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Please check the attached documents to see the questions
– 2 years ago
**files (12)**
– ## Practiceexercises2024.docx.pdf
– ## guidancetohomework1.docx.pdf
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**## Practiceexercises2024.docx.pdf**
Practice exercise multifactor productivity – chapter 1
Gibson Products produces cast bronze valves for use in offshore oil platforms. Currently, Gibson
produces 1600 valves per day. The 20 workers at Gibson work from 7 a.m. until 4 p.m., with 30 minutes off for
lunch and a 15-minute break during the morning work session and another at the afternoon work session. Gibson is
in a competitive industry, and needs to increase productivity to stay competitive. They feel that a 20 percent increase
is needed.
Gibson's management believes that the 20 percent increase will not be possible without a change in working
conditions, so they change work hours. The new schedule calls on workers to work from 7:30 a.m. until 4:30 p.m.,
during which workers can take one hour off at any time of their choosing. Obviously, the number of paid hours is the
same as before, but production increases, perhaps because workers are given a bit more control over their workday.
After this change, valve production increased to 1800 units per day.
a. Calculate labor productivity for the initial situation
b. Calculate labor productivity for the hypothetical 20 percent increase
c. What is the productivity after the change in work rules?
d. Write a short paragraph analyzing these results.
Answer:
(a) Workers are active for eight hours per day; labor productivity is 10 valves/hour
(b) If Productivity rises by 20 percent, to 12 valves/hour; output would be 12 × 8 × 20 = 1920
(c) New productivity is 1800 / (20 × 8) = 11.25 valves/hour
(d) Gibson did not gain the desired 20 percent increase in productivity, but they did gain over 11 percent, without
extra equipment or energy, and without increasing the labor cost.
Practice exercise project management – chapter 3
A network consists of the activities in the following list. Times are given in weeks.
Activity PrecedingTime
A — 8
B — 3
C A 7
D A, B 3
E C 4
F D 6
a. Draw the network diagram.
b. Calculate the ES, EF, LS, LF, and Slack for each activity.
c. What is project completion time?
Answer: (a)
(b ,c)
Results
Task Early StartEarly FinishLate StartLate FinishSlack
A 0 8 0 8 0
B 0 3 7 10 7
C 8 15 8 15 0
D 8 11 10 13 2
E 15 19 15 19 0
F 11 17 13 19 2
Project 19
Practice exercise forecasting MAD chapter 4
The department manager using a combination of methods has forecast sales of toasters at a local department store.
Calculate the MAD for the manager's forecast. Compare the manager's forecast against a naive forecast covering the
same time period. Which is better?
Month Unit Sales
Manager's
Forecast
January 52
February 61
March 73
April 79
May 66
June 51
July 47 50
August 44 55
September 30 52
October 55 42
November 74 60
December 125 75
Answer:
Month ActualManager'sAbs. Error NaiveAbs. Error
January 52
February 61
March 73
April 79
May 66
June 51
July 47 50 3 51 4
August 44 55 11 47 3
September 30 52 22 44 14
October 55 42 13 30 25
November 74 60 14 55 19
December 125 75 50 74 51
The manager's forecast has a MAD of 18.83, while the naive is 19.33. Therefore, the manager's forecast is slightly
better than the naive.
Practice exercise exponential smoothing – chapter 4
Given an actual demand this period of 103, a forecast value for this period of 99, and an alpha of .4, what is the
exponential smoothing forecast for next period?
A) 94.6
B) 97.4
C) 100.6
D) 101.6
E) 103.0
Answer: C
Practice exercise seasonal factor forecasting chapter 4
The quarterly sales for specific educational software over the past three years are given in the following table.
Compute the four seasonal factors.
YEAR 1YEAR 2YEAR 3
Quarter 1 1710 1820 1830
Quarter 2 960 910 1090
Quarter 3 2720 2840 2900
Quarter 4 2430 2200 2590
Answer:
Avg. Sea. Fact.
Quarter 1 1786.670.8933
Quarter 2 986.67 0.4933
Quarter 3 2820.001.4100
Quarter 4 2406.671.2033
Grand Average 2000.00
Practice exercise – process management
17) Consider the assembly line below. The three fabrication operations run in parallel, such that each batch of 20
units only needs to go through one of the three fabrication operations. After that, each batch needs to go through
both assembly operations, which occur simultaneously (specifically, 10 components are made for each unit in the
fabrication stage—some components are then assembled in the Assembly 1 area while others are assembled in the
Assembly 2 area). The units are packaged and made ready for shipment in the final stage.
What is the bottleneck time per batch of this operation?
A) 24 min.
B) 8 min.
C) 88 min.
D) 34 min.
E) 3 min.
Answer: B
Practice exercise bullwhip effect – chapter 11
Consider a supply chain where a manufacturer sells to a distributor who sells to a wholesaler who sells to a retailer.
Last year, the retailer's weekly variance of demand was 4000 units. The weekly variance of orders was 5000; 8000;
12,000; and 17,000 units for the retailer, wholesaler, distributor, and manufacturer, respectively. (Note that the
variance of orders equals the variance of demand for that firm's supplier.)
(a) Calculate the bullwhip measure for the retailer.
(b) Calculate the bullwhip measure for the wholesaler.
(c) Calculate the bullwhip measure for the distributor.
(d) Calculate the bullwhip measure for the manufacturer.
(e) Which firm appears to be contributing the most to the bullwhip effect in this supply chain?
Answer:
(a) 5000 / 4000 = 1.25; (b) 8000 / 5000 = 1.6; (c) 12,000 / 8000 = 1.5; (d) 17,000 / 12,000 = 1.42
(e) With the highest bullwhip measure value of 1.6, the wholesaler appears to be contributing the most to the
bullwhip effect in this supply chain.
Practice exercise supply chain – chapter 11
The following data are pulled from a recent Walsh Manufacturing annual report.
