Practice Problems: Chapter 1, Operations and Productivity
Problem 1:
Mance Fraily, the Production Manager at Ralts Mills, can currently expect his operation to
produce 1000 square yards of fabric for each ton of raw cotton. Each ton of raw cotton requires 5
labor hours to process. He believes that he can buy a better quality raw cotton, which will enable
him to produce 1200 square yards per ton of raw cotton with the same labor hours.
What will be the impact on productivity (measured in square yards per labor-hour) if he
purchases the higher quality raw cotton?
Problem 2:
C. A. Ratchet, the local auto mechanic, finds that it usually takes him 2 hours to diagnose and fix
a typical problem. What is his daily productivity (assume an 8 hour day)?
Mr. Ratchet believes he can purchase a small computer trouble-shooting device, which will
allow him to find and fix a problem in the incredible (at least to his customers!) time of 1 hour.
He will, however, have to spend an extra hour each morning adjusting the computerized
diagnostic device. What will be the impact on his productivity if he purchases the device?
Problem 3:
Joanna French is currently working a total of 12 hours per day to produce 240 dolls. She thinks
that by changing the paint used for the facial features and fingernails that she can increase her
rate to 360 dolls per day. Total material cost for each doll is approximately $3.50; she has to
invest $20 in the necessary supplies (expendables) per day; energy costs are assumed to be only
$4.00 per day; and she thinks she should be making $10 per hour for her time. Viewing this from
a total (multifactor) productivity perspective, what is her productivity at present and with the
new paint?
Problem 4:
How would total (multifactor) productivity change if using the new paint raised Ms. French’s
material costs by $0.50 per doll?
Problem 5:
If she uses the new paint, by what amount could Ms. French’s material costs increase without
reducing total (multifactor) productivity?
1
ANSWERS:
Problem 1:
Current labor productivity =
New labor productivity =
1000 sq yds
= 200 sq yds per hour
1 ton*5 hours
1200 sq yds
= 240 sq yds per hour
1 ton * 5 hours
Productivity improves 20% = ( 240 - 200 ) / 200 = .2
Problem 2:
Current productivity =
8 hours per day
= 4 problems per day
2 hours per problem
Productivity with computer =
7 hours per day
= 7 problems per day
1 hour per problem
⎛ 7−4 3
⎞
Productivity improves 75% ⎜
= = .75 ⎟
4
⎝ 4
⎠
Problem 3:
Currently
Using the new paint
Labor
12 hrs * $10 = $120
12 hrs * $10 = $ 120
Material
240 * $3.50 = $840
360 * $3.50 = $1260
Supplies
= $ 20
= $ 20
Energy
=$ 4
=$
Total Inputs
= $984
= $1404
Productivity
240/984 = 0.24
360/1404 = .26
2
4
Problem 4:
If the material costs increase by $0.50 per doll:
Using the new paint
Labor
12 hrs * $10 = $ 120
Material
360 * $4.00 = $1440
Supplies
=$
20
Energy
=$
4
Total Inputs
= $1584
Productivity
360/1584 = 0.23
Problem 5:
From the answer to Problem 3 we know the following:
Currently
Using the new paint
Labor
12 hrs * $10 = $120
12 hrs * $10 = $ 120
Material
240 * $3.50 = $840
360 * $3.50 = $1260
Supplies
= $ 20
= $ 20
Energy
=$ 4
=$
Total Inputs
= $984
= $1404
Productivity
240/984 = 0.24
4
360/1404 = .26
We want to know how high the material cost could go, using the new paint, before the
productivity drops to the current level of 0.24. In mathematical terms we make the material cost
a variable (X), set the new multifactor productivity value to the current level, 0.24, and solve for
X.
3
360/(($12x10) + 360 $(X) + $20 + $4) = 0.24
360 = 0.24($120 + 360$(X) + $20 + $4)
360 = $28.8 + 86.4$(X) + $4.8 + $.96
325.44 = 86.4$(X)
$(X)= 325.44/86.4 = $3.7666 ≅ $3.77
It follows then that the new paint could raise Materials cost by no more than approximately
$0.27 (the difference between $3.77 and $3.50) before Ms. French would experience a decrease
in multifactor productivity.
4
Practice Problems: Chapter 2, Operations Strategy in a Global
Environment
Problem 1:
Identify how changes in the external environment may affect the OM strategy for a
company. For example, what impact are the following factors likely to have on OM
strategy?
a. The occurrence of a major storm or hurricane.
b. Terrorist attacks of 9/11/01.
c. The much discussed decrease in the quality of American primary and secondary
school systems.
d. Trade Legislation such as WTO and NAFTA and changes in tariffs and quotas.
e. The rapid rate at which the cost of health insurance is increasing.
f. The Internet.
