.25) 7,500 (14,000)(0.35) 17,500 12,400 $5, a) $1,700 $12 per pupil p $75 1, or 27 pupils b) Z p $5,000 (75) $1,700 (12) 63 6, pupils ) Z p $5,000 60p $1,700 60(12) 60p 7,420 p $ a) $350,000 $12,000 p $18,000 p 350,000 18,000 12, or 59 students b) Z (75)(18,000) 350,000 (75)(12,000) $100,000 ) Z (35)(25,000) 350,000 (35)(12,000) 105,000 This is approximately the same as the proit or 75 students and a lower tuition in part (b). 23. p $400 $8,000 $75 Z $60,000 Z + p 60, , teams 24. Fixed ost ( ) 875,000 Variable ost ( ) $200 Prie (p) (225)(12) $2,700 /(p ) 875,000/(2, ) 350 With olume doubled to 700: Proit (Z) (2,700)(700) 875,000 (700)(200) $875, Fixed ost ( ) 100,000 Variable ost ( ) $(.50)(.35) + (.35)(.50) + (.15)(2.30) $0.695 Prie (p) $6 Proit (Z) (6)(45,000) 100,000 (45,000)(0.695) $138,725 This is not the inanial proit goal o $150,000. The prie to ahiee the goal o $150,000 is, p (Z + + )/ (150, ,000 + (45,000)(.695))/45,000 $

4 The olume to ahiee the goal o $150,000 is, (Z + )/(p ) (150, ,000)/(6.695) 47, a) Monthly ixed ost ( ) ost o an/60 months + labor (drier)/month (21,500/60) + (30.42 days/month)($8/hr) (5 hr/day) , $1, Variable ost ( ) $ $16.35 Prie (p) $34 /(p ) (1,575.13)/( ) orders/month b) 89.24/ orders/day Monday through Thursday Double or weekend 5.86 orders/day Friday through Sunday Orders per month approximately (18 days) (2.93 orders) + (12.4 days)(5.86 orders) deliery orders per month Proit total reenue total ost p ( + ) (125.4)(34) 1, (125.4)(16.35) a) jobs b) (8 weeks)(6 days/week)(3 lawns/day) 144 lawns Z (144)(30) 500 (144)(14) Z $1,804 ) (8 weeks)(6 days/week)(4 lawns/day) 192 lawns Z (192)(25) 500 (192)(14) Z $1,612 No, she would make less money than (b) a) jobs b) (6 snows)(2 days/snow)(10 jobs/day) 120 jobs Z (120)(35) 700 (120)(3) Z $3,140 ) (6 snows)(2 days/snow)(4 jobs/day) 48 jobs Z (48)(150) 1800 (48)(28) Z $4,056 Yes, better than (b) d) Z (120)(35) 700 (120)(18) Z $1,340 Yes, still a proit with one more person 29. There are two possible answers, or solution points: x 25, y 0 or x 0, y 50 Substituting these alues in the objetie untion: Z 15(25) + 10(0) 375 Z 15(0) + 10(50) 500 Thus, the solution is x 0 and y 50 This is a simple linear programming model, the subjet o the next seeral hapters. The student should reognize that there are only two possible solutions, whih are the orner points o the easible solution spae, only one o whih is optimal. 30. The solution is omputed by soling simultaneous equations, x 30, y 10, Z $1,400 It is the only, i.e., optimal solution beause there is only one set o alues or x and y that satisy both onstraints simultaneously. 1-4

5 31. Labor usage Clay usage Proit Possible # bowls # mugs 12x + 15y < 60 9x+5y < x + 250y solution? yes yes yes yes yes yes yes yes, best solution yes yes yes no no no no no yes no no no no y ~9x + 5y ~12x + 15y x 1-5