Assets
Raw material inventory $120,000
Work-in-process inventory $50,000
Finished goods inventory $300,000
Property, plant & equipment $500,000
Other assets $200,000
Total assets $1,170,000
Condensed Income Statement
Revenue $2,000,000
Cost of goods sold $600,000
Other expenses $1,000,000
Net income $400,000
Calculate: (a) Percent invested in inventory, (b) Inventory turnover, and (c) Weeks of supply.
Answer:
(a) Percent invested in inventory = (120,000 + 50,000 + 300,000)/1,170,000 = 40.17%
(b) Inventory turnover = 600,000/(120,000 + 50,000 + 300,000) = 1.28
(c) Weeks of supply = (120,000 + 50,000 + 300,000)/(600,000/52) = 40.73
Practice exercise ABC analysis chapter 12
Perform an ABC analysis on the following set of products.
Item Annual DemandUnit Cost
A211 1200 $9
B390 100 $90
C003 4500 $6
D100 400 $150
E707 35 $2000
F660 250 $120
G473 1000 $90
H921 100 $75
Answer: The table below details the contribution of each of the eight products. Item G473 is clearly an A item, and
items A211, B390, and H921 are all C items. Other classifications are somewhat subjective, but one choice is to
label E707 and D100 as A items, and F660 and C003 as B items.
Item
Annual
DemandUnit Cost Volume
Cumulative
volume
Cumulative
percent
G473 1000 $90 $90,000 $90,000 29.6%
E707 35 $2,000 $70,000 $160,000 52.6%
D100 400 $150 $60,000 $220,000 72.3%
F660 250 $120 $30,000 $250,000 82.2%
C003 4500 $6 $27,000 $277,000 91.0%
A211 1200 $9 $10,800 $287,800 94.6%
B390 100 $90 $9,000 $296,800 97.5%
H921 100 $75 $7,500 $304,300 100.0%
$304,300
Practice exercise safety stock – chapter 12
Thomas' Bike Shop stocks a high volume item that has a normally distributed demand during lead time. The average
daily demand is 70 units, the lead time is 4 days, and the standard deviation of demand during lead time is 15.
(a) How much safety stock provides a 95% service level to Thomas?
(b) What should the reorder point be?
Answer:
(a) SS = 1.65 × 15 = 24.75 units or 25 units
(b) ROP = (70)(4) + 25 = 305 units.
Practice exercise production order quantity model – chapter 12
Montegut Manufacturing produces a product for which the annual demand is 10,000 units. Production averages 100
units per day, while demand is 40 units per day. Holding costs are $2.00 per unit per year, and setup cost is $200.00.
(a) If the firm wishes to produce this product in economic batches, what size batch should be used? (b) What is the
maximum inventory level? (c) How many order cycles are there per year? (d) What are the total annual holding and
setup costs?
Answer: This is a production order quantity problem.
(a) Q*p = = = 1825.7 or 1826 units
(b) The maximum inventory level is Q = 1825.7 = 1095.45 or 1095 units
(c) There are approximately N = = = 5.48 cycles per year
(d) Total annual costs = (5.48)($200) + (1095.45/2)$2 = $2,190.89 or $2,191
Practice exercise of EOQ model – chapter 12
A printing company estimates that it will require 1,000 reams of a certain type of paper in a given period. The cost
of carrying one unit in inventory for that period is 50 cents. The company buys the paper from a wholesaler in the
same town, sending its own truck to pick up the orders at a fixed cost of $20.00 per trip. Treating this cost as the
order cost, (a) what is the optimum number of reams to buy at one time? (b) How many times should lots of this size
be bought during this period? (c) What is the minimum cost (holding and setup) of maintaining inventory on this
item for the period? (d) Of this total cost, how much is carrying cost and how much is ordering cost?
Answer: This is an EOQ problem, even though the time period is not a year. All that is required is that the demand
value and the carrying cost share the same time reference. This will require approximately 3.5 orders per period.
Setup costs and carrying costs are each $70.71, and the annual total is $141.42.
(a) EOQ = = 283; (b) N = = 3.54
(d) Carrying cost = ($0.50) = $70.71; setup cost = ($20) = $70.71
(c) Total cost = $70.71 + $70.71 = $141.42
Practice exercise of production order quantity model – chapter 12
Holstein Computing manufactures an inexpensive audio card (Audio Max) for assembly into several models of its
microcomputers. The annual demand for this part is 100,000 units. The annual inventory carrying cost is $5 per unit
and the cost of preparing an order and making production setup for the order is $750. The company operates 250
days per year. The machine used to manufacture this part has a production rate of 2000 units per day.
(a) Calculate the optimum lot size.
(b) How many lots are produced in a year?
(c) What is the average inventory for Audio Max?
(d) What is the annual holding and setup cost for Audio Max?
Answer:
(a) Q*p = = = 6123.7 or 6124 units.
(b) There are approximately N = = = 16.33 cycles per year.
(c) The maximum inventory is Q = 6123.7 = 4899 units; average inventory is 4899 / 2 = 2449.5
units.
(d) Annual inventory management costs are 16.33($750) + 2449.5($5) = $12,247.50 + $12,247.50 = $24,495.
Practice exercise Quantity Discount – chapter 12
The annual demand for an item is 10,000 units. The cost to process an order is $75 and the annual inventory holding
cost is 20% of item cost.
(a) What is the optimal order quantity, given the following price breaks for purchasing the item?
(b) What price should the firm pay per unit?
(c) What is the total annual cost at the optimal behavior?
Quantity Price
1-9 $2.95 per unit
10 – 999 $2.50 per unit
1,000 – 4,999 $2.30 per unit
5,000 or more $1.85 per unit
Answer:
STEP 1: EOQ at $1.85 = 2013.47 → infeasible
EOQ at $2.30 = 1805.79 → feasible
So the possible best answers are 1806 and 5000
STEP 2: TC for 1806 units = $23,830.66
TC for 5000 units = $19,575.00
(a) Order 5000 units at a time.
(b) The price is $1.85 per unit.
(c) Total annual purchasing, holding, and setup costs = $19,575.00.