Problem 2:
Identify how the changes in the internal environment affect the OM strategy for a
company. For example, what impact are the following factors likely to have on OM
strategy?
a. The increased use of Local and Wide Area Networks (LANs and WANs)
b. An increased emphasis on service
c. The increased role of women in the workplace
d. The seemingly increasing rate at which both internal and external environments
change.
Problem 3:
Operations managers are called upon to support the organization's strategy. OM does this
with some combination of one of three strategies. What are these three strategies?
1
ANSWERS:
Problem 1:
a. A major storm or hurricane may have considerable impact on a company’s facilities
and scheduling. Flooding and wind damage can make a facility unusable or
significantly reduce its capacity. Stocks of raw materials, especially agricultural
products, might be damaged or in short supply. The long-term availability of some
materials might be significantly reduced. There may be a shortage of important
services during the recovery. For example, the demand for roofers and builders is high
after a major storm and they would like to be able to rapidly increase their capacity to
handle the higher demand.
b. Terrorist activity has forced organizations to rethink, and in many cases expand, their
security systems. Firms have also had to reevaluate their supply networks and consider
increasing their inventory safety stock. They may also reassess the risks of foreign
locations and expansion.
c. A decrease in the skill levels of Americans entering the labor market requires that
organizations place more emphasis on training, turn to automation to obviate the need
for human labor, and hire from outside the United States.
d. WTO and NAFTA changed the rules for trading, opened new markets, and in some
instances, changed the role of labor versus capital (where labor is especially low cost,
emphasis often shifts from the use of capital to the use of labor).
e. The increasing cost of health insurance adds significantly to the cost of labor. Some
large US organizations are passing on this increased cost to the employees or reducing
other parts of the benefit package in response to these pressures.
f. The Internet has promoted globalization of markets, and eliminated barriers of
geography and time.
Problem 2:
a. The increased use of LANs and WANs has, among other things, enabled new
organizational structures, the movement of the locus of responsibility further down the
organizational hierarchy (elimination of middle management), and the increasing
practicality of JIT operations, mass customization, etc..
b. The increased emphasis on service has, among other things, fostered an increased
information or information technology content of many products. Firms are also
increasing training because so much of the service economy is dependent upon
individual competence.
c. The increased role of women in the workplace is requiring an increased emphasis on
the creation and communication of appropriate human resource policies. It may also
be fostering the creation of flexible work schedules and, to a lesser degree,
telecommuting.
2
d. Some companies seem to be adopting the perspective that their main problem is now
the “management of change” as opposed to the management of a specific process or
product. If nothing else, the management of change is becoming a formal part of the
manager’s responsibility.
Problem 3:
OM managers support the firm's strategy by achieving a competitive advantage through
some combination of differentiation, low-cost leadership, and response.
3
Practice Problem: Chapter 3, Project Management
Problem 1:
The following represent activities in a major construction project. Draw the network to
represent this project.
Activity
Immediate Predecessor
A
-
B
-
C
A
D
B
E
B
F
C, E
G
D
H
F, G
1
Problem 2:
Given the following Time Chart and Network Diagram, find the Critical Path.
Activity
a
m
b
t
Variance
A
2
3
4
3
1/9
B
1
2
3
2
1/9
C
4
5
12
6
16/9
D
1
3
5
3
4/9
E
1
2
3
2
1/9
Problem 3:
What is the variance in completion time for the critical path found in Problem 2?
Problem 4:
A project has an expected completion time of 40 weeks and a standard deviation of 5
weeks. It is assumed that the project completion time is normally distributed.
(a) What is the probability of finishing the project in 50 weeks or less?
(b) What is the probability of finishing the project in 38 weeks or less?
(c) The due date for the project is set so that there is a 90% chance that the project will
be finished by this date. What is the date?
2
Problem 5:
Development of a new deluxe version of a particular software product is being
considered. The activities necessary for the completion of this project are listed in the
table below along with their costs and completion times in weeks.
Activity
Normal
Time
Crash
Time
Normal
Cost
Crash
Cost
Immediate
Predecessor
A
4
3
2,000
2,600
-
B
2
1
2,200
2,800
A
C
3
3
500
500
A
D
8
4
2,300
2,600
A
E
6
3
900
1,200
B, D
F
3
2
3,000
4,200
C, E
G
4
2
1,400
2,000
F
(a) What is the project expected completion date?
(b) What is the total cost required for completing this project on normal time?
(c) If you wish to reduce the time required to complete this project by 1 week, which
activity should be crashed, and how much will this increase the total cost?
