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Organization Management

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Submitted By dabbang
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CHAPTER = 11 Q-1
| | |Factor |Factor Score for Each Location |
| |Location Factor |Weight |A |B |C |D |
| |1. Labor climate | 5 |5 |
| |Location Factor |Weight |A |B |C |D |
| |1. Rent |25 |3 |75 |1 |25 |2 |50 |5 |125 |
| |2. Quality of life |20 |2 |40 |5 |100 |5 |100 |4 |80 |
| |3. Schools |5 |3 |15 |5 |25 |3 |15 |1 |5 |
| |4. Proximity to work |10 |5 |50 |3 |30 |4 |40 |3 |30 |
| |5. Proximity to recreation |15 |4 |60 |4 |60 |5 |75 |2 |30 |
| |6. Neighborhood security |15 |2 |30 |4 |60 |4 |60 |4 |60 |
| |7. Utilities |10 |4 |40 |2 |20 |3 |30 |5 |50 |
| |Total |100 | |310 | |320 | |370 | |380 |

Location D, the in-laws’ downstairs apartment, is indicated by the highest score. This points out a criticism of the technique: the Darlings did not include or give weight to a relevant factor.

3. Jackson or Dayton locations Jackson — [pic] Dayton — [pic] Jackson yields higher total profit contribution per year.

4. Fall-Line, Inc. a. Plot of total costs (in $ millions) versus volume (in thousands)

[pic]

b. Medicine Lodge is the lowest-cost location for volumes up to 25,000 pairs per year. Broken Bow is the best choice over the range of 25,000 to 44,000 pairs per year. Wounded Knee is the lowest-cost location for volumes over 44,000 pairs per year. Aspen is not the low-cost location at any volume.

c. Aspen — [pic] Medicine Lodge — [pic] Broken Bow — [pic] Wounded Knee— [pic]

d. Aspen would surpass Broken Bow when the Aspen profit is $7,780,000. $500Q ($8,000,000 + ($250Q)} = $7,780,000

$250Q = 15,780,000

Q = 63,120 Aspen would be the best location if sales would exceed 63,120 pairs per year.

5. Wiebe Trucking, Inc. a. Plot of total costs (in $ millions) versus volume (in thousands)

[pic]

b. For up to 576,923 shipments per year, Salt Lake City is the best location. Beyond that, Denver is the best location.

6. Sam’s Bagels Expected annual profits from “Downtown” location: 30,000(3.25 – 1.50) – 12,000 = $40,500 Expected annual profits from “Suburban” location: 25,000(2.85 – 1.00) – 8,000 = $38,250 Recommend “Downtown” location.

7. Distance between three points Point A = (20, 20) Point B = (50, 10) Point C = (50, 60) a. Euclidean distance [pic]

[pic] [pic] [pic]

b. Rectilinear distances [pic] [pic]

8. Centura High School

|Inputs |
|Solver - Center of Gravity |
|Enter data in yellow shaded areas. |
| | | | | | | | |
|Enter the names of the towns and the coordinates (x and y) and population (or load, l) of each town. |
| | | | | | | | |
| | | | | | | | |
| | | | | | | | |
| | | | | | | | |
| | | | | | | | |
|City/Town Name |x |y |l |lx |ly | | |
|Boelus |106.72 |46.31 |228 |24332.16 |10558.68 | | |
|Center-of-Gravity Coordinates | |x* |106.71 | | | |
| | | |y* |46.35 | | | |

1. A part a. Average cycle inventory [pic] [pic] Value of cycle inventory = (500 units) ($50+$60) = $55,000

b. Pipeline inventory = dL [(3800 units/year)/(50wks/yr)](6 weeks) = 456 units Value of the pipeline inventory = (456 units)($50+$30) = $36,480

2. Prince Electronics a. Value of each DC’s pipeline inventory = (75 units/wk)(2 wk)($350/unit) = $52,500

b. Total inventory = cycle + safety + pipeline = 5[(400/2) + (2*75) + (2*75)] = 2,500 units

3. Lockwood Industries First we rank the items from top to bottom on the basis of their dollar usage. Then we partition them into classes. The analysis was done using OM Explorer Tutor12.2—ABC Analysis.

| | | | | | | |Cumulative % |Cumulative % | |
| |Part # |Description |Qty Used/Year |Value |Dollar Usage |Pct of Total |of Dollar Value |of Items |Class |
| |4 | |44,000 |$1.00 |
| |D205 |9,690 |9,690 | |
| |U404 |6,075 |15,765 |A: 55% |
| | | | | |
| |A104 |3,220 |18,985 | |
| |L205 |3,035 |22,020 |B: 22% |
| | | | | |
| |L104 |2,005 |24,025 | |
| |S104 |1,604 |25,629 | |
| |X205 |1,603 |27,232 |C: 23% |
| |X104 |1,500 |28,732 | |

One classification might be to group the top two items (i.e., 25% of the items) in A class accounting for 55% of the total value. The next two items would be classified as B and the last four as C. The dollar usage percentages don’t exactly match the predictions of ABC analysis. For example, Class A items account for only 55% of the total, rather than 80%. Nonetheless, the important finding is that ABC analysis did find the “significant few.” For the items sampled, particularly close inventory management is needed for items D205 and U404.

6. Yellow Press, Inc. a. Economic order quantity [pic]

b. Time between orders [pic]

7. Babble Inc. a. [pic] tapes/month [pic] tapes/year [pic] [pic]

b. Time between orders [pic] years or 2.5 months

8. Dot Com a. [pic]

b. Optimal number of orders/year = (32,000)/400 = 80 orders c. Optimal interval between orders = 300/80 = 3.75 days d. Demand during lead time = dL = (5 days)(32,000/300) = 533 books

e. Reorder point = dL + safety stock = 533 + 0 = 533 books f. Inventory position = OH + SR – BO = 533 + 400 – 0 = 933 books

9. Leaky Pipe Inc. a. [pic]

b. Optimal number of orders = (30,000)/(775) = 38.7 or 39 c. Optimal interval between orders = (300)/(39) = 7.69 days d. Demand during lead time = dL = (4 days)(30,000/300) = 400 units e. Reorder point = dL + safety stock = 400 + 0 = 400 units f. Inventory position = OH + SR – BO = 400 +775 – 0 = 1175 units

10. Sam’s Cat Hotel a. Economic order quantity [pic] Time between orders, in weeks [pic]

b. Reorder point, R R = demand during protection interval + safety stock Demand during protection interval = dL = 90 * 3 = 270 bags Safety stock [pic] When the desired cycle-service level is 80%, [pic].

