《运营管理》课后习题答案

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Solutions_Problems_OM_11e_Stevenson Chapter 02 - Competitiveness, Strategy, and Productivity 3. (1) (2) (3) (4) (5) (6) (7) Week Output Worker Cost@ $12x40 Overhead Cost @1.5 Material Cost@$6 Total Cost MFP (2) (6) 1 30,000 2,880 4,320 2,700 9,900 3.03 2 33,600 3,360 5,040 2,820 11,220 2.99 3 32,200 3,360 5,040 2,760 11,160 2.89 4 35,400 3,840 5,760 2,880 12,480 2.84 *refer to solved problem #2 Multifactor productivity dropped steadily from a high of 3.03 to about 2.84. 4. a. Before: 80 5 = 16 carts per worker per hour. After: 84 4 = 21 carts per worker per hour. b. Before: ($10 x 5 = $50) + $40 = $90; hence 80 $90 = .89 carts/$1. After: ($10 x 4 = $40) + $50 = $90; hence 84 $90 = .93 carts/$1. c. Labor productivity increased by 31.25% ((21-16)/16). Multifactor productivity increased by 4.5% ((.93-.89)/.89). *Machine Productivity Before: 80 40 = 2 carts/$1. After: 84 50 = 1.68 carts/$1. Productivity increased by -16% ((1.68-2)/2) Chapter 03 - Product and Service Design 6. Steps for Making Cash Withdrawal from an ATM 1. Insert Card: Magnetic Strip Should be Facing Down 2. Watch Screen for Instructions 3. Select Transaction Options: 1) Deposit 2) Withdrawal 3) Transfer 4) Other 4. Enter Information: 1) PIN Number 2) Select a Transaction and Account 3) Enter Amount of Transaction 5. Deposit/Withdrawal: 1) Deposit—place in an envelope (which you’ll find near or in the ATM) and insert it into the deposit slot 2) Withdrawal—lift the “Withdrawal Door,” being careful to remove all cash 6. Remove card and receipt (which serves as the transaction record) 8. Technical Requirements Ingredients Handling Preparation Customer Requirements Taste √ √ Appearance √ √ √ Texture/consistency √ √ Chapter 04 - Strategic Capacity Planning for Products and Services 2. Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4 10. a. Given: 10 hrs. or 600 min. of operating time per day. 250 days x 600 min. = 150,000 min. per year operating time. Total processing time by machine Product A B C 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 4 60,000 60,000 30,000 Total 186,000 208,000 122,000 You would have to buy two “A” machines at a total cost of $80,000, or two “B” machines at a total cost of $60,000, or one “C” machine at $80,000. b. Total cost for each type of machine: A (2): 186,000 min 60 = 3,100 hrs. x $10 = $31,000 + $80,000 = $111,000 B (2) : 208,000 60 = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133 C(1): 122,000 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400 Buy 2 Bs—these have the lowest total cost. Chapter 05 - Process Selection and Facility Layout 3 a d f 7 5 2 b 4 c 4 e 9 h 5 i 6 g 3. Desired output = 4 Operating time = 56 minutes Task # of Following tasks Positional Weight A 4 23 B 3 20 C 2 18 D 3 25 E 2 18 F 4 29 G 3 24 H 1 14 I 0 5 a. First rule: most followers. Second rule: largest positional weight. Assembly Line Balancing Table (CT = 14) Work Station Task Task Time Time Remaining Feasible tasks Remaining I F 5 9 A,D,G A 3 6 B,G G 6 – – II D 7 7 B, E B 2 5 C C 4 1 – III E 4 10 H H 9 1 – IV I 5 9 – b. First rule: Largest positional weight. Assembly Line Balancing Table (CT = 14) Work Station Task Task Time Time Remaining Feasible tasks Remaining I F 5 9 A,D,G D 7 2 – II G 6 8 A, E A 3 5 B,E B 2 3 – III C 4 10 E E 4 6 – IV H 9 5 I I 5 – c. a b d c f 1. e g h 1. 4. a. l. 2. Minimum Ct = 1.3 minutes Task Following tasks a 4 b 3 c 3 d 2 e 3 f 2 g 1 h 0 Work Station Eligible Assign Time Remaining Idle Time I a A 1.1 b,c,e, (tie) B 0.7 C 0.4 E 0.3 0.3 II d D 0.0 0.0 III f,g F 0.5 G 0.2 0.2 IV h H 0.1 0.1 0.6 3. 4. b. 1. 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time] Assign f, g, and h to station 2: 2.3 minutes 3. 