3
ANSWERS:
Problem 1:
Problem 2:
Critical path: ACDE = 14
Problem 3:
Total variance = ∑ variances of activities on critical path
Total variance = 1/9 + 16/9 + 4/9 + 1/9 = 2 2/9 = 2.44
And σ =
2.44 = 1.6
4
Problem 4:
(a) Z =
X −μ
σ
=
50 − 40
=2
5
Therefore: P (X ≤ 50) = P(Z ≤ 2) = 0.97725
(b) Z =
X−μ
σ
=
−2
= −0.4
5
Therefore: P (X ≤ 38) = P(Z ≤ −0.4) = 0.34458
(c) 90% ≥ Z = 1.28 = (χ - μ ) / σ = χ − 40 / 5
Therefore: χ = 1.28*5 + 40 = 46.4weeks
Problem 5:
(a)
Project completion time is therefore t A + t D + t E + t F + t G = 4 + 8 + 6 + 3 + 4 = 25
(b) Total cost = $2, 000 + $2200 + $500 + $2,300 + $900 + $3, 000 + $1, 400 = $12,300
(c) Crash D 1 week at an additional cost of
$2, 600 − $2,300 $300
=
= $75
8−4
4
5
Practice Problems: Chapter 6, Managing Quality
Problem 1:
The accounts receivable department has documented the following defects over a 30-day period:
Category
Frequency
Invoice amount does not agree with the check amount
108
Invoice not on record (not found)
24
No formal invoice issued
18
Check (payment) not received on time
30
Check not signed
8
Invoice number and invoice referenced do not agree
12
What techniques would you use and what conclusions can you draw about defects in the accounts
receivable department?
Problem 2:
Prepare a flow chart for purchasing a Big Mac at the drive-through window at McDonalds.
Problem 3:
Draw a fishbone chart detailing reasons why a part might not be correctly machined.
1
ANSWERS:
Problem 1:
Category
Frequency
Percent
Invoice amount does not agree with the check amount
108
54
Invoice not on record (not found)
24
12
No formal invoice issued
18
9
Check (payment) not received on time
30
15
Check not signed
8
4
Invoice number and invoice referenced do not agree
12
6
200
100
Σ=
Use a Pareto chart to organize the defects and conclude that the obvious problem (about half the
defects) is the failure of the check to agree with the company’s records as to the correct amount.
Other problems are late payments and an apparent invoice-filing problem in the office. Notice that
27% of these common errors appear to be the result of procedural problems within accounts
receivable (invoice not on record, no invoice issued, and invoice numbering problems). This value
could be considerably higher depending on how much of the problem of disagreement between
invoice and check amounts is the result of accounts receivable process problems.
2
3
Problem 2:
Distance
Symbol
Activity
--
Pull up to speaker
--
Press button
--
Wait for response
--
Verbalize order
--
Get confirmation of order and cost
20
Move car up in line
--
Wait
20
Move car up in line
--
Wait
--
Verify order and cost
--
Pay and receive order
--
Leave
--
Realize they forgot the extra catsup!
4
Problem 3
5
Practice Problems: Chapter 7, Process Strategy
Problem 1:
Jackson Custom Machine Shop has a contract for 130,000 units of a new product. Sam Jumper, the
owner, has calculated the cost for three process alternatives. Fixed costs will be: for generalpurpose equipment (GPE), $150,000; flexible manufacturing (FMS), $350,000; and dedicated
automation (DA), $950,000. Variable costs will be: GPE, $10; FMS, $8; and DA, $6. Which should
he choose?
Problem 2:
Solve Problem 1 graphically
Problem 3:
Using either your analytical solution found in Problem 1, or the graphical solution found in
Problem 2, identify the volume ranges where each process should be used.
Problem 4:
If Jackson Custom Machine is able to convince the customer to renew the contract for another one
or two years, what implications does this have for his decision?
ANSWERS:
Problem 1:
Solve for the crossover between GPE and FMS:
10X + 150000 = 8X + 350000
or
2X = 200000
x = 100,000 units
Solve for the crossover between FMS and DA:
8X + 350000 = 6X + 950000
or
2X = 600000
X = 300000
Therefore, at a volume of 130,000 units, FMS is the appropriate strategy.
1
Problem 2 & 3:
Below 100,000 units use GPE, between 100,000 and 300,000 use FMS, above 300,000 use DA
Problem 4:
If Jackson Custom Machine is able to get the customer to extend the contract for another two years,
the owner would certainly wish to take advantage of the savings using Dedicated Automation.
2
Practice Problems: Chapter 8, Location Strategies
Problem 1:
A major drug store chain wishes to build a new warehouse to serve the whole Midwest. At the
moment, it is looking at three possible locations. The factors, weights, and ratings being considered
are given below:
Ratings
Factor
Weights
Peoria
Des Moines
Chicago
Nearness to markets
20
4
7
5
Labor cost
5
8
8
4
Taxes
15
8
9
7
Nearness to suppliers
10
10
6
10
Which city should they choose?
Problem 2:
Balfour’s is considering building a plant in one of three possible locations. They have estimated the
following parameters for each location:
Location
Fixed Cost
Variable Cost
Waco, Texas
$300,000
$5.75
Tijuana, Mexico
$800,000
$2.75
Fayetteville, Arkansas
$100,000
$8.00
For what unit sales volume should they choose each location?