[pic] = 25.98 or 26

Safety stock = 0.84 * 26 = 21.82 or about 22 bags [pic]

c. Initial inventory position = OH + SR – BO = 320 + 0 – 0 320 – 10 = 310. Because inventory position remains above 292, it is not yet time to place an order.

d. Annual holding cost Annual ordering cost [pic] [pic]

At the EOQ, these two costs are equal. When [pic], the annual holding cost is larger than the ordering cost, therefore Q is too large. Total costs are $789.75 + $505.44 = $1,295.19.

e. Annual holding cost Annual ordering cost [pic] [pic]

Total costs at EOQ: = $1,263.60, which is $31.59 less than when order quantity is 500 bags.

11. Sam’s Cat Hotel, revisited a. If the demand is only 60 bags per week, the correct EOQ is: D = (60 units/wk)(52 wk/yr) = 3,120 bags

[pic] or 327 bags

If the demand is incorrectly estimated at 90 bags, the EOQ would be incorrectly calculated (from problem 10) as 400 bags: The total cost, working with the actual demand, is: [pic]

[pic]

[pic]

We can see clearly now that the cost penalty of Sam’s difficulty in foreseeing demand for kitty litter is $21.27 ($1,053.00 – $1,031.73).

b. If S = $6, and [pic], the correct EOQ is:

[pic] or 109 bags

The total cost, working with the actual ordering cost, is

[pic] If the reduced ordering cost continues to be unseen, the cost penalty for not updating the EOQ is (573.74 – 343.91) = $229.83.

12. A Q system (also known as a reorder point system) [pic]

a. Standard deviation of demand during the protection interval: [pic]

b. Average demand during the protection interval: [pic]

c. Reorder point R = demand during protection interval + safety stock – backorders Safety stock [pic] When the desired cycle-service level is 99%, z = 2.33. Safety stock = 2.33 * 45 = 104.85 or 105 gizmos R = 2,700 + 105 – 0 = 2,805

13. Petromax Enterprises a. [pic]

b. Safety stock [pic] or 280 units Reorder point = average lead time demand + safety stock = (3)(50,000/50) + 278 = 3,278 units

14. A perpetual system (also known as a continuous review system). Find the safety stock reduction when lead time is reduced from five weeks to one week, given: Standard deviation of demand during the (five-week) protection interval is 85 dohickies. Desired cycle service level is 99% (therefore z = 2.33). Safety stock required for five-week protection interval Safety stock = [pic] = 2.33(85) = 198 dohickies Safety stock required for one-week protection interval [pic] Safety stock = [pic] = 2.33(38) = 88.54 or 89 dohickies

15. A two-bin system. “The two-bin system is really a Q system, with the normal level in the second bin being the reorder point R.” Find cycle-service level, given: [pic] Safety stock = R – dL = 120 – (53*2) = 14 whatchamacallits Safety stock = [pic] = 14 whatchamacallits [pic] or 7 whatchamacallits [pic] When z = 2, the cycle-service level is 97.72%.

16. Nationwide Auto Parts a. Protection interval (PI) = P + L = 6 +3 = 9 weeks Average demand during PI = 9 (100) = 900 units Standard deviation during PI = [pic] = 60 units b. Target inventory = d(P+L)+z(P+L = 900 + (1.96)(60) = 1,018 c. Order quantity = Target inventory – IP = 1,018 – 350 = 668 units presuming no SR or BO

17. A P system (also known as a periodic review system). Find cycle-service level, given: L = 2 weeks P = 1 week d(P + L) = 218 gadgets [pic] = 40 gadgets T = 300 gadgets T = Average demand during protection interval + Safety stock T = 218 + z(40) = 300 z = (300 – 218)/40 = 2.05 When z = 2.05, cycle-service level is 97.98 or 98%.

18. A Successful Product Annual Demand, D = (200)(50) = 10,000 units, H = ((0.20)(12.50)) = 2.50 a. Optimal ordering quantity [pic]

b. Safety stock [pic] = (2.33)(16)[pic] = 74.56 or 75 units c. Safety stock will now be = (2.33)(16)[pic] = 52.72 or 53 units % reduction in safety stock = (75 – 53)/75 = 29.33% d. Safety stock will be = (2.33)(8)[pic] = 37.28 or 38 units % reduction in safety stock = (75 – 38)/75 = 49.33%

19. Sam’s Cat Hotel with a P system a. Referring to Problem 10, the EOQ is 400 bags. When the demand rate is 15 per day, the average time between orders is (400/15) = 26.67 or about 27 days. The lead time is 3 weeks 6 days per week = 18 days. If the review period is set equal to the EOQs average time between orders (27 days), then the protection interval (P + L) = (27 + 18) = 45 days. For an 80% cycle-service level [pic] Safety stock = [pic] = 0.84(41.08) = 34.51 or 35 bags T = Average demand during the protection interval + Safety stock T = (15*45) + 35 = 710

b. In Problem 10, the Q system required a safety stock of 22 bags to achieve an 80% cycle-service level. Therefore, the P system requires a safety stock that is larger by (35 – 22) 13 bags.

c. From Problem 10, inventory position, IP = 310. The amount to reorder is T – IP = 710 – 310 = 400.