Practice exercise about MTBF – chapter 17
Ten high-intensity bulbs are tested for 100 hours each. One failed at 10 hours; all others completed the test.
Calculate FR(%), FR(N) and MTBF.
Answer: FR(%) = 1/10 or 10%; operating hours = 9 × 100 + 10 = 910; FR(N) = 1/910 or .0011, and
MTBF = 910 hours
Practice exercise about reliability – chapter 17
The academic service commonly referred to as "registration" consists of several smaller components: advising,
registration for courses, fee assessment, financial aid calculations, and fee payment. Each of these modules operates
independently and has some probability of failure for each student. If the five probabilities that accompany these
services are 95%, 90%, 99%, 98%, and 99%, respectively, what is the "reliability" of the entire product from the
student's perspective, i.e., the probability that all five will work according to plan?
Answer: Reliability is (.95)(.90)(.99)(.98)(.99) = .8212
Practice exercise about maintenance – chapter 17
Great Southern Consultants Group's computer system has been down several times over the past few months, as
shown below.
Number of breakdowns 01234
Monthly frequency 92441
Each time the system is down, the firm loses an average of $400 in time and service expenses. They are considering
signing a contract for preventive maintenance. With preventive maintenance, the system would be down on average
only 0.5 per month. The monthly cost of preventive maintenance would be $200 a month. Which is cheaper,
breakdown or preventive maintenance?
Answer:
Number of breakdowns 0 1 2 3 4 Total
Monthly probability 0.450.100.200.200.05 1.00
Expected # of breakdowns per month = (0)(.45) + (1)(.10) + (2)(.20) + (3)(.20) + (4)(.05) = 1.30 Expected cost of
breakdowns per month with breakdown maintenance = (1.30)($400) = $520
Preventive maintenance cost per month = (.5)($400) + $200 = $400
Preventive maintenance is more cost-effective.
**## guidancetohomework1.docx.pdf**
3.9 (a)
(b, c) There are four paths:
Path Time (hours)
A–C–E–G 19.5
B–D–F–G 24.9
A–C–D–F–G 28.7 (critical)
B–E–G 15.7
3.20 (a) Activity a m b te Variance
A 9 10 11 10 0.11
B 4 10 16 10 4
C 9 10 11 10 0.11
D 5 8 11 8 1
(b) Critical path is A–C with mean (te) completion time
of 20 weeks. The other path is B–D, with mean completion time of 18 weeks.
(c) Variance of A–C = (Variance of A) + (Variance of C)
= 0.11 + 0.11 = 0.22
Variance of B–D = (Variance of B) + (Variance of D)
= 4 + 1 = 5
(d) Probability A–C is finished in 22 weeks or less =
−⎛ ⎞
≤ = ≤ ≅⎜ ⎟
⎝ ⎠
2 2 2 0 ( 4 . 2 6 ) 1 . 0 0
0 . 2 2
P Z P Z
(e) Probability B–D is finished in 22 weeks or less =
−⎛ ⎞
≤ = ≤ =⎜ ⎟⎝ ⎠
2 2 1 8 ( 1 . 7 9 ) 0 . 9 6 3
5
P Z P Z
(f) The critical path has a relatively small variance and will almost certainly be finished in 22 weeks or less. Path
B–D has a relatively high variance. Due to this, the probability B–D is finished in 22 weeks or less is only about 0.96.
Since the project is not finished until all activities (and paths) are finished, the probability that the project will be
finished in 22 weeks or less is not 1.00 but is approximately 0.96.
3.32 (a)
Project completion time = 14 weeks
Task Time ES EF LS LF Slack
A 3 0 3 0 3 0
B 2 0 2 2 4 2
C 1 0 1 11 12 11
D 7 3 10 3 10 0
E 6 2 8 4 10 2
F 2 1 3 12 14 11
G 4 10 14 10 14 0
(b) To crash to 10 weeks, we follow 2 steps:
1. Crash D by 2 weeks ($150).
2. Crash D and E by 2 weeks each ($150 + $100).
Total crash cost = $400 additional
**28E6EE0E-86CE-495A-A42D-B7419999F357.jpeg**
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**## Practiceexercises2024.docx.pdf**
Practice exercise multifactor productivity – chapter 1
Gibson Products produces cast bronze valves for use in offshore oil platforms. Currently, Gibson
produces 1600 valves per day. The 20 workers at Gibson work from 7 a.m. until 4 p.m., with 30 minutes off for
lunch and a 15-minute break during the morning work session and another at the afternoon work session. Gibson is
in a competitive industry, and needs to increase productivity to stay competitive. They feel that a 20 percent increase
is needed.
Gibson's management believes that the 20 percent increase will not be possible without a change in working
conditions, so they change work hours. The new schedule calls on workers to work from 7:30 a.m. until 4:30 p.m.,
during which workers can take one hour off at any time of their choosing. Obviously, the number of paid hours is the
same as before, but production increases, perhaps because workers are given a bit more control over their workday.
After this change, valve production increased to 1800 units per day.
a. Calculate labor productivity for the initial situation
b. Calculate labor productivity for the hypothetical 20 percent increase
c. What is the productivity after the change in work rules?
d. Write a short paragraph analyzing these results.
Answer:
(a) Workers are active for eight hours per day; labor productivity is 10 valves/hour
(b) If Productivity rises by 20 percent, to 12 valves/hour; output would be 12 × 8 × 20 = 1920
(c) New productivity is 1800 / (20 × 8) = 11.25 valves/hour
(d) Gibson did not gain the desired 20 percent increase in productivity, but they did gain over 11 percent, without
extra equipment or energy, and without increasing the labor cost.
Practice exercise project management – chapter 3
A network consists of the activities in the following list. Times are given in weeks.
Activity PrecedingTime
A — 8
B — 3
C A 7
D A, B 3
E C 4
F D 6
a. Draw the network diagram.
b. Calculate the ES, EF, LS, LF, and Slack for each activity.
c. What is project completion time?