1
Problem 3:
Our main distribution center in Phoenix, AZ is due to be replaced with a much larger, more modern
facility that can handle the tremendous needs that have developed with the city’s growth. Fresh
produce travels to the seven store locations several times a day making site selection critical for
efficient distribution. Using the data in the following table, determine the map coordinates for the
proposed new distribution center.
Store Locations
Map Coordinates (x,y)
Truck Round Trips per Day
Mesa
(10,5)
3
Glendale
(3,8)
3
Camelback
(4,7)
2
Scottsdale
(15,10)
6
Apache Junction
(13,3)
5
Sun City
(1,12)
3
Pima
(5,5)
10
2
Problem 4:
A company is planning on expanding and building a new plant in one of three countries in Middle
or Eastern Europe. The general manager, Patricia Donegal, has decided to base her decision on six
critical success factors: technology availability and support, availability and quality of public
education, legal and regulatory aspects, social and cultural aspects, economic factors, and political
stability.
Using a rating system of 1 (least desirable) to 5 (most desirable) she has arrived at the following
ratings (you may, of course, have different opinions). In which country should the plant be built?
Critical Success Factor
Turkey
Serbia
Slovakia
Technology availability and support
4
3
4
Availability and quality of public education
4
4
3
Legal and regulatory aspects
2
4
5
Social and cultural aspects
5
3
4
Economic factors
4
3
3
Political stability
4
2
3
Problem 5:
Assume that Patricia decides to use the following weights for the critical success factors:
Technology availability and support
0.3
Availability and quality of public education
0.2
Legal and regulatory aspects
0.1
Social and cultural aspects
0.1
Economic factors
0.1
Political stability
0.2
Would this change her decision?
3
Problem 6:
Patricia’s advisors have suggested that Turkey and Slovakia might be better differentiated by either
(a) doubling the number of critical success factors, or (b) breaking down each of the existing
critical success factors into smaller, more narrowly defined items, e.g., Availability and quality of
public education might be broken into primary, secondary, and post-secondary education. How
would you advise Ms. Donegal?
4
ANSWERS:
Problem 1:
Ratings
Weighted Ratings
Weights
Peoria
Des
Moines
Chicago
Peoria
Des
Moines
Chicago
Nearness to
markets
20
4
7
5
80
140
100
Labor cost
5
8
8
4
40
40
20
Taxes
15
8
9
7
120
135
105
Nearness to
suppliers
10
10
6
10
100
60
100
Sum of Weighted ratings:
340
375
325
Factor
Therefore, it appears that based upon the weights and rating, Des Moines should be chosen.
5
Problem 2:
Transition between Waco and Tijuana:
300, 000 + (5.75 x) = 800, 000 + (2.75 x)
3 x = 500, 000
x = 166, 000
Transition between Waco and Fayetteville:
300, 000 + (5.75 x) = 100, 000 + (8.00 x)
200, 000 = 2.25 x
88,888 = x
6
Problem 3:
New Distribution Center should be located at:
Cx =
(10*3) + (3*3) + (4* 2) + (15*6) + (13*5) + (1*3) + (5*10) 255
=
= 7.97
3 + 3 + 2 + 6 + 5 + 3 + 10
32
Cy =
(5*3) + (8*3) + (7 * 2) + (10*6) + (3*5) + (12*3) + (5*10) 214
=
= 6.69
3 + 3 + 2 + 6 + 5 + 3 + 10
32
7
Problem 4:
Critical Success Factor
Turkey
Serbia
Slovakia
Technology availability and support
4
3
4
Availability and quality of public education
4
4
3
Legal and regulatory aspects
2
4
5
Social and cultural aspects
5
3
4
Economic factors
4
3
3
Political stability
4
2
3
23
19
22
Σ =
Based upon her ratings of the critical success factors, Patricia should choose Turkey. From a
practical perspective, given the small difference between the scores for Turkey and Slovakia, and
the subjectivity of the ratings themselves, Patricia would be better advised to develop additional
critical success factors, more carefully weigh the individual factors; or, in general, to acquire more
information before making her decisions.
8
Problem 5:
Critical Success Factor
Wgt
Turkey
Serbia
Technology availability and support
0.3
4
1.2
3
0.9
4
1.2
Availability and quality of public
education
0.2
4
0.8
4
0.8
3
0.6
Legal and regulatory aspects
0.1
2
0.2
4
0.4
5
.5
Social and cultural aspects
0.1
5
0.5
3
0.3
4
0.4
Economic factors
0.1
4
0.4
3
0.3
3
0.3
Political stability
0.2
4
0.8
2
0.4
3
0.6
Σ = 3.9
Slovakia
3.1
3.6
No, in this case, use of the weighting factors does not change the recommendation. One might
again suggest that additional information be considered in making the decision.