1. Printer rentals a. The forecast for week 11 is 29 rentals.

| | | | |Forecast for Following |
| | |Week Forecast Calculated |[pic] |Week ([pic]) |
| | |5 |[pic] |= 27.2 or 27 |
| | |6 |[pic] |= 28.2 or 28 |
| | |7 |[pic] |= 29.8 or 30 |
| | |8 |[pic] |= 30.4 or 30 |
| | |9 |[pic] |= 30.4 or 30 |
| | |10 |[pic] |= 29.0 or 29 |

b. The Mean Absolute Deviation is 4 rentals.

| | |Week |Actual |Forecast | |Absolute Error |
| | |6 |28 |27 | |1 | |
| | | |(Thousands) |Moving Average |Error |% Error |Error |
| | | | |Forecast | | | |
| | |Jan. |20 | | | | |
| | |Feb. |24 | | | | |
| | |Mar. |27 | | | | |
| | |Apr. |31 |(20+24+27)/3 = 23.67 |7.33 |23.65 |53.73 |
| | |May |37 |(24+27+31)/3 = 27.33 |9.67 |26.14 |93.51 |
| | |June |47 |(27+31+37)/3 = 31.67 |15.33 |32.62 |235.01 |
| | |July |53 |(31+37+47)/3 = 38.33 |14.67 |27.68 |215.21 |
| | |Aug. |62 |(37+47+53)/3 = 45.67 |16.33 |26.34 |266.67 |
| | |Sept. |54 |(47+53+62)/3 = 54.00 |0.00 |0.00 |0.00 |
| | |Oct. |36 |(53+62+54)/3 = 56.33 |20.33 |56.47 |413.31 |
| | |Nov. |32 |(62+54+36)/3 = 50.67 |18.67 |58.34 |348.57 |
| | |Dec. |29 |(54+36+32)/3 = 40.67 |11.67 |40.24 |136.19 |
| | |Total | | |114.00 |291.48 |1,762.20 |
| | |Average | | |12.67 |32.39 |195.80 |

Such results also can be obtained from the Time Series Forecasting Solver:

[pic]

[pic]

b. Four-month simple moving average

| | |Month |Actual Sales |Four-Month Simple |Absolute |Absolute |Squared |
| | | |(Thousands) |Moving Average |Error |% Error |Error |
| | | | |Forecast | | | |
| | |Apr. |31 | | | | |
| | |May |37 |(20+24+27+31)/4 = 25.5 |11.50 |31.08 |132.25 |
| | |June |47 |(24+27+31+37)/4 = 29.75 |17.25 |36.70 |297.56 |
| | |July |53 |(27+31+37+47)/4 = 35.5 |17.50 |33.02 |306.25 |
| | |Aug. |62 |(31+37+47+53)/4 = 42.00 |20.00 |32.26 |400.00 |
| | |Sept. |54 |(37+47+53+62)/4 = 49.75 |4.25 |7.87 |18.06 |
| | |Oct. |36 |(47+53+62+54)/4 = 54.00 |18.00 |50.00 |324.00 |
| | |Nov. |32 |(53+62+54+36)/4 = 51.25 |19.25 |60.16 |370.56 |
| | |Dec. |29 |(62+54+36+32)/4 = 46.00 |17.00 |58.62 |289.00 |
| | |Total | | |124.75 |309.71 |2,137.68 |
| | |Average | | |15.59 |38.71 |267.21 |

Similarly, using Time Series Forecasting Solver, we get:

[pic]

[pic]

c.e. Comparison of performance

| | |Question |Measure |3-Month |4-Month |Recommendation |
| | | | |SMA |SMA | |
| | |c. |MAD |12.67 |15.59 |3-month SMA |
| | |d. |MAPE |32.39 |38.71 |3-month SMA |
| | |e. |MSE |195.80 |267.21 |3-month SMA |

3. Karl’s Copiers

| |Week Forecast Calculated|[pic] |[pic] |
| |7/3 |0.20(24) + 0.80(24) |= 24 |
| |7/10 |0.20(32) + 0.80(24) |= 25.6 or 26 |
| |7/17 |0.20(36) + 0.80(25.6) |= 27.68 or 28 |
| |7/24 |0.20(23) + 0.80(27.68) |= 26.744 or 27 |
| |7/31 |0.20(25) + 0.80(26.744) |= 26.3952 or 26 |

The forecast for the week of August 7 is 26 calls. Similarly, using Time Series Forecasting Solver, we get:

[pic]

[pic]

4. Dalworth Company (continued) a. Three-month weighted moving average (weights of 3/6, 2/6, and 1/6)

| | |Month |Actual Sales |Three-Month Weighted |Absolute |Absolute % |Squared |
| | | |(000s) |Moving Average Forecast |Error |Error |Error |
| | |Jan. |20 | | | | |
| | |Feb. |24 | | | | |
| | |Mar. |27 | | | | |
| | |Apr. |31 |[(3[pic]27)+(2[pic]24)+(l [pic]20)]/6 = 24.83|6.17 |19.90 |38.07 |
| | |May |37 |[(3[pic]31)+(2[pic]27)+(l [pic]24)]/6 = 28.50|8.50 |22.97 |72.25 |
| | |June |47 |[(3[pic]37)+(2[pic]31)+(l [pic]27)]/6 = 33.33|13.67 |29.09 |186.87 |
| | |July |53 |[(3[pic]47)+2[pic]37)+(l [pic]31)]/6 = 41.00 |12.00 |22.64 |144.00 |
| | |Aug. |62 |[(3[pic]53)+(2[pic]47)+(l [pic]37)]/6 = 48.33|13.67 |22.05 |186.87 |
| | |Sept. |54 |[(3[pic]62)+(2[pic]53)+(l [pic]47)]/6 = 56.50|2.50 |4.63 |6.25 |
| | |Oct. |36 |[(3[pic]54)+(2[pic]62)+(l [pic]53)]/6 = 56.50|20.50 |56.94 |420.25 |
| | |Nov. |32 |[(3[pic]36)+(2[pic]54)+(l[pic]62)]/6 = 46.33 |14.33 |44.78 |205.35 |
| | |Dec. |29 |[(3[pic]32)+(2[pic]36)+(l [pic]54)]/6 = 37.00|8.00 |27.59 |64.00 |
| | |Total | | |99.34 |250.59 |1,323.91 |
| | |Average | | |11.04 |27.84 |147.09 |