4. 7. 1 5 4 3 8 7 6 2 Chapter 06 - Work Design and Measurement 3. Element PR OT NT AFjob ST 1 .90 .46 .414 1.15 .476 2 .85 1.505 1.280 1.15 1.472 3 1.10 .83 .913 1.15 1.050 4 1.00 1.16 1.160 1.15 1.334 Total 4.332 8. A = 24 + 10 + 14 = 48 minutes per 4 hours 9. a. Element PR OT NT A ST 1 1.10 1.19 1.309 1.15 1.505 2 1.15 .83 .955 1.15 1.098 3 1.05 .56 .588 1.15 .676 b. c. e = .01 minutes Chapter 07- Location Planning and Analysis 1. Factor Local bank Steel mill Food warehouse Public school 1. Convenience for customers H L M–H M–H 2. Attractiveness of building H L M M–H 3. Nearness to raw materials L H L M 4. Large amounts of power L H L L 5. Pollution controls L H L L 6. Labor cost and availability L M L L 7. Transportation costs L M–H M–H M 8. Construction costs M H M M–H Location (a) Location (b) 4. Factor A B C Weight A B C 1. Business Services 9 5 5 2/9 18/9 10/9 10/9 2. Community Services 7 6 7 1/9 7/9 6/9 7/9 3. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/9 4. Construction Costs 5 6 5 2/9 10/9 12/9 10/9 5. Cost of Living 4 7 8 1/9 4/9 7/9 8/9 6. Taxes 5 5 5 1/9 5/9 5/9 4/9 7. Transportation 6 7 8 1/9 6/9 7/9 8/9 Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7. a. Composite Scores 39 44 45 7 7 7 B or C is the best and A is least desirable. b. Business Services and Construction Costs both have a weight of 2/9; the other factors each have a weight of 1/9. 5 x + 2 x + 2 x = 1 x = 1/9 c. Composite Scores A B C 53/9 55/9 54/9 B is the best followed by C and then A. 5. Location x y A 3 7 B 8 2 C 4 6 D 4 1 E 6 4 Totals 25 20 = xi = 25 = 5.0 = yi = 20 = 4.0 n 5 n 5 Hence, the center of gravity is at (5,4) and therefore the optimal location. Chapter 08 - Management of Quality 1. Checksheet Work Type Frequency Lube and Oil 12 Brakes 7 Tires 6 Battery 4 Transmission 1 Total 30 12 7 6 4 1 Lube & Oil Brakes Tires Battery Trans. Pareto break lunch break 3 2 1 0 2 . The run charts seems to show a pattern of errors possibly linked to break times or the end of the shift. Perhaps workers are becoming fatigued. If so, perhaps two 10 minute breaks in the morning and again in the afternoon instead of one 20 minute break could reduce some errors. Also, errors are occurring during the last few minutes before noon and the end of the shift, and those periods should also be given management’s attention. Power off Person Lamp Missing Didn’t turn completely on Not plugged in Outlet defective Defective Burned out Loose Lamp fails to light Other Cord 4 Chapter 9 - Quality Control 4. Sample Mean Range 1 79.48 2.6 Mean Chart: A2= 79.96 0.58(1.87) 2 80.14 2.3 = 79.96 1.08 3 80.14 1.2 UCL = 81.04, LCL = 78.88 4 79.60 1.7 Range Chart: UCL = D4= 2.11(1.87) = 3.95 5 80.02 2.0 LCL = D3= 0(1.87) = 0 6 80.38 1.4 [Both charts suggest the process is in control: Neither has any points outside the limits.] 6. n = 200 Control Limits = Thus, UCL is .0234 and LCL becomes 0. Since n = 200, the fraction represented by each data point is half the amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc. Sample 10 is too large. 7. Control limits: UCL is 16.266, LCL becomes 0. All values are within the limits. 