Problem 6:
(a) Doubling the number of critical success factors. There are two issues here. First, from a
practical perspective there are a limited number of truly “critical” success factors – and these
should be the ones presently being considered. Any additional factors should be of secondary
or tertiary importance. Second, given the subjective nature of the rating process, adding
additional factors would also increase the overall margin of error of the final ratings to a degree
that may eliminate any gain in differentiation arising from the use of the additional factors. The
use of a maximum of seven to nine critical success factors is usually appropriate.
(b) Given that one’s ability to estimate or rate an aggregate is usually better than one’s ability to
estimate or rate the individual components of the aggregate, this approach is unlikely to
provide much help.
9
Practice Problems: Chapter 9, Layout Strategy
Problem 1:
As in most kitchens, the baking ovens in Lori’s Kitchen in New Orleans are located in one area
near the cooking burners. The refrigerators are located next to each other as are the dishwashing
facilities. A work area of tabletops is set aside for cutting, mixing, dough rolling, and assembling of
final servings, although different table areas may be reserved for each of these functions.
Given the following Interdepartmental Activity Matrix, develop an appropriate layout for Lori’s
Kitchen.
Interdepartmental Activity Matrix
Cooking burners (A)
Cooking
Burners (A)
Refrigerators (B)
Dishwashing (C)
Work Area (D)
-
7
193
12
-
4
82
-
222
Refrigerator (B)
Dishwashing (C)
Work Area (D)
-
The present layout is:
A
B
C
with a distance of 10 feet between adjacent areas.
1
D
Computing the Load * Distance measure:
Load * Distance
A to B
7 * 10
A to C
193*20
A to D
12*30
B to C
4*10
B to D
82*20
C to D
222*10
Total
70
3860
360
40
1640
2220
8190
Develop a preferred layout. What is the sum of the loads * distance of your new layout?
2
Problem 2:
A firm must produce 40 units/day during an 8-hour workday. Tasks, times, and predecessor
activities are given below.
Task
Time (Minutes)
Predecessor(s)
A
2
-
B
2
A
C
8
-
D
6
C
E
3
B
F
10
D, E
G
4
F
H
3
G
Total
38 minutes
Determine the cycle time and the appropriate number of workstations to produce the 40 units per
day.
3
ANSWERS
Problem 1:
From the Activity Matrix, C and D should be next to each other and A should be next to C. The
other relationships are minor by comparison. One possible solution is:
B
A
C
D
with a distance of 10 feet between adjacent areas.
Computing the Load * Distance measure:
Load * Distance
A to B
7 * 10
70
A to C
193*10
1930
A to D
12*20
240
B to C
4*20
80
B to D
82*30
2460
C to D
222*10
2220
Total
7000
Further improvement is possible. Try analyzing the following layouts.
A
C
B
D
A
C
D
B
4
Problem 2:
Cycle time =
Production time available 8 hrs *60 minutes/hr 480
=
=
= 12 minutes/cycle
Units required
40 units
40
Minimum number of workstations =
=
∑t
i
Cycle time
=
Work time required
Cycle time
38 minutes
= 3.17 station
12 minutes/cycle
3.17 workstations must be rounded up to 4 as 3 workstations would not be able to produce the
required output.
One layout – not necessarily optimal
5
Practice Problems: Chapter 11, Supply-Chain Management
Problem 1:
Determine the sales necessary to equal a dollar of savings on purchases for a company
that has a net profit of 6% and spends 70% of its revenues on purchases.
Problem 2:
Determine the sales necessary to equal a dollar of savings on purchases for a company
that has a net profit of 8% and spends 40% of its revenues on purchases.
Problem 3
Phil Carter, President of Carter Computer Components, Corp. has the option of shipping
computer transformers from its Singapore plant via container ship or airfreight. The
typical shipment has a value of $75,000. A container ship takes 24 days and costs $5,000;
airfreight takes 1 day and costs $8,000. Holding cost is estimated to be 40% in either
case. How should shipments be made?
Problem 4
Carol King is evaluating the inventory performance of Johnston Systems. A recent annual
report (all figures in millions) indicates assets of $16,000, inventory of $1,000, and cost
of goods sold of $24,000. What is the inventory turnover and what percent of assets are
tied up in inventory?
1
ANSWERS
Problem 1:
From Table 11.3, we see that this company would have to increase sales by
approximately $5.56
Problem 2:
From Table 11.3, we see that this company would have to increase sales by
approximately $2.94
Problem 3:
Cost via container ship:
[24 *
(.40 * 75,000)
] + 5,000 = (24 * 82.19) + 5,000 = 1,972.56 + 5,000 = $6,972.56
365
Cost via airfreight:
[1*
(.40 * 75,000)
] + 8,000 = (1* 82.19) + 8,000 = 82.19 + 8,000 = $8,082.19
365
Therefore, use the container ship as it has a lower total cost.