The results from Time Series Forecasting Solver give the same results:

[pic]

[pic]

b. Exponential smoothing ( = 0.6)

| | |Month |Dt |Ft |Ft+1 = Ft + (Dt Ft) |Absolute |Absolute |Squared |
| | |(t) | (millions) | | |Error |% Error |Error |
| | |Jan. |20 |22.00 |20.80 | | | |
| | |Feb. |24 |20.80 |22.72 | | | |
| | |Mar. |27 |22.72 |25.29 | | | |
| | |Apr. |31 |25.29 |28.72 |5.71 |18.41 |32.60 |
| | |May |37 |28.72 |33.69 |8.28 |22.38 |68.56 |
| | |June |47 |33.69 |41.67 |13.31 |28.32 |177.16 |
| | |July |53 |41.67 |48.47 |11.33 |21.38 |128.37 |
| | |Aug. |62 |48.47 |56.59 |13.53 |21.82 |183.06 |
| | |Sept. |54 |56.59 |55.04 |2.59 |4.80 |6.71 |
| | |Oct. |36 |55.04 |43.62 |19.04 |52.88 |362.52 |
| | |Nov. |32 |43.62 |36.64 |11.61 |36.28 |134.79 |
| | |Dec. |29 |36.64 |32.06 |7.65 |26.38 |58.52 |
| | |Total | | | |93.05 |232.65 |1,152.29 |
| | |Average | | | | 10.34 |25.85 |128.03 |

c.e. Comparison of performance

| | |Question |Measure |3-Month |Exponential |Recommendation |
| | | | |WMA |Smoothing | |
| | |c. |MAD | 11.04 |10.34 |Exponential smoothing |
| | |d. |MAPE | 27.84 |25.85 |Exponential smoothing |
| | |e. |MSE |147.10 |128.03 |Exponential smoothing |

5. Convenience Store [pic]

May [pic]

June [pic]

July [pic]

6. Community Federal [pic]

June [pic]

July [pic]

7. Heartville General Hospital i. Exponential smoothing, [pic]

| | |Year |Demand |Exponential Smoothing |Absolute |Absolute % |Square |
| | | | | |Deviation |Deviation |Error |
| | |1 |45 |41 | | | |
| | |2 |50 |41 + .6(45 – 41) = 43.4 | 6.60 |13.20 | 43.56 |
| | |3 |52 |43.4 + .6(50 – 43.4) = 47.4 | 4.60 | 8.85 | 21.16 |
| | |4 |56 |47.4 + .6(52 – 47.4) = 50.2 | 5.80 |10.36 | 33.64 |
| | |5 |58 |50.2 + .6(56 – 50.2) = 53.7 | 4.30 | 7.41 | 18.49 |
| | | | |Total |21.30 |39.82 | 116.85 |
| | | | |Average | 5.33 | 9.96 |29.2125 |

ii. Exponential smoothing, = 0.9

| | |Year |Demand |Exponential Smoothing |Absolute |Absolute % |Squared |
| | | | | |Deviation |Deviation |Error |
| | |1 |45 |41 | | | |
| | |2 |50 |41 + .9(45 – 41) = 44.6 | 5.40 |10.80 |29.16 |
| | |3 |52 |44.6 + .9(50 – 44.6) = 49.5 | 2.50 | 4.81 |6.25 |
| | |4 |56 |49.5 + .9(52 – 49.5) = 51.8 | 4.20 | 7.50 |17.64 |
| | |5 |58 |51.8 + .9(56 – 51.8) = 55.6 | 2.40 | 4.14 |5.76 |
| | | | |Total |14.50 |27.25 |58.81 |
| | | | |Average | 3.63 | 6.81 |14.7025 |

iii. Trend-adjusted exponential smoothing ([pic])

| | |Year |Demand |[pic] |[pic] |[pic] |Absolute |Absolute % |Squared |
| | | | | | | |Deviation |Deviation |Error |
| | |1 |45 |43.40 |2.24 |41.00 | 4.00 | 8.89 |16.00 |
| | |2 |50 |48.26 |2.50 |45.64 | 4.36 | 8.72 |19.01 |
| | |3 |52 |51.50 |2.58 |50.76 | 1.24 | 2.38 | 1.54 |
| | |4 |56 |55.23 |2.69 |54.08 | 1.92 | 3.43 | 3.69 |
| | |5 |58 |57.97 |2.70 |57.92 | 0.08 | 0.14 | 0.01 |
| | | | | | |Total |11.60 |23.56 |40.24 |
| | | | | | |Average | 2.32 | 4.71 | 8.05 |

Calculations by year: Year 1 [pic]

Year 2 [pic]

Year 3 [pic]

Year 4 [pic]

Year 5 [pic]

iv. Three-year moving average

| | |Year |Demand |3-Year Moving |Absolute |Absolute % |Square |
| | | | |Average |Deviation |Deviation |Error |
| | |1 |45 | | | | |
| | |2 |50 | | | | |
| | |3 |52 | | | | |
| | |4 |56 |(45 + 50 + 52)/3 = 49 | 7.00 |12.50 |49.00 |
| | |5 |58 |(50 + 52 + 56)/3 = 52.7 | 5.30 | 9.14 |28.09 |
| | | | |Total |12.30 |21.64 |77.09 |
| | | | |Average | 6.15 |10.82 |38.55 |

v. Three-year weighted moving average

| | |Year |Demand |3-Year Weighted |Absolute |Absolute % |Squared |
| | | | |Moving Average |Deviation |Deviation |Error |
| | |1 |45 | | | | |
| | |2 |50 | | | | |
| | |3 |52 | | | | |
| | |4 |56 |(45(1/6) + 50(2/6) + 52(3/6)) = 50 |6 |10.71 |36 |
| | |5 |58 |(50(1/6) + 52(2/6) + 56(3/6)) = 54 |4 | 6.90 |16 |
| | | | |Total |10 |17.61 |52 |
| | | | |Average |5 | 8.81 |26 |

vi. Regression model [pic]