6 32. Maximize Z $30x AN + 70x AJ + 40x BN + 60x BJ subjet to x AN + x AJ 400 x BN + x BJ 400 x AN + x BN 500 x AJ + x BJ 300 The solution is x AN 400, x BN 100, x BJ 300, and Z 34,000 This problem an be soled by alloating as muh as possible to the lowest ost ariable, x AN 400, then repeating this step until all the demand has been met. This is a similar logi to the minimum ell ost method. 33. This is irtually a straight linear relationship between time and site isits; thus, a simple linear graph would result in a oreast o approximately 34,500 site isits. 34. Determine logial solutions: Cakes Bread Total Sales $ $ $ $40 Eah solution must be heked to see i it iolates the onstraints or baking time and lour. Some possible solutions an be logially disarded beause they are obiously inerior. For example, 0 akes and 1 loa o bread is learly inerior to 0 akes and 2 loaes o bread. 0 akes and 3 loaes o bread is not possible beause there is not enough lour or 3 loaes o bread. Using this logi, there are our possible solutions as shown. The best one, 4 akes and 0 loaes o bread, results in the highest total sales o $ This problem demonstrates the ost trade-o inherent in queuing analysis, the topi o Chapter 13. In this problem the ost o serie, i.e., the ost o staing registers, is added to the ost o ustomers waiting, i.e., the ost o lost sales and ill will, as shown in the ollowing table. 36. The shortest route problem is one o the topis o Chapter 7. At this point, the most logial trial and error way that most students will probably approah this problem is to identiy all the easible routes and ompute the total distane or eah, as ollows: Obiously inerior routes like and that inlude additional segments to the routes listed aboe an be logially eliminated rom onsideration. As a result, the route is the shortest. An additional aspet to this problem ould be to hae the students look at these routes on a real map and indiate whih they think might pratially be the best route. In this ase, would likely be a better route, beause een though it s two miles arther it is Interstate highway the whole way, whereas enompasses U.S. 4-lane highways and state roads. Registers staed Waiting time (mins) Cost o serie ($) Cost o waiting ($) Total ost ($) The total minimum ost o $290 ours with 4 registers staed 1-6

7 CASE SOLUTION: CLEAN CLOTHES CORNER LAUNDRY 1, 700 a) 2,000 items per month b) Solution depends on number o months; 36 used here. $16, $450 per month, thus monthly ixed ost is $2,150 2,150 2,529.4 items per month additional items per month ) Z p 4,300(1.10) 2,150 4,300(.25) $1,505 per month Ater 3 years, Z $1,955 per month 1, 700 d) 2, Z p 3,800(.99) 1, 700 3,800(.25) $1,112 per month e) With both options: Z p 4,700(.99) 2,150 4,700(.25) $1,328 She should purhase the new equipment but not derease pries. CASE SOLUTION: OCOBEE RIVER RAFTING COMPANY Alternatie 1: $3,000 p $20 $12 3, rats Alternatie 2: $10,000 p $20 $8 10, I demand is less than 375 rats, the students should not start the business. I demand is less than 833 rats, alternatie 2 should not be seleted, and alternatie 1 should be used i demand is expeted to be between 375 and rats. I demand is greater than rats, whih alternatie is best? To determine the answer, equate the two ost untions. 3, , ,000 1,750 This is reerred to as the point o indierene between the two alternaties. In general, or demand lower than this point (1,750) the alternatie should be seleted with the lowest ixed ost; or demand greater than this point the alternatie with the lowest ariable ost should be seleted. (This general relationship an be obsered by graphing the two ost equations and seeing where they interset.) Thus, or the Oobee Rier Rating Company, the ollowing guidelines should be used: demand < 375, do not start business; 375 < demand < 1,750, selet alternatie 1; demand > 1,750, selet alternatie 2 Sine Penny estimates demand will be approximately 1,000 rats, alternatie 1 should be seleted. Z p (1,000)(20) 3,000 (1,000)(12) Z $5,000 CASE SOLUTION: CONSTRUCTING A DOWNTOWN PARKING LOT IN DRAPER a) The annual apital reoery payment or a apital expenditure o $4.5 million oer 30 years at 8% is, (4,500,000)[0.08(1 +.08) 30 ] / (1 +.08) 30 1 $399,

8 This is part o the annual ixed ost. The other part o the ixed ost is the employee annual salaries o $140,000. Thus, total ixed osts are, $399, ,000 $539, p 539, , parked ars per year b) I 365 days per year are used, then the daily usage is, 207, or approximately 569 ars 365 per day This seems like a reahable goal gien the size o the town and the student population. 1-8