Problem 4
Cost of good sold / inventory investment = 24,000 / 1,000
= 24 (inventory turnover)
Total inventory investment/ Total assets = 1,000 / 16,000
= .0625 = 6.25% (percent of assets in inventory)
2
Practice Problems: Chapter 12, Inventory Management
Problem 1:
ABC Analysis
Stock Number
Annual
$
Volume
Percent of Annual $
Volume
J24
12,500
46.2
R26
9,000
33.3
L02
3,200
11.8
M12
1,550
5.8
P33
620
2.3
T72
65
0.2
S67
53
0.2
Q47
32
0.1
V20
30
0.1
Σ = 100.0
What are the appropriate ABC groups of inventory items?
Problem 2:
A firm has 1,000 “A” items (which it counts every week, i.e., 5 days), 4,000 “B” items (counted
every 40 days), and 8,000 “C” items (counted every 100 days). How many items should be counted
per day?
Problem 3:
Assume you have a product with the following parameters:
Demand = 360
Holding cost per year = $1.00 per unit
Order cos t: = $100 per order
What is the EOQ?
Problem 4:
Given the data from Problem 3, and assuming a 300-day work year; how many orders should be
processed per year? What is the expected time between orders?
Problem 5:
What is the total cost for the inventory policy used in Problem 3?
Problem 6:
Assume that the demand was actually higher than estimated (i.e., 500 units instead of 360 units).
What will be the actual annual total cost?
Problem 7:
If demand for an item is 3 units per day, and delivery lead-time is 15 days, what should we use for
a re-order point?
Problem 8:
Assume that our firm produces type C fire extinguishers. We make 30,000 of these fire
extinguishers per year. Each extinguisher requires one handle (assume a 300 day work year for
daily usage rate purposes). Assume an annual carrying cost of $1.50 per handle; production setup
cost of $150, and a daily production rate of 300. What is the optimal production order quantity?
Problem 9:
We need 1,000 electric drills per year. The ordering cost for these is $100 per order and the
carrying cost is assumed to be 40% of the per unit cost. In orders of less than 120, drills cost $78;
for orders of 120 or more, the cost drops to $50 per unit.
Should we take advantage of the quantity discount?
Problem 10:
Litely Corp sells 1,350 of its special decorator light switch per year, and places orders for 300 of
these switches at a time. Assuming no safety stocks, Litely estimates a 50% chance of no shortages
in each cycle, and the probability of shortages of 5, 10, and 15 units as 0.2, 0.15, and 0.15
respectively. The carrying cost per unit per year is calculated as $5 and the stockout cost is
estimated at $6 ($3 lost profit per switch and another $3 lost in goodwill, or future sales loss). What
level of safety stock should Litely use for this product? (Consider safety stock of 0, 5, 10, and 15
units)
Problem 11:
Presume that Litely carries a modern white kitchen ceiling lamp that is quite popular. The
anticipated demand during lead time can be approximated by a normal curve having a mean of 180
units and a standard deviation of 40 units. What safety stock should Litely carry to achieve a 95%
service level?
ANSWERS
Problem 1:
ABC Groups
Class
Items
Annual
Volume
Percent of $
Volume
A
J24, R26
21,500
79.5
B
L02, M12
4,750
17.6
C
P33, T72, S67, Q47, V20
800
2.9
Σ = 100.0
Problem 2:
Item
Class
Quantity
Policy
Number of Items to
Count Per Day
A
1,000
Every 5 days
1000/5 = 200/day
B
4,000
Every
days
40 4000/40=100/day
C
8,000
Every
days
100 8000/100=80/day
Total items to count: 380/day
Problem 3:
EOQ =
2 * Demand * Order cost
2 * 360 * 100
=
= 72000 = 268 items
1
Holding cost
Problem 4:
N=
Demand 360
=
= 134
. orders per year
268
Q
T=
Working days
= 300 / 134
. = 224 days between orders
Expected number of orders
Problem 5:
Demand * Order Cost (Quantity of Items) * ( Holding Cost ) 360 * 100 268 * 1
+
=
+
= 134 + 134 = $268
Q
2
268
2
TC =
Problem 6:
TC =
Demand * Order Cost (Quantity of Items) * ( Holding Cost ) 500 * 100 268 * 1
+
=
+
= 186.57 + 134 = $320.57
268
2
Q
2
Note that while demand was underestimated by nearly 50%, annual cost increases by only 20%
(320 / 268 = 1.20) an illustration of the degree to which the EOQ model is relatively insensitive to
small errors in estimation of demand.
Problem 7:
ROP = Demand during lead - time = 3 * 15 = 45 units
Problem 8:
Q*p =
2 * Demand * Order Cost
=
⎛
Daily Usage Rate ⎞
Holding Cost ⎜ 1⎟
Daily Production Rate ⎠
⎝
(2)(30,000)(150)
= 3000 units
⎛ 100 ⎞
1.50⎜1 −
⎟
⎝
300 ⎠
Problem 9:
Q*p ($78) =
(2)(1000)(100)
= 80 units
(0.4)(78)
Q*p ($50) =
(2)(1000)(100)
= 100 units = 120 to take advantage of quantity discount.