| | |Year |Demand |Trend Projection |Absolute |Absolute % |Squared |
| | | | | |Deviation |Deviation |Error |
| | |1 |45 |42.6 + 3.2[pic]1 = 45.8 |0.80 |1.78 |0.64 |
| | |2 |50 |42.6 + 3.2[pic]2 = 49.0 |1.00 |2.00 |1.00 |
| | |3 |52 |42.6 + 3.2[pic]3 = 52.2 |0.20 |0.38 |0.04 |
| | |4 |56 |42.6 + 3.2[pic]4 = 55.4 |0.60 |1.07 |0.36 |
| | |5 |58 |42.6 + 3.2[pic]5 = 58.6 |0.60 |1.03 |0.36 |
| | | | |Total |3.20 |6.26 |2.40 |
| | | | |Average |0.64 |1.25 |0.48 |

a.-c. Comparison of the forecasting methodologies

| | |Forecast |MAD |MAPE |MSE |
| | |Methodology | | | |
| | |Exponential smoothing ( = .6 |5.33 |9.96 |29.21 |
| | |Exponential smoothing ( = .9 |3.63 |6.81 |14.70 |
| | |Trend-adjusted exp. smoothing |2.32 |4.71 |8.05 |
| | |Three-year moving average |6.15 |10.82 |38.55 |
| | |Three-year weighted moving average |5.00 |8.81 |26.00 |
| | |Regression model |0.64 |1.25 |0.48 |

Regression model methodology works best in this case under all performance criteria.

8. Calculator sales

[pic]

[pic]

[pic]

[pic]

[pic]

9. Forrest’s boxes of chocolates a. One possible estimated forecast for Year 4:

| | |Quarter |Forecast |
| | |1 |3,700 |
| | |2 |2,700 |
| | |3 |1,900 |
| | |4 |6,500 |
| | | |14,800 |

b. Multiplicative seasonal method

| | | | | | | | | |Average |
| | | | |Seasonal | |Seasonal | |Seasonal |Seasonal |
| | |Quarter |Year 1 |Factor |Year 2 |Factor |Year 3 |Factor |Factor |
| | |1 |3,000 |1.20 |3,300 |1.1 |3502 |1.03 |1.11 |
| | |2 |1,700 |0.68 |2,100 |0.7 |2448 |0.72 |0.70 |
| | |3 |900 |0.36 |1,500 |0.5 |1768 |0.52 |0.46 |
| | |4 |4,400 |1.76 |5,100 |1.7 |5882 |1.73 |1.73 |
| | |Total |10,000 | |12,000 | |13,600 | | |
| | |Average |2,500 | |3,000 | |3,400 | | |

Forecast for year 4, 14,800. Average = 3,700.

| | |Quarter |Average |Factor |Forecast |
| | |1 |3,700 |1.11 |4,107 |
| | |2 |3,700 |0.70 |2,590 |
| | |3 |3,700 |0.46 |1,702 |
| | |4 |3,700 |1.73 |6,401 |
| | | | | |14,800 |

This technique forecasts that the third-quarter sales will decrease compared to sales for the third quarter of the third year. Betcha thought it would increase. Mamma always said: “Life is full of surprises!” Just to make sure, we find confirmation of our calculations using the Seasonal

Forecasting Solver:

[pic]

10. Snyder’s Garden Center

| | | |Seasonal | |Seasonal |Average |
| |Quarter |Year 1 |Factor |Year 2 |Factor |Seasonal Factor |
| |1 |40 |0.179 |60 |0.218 |0.199 |
| |2 |350 |1.573 |440 |1.600 |1.587 |
| |3 |290 |1.303 |320 |1.164 |1.234 |
| |4 |210 |0.944 |280 |1.018 |0.981 |
| |Total |890 | |1,100 | | |
| |Average |222.50 | |275.00 | | |

Average quarterly sales in year 3 are expected to be 287.50 (1,150/4). Using the average seasonal factors, the forecasts for year 3 are:

| |Quarter | |Forecast |
| |1 |0.199(287.50) | 57 |
| |2 |1.587(287.50) |456 |
| |3 |1.234(287.50) |355 |
| |4 |0.981(287.50) |282 |

With the Seasonal Forecasting Solver, we get the same results

[pic]
11. Utility company

| |Quarter |Year 1 |Year 2 |Year 3 |Year 4 |
| |1 |103.5 |94.7 |118.6 |109.3 |
| |2 |126.1 |116.0 |141.2 |131.6 |
| |3 |144.5 |137.1 |159.0 |149.5 |
| |4 |166.1 |152.5 |178.2 |169.0 |
| |Total |540.2 |500.3 |597.0 |559.4 |
| |Average |135.05 |125.075 |149.25 |139.85 |

| |Quarter |Year 1 |Year 2 |Year 3 |Year 4 |Average |
| | | | | | |Seasonal Index |
| |1 |0.7664 |0.7571 |0.7946 |0.7816 |0.7749 |
| |2 |0.9337 |0.9274 |0.9410 |0.9410 |0.9371 |
| |3 |1.0700 |1.0961 |1.0653 |1.0690 |1.0751 |
| |4 |1.2299 |1.2193 |1.1940 |1.2084 |1.2129 |
| |Total |4.0 |4.0 |4.0 |4.0 |4.0 |

Forecast for Year 5

| |Quarter |Average Demand |Adjusted | | |
| | |per Quarter |Demand | | |
| |1 |150 |116.235 |= |116 |
| |2 |150 |140.565 |= |141 |
| |3 |150 |161.265 |= |161 |
| |4 |150 |181.935 |= |182 |
| | |600 |600 | | |

Turning to the Seasonal Forecasting Solver, we get the same results:

[pic]

Chapter 15

1. Bill of materials, Fig. 15.26 a. Item I has only one parent (E). However, item E has two parents (B and C). b. Item A has 10 unique components (B, C, D, E, F, G, H, I, J, and K). c. Item A has five purchased items (I, F, G, H, and K). These are the items without components. d. Item A has five intermediate items (B, C, D, E, and J). These items have both parents and components. e. The longest path is I–E–C–A at 11 weeks.