Answer: (a)
(b ,c)
Results
Task Early StartEarly FinishLate StartLate FinishSlack
A 0 8 0 8 0
B 0 3 7 10 7
C 8 15 8 15 0
D 8 11 10 13 2
E 15 19 15 19 0
F 11 17 13 19 2
Project 19
Practice exercise forecasting MAD chapter 4
The department manager using a combination of methods has forecast sales of toasters at a local department store.
Calculate the MAD for the manager's forecast. Compare the manager's forecast against a naive forecast covering the
same time period. Which is better?
Month Unit Sales
Manager's
Forecast
January 52
February 61
March 73
April 79
May 66
June 51
July 47 50
August 44 55
September 30 52
October 55 42
November 74 60
December 125 75
Answer:
Month ActualManager'sAbs. Error NaiveAbs. Error
January 52
February 61
March 73
April 79
May 66
June 51
July 47 50 3 51 4
August 44 55 11 47 3
September 30 52 22 44 14
October 55 42 13 30 25
November 74 60 14 55 19
December 125 75 50 74 51
The manager's forecast has a MAD of 18.83, while the naive is 19.33. Therefore, the manager's forecast is slightly
better than the naive.
Practice exercise exponential smoothing – chapter 4
Given an actual demand this period of 103, a forecast value for this period of 99, and an alpha of .4, what is the
exponential smoothing forecast for next period?
A) 94.6
B) 97.4
C) 100.6
D) 101.6
E) 103.0
Answer: C
Practice exercise seasonal factor forecasting chapter 4
The quarterly sales for specific educational software over the past three years are given in the following table.
Compute the four seasonal factors.
YEAR 1YEAR 2YEAR 3
Quarter 1 1710 1820 1830
Quarter 2 960 910 1090
Quarter 3 2720 2840 2900
Quarter 4 2430 2200 2590
Answer:
Avg. Sea. Fact.
Quarter 1 1786.670.8933
Quarter 2 986.67 0.4933
Quarter 3 2820.001.4100
Quarter 4 2406.671.2033
Grand Average 2000.00
Practice exercise – process management
17) Consider the assembly line below. The three fabrication operations run in parallel, such that each batch of 20
units only needs to go through one of the three fabrication operations. After that, each batch needs to go through
both assembly operations, which occur simultaneously (specifically, 10 components are made for each unit in the
fabrication stage—some components are then assembled in the Assembly 1 area while others are assembled in the
Assembly 2 area). The units are packaged and made ready for shipment in the final stage.
What is the bottleneck time per batch of this operation?
A) 24 min.
B) 8 min.
C) 88 min.
D) 34 min.
E) 3 min.
Answer: B
Practice exercise bullwhip effect – chapter 11
Consider a supply chain where a manufacturer sells to a distributor who sells to a wholesaler who sells to a retailer.
Last year, the retailer's weekly variance of demand was 4000 units. The weekly variance of orders was 5000; 8000;
12,000; and 17,000 units for the retailer, wholesaler, distributor, and manufacturer, respectively. (Note that the
variance of orders equals the variance of demand for that firm's supplier.)
(a) Calculate the bullwhip measure for the retailer.
(b) Calculate the bullwhip measure for the wholesaler.
(c) Calculate the bullwhip measure for the distributor.
(d) Calculate the bullwhip measure for the manufacturer.
(e) Which firm appears to be contributing the most to the bullwhip effect in this supply chain?
Answer:
(a) 5000 / 4000 = 1.25; (b) 8000 / 5000 = 1.6; (c) 12,000 / 8000 = 1.5; (d) 17,000 / 12,000 = 1.42
(e) With the highest bullwhip measure value of 1.6, the wholesaler appears to be contributing the most to the
bullwhip effect in this supply chain.
Practice exercise supply chain – chapter 11
The following data are pulled from a recent Walsh Manufacturing annual report.
Assets
Raw material inventory $120,000
Work-in-process inventory $50,000
Finished goods inventory $300,000
Property, plant & equipment $500,000
Other assets $200,000
Total assets $1,170,000
Condensed Income Statement
Revenue $2,000,000
Cost of goods sold $600,000
Other expenses $1,000,000
Net income $400,000
Calculate: (a) Percent invested in inventory, (b) Inventory turnover, and (c) Weeks of supply.
Answer:
(a) Percent invested in inventory = (120,000 + 50,000 + 300,000)/1,170,000 = 40.17%
(b) Inventory turnover = 600,000/(120,000 + 50,000 + 300,000) = 1.28
(c) Weeks of supply = (120,000 + 50,000 + 300,000)/(600,000/52) = 40.73
Practice exercise ABC analysis chapter 12
Perform an ABC analysis on the following set of products.
Item Annual DemandUnit Cost
A211 1200 $9
B390 100 $90
C003 4500 $6
D100 400 $150
E707 35 $2000
F660 250 $120
G473 1000 $90
H921 100 $75
Answer: The table below details the contribution of each of the eight products. Item G473 is clearly an A item, and
items A211, B390, and H921 are all C items. Other classifications are somewhat subjective, but one choice is to
label E707 and D100 as A items, and F660 and C003 as B items.
Item
Annual
DemandUnit Cost Volume
Cumulative
volume
Cumulative
percent
G473 1000 $90 $90,000 $90,000 29.6%
E707 35 $2,000 $70,000 $160,000 52.6%
D100 400 $150 $60,000 $220,000 72.3%
F660 250 $120 $30,000 $250,000 82.2%
C003 4500 $6 $27,000 $277,000 91.0%
A211 1200 $9 $10,800 $287,800 94.6%
B390 100 $90 $9,000 $296,800 97.5%
H921 100 $75 $7,500 $304,300 100.0%
$304,300
Practice exercise safety stock – chapter 12
Thomas' Bike Shop stocks a high volume item that has a normally distributed demand during lead time. The average
daily demand is 70 units, the lead time is 4 days, and the standard deviation of demand during lead time is 15.