(0.4)(50)
Ordering 100 units at $50 per unit is not possible; the discount does not apply until 120 the order
equals 120 units. Therefore, we need to compare the total costs for the two alternatives.
Total cos t = Demand * Cost +
Demand * Order Cost (Quantity of Items) * ( Holding cos t )
+
Q
2
Total cos t ($78) = (1000)(78) +
(1000)(100) (80)(0.4)(78)
+
= $80,498
80
2
Total cos t ($50) = (1000)(50) +
(1000)(100) (120)(0.4)(50)
+
= $52,033
120
2
Therefore, we should order 120 each time at a unit cost of $50 and a total cost of $52,033.
Problem 10:
Safety stock = 0 units:
Carrying cost equals zero.
Stockout cos ts = (Stockout cos t * possible units of shortage * probability of shortage * number of orders per year )
S0 = 6 * 5* 0.2 *
1350
1350
1350
+ 6 * 10 * 015
. *
+ 6 * 15* 015
. *
= $128.25
300
300
300
Safety stock = 5 units:
Carrying cos t = $5 per unit * 5 units = $25
Stockout cost: S5 = 6* 5* 015
. *
1350
1350
+ 6* 10* 015
. *
= $60.75
300
300
Total cos t = carrying cos t + stockout cos t = $25 + $60.75 = $85.75
Safety stock = 10 units:
Carrying cos t = 10 * 5 = $50.00
. *
Stockout cost: S10 = 6* 5* 015
1350
= $20.25
300
Total cos t = carrying cos t + plus stockout cos t = $50.00 + $20.25 = $70.25
Safety stock = 15:
Carrying cos t = 15* 5 = $75.00
Stockout cos ts = 0 (there is no shortage if 15 units are maintained)
Total cos t = carrying cos t + stockout cos t = $75.00 + $0 = $75.00
Therefore: Minimum cost comes from carrying a 10-unit safety stock.
Problem 11:
To find the safety stock for a 95% service level it is necessary to calculate the 95th percentile on
the normal curve. Using the standard Normal table from the text, we find the Z value for 0.95 is
1.65 standard units. The safety stock is then given by:
(165
. * 40) + 180 = 66 + 180 = 246 Ceiling Lamps
Practice Problem: Chapter 13, Aggregate Planning
Problem 1:
Set the following problem up in transportation format and solve for the minimum cost
plan.
Period
Feb
Mar
Apr
55
70
75
Regular
50
50
50
Overtime
5
5
5
Subcontract
12
12
10
Beginning Inventory
10
Demand
Capacity
Costs
Regular time
$60 per unit
Overtime
$80 per unit
Subcontract
$90 per unit
Inventory carrying cost
$1 per unit per month
Back order cost
$3 per unit per month
1
ANSWERS
Problem 1:
2
Practice Problems: Chapter 14, Material Requirements Planning
(MRP) and ERP
Problem 1:
The Hunicut and Hallock Corporation makes two versions of the same basic file cabinet,
the TOL (Top-of-the-line) five drawer file cabinet and the HQ (High-quality) five drawer
filing cabinet.
The TOL and HQ use the same cabinet frame and locking mechanism. The drawer
assemblies are different although both use the same drawer frame assembly. The drawer
assemblies for the TOL cabinet use a sliding assembly that requires four bearings per side
whereas the HQ sliding assembly requires only two bearings per side. (These bearings are
identical for both cabinet types.) 100 TOL and 300 HQ file cabinets need to be assembled
in week #10. No current stock exists.
Develop a material structure tree for the TOL and the HQ file cabinets.
Problem 2:
Develop a gross material requirements plan for the TOL and HQ cabinets in the previous
example.
Problem 3:
Develop a net material requirements plan for the TOL and HQ file cabinets in the
previous problems assuming a current on-hand finished goods inventory of 100 TOL
cabinets. The lead times are given below.
Painting and final assembly of both HQ and TOL requires 2 weeks.
Both cabinet frames and lock assembly require 1 week for manufacturing.
Both drawer assemblies require 2 weeks for assembly.
Both sliding assemblies require 2 weeks for manufacturing.
Bearings require 2 week to arrive from the supplier.
1
Problem 4:
If the TOL file cabinet has a gross material requirements plan as shown below, no
inventory, and 2 weeks lead time is required for assembly, what are the order release
dates and lot sizes when lot sizing is determined using lot-for-lot? Use a holding cost of
$2.00 and a setup cost of $20.00, and assume no initial inventory.
Gross Material Requirements Plan
Week
1
2
3
TOL
4
50
5
6
100
7
8
9
50
10
100
Problem 5:
If the TOL file cabinet has a gross material requirements plan as shown below, no
inventory, and 2 weeks of lead time is required for assembly, what are the order release
dates and lot sizes when lot sizing is determined by EOQ (Economic Order Quantity)?