2. Item A. The bill of materials for item A is shown following.

[pic]

3. Lead time is determined by the longest path, C–B–A, at 13 weeks.

4. The bill of materials for item A with lead times is shown following. a. Lead time is determined by the longest path G-E-B-A = 12 weeks. b. If purchased items D, F, G, and H are already in inventory, the lead time is reduced to: A–B–E = 8 weeks. c. Item G with the longest lead time.

[pic]

5. Refer to Figure 15.21 and Solved Problem 1.

| |FIGURE 15.21 | |
| | | |

[pic]

Material requirement to produce 5 units of end-item A: [pic] on hand [pic] [pic] Material requirement to produce 13B: [pic]13D [pic]26E Material requirement to produce 5C:

[pic]5D

[pic] on hand [pic] Material requirement to produce 4F: [pic]4G – 3G on hand = 1G Purchased material requirements net of on-hand inventory: [pic], 26E, and 1G.

6. MPS record in Fig. 15.28. The following table is from the Master Production Scheduling Solver in OM Explorer. The ATP row is not required for this problem.

| |Solver Master Production Scheduling |
| |Enter data in yellow shaded areas. |
| | |
| |Lot Size |

1. Michaels Distribution Center

| |Day |M |T |W |Th |F |S |Su |

|
| |Requirements |6 |3 |5 |3 |7 |2 |3 |

|
| |M |T |W |Th |F |S |Su |Employee |

|
| | | | | | | | | |

|
| |6 |3 |5 |3 |7 |2 |3 |1 |

|
| | | | | | | | | |

|
| |5 |2 |4 |2 |6 |2 |3 |2 |

|
| | | | | | | | | |

|
| |4 |1 |3 |1 |5 |2 |3 |3 |

|
| | | | | | | | | |

|
| |3 |1 |3 |0 |4 |1 |2 |4 |

|
| | | | | | | | | |

|
| |2 |0 |2 |0 |3 |1 |2 |5 |

|
| | | | | | | | | |

|
| |1 |0 |2 |0 |2 |0 |1 |6 |

|
| | | | | | | | | |

|
| |0 |0 |1 |0 |1 |0 |1 |7 |

|
| | | | | | | | | |

| The number of employees is 7. They are scheduled to take the boxed days off.

2. Cara Ryder’s ski school needs 11 instructors. a. Alternative 1. The heuristic does have a number of different solutions.

| | |M |T |W |Th |F |S |Su |Instructor |

|
| | | | | | | | | | |

|
| | |7 |5 |4 |5 |6 |9 |8 |1 |

|
| | | | | | | | | | |

|
| | |6 |5 |4 |5 |5 |8 |7 |2 |

|
| | | | | | | | | | |

|
| | |5 |4 |4 |5 |5 |7 |6 |3 |

|
| | | | | | | | | | |

|
| | |5 |4 |4 |4 |4 |6 |5 |4 |

|
| | | | | | | | | | |

|
| | |4 |4 |4 |4 |3 |5 |4 |5 |

|
| | | | | | | | | | |

|
| | |3 |3 |4 |4 |3 |4 |3 |6 |

|
| | | | | | | | | | |

|
| | |3 |3 |3 |3 |2 |3 |3 |7 |

|
| | | | | | | | | | |

|
| | |2 |2 |2 |2 |2 |3 |3 |8 |

|
| | | | | | | | | | |

|
| | |2 |2 |2 |1 |1 |2 |2 |9 |

|
| | | | | | | | | | |

|
| | |1 |1 |1 |1 |1 |2 |1 |10 |

|
| | | | | | | | | | |

|
| | |1 |1 |0 |0 |0 |1 |0 |11 |

|
| | | | | | | | | | |

| b. Instructors are scheduled to take the boxed days off in the solution shown in part (a).

| | | |M |T |W |Th |F |S |Su |

|
| | |On-duty |7 |5 |4 |5 |6 |9 |8 |

|
| | |Requirements |7 |5 |4 |5 |6 |9 |8 |

|
| | |Slack |0 |0 |0 |0 |0 |0 |0 |

|

Alternative 2 (Optional)

| | |M |T |W |Th |F |S |Su |Instructor |

|
| | | | | | | | | | |

|
| | |7 |5 |4 |5 |6 |9 |8 |1 |

|
| | | | | | | | | | |

|
| | |6 |5 |4 |5 |5 |8 |7 |2 |

|
| | | | | | | | | | |

|
| | |5 |4 |4 |5 |5 |7 |6 |3 |

|
| | | | | | | | | | |

|
| | |5 |4 |4 |4 |4 |6 |5 |4 |

|
| | | | | | | | | | |

|
| | |4 |3 |4 |4 |4 |5 |4 |5 |

|
| | | | | | | | | | |

|
| | |4 |3 |3 |3 |3 |4 |4 |6 |

|
| | | | | | | | | | |

|
| | |3 |3 |3 |3 |2 |3 |3 |7 |

|
| | | | | | | | | | |

|
| | |2 |2 |2 |2 |2 |3 |3 |8 |

|
| | | | | | | | | | |

|
| | |2 |2 |2 |1 |1 |2 |2 |9 |

|
| | | | | | | | | | |

|
| | |1 |1 |1 |1 |1 |2 |1 |10 |

|
| | | | | | | | | | |

|
| | |1 |1 |0 |0 |0 |1 |1 |11 |

|
| | | | | | | | | | |

|

4. Hickory Company

| |a. |FCFS: |SPT: |EDD: |

|
| | |Job |Start Time |Flow Time |Job |Start Time |Flow Time |Job |Start Time |Flow Time |

|
| | |1 | 0 |10 |2 | 0 | 3 |2 | 0 | 3 |

|
| | |2 |10 |13 |5 | 3 |10 |1 | 3 |13 |

|
| | |3 |13 |28 |4 |10 |19 |3 |13 |28 |

|
| | |4 |28 |37 |1 |19 |29 |4 |28 |37 |

|
| | |5 |37 |44 |3 |29 |44 |5 |37 |44 |

|

| |b. | |FCFS: |SPT: |EDD: |

|
| | |Average flow times |26.4 |21.0 |25.0 |

|
| | |Average early time |0.4 |3.4 |1.0 |

|
| | |Average past due |11.0 |8.6 |10.2 |

|
| | |Average WIP inv. |3.0 |2.4 |2.8 |

|
| | |Average total inv. |3.1 |2.8 |3.0 |

| c. The rules perform as expected, except for SPT on the average past due measure. Typically EDD will do better here. Nonetheless, SPT does well on flow times, WIP, and inventory levels.