14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit, = Process mean, s = Process standard deviation For process H: For process K: Assuming the minimum acceptable is 1.33, since 1.0 < 1.33, the process is not capable. For process T: Since 1.33 = 1.33, the process is capable. Chapter 10 - Aggregate Planning and Master Scheduling 7. a. No backlogs are allowed Period Mar. Apr. May Jun. July Aug. Sep. Total Forecast 50 44 55 60 50 40 51 350 Output Regular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 3 8 51 Subcontract 2 0 3 12 2 0 0 19 Output - Forecast 0 4 –4 0 0 3 –3 Inventory Beginning 0 0 4 0 0 0 3 Ending 0 4 0 0 0 3 0 Average 0 2 2 0 0 1.5 1.5 7 Backlog 0 0 0 0 0 0 0 0 Costs: Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 360 960 6,120 Subcontract 280 0 420 1,680 280 0 0 2,660 Inventory 0 20 20 0 0 15 15 70 Total 4,440 4,180 4,600 5,840 4,440 3,575 4,175 31,250 b. Level strategy Period Mar. Apr. May Jun. July Aug. Sep. Total Forecast 50 44 55 60 50 40 51 350 Output Regular 40 40 40 40 40 40 40 280 Overtime 8 8 8 8 8 8 8 56 Subcontract 2 2 2 2 2 2 2 14 Output - Forecast 0 6 –5 –10 0 10 –1 Inventory Beginning 0 0 6 1 0 0 1 Ending 0 6 1 0 0 1 0 Average 0 3 3.5 .5 0 .5 .5 8 Backlog 0 0 0 9 9 0 0 18 Costs: Regular 3,200 3,200 3,200 3,200 3,200 3,200 3,200 22,400 Overtime 960 960 960 960 960 960 960 6,720 Subcontract 280 280 280 280 280 280 280 1,960 Inventory 30 35 5 0 5 5 80 Backlog 180 180 360 Total 4,440 4,470 4,475 4,625 4,620 4,445 4,445 31,520 8. Period 1 2 3 4 5 6 Total Forecast 160 150 160 180 170 140 960 Output Regular 150 150 150 150 160 160 920 Overtime 10 10 0 10 10 10 50 Subcontract 0 0 10 10 0 0 20 Output- Forecast 0 10 0 –10 0 0 Inventory Beginning 0 0 10 10 0 0 Ending 0 10 10 0 0 0 Average 0 5 10 5 0 0 20 Backlog 0 0 0 0 0 0 0 Costs: Regular 7,500 7,500 7,500 7,500 8,000 8,000 46,000 Overtime 750 750 0 750 750 750 3,750 Subcontract 0 0 800 800 0 0 1,600 Inventory 20 40 20 80 Backlog 0 0 0 0 0 0 Total 8,250 8,270 8,340 9,070 9,050 8,750 51,430 Chapter 11 - MRP and ERP 1. a. F: 2 G: 1 H: 1 J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4 D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2 Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 4 Stapler Top Assembly Base Assembly Cover Spring Slide Assembly Base Strike Pad Rubber Pad 2 Slide Spring b. 4. Master Schedule Day Beg. Inv. 1 2 3 4 5 6 7 Quantity 100 150 200 Table Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 100 150 200 Scheduled receipts Projected on hand Net requirements 100 150 200 Planned-order receipts 100 150 200 Planned-order releases 100 150 200 Wood Sections Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 200 300 400 Scheduled receipts 100 Projected on hand 100 100 Net requirements 100 300 400 Planned-order receipts 100 300 400 Planned-order releases 400 400 Braces Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 300 450 600 Scheduled receipts Projected on hand 60 60 60 60 Net requirements 240 450 600 Planned-order receipts 240 450 600 Planned-order releases 240 450 600 Legs Beg. Inv. 1 2 3 4 5 6 7 Gross requirements 400 600 800 Scheduled receipts Projected on hand 120 120 120 120 88 88 71 Net requirements 280 600 800 Planned-order receipts 308 660 880 Planned-order releases 968 880 10. Week 1 2 3 4 Material 40 80 60 70 Week 1 2 3 4 Labor hr. 160 320 240 280 Mach. hr. 120 240 180 210 a. Capacity utilization Week 1 2 3 4 Labor 53.3% 106.7% 80% 93.3% Machine 60% 120% 90% 105% b. Capacity utilization exceeds 100% for both labor and machine in week 2, and for machine alone in week 4. Production could be shifted to earlier or later weeks in which capacity is underutilized. Shifting to an earlier week would result in added carrying costs; shifting to later weeks would mean backorder costs. Another option would be to work overtime. Labor cost would increase due to overtime premium, a probable decrease in productivity, and possible increase in accidents. Chapter 12 - Inventory Management 2. The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items. Dollar Item Unit Cost Usage Usage Category K34 10 200 2,000 C K35 25 600 15,000 A K36 36 150 5,400 B M10 16 25 400 C M20 20 80 1,600 C Z45 80 250 16,000 A F14 20 300 6,000 B F95 30 800 24,000 A F99 