Use a holding cost of $2.00 and a setup cost of $20.00, and assume no initial inventory.
Gross Material Requirements Plan
Week
TOL
1
2
3
50
4
5
100
2
6
7
50
8
9
10
100
Problem 6:
If the TOL file cabinet has a gross materials requirements plan as shown below, no
inventory, and 2 weeks of lead time is required for assembly, what are the order release
dates and lot size when lot sizing is determined using PPB (part period balancing)? Use a
holding cost of $2.00 and a setup cost of $20,000, and no initial inventory.
Gross Material Requirements Plan
Week
TOL
1
2
3
50
4
5
100
3
6
7
50
8
9
10
100
ANSWERS
Problem 1:
Problem 2:
Gross Requirements Plan
Week
1
2
3
4
5
6
7
8
9
10
TOL
100
HQ
300
4
Problem 3:
Week
1
2
3
4
5
6
7
8
Required date
9
10
Lead
Time
100
2 weeks
300
2 weeks
TOL
Order release date
Required date
HQ
Cabinet
frame and
lock
HQ drawer
assembly
Drawer
frame
assembly
HQ sliding
assembly
Order release date
300
Required date
300
1 week
1500
2 weeks
Order release date
300
Required date
Order release date
1500
Required date
1500
2 weeks
1500
2 weeks
Order release date
1500
Required date
Order release date
1500
Required date
6000
Bearings
Order release date
6000
Receipts:
300 cabinet frames and locks in week 8
1500 HQ drawer assemblies in week 8
1500 drawer frame assemblies in week 6
1500 HQ sliding assemblies in week 6
6000 bearings in week 4
5
2 weeks
Problem 4:
Gross Material Requirements Plan
Week
1
2
TOL
Release dates
and lot sizes
50
3
4
5
50
100
100
50
6
7
8
9
50
10
100
100
Holding cost = $0
Setup cost = 4 * $20 = $80
Total cost = $80
Problem 5:
Solution using POM for Windows:
Gross Material Requirements Plan
Week
1
TOL
Release dates
and lot sizes
72
2
3
4
5
50
100
96
48
Holding cost = $280
Setup cost = 4 * $20 = $80
Total cost = $360
6
6
7
8
50
9
10
100
96
Problem 6:
Solution using POM for Windows:
Gross Material Requirements Plan
Week
1
TOL
Release dates
and lot sizes
50
2
3
4
5
50
100
100
50
Holding cost = $0
Setup cost = 4 * $20 = $80
Total cost = $80
7
6
7
8
50
9
10
100
100
Practice Problems: Chapter 17, Maintenance and Reliability
Problem 1:
California Instruments, Inc., produces 3,000 computer chips per day. Three hundred are tested for a
period of 500 operating hours each. During the test, six failed: two after 50 hours, two at 100 hours,
one at 300 hours, and one at 400 hours.
Find FR(%) and FR(N).
Problem 2:
If 300 of these chips are used in building a mainframe computer, how many failures of the
computer can be expected per month?
Problem 3:
Find the reliability of this system:
1
Problem 4:
Given the probabilities below, calculate the expected breakdown cost.
Number of Breakdowns
Daily Frequency
0
3
1
2
2
2
3
3
Assume a cost of $10 per breakdown.
2
ANSWERS
Problem 1:
FR(%) = failures per number tested = 6/300 = 0.02 = 2%
FR(N) = failures per operating time:
Total time = 300 * 500 = 150,000 hours
Downtime = 2(450) + 2(400) + 1(200) + 1(100) = 2,000 hours
Operating time = Total time – Downtime = 150,000 – 2,000 = 148,000
Therefore: FR(N) = 6/148,000 = 0.0000405 failures/hour
MTBF = 1/FR(N) = 24,691 hours
Problem 2:
Converting the units of FR(N) to months:
FR(N) = 0.0000405 * 24 hours/day * 30 days/month = 0.029 failures/month
FR(N) for the 300 units:
FR(N) = 0.029 failures/month * 300 units = 8.75 failures/month
MTBF for the mainframe:
MTBF = 1/FR(N) = 1/8.75 = 0.11 month = 0.11 * 30 = 3.4 days
Calculation for MTBF assumes that failure of any one chip brings down entire system.
Problem 3:
R = [0.95 + 0.92(1 − 0.95)] * [0.98] * [0.90 + 0.90(1 − 0.90)]
= 0.996 * 0.98 * 0.99 = 96.6%
3
Problem 4:
Number of Breakdowns
Daily Frequency
Probability
0
3
0.3
1
2
0.2
2
2
0.2
3
3
0.3
Expected number of breakdowns = (0)(0.3) + (1)(0.2) + (2)(0.2) + (3)(0.3)
= 0 + 0.2 + 0.4 + 0.9
= 1.5 breakdowns/day
Expected breakdown cost = Expected number of breakdowns * Cost per breakdown
= 1.5 * $10
= $15/day
4