(a) How much safety stock provides a 95% service level to Thomas?
(b) What should the reorder point be?
Answer:
(a) SS = 1.65 × 15 = 24.75 units or 25 units
(b) ROP = (70)(4) + 25 = 305 units.
Practice exercise production order quantity model – chapter 12
Montegut Manufacturing produces a product for which the annual demand is 10,000 units. Production averages 100
units per day, while demand is 40 units per day. Holding costs are $2.00 per unit per year, and setup cost is $200.00.
(a) If the firm wishes to produce this product in economic batches, what size batch should be used? (b) What is the
maximum inventory level? (c) How many order cycles are there per year? (d) What are the total annual holding and
setup costs?
Answer: This is a production order quantity problem.
(a) Q*p = = = 1825.7 or 1826 units
(b) The maximum inventory level is Q = 1825.7 = 1095.45 or 1095 units
(c) There are approximately N = = = 5.48 cycles per year
(d) Total annual costs = (5.48)($200) + (1095.45/2)$2 = $2,190.89 or $2,191
Practice exercise of EOQ model – chapter 12
A printing company estimates that it will require 1,000 reams of a certain type of paper in a given period. The cost
of carrying one unit in inventory for that period is 50 cents. The company buys the paper from a wholesaler in the
same town, sending its own truck to pick up the orders at a fixed cost of $20.00 per trip. Treating this cost as the
order cost, (a) what is the optimum number of reams to buy at one time? (b) How many times should lots of this size
be bought during this period? (c) What is the minimum cost (holding and setup) of maintaining inventory on this
item for the period? (d) Of this total cost, how much is carrying cost and how much is ordering cost?
Answer: This is an EOQ problem, even though the time period is not a year. All that is required is that the demand
value and the carrying cost share the same time reference. This will require approximately 3.5 orders per period.
Setup costs and carrying costs are each $70.71, and the annual total is $141.42.
(a) EOQ = = 283; (b) N = = 3.54
(d) Carrying cost = ($0.50) = $70.71; setup cost = ($20) = $70.71
(c) Total cost = $70.71 + $70.71 = $141.42
Practice exercise of production order quantity model – chapter 12
Holstein Computing manufactures an inexpensive audio card (Audio Max) for assembly into several models of its
microcomputers. The annual demand for this part is 100,000 units. The annual inventory carrying cost is $5 per unit
and the cost of preparing an order and making production setup for the order is $750. The company operates 250
days per year. The machine used to manufacture this part has a production rate of 2000 units per day.
(a) Calculate the optimum lot size.
(b) How many lots are produced in a year?
(c) What is the average inventory for Audio Max?
(d) What is the annual holding and setup cost for Audio Max?
Answer:
(a) Q*p = = = 6123.7 or 6124 units.
(b) There are approximately N = = = 16.33 cycles per year.
(c) The maximum inventory is Q = 6123.7 = 4899 units; average inventory is 4899 / 2 = 2449.5
units.
(d) Annual inventory management costs are 16.33($750) + 2449.5($5) = $12,247.50 + $12,247.50 = $24,495.
Practice exercise Quantity Discount – chapter 12
The annual demand for an item is 10,000 units. The cost to process an order is $75 and the annual inventory holding
cost is 20% of item cost.
(a) What is the optimal order quantity, given the following price breaks for purchasing the item?
(b) What price should the firm pay per unit?
(c) What is the total annual cost at the optimal behavior?
Quantity Price
1-9 $2.95 per unit
10 – 999 $2.50 per unit
1,000 – 4,999 $2.30 per unit
5,000 or more $1.85 per unit
Answer:
STEP 1: EOQ at $1.85 = 2013.47 → infeasible
EOQ at $2.30 = 1805.79 → feasible
So the possible best answers are 1806 and 5000
STEP 2: TC for 1806 units = $23,830.66
TC for 5000 units = $19,575.00
(a) Order 5000 units at a time.
(b) The price is $1.85 per unit.
(c) Total annual purchasing, holding, and setup costs = $19,575.00.
Practice exercise about MTBF – chapter 17
Ten high-intensity bulbs are tested for 100 hours each. One failed at 10 hours; all others completed the test.
Calculate FR(%), FR(N) and MTBF.
Answer: FR(%) = 1/10 or 10%; operating hours = 9 × 100 + 10 = 910; FR(N) = 1/910 or .0011, and
MTBF = 910 hours
Practice exercise about reliability – chapter 17
The academic service commonly referred to as "registration" consists of several smaller components: advising,
registration for courses, fee assessment, financial aid calculations, and fee payment. Each of these modules operates
independently and has some probability of failure for each student. If the five probabilities that accompany these
services are 95%, 90%, 99%, 98%, and 99%, respectively, what is the "reliability" of the entire product from the
student's perspective, i.e., the probability that all five will work according to plan?
Answer: Reliability is (.95)(.90)(.99)(.98)(.99) = .8212
Practice exercise about maintenance – chapter 17
Great Southern Consultants Group's computer system has been down several times over the past few months, as
shown below.
Number of breakdowns 01234
Monthly frequency 92441
Each time the system is down, the firm loses an average of $400 in time and service expenses. They are considering
signing a contract for preventive maintenance. With preventive maintenance, the system would be down on average
only 0.5 per month. The monthly cost of preventive maintenance would be $200 a month. Which is cheaper,
breakdown or preventive maintenance?
Answer:
Number of breakdowns 0 1 2 3 4 Total
Monthly probability 0.450.100.200.200.05 1.00
Expected # of breakdowns per month = (0)(.45) + (1)(.10) + (2)(.20) + (3)(.20) + (4)(.05) = 1.30 Expected cost of
breakdowns per month with breakdown maintenance = (1.30)($400) = $520
Preventive maintenance cost per month = (.5)($400) + $200 = $400
Preventive maintenance is more cost-effective.