20 60 1,200 C D45 10 550 5,500 B D48 12 90 1,080 C D52 15 110 1,650 C D57 40 120 4,800 B N08 30 40 1,200 C P05 16 500 8,000 B P09 10 30 300 C a. See table. b. To allocate control efforts. c. It might be important for some reason other than dollar usage, such as cost of a stockout, usage highly correlated to an A item, etc. 3. D = 1,215 bags/yr. S = $10 H = $75 a. b. Q/2 = 18/2 = 9 bags c. d. e. Assuming that holding cost per bag increases by $9/bag/year Q = 17 bags Increase by [$1,428.71 – $1,350] = $78.71 4. D = 40/day x 260 days/yr. = 10,400 packages S = $60 H = $30 a. b. c. Yes d. TC200 = 3,000 + 3,120 = $6,120 6,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.) 7. H = $2/month S = $55 D1 = 100/month (months 1–6) D2 = 150/month (months 7–12) a. b. The EOQ model requires this. c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost) 1–6 TC74 = $148.32 7–12 TC91 = $181.66 10. p = 50/ton/day D= 20 tons/day x 200 days/yr. = 4,000 tons/yr. u = 20 tons/day 200 days/yr. S = $100 H = $5/ton per yr. a. b. Average is tons [approx. 3,098 bags] c. Run length = d. Runs per year = e. Q = 258.2 TC = TCorig. = $1,549.00 TCrev. = $ 774.50 Savings would be $774.50 15. Range P H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+ 4.85 1.94 503 NF Compare TC495 with TC for all lower price breaks: TC495 = 495 ($2) + 4,900 ($50) + $5.00(4,900) = $25,490 2 495 TC1,000 = 1,000 ($1.98) + 4,900 ($50) + $4.95(4,900) = $25,490 2 1,000 TC4,000 = 4,000 ($1.96) + 4,900 ($50) + $4.90(4,900) = $27,991 2 4,000 TC6,000 = 6,000 ($1.94) + 4,900 ($50) + $4.85(4,900) = $29,626 2 6,000 495 497 500 503 1,000 4,000 6,000 Quantity TC Hence, one would be indifferent between 495 or 1,000 units 22. d = 30 gal./day ROP = 170 gal. LT = 4 days, ss = ZsLT = 50 gal Risk = 9% Z = 1.34 Solving, sLT = 37.31 3% Z = 1.88, ss=1.88 x 37.31 = 70.14 gal. Chapter 13 - JIT and Lean Operations 1. N = ? N = DT(1 + X) D = 80 pieces per hour C T = 75 min. = 1.25 hr. = 80(1.25) (1.35) = 3 C = 45 45 X = .35 4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, use three cycles. Product Daily quantity Units per cycle A 21 21/3 = 7 B 12 12/3 = 4 C 3 3/3 = 1 D 15 15/3 = 5 5. a. Cycle 1 2 3 4 A 6 6 5 5 B 3 3 3 3 C 1 1 1 1 D 4 4 5 5 E 2 2 2 2 b. Cycle 1 2 A 11 11 B 6 6 C 2 2 D 8 8 E 4 4 c. 4 cycles = lower inventory, more flexibility 2 cycles = fewer changeovers 7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling 6. a. FCFS: A–B–C–D SPT: D–C–B–A EDD: C–B–D–A CR: A–C–D–B FCFS: Job time Flow time Due date Days Job (days) (days) (days) tardy A 14 14 20 0 B 10 24 16 8 C 7 31 15 16 D 6 37 17 20 37 106 44 SPT: Job time Flow time Due date Days Job (days) (days) (days) tardy D 6 6 17 0 C 7 13 15 0 B 10 23 16 7 A 14 37 20 17 37 79 24 EDD: Job time Flow time Due date Days Job (days) (days) (days) tardy C 7 7 15 0 B 10 17 16 1 D 6 23 17 6 A 14 37 20 17 84 24 Critical Ratio Job Processing Time (Days) Due Date Critical Ratio Calculation A 14 20 (20 – 0) / 14 = 1.43 B 10 16 (16 – 0) /10 = 1.60 C 7 15 (15 – 0) / 7 = 2.14 D 6 17 (17 – 0) / 6 = 2.83 Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After the completion of Job A, the revised critical ratios are: Job Processing Time (Days) Due Date Critical Ratio Calculation A – – – B 10 16 (16 – 14) /10 = 0.20 C 7 15 (15 – 14) / 7 = 0.14 D 6 17 (17 – 14) / 6 = 0.50 Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After the completion of Job C, the revised critical ratios are: Job Processing Time (Days) Due Date Critical Ratio Calculation A – – – B 10 16 (16 – 21) /10 = –0.50 C – – – D 6 17 (17 – 21) / 6 = –0.67 Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27. The critical ratio sequence is A–C–D–B and the makespan is 37 days. Critical Ratio sequence P