Solver—Singe Machine Scheduler

| | | | | |

|
|Job |StartTime |Flow Time |Job |Start Time |Flow Time |

|
|Average Flow TImes |21.0 |14.6 |16.8 |20.2 |20.0 |

|
|Average Early TIme |0.2 |2.2 |0.4 |1.8 |0.2 |

|
|Average Past Due |9.8 |5.4 |5.8 |10.6 |8.8 |

|
|Average WIP Inv. |3.2 |2.2 |2.5 |3.1 |3.0 |

|
|Average Total Inv. |3.2 |2.5 |2.6 |3.3 |3.1 |

|

c. Priority planning with an MRP system relies on proper timing of materials. Planners manipulate scheduled due dates to match material need dates with order due dates. Consequently, priority rules incorporating due dates would be most useful in communicating these changes to the shop floor. Of those listed in this problem, EDD, S/RO, and CR would work best.

6. Bycraft Enterprises

| |Job |Total Processing Time (hours) |

|
| |1 |50(.06) + 4 = 7.0 |

|
| |2 |120(.05) + 3 = 9.0 |

|
| |3 |260(.03) + 5 = 12.8 |

|
| |4 |200(.04) + 2 = 10.0 |

| a. Using SPT

| | |Job |Arrival |Start |Finish |Flow (hr) |Past Due (hr) |

|
| | |1 |9:00 a.m.(M) |9:00 a.m.(M) |4:00 p.m.(M) | 7.0 | 0.0 |

|
| | |2 |10:00 a.m.(M) |4:00 p.m.(M) |1:00 a.m.(T) |15.0 | 3.0 |

|
| | |4 |12:00 p.m. (M) |1:00 a.m.(T) |11:00 a.m.(T) |23.0 | 9.0 |

|
| | |3 |11:00 a.m.(M) |11:00 a.m.(T) |11:48 p.m.(T) |36.8 |24.8 |

|
| | |Total | | | | 81.80 |36.8 |

|

| | |Monday |Tuesday |

|
| | |8–|12–4 |4–8 |8–12 |12–4 |
| | |12| | | | |

|
| | | |7 hours |9 hours |10 hours |12.8 hours |

|

Using EDD

| | |Job |Arrival |Start |Finish |Flow |Past Due |

|
| | | | | | |(hr) |(hr) |

|
| | |1 |9:00 a.m.(M) |9:00 a.m.(M) |4:00 p.m.(M) |7.0 |0.0 |

|
| | |2 |10:00 a.m.(M) |4:00 p.m.(M) |1:00 a.m.(T) |15.0 |3.0 |

|
| | |3 |11:00 a.m.(M) |1:00 a.m.(T) |1:48 p.m.(T) |26.8 |14.8 |

|
| | |4 |12:00 p.m.(M) |1:48 p.m.(T) |11:48 p.m.(T) |35.8 |21.8 |

|
| | |Total | | | |84.6 |39.6 |

|

| | |Monday |Tuesday |

|
| | |8–|12–4 |4–8 |8–12 |12–4 |4–8|
| | |12| | | | | |

|
| | | |7 hours |9 hours |12.8 hours |10 hours | |

|

| |b. | |SPT |EDD |

|
| | |Average flow time (hours) |20.45 |21.15 |

|
| | |Average hours past due |9.20 |9.90 |

| c. EDD minimizes the maximum number of past-due hours and the variance of the past-due hours; however, EDD does worse with regard to average flow times and average hours past due. Consequently, in this example EDD does better with respect to some customer service measures but does worse with respect to inventory. SPT processes some jobs and gets them out of inventory quickly, assuming jobs can be shipped on completion whether or not they are due. Typical trade-offs involve customer service and inventory investment.

7. Refer to Gantt chart in Fig. 16.8

[pic]

a. To minimize the makespan if each job must be processed on machine A first, we can use Johnson’s rule:

| | | |Process Time (hr) |

|
| | |Job |Machine A |Machine B |

|
| | |1 |2 |1 |

|
| | |2 |1 |4 |

|
| | |3 |3 |2 |

| The optimal sequence would be 2–3–1. The revised Gantt chart is:

[pic]

The makespan is now 8 hours, which is an improvement of 1 hour.

b. Now suppose that the only restriction is that no job may be processed on different machines at the same time. One of several schedules that yield a makespan of 7 hours is given following:

[pic]

With the restriction of flow from machine A to machine B removed, we are able to utilize the first hour on machine B. This is why we could beat the schedule in part (a).
[pic]

1. Bold Vision Inc. [pic]

2. Sharpe Cutter
| |a. |[pic] |

b. Total annual cost [pic]

c. [pic] days

d. Production time/lot [pic] days

3. Sud’s Bottling Company [pic]

4. Bucks Grande major-league baseball In this problem we assume the annual demand is based on a 52-week year. Step 1: Calculate the EOQ at the lowest price ($48.50):

[pic] or 40 bats

This solution is infeasible. We cannot buy 40 bats at a price of $48.50 each. Therefore we calculate the EOQ at the next lowest price ($51.00):

[pic] or 39 bats

This solution is feasible.