**## guidancetohomework1.docx.pdf**
3.9 (a)
(b, c) There are four paths:
Path Time (hours)
A–C–E–G 19.5
B–D–F–G 24.9
A–C–D–F–G 28.7 (critical)
B–E–G 15.7
3.20 (a) Activity a m b te Variance
A 9 10 11 10 0.11
B 4 10 16 10 4
C 9 10 11 10 0.11
D 5 8 11 8 1
(b) Critical path is A–C with mean (te) completion time
of 20 weeks. The other path is B–D, with mean completion time of 18 weeks.
(c) Variance of A–C = (Variance of A) + (Variance of C)
= 0.11 + 0.11 = 0.22
Variance of B–D = (Variance of B) + (Variance of D)
= 4 + 1 = 5
(d) Probability A–C is finished in 22 weeks or less =
−⎛ ⎞
≤ = ≤ ≅⎜ ⎟
⎝ ⎠
2 2 2 0 ( 4 . 2 6 ) 1 . 0 0
0 . 2 2
P Z P Z
(e) Probability B–D is finished in 22 weeks or less =
−⎛ ⎞
≤ = ≤ =⎜ ⎟⎝ ⎠
2 2 1 8 ( 1 . 7 9 ) 0 . 9 6 3
5
P Z P Z
(f) The critical path has a relatively small variance and will almost certainly be finished in 22 weeks or less. Path
B–D has a relatively high variance. Due to this, the probability B–D is finished in 22 weeks or less is only about 0.96.
Since the project is not finished until all activities (and paths) are finished, the probability that the project will be
finished in 22 weeks or less is not 1.00 but is approximately 0.96.
3.32 (a)
Project completion time = 14 weeks
Task Time ES EF LS LF Slack
A 3 0 3 0 3 0
B 2 0 2 2 4 2
C 1 0 1 11 12 11
D 7 3 10 3 10 0
E 6 2 8 4 10 2
F 2 1 3 12 14 11
G 4 10 14 10 14 0
(b) To crash to 10 weeks, we follow 2 steps:
1. Crash D by 2 weeks ($150).
2. Crash D and E by 2 weeks each ($150 + $100).
Total crash cost = $400 additional
**28E6EE0E-86CE-495A-A42D-B7419999F357.jpeg**
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**## Practiceexercises2024.docx.pdf**
Practice exercise multifactor productivity – chapter 1
Gibson Products produces cast bronze valves for use in offshore oil platforms. Currently, Gibson
produces 1600 valves per day. The 20 workers at Gibson work from 7 a.m. until 4 p.m., with 30 minutes off for
lunch and a 15-minute break during the morning work session and another at the afternoon work session. Gibson is
in a competitive industry, and needs to increase productivity to stay competitive. They feel that a 20 percent increase
is needed.
Gibson's management believes that the 20 percent increase will not be possible without a change in working
conditions, so they change work hours. The new schedule calls on workers to work from 7:30 a.m. until 4:30 p.m.,
during which workers can take one hour off at any time of their choosing. Obviously, the number of paid hours is the
same as before, but production increases, perhaps because workers are given a bit more control over their workday.
After this change, valve production increased to 1800 units per day.
a. Calculate labor productivity for the initial situation
b. Calculate labor productivity for the hypothetical 20 percent increase
c. What is the productivity after the change in work rules?
d. Write a short paragraph analyzing these results.
Answer:
(a) Workers are active for eight hours per day; labor productivity is 10 valves/hour
(b) If Productivity rises by 20 percent, to 12 valves/hour; output would be 12 × 8 × 20 = 1920
(c) New productivity is 1800 / (20 × 8) = 11.25 valves/hour
(d) Gibson did not gain the desired 20 percent increase in productivity, but they did gain over 11 percent, without
extra equipment or energy, and without increasing the labor cost.
Practice exercise project management – chapter 3
A network consists of the activities in the following list. Times are given in weeks.
Activity PrecedingTime
A — 8
B — 3
C A 7
D A, B 3
E C 4
F D 6
a. Draw the network diagram.
b. Calculate the ES, EF, LS, LF, and Slack for each activity.
c. What is project completion time?
Answer: (a)
(b ,c)
Results
Task Early StartEarly FinishLate StartLate FinishSlack
A 0 8 0 8 0
B 0 3 7 10 7
C 8 15 8 15 0
D 8 11 10 13 2
E 15 19 15 19 0
F 11 17 13 19 2
Project 19
Practice exercise forecasting MAD chapter 4
The department manager using a combination of methods has forecast sales of toasters at a local department store.
Calculate the MAD for the manager's forecast. Compare the manager's forecast against a naive forecast covering the
same time period. Which is better?
Month Unit Sales
Manager's
Forecast
January 52
February 61
March 73
April 79
May 66
June 51
July 47 50
August 44 55
September 30 52
October 55 42
November 74 60
December 125 75
Answer:
Month ActualManager'sAbs. Error NaiveAbs. Error
January 52
February 61
March 73
April 79
May 66
June 51
July 47 50 3 51 4
August 44 55 11 47 3
September 30 52 22 44 14
October 55 42 13 30 25
November 74 60 14 55 19
December 125 75 50 74 51
The manager's forecast has a MAD of 18.83, while the naive is 19.33. Therefore, the manager's forecast is slightly
better than the naive.
Practice exercise exponential smoothing – chapter 4
Given an actual demand this period of 103, a forecast value for this period of 99, and an alpha of .4, what is the
exponential smoothing forecast for next period?
A) 94.6
B) 97.4
C) 100.6
D) 101.6
E) 103.0
Answer: C
Practice exercise seasonal factor forecasting chapter 4
The quarterly sales for specific educational software over the past three years are given in the following table.
Compute the four seasonal factors.
YEAR 1YEAR 2YEAR 3
Quarter 1 1710 1820 1830
Quarter 2 960 910 1090
Quarter 3 2720 2840 2900
Quarter 4 2430 2200 2590
Answer:
Avg. Sea. Fact.
Quarter 1 1786.670.8933
Quarter 2 986.67 0.4933
Quarter 3 2820.001.4100
Quarter 4 2406.671.2033
Grand Average 2000.00
Practice exercise – process management
17) Consider the assembly line below. The three fabrication operations run in parallel, such that each batch of 20
units only needs to go through one of the three fabrication operations. After that, each batch needs to go through
both assembly operations, which occur simultaneously (specifically, 10 components are made for each unit in the
fabrication stage—some components are then assembled in the Assembly 1 area while others are assembled in the
Assembly 2 area). The units are packaged and made ready for shipment in the final stage.
What is the bottleneck time per batch of this operation?
A) 24 min.
B) 8 min.
C) 88 min.
D) 34 min.
E) 3 min.
Answer: B
Practice exercise bullwhip effect – chapter 11
Consider a supply chain where a manufacturer sells to a distributor who sells to a wholesaler who sells to a retailer.
Last year, the retailer's weekly variance of demand was 4000 units. The weekly variance of orders was 5000; 8000;
12,000; and 17,000 units for the retailer, wholesaler, distributor, and manufacturer, respectively. (Note that the
variance of orders equals the variance of demand for that firm's supplier.)
(a) Calculate the bullwhip measure for the retailer.
(b) Calculate the bullwhip measure for the wholesaler.
(c) Calculate the bullwhip measure for the distributor.
(d) Calculate the bullwhip measure for the manufacturer.
(e) Which firm appears to be contributing the most to the bullwhip effect in this supply chain?
Answer:
(a) 5000 / 4000 = 1.25; (b) 8000 / 5000 = 1.6; (c) 12,000 / 8000 = 1.5; (d) 17,000 / 12,000 = 1.42
(e) With the highest bullwhip measure value of 1.6, the wholesaler appears to be contributing the most to the
bullwhip effect in this supply chain.
Practice exercise supply chain – chapter 11
The following data are pulled from a recent Walsh Manufacturing annual report.
Assets
Raw material inventory $120,000
Work-in-process inventory $50,000
Finished goods inventory $300,000
Property, plant & equipment $500,000
Other assets $200,000
Total assets $1,170,000
Condensed Income Statement
Revenue $2,000,000
Cost of goods sold $600,000
Other expenses $1,000,000
Net income $400,000
Calculate: (a) Percent invested in inventory, (b) Inventory turnover, and (c) Weeks of supply.
Answer:
(a) Percent invested in inventory = (120,000 + 50,000 + 300,000)/1,170,000 = 40.17%
(b) Inventory turnover = 600,000/(120,000 + 50,000 + 300,000) = 1.28
(c) Weeks of supply = (120,000 + 50,000 + 300,000)/(600,000/52) = 40.73
Practice exercise ABC analysis chapter 12
Perform an ABC analysis on the following set of products.
Item Annual DemandUnit Cost
A211 1200 $9
B390 100 $90
C003 4500 $6
D100 400 $150
E707 35 $2000
F660 250 $120
G473 1000 $90
H921 100 $75
Answer: The table below details the contribution of each of the eight products. Item G473 is clearly an A item, and
items A211, B390, and H921 are all C items. Other classifications are somewhat subjective, but one choice is to
label E707 and D100 as A items, and F660 and C003 as B items.
Item
Annual
DemandUnit Cost Volume
Cumulative
volume
Cumulative
percent
G473 1000 $90 $90,000 $90,000 29.6%
E707 35 $2,000 $70,000 $160,000 52.6%
D100 400 $150 $60,000 $220,000 72.3%
F660 250 $120 $30,000 $250,000 82.2%
C003 4500 $6 $27,000 $277,000 91.0%
A211 1200 $9 $10,800 $287,800 94.6%
B390 100 $90 $9,000 $296,800 97.5%
H921 100 $75 $7,500 $304,300 100.0%
$304,300
Practice exercise safety stock – chapter 12
Thomas' Bike Shop stocks a high volume item that has a normally distributed demand during lead time. The average
daily demand is 70 units, the lead time is 4 days, and the standard deviation of demand during lead time is 15.
(a) How much safety stock provides a 95% service level to Thomas?
(b) What should the reorder point be?
Answer:
(a) SS = 1.65 × 15 = 24.75 units or 25 units
(b) ROP = (70)(4) + 25 = 305 units.
Practice exercise production order quantity model – chapter 12
Montegut Manufacturing produces a product for which the annual demand is 10,000 units. Production averages 100
units per day, while demand is 40 units per day. Holding costs are $2.00 per unit per year, and setup cost is $200.00.
(a) If the firm wishes to produce this product in economic batches, what size batch should be used? (b) What is the
maximum inventory level? (c) How many order cycles are there per year? (d) What are the total annual holding and
setup costs?
Answer: This is a production order quantity problem.
(a) Q*p = = = 1825.7 or 1826 units
(b) The maximum inventory level is Q = 1825.7 = 1095.45 or 1095 units
(c) There are approximately N = = = 5.48 cycles per year
(d) Total annual costs = (5.48)($200) + (1095.45/2)$2 = $2,190.89 or $2,191
Practice exercise of EOQ model – chapter 12
A printing company estimates that it will require 1,000 reams of a certain type of paper in a given period. The cost
of carrying one unit in inventory for that period is 50 cents. The company buys the paper from a wholesaler in the
same town, sending its own truck to pick up the orders at a fixed cost of $20.00 per trip. Treating this cost as the
order cost, (a) what is the optimum number of reams to buy at one time? (b) How many times should lots of this size
be bought during this period? (c) What is the minimum cost (holding and setup) of maintaining inventory on this
item for the period? (d) Of this total cost, how much is carrying cost and how much is ordering cost?
Answer: This is an EOQ problem, even though the time period is not a year. All that is required is that the demand
value and the carrying cost share the same time reference. This will require approximately 3.5 orders per period.
Setup costs and carrying costs are each $70.71, and the annual total is $141.42.