Step 2: Calculate total costs at the feasible EOQ and at higher discount quantities: [pic]

a. It is less costly on an annual basis to buy 39 bats at a time. b. The total annual costs associated with buying 39 bats at a time are $11,359.24. c. Yes, if the price is reduced to $45 at an order quantity of 180 bats. The Quantity Discount Solver from OM Explorer is used to find the solution.

| |Solver Evaluating Quantity Discounts | |
| |Enter data in yellow shaded areas. | |
| | | | | | | | |
| | | | | | | | |
| | | | | | | | |
| |Min. Amount Req'd for Price Point |Lot Sizes |Price/Unit | | | | |
| |--- |0 - 11 |$54.00 | | | | |
| |12 |12 -143 |$51.00 | | | | |
| |144 |144 -179 |$48.50 | | | | |
| |180 |180 or more |$45.00 | | | | |
| | | | | | | | |
| |Annual demand |208 | | | | | |
| |Order cost |$70 | | | | | |
| |Holding cost (% of Price) |38% | | | | | |
| | | | | | | | |
| |Best order quantity | |180 | | | | |
| | | | | | | | |
| | |EOQ or Req'd | | | | | |
| | |Order for Price |Inventory Cost |Order |Purchase Cost |Total | |
| |Price Point |Point | |Cost | |Cost | |
| |$54.00 |38 |$389.88 |$383.16 |$11,232 |$12,005 | |

Therefore, Bucks Grande should purchase 180 bats at a time.

5. Pfisher. The EOQ at the lowest price ($49.00) remains infeasible and the EOQ at the next lowest price ($50.25) remains at 79 packages. The total annual cost of buying disposable surgical packages also remains at $25,416.44 per year. Now we calculate the annual cost associated with ordering 500 at a time:

[pic]

The quantity discount is not sufficient to cause Pfisher to buy the larger order quantity.

6. University Bookstore Step 1: Calculate the EOQ at the lowest price ($3.25)

[pic]

This solution is infeasible; we cannot buy 226 pencils at $3.25 a piece. Therefore, we calculate the EOQ at the next lowest price ($3.50).

[pic] This solution is feasible.

Step 2: Calculate total costs at the feasible EOQ and at higher discount quantities.

[pic]

The best order quantity is 218 units.

7. Mac-in-the-Box, Inc. Step 1: Calculate the EOQ at the lowest price ($400):

[pic] or 106 scanners

This solution is infeasible. We cannot buy 106 scanners at a price of $400 each. Therefore, we calculate the EOQ at the next lowest price ($500).

[pic] or 95 scanners This solution is feasible.

Step 2: Calculate total costs at the feasible EOQ and at higher discount quantities.

[pic]

The order quantity should be 144 scanners.

8. Order quantity for “an item” Step 1: Calculate the EOQ at the lowest price ($2.00).

[pic] or 447 units

This solution is infeasible. We cannot buy 447 units at a price of $2.00 each. Therefore, we calculate the EOQ at the next lowest price ($2.25).

[pic] or 422 units

This solution is feasible.

Step 2: Calculate total costs at the feasible EOQ and at higher discount quantities.

[pic]

The order quantity should be 1,000 units.

9. National Printing Company The payoff matrix is constructed, using

Payoff = [pic]

The following table comes from the One-Period Inventory Decision Solver in OM Explorer.

| | | | | | | | | | |
| |Demand | |2000 |3000 |4000 |5000 |6000 | | |
|Payoff Table | |
| | | | | | | | | | |
| | | | |Demand | | | | |
| | | |2000 |3000 |4000 |5000 |6000 | | |
| |Quantity |2000 |12000 |12000 |12000 |12000 |12000 | | |
| | | | | | | | | | |
| | |Weighted Payoffs | | | | | |
| | | | | | | | | | |
| | |Order |Expected | | | | | |
| | |Quantity |Payoff | | | | | | |
| | |2000 |12000 | |Greatest |24400 | | | |
| | | | | |Expected | | | | |
| | | | | |Payoff | | | | |

The best order quantity is 5,000 units.

10. Dorothy’s Pastries The following payoff matrix was constructed, where

[pic]

| | |D |
| |Q |50 |150 |200 |
| |50 |$20 |$20 |$20 |
| |150 |($10) |$60 |$60 |
| |200 |($25) |$45 |$80 |

Now we can compute the expected payoff for each baking quantity Q.

| |Order Quantity |Expected Payoff |
| | 50 |0.25(20) |+ |0.50(20) |+ |0.25(20) |= $20.00 |
| |150 |0.25(–10) |+ |0.50(60) |+ |0.25(60) |= $42.50 |
| |200 |0.25(–25) |+ |0.50(45) |+ |0.25(80) |= $36.25 |

Therefore, 150 pastries should be baked each day.

11. Aggies versus Tech The following payoff matrix was constructed, where

[pic]

| | |D |
| |Q |2,000 |3,000 |4,000 |5,000 |6,000 |
| |2,000 |$3,000 |$3,000 |$3,000 |$3,000 |$3,000 |
| |3,000 |$2,000 |$4,500 |$4,500 |$4,500 |$4,500 |
| |4,000 |$1,000 |$3,500 |$6,000 |$6,000 |$6,000 |
| |5,000 |$ 0 |$2,500 |$5,000 |$7,500 |$7,500 |
| |6,000 |($1,000) |$1,500 |$4,000 |$6,500 |$9,000 |

Now we can compute the expected payoff for each baking quantity Q.

| |Order Quantity |Expected Payoff |

|2,000 |0.10(3,000) |+ |0.30(3,000) |+ |0.30(3,000) |+ |0.20(3,000) |+ |0.10(3,000) |= |$3,000 | | |3,000 |0.10(2,000) |+ |0.30(4,500) |+ |0.30(4,500) |+ |0.20(4,500) |+ |0.10(4,500) |= |$4,250 | | |4,000 |0.10(1,000) |+ |0.30(3,500) |+ |0.30(6,000) |+ |0.20(6,000) |+ |0.10(6,000) |= |$4,750 | | |5,000 |0.10(0) |+ |0.30(2,500) |+ |0.30(5,000) |+ |0.20(7,500) |+ |0.10(7,500) |= |$4,500 | | |6,000 |0.10(–1,000) |+ |0.30(1,500) |+ |0.30(4,000) |+ |0.20(6,500) |+ |0.10(9,000) |= |$3,750 | | Order 4,000 hot dogs.

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