(a) EOQ = = 283; (b) N = = 3.54
(d) Carrying cost = ($0.50) = $70.71; setup cost = ($20) = $70.71
(c) Total cost = $70.71 + $70.71 = $141.42
Practice exercise of production order quantity model – chapter 12
Holstein Computing manufactures an inexpensive audio card (Audio Max) for assembly into several models of its
microcomputers. The annual demand for this part is 100,000 units. The annual inventory carrying cost is $5 per unit
and the cost of preparing an order and making production setup for the order is $750. The company operates 250
days per year. The machine used to manufacture this part has a production rate of 2000 units per day.
(a) Calculate the optimum lot size.
(b) How many lots are produced in a year?
(c) What is the average inventory for Audio Max?
(d) What is the annual holding and setup cost for Audio Max?
Answer:
(a) Q*p = = = 6123.7 or 6124 units.
(b) There are approximately N = = = 16.33 cycles per year.
(c) The maximum inventory is Q = 6123.7 = 4899 units; average inventory is 4899 / 2 = 2449.5
units.
(d) Annual inventory management costs are 16.33($750) + 2449.5($5) = $12,247.50 + $12,247.50 = $24,495.
Practice exercise Quantity Discount – chapter 12
The annual demand for an item is 10,000 units. The cost to process an order is $75 and the annual inventory holding
cost is 20% of item cost.
(a) What is the optimal order quantity, given the following price breaks for purchasing the item?
(b) What price should the firm pay per unit?
(c) What is the total annual cost at the optimal behavior?
Quantity Price
1-9 $2.95 per unit
10 – 999 $2.50 per unit
1,000 – 4,999 $2.30 per unit
5,000 or more $1.85 per unit
Answer:
STEP 1: EOQ at $1.85 = 2013.47 → infeasible
EOQ at $2.30 = 1805.79 → feasible
So the possible best answers are 1806 and 5000
STEP 2: TC for 1806 units = $23,830.66
TC for 5000 units = $19,575.00
(a) Order 5000 units at a time.
(b) The price is $1.85 per unit.
(c) Total annual purchasing, holding, and setup costs = $19,575.00.
Practice exercise about MTBF – chapter 17
Ten high-intensity bulbs are tested for 100 hours each. One failed at 10 hours; all others completed the test.
Calculate FR(%), FR(N) and MTBF.
Answer: FR(%) = 1/10 or 10%; operating hours = 9 × 100 + 10 = 910; FR(N) = 1/910 or .0011, and
MTBF = 910 hours
Practice exercise about reliability – chapter 17
The academic service commonly referred to as "registration" consists of several smaller components: advising,
registration for courses, fee assessment, financial aid calculations, and fee payment. Each of these modules operates
independently and has some probability of failure for each student. If the five probabilities that accompany these
services are 95%, 90%, 99%, 98%, and 99%, respectively, what is the "reliability" of the entire product from the
student's perspective, i.e., the probability that all five will work according to plan?
Answer: Reliability is (.95)(.90)(.99)(.98)(.99) = .8212
Practice exercise about maintenance – chapter 17
Great Southern Consultants Group's computer system has been down several times over the past few months, as
shown below.
Number of breakdowns 01234
Monthly frequency 92441
Each time the system is down, the firm loses an average of $400 in time and service expenses. They are considering
signing a contract for preventive maintenance. With preventive maintenance, the system would be down on average
only 0.5 per month. The monthly cost of preventive maintenance would be $200 a month. Which is cheaper,
breakdown or preventive maintenance?
Answer:
Number of breakdowns 0 1 2 3 4 Total
Monthly probability 0.450.100.200.200.05 1.00
Expected # of breakdowns per month = (0)(.45) + (1)(.10) + (2)(.20) + (3)(.20) + (4)(.05) = 1.30 Expected cost of
breakdowns per month with breakdown maintenance = (1.30)($400) = $520
Preventive maintenance cost per month = (.5)($400) + $200 = $400
Preventive maintenance is more cost-effective.
**## guidancetohomework1.docx.pdf**
3.9 (a)
(b, c) There are four paths:
Path Time (hours)
A–C–E–G 19.5
B–D–F–G 24.9
A–C–D–F–G 28.7 (critical)
B–E–G 15.7
3.20 (a) Activity a m b te Variance
A 9 10 11 10 0.11
B 4 10 16 10 4
C 9 10 11 10 0.11
D 5 8 11 8 1
(b) Critical path is A–C with mean (te) completion time
of 20 weeks. The other path is B–D, with mean completion time of 18 weeks.
(c) Variance of A–C = (Variance of A) + (Variance of C)
= 0.11 + 0.11 = 0.22
Variance of B–D = (Variance of B) + (Variance of D)
= 4 + 1 = 5
(d) Probability A–C is finished in 22 weeks or less =
−⎛ ⎞
≤ = ≤ ≅⎜ ⎟
⎝ ⎠
2 2 2 0 ( 4 . 2 6 ) 1 . 0 0
0 . 2 2
P Z P Z
(e) Probability B–D is finished in 22 weeks or less =
−⎛ ⎞
≤ = ≤ =⎜ ⎟⎝ ⎠
2 2 1 8 ( 1 . 7 9 ) 0 . 9 6 3
5
P Z P Z
(f) The critical path has a relatively small variance and will almost certainly be finished in 22 weeks or less. Path
B–D has a relatively high variance. Due to this, the probability B–D is finished in 22 weeks or less is only about 0.96.
Since the project is not finished until all activities (and paths) are finished, the probability that the project will be
finished in 22 weeks or less is not 1.00 but is approximately 0.96.
3.32 (a)
Project completion time = 14 weeks
Task Time ES EF LS LF Slack
A 3 0 3 0 3 0
B 2 0 2 2 4 2
C 1 0 1 11 12 11
D 7 3 10 3 10 0
E 6 2 8 4 10 2
F 2 1 3 12 14 11
G 4 10 14 10 14 0
(b) To crash to 10 weeks, we follow 2 steps:
1. Crash D by 2 weeks ($150).
2. Crash D and E by 2 weeks each ($150 + $100).
Total crash cost = $400 additional
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