Exam 1 Study Guide
PROBLEMS
1.
Mabel's Ceramics spent $3000 on a new kiln last year, in the belief that it would cut energy usage 25%
over the old kiln. This kiln is an oven that turns "greenware" into finished pottery. Mabel is concerned
that the new kiln requires extra labor hours for its operation. Mabel wants to check the energy savings of
the new oven, and also to look over other measures of their productivity to see if the change really was
beneficial. Mabel has the following data to work with:
Production (finished units)
Labor (hrs)
Capital ($)
Energy (kWh)
The year before
4000
350
15000
3000
Year just ended
4100
375
18000
2600
Also, suppose that the average labor cost is $12 per hour, cost of capital is 20%, and cost of energy is
$0.40 per kwh.
a. Were the modifications beneficial? (Compute labor, energy, and capital productivity for the two
years and compare.)
b. Compute percentage change in multi-factor productivity of the year just ended from that of year
before.
c. If the multifactor productivity for next year must be restored to what it was the year before, assuming
the same output next year as the year just ended, by how the input must be reduced from what it is
this year?
2.
An Appliance Service company made house calls and repaired 10 lawn-mowers, 2 refrigerators, and 3
washers in an 8-hour day with his standard crew of 3 workers. The retail price for each respective service
is $50, $200, and $120. The average wage for the workers is $12 per hour. The materials cost for a day
was $200 while the overhead cost was $50.
a. What is the company’s labor productivity?
b. What is the multifactor productivity?
c. How much of a reduction in input is necessary for a 5% increase in multifactor productivity?
3.
Consider the tasks, durations, and predecessor relationships in the following network. Draw the AON
network and answer the questions that follow.
Activity Description
A
B
C
D
E
F
G
H
I
J
Immediate
Predecessor(s)
--A
A
B
D, C
C
F
F
E, G
I
Optimistic
(Weeks)
4
2
8
1
6
2
2
6
4
1
Most Likely
(Weeks)
7
8
12
2
8
3
2
8
8
2
Pessimistic
(Weeks)
10
20
16
3
22
4
2
10
12
3
a.
b.
c.
d.
4.
Schedule the activities of this project and determine (i) the expected project completion time, (ii)
the earliest and latest start and finish times, and the slack for all the activities, and (iii) all the
critical paths.
What is the probability of completion of the project before week 42?
What is the probability of completion of the project before week 35?
With 99% confidence what is your estimate for the project completion time.
Consider the following project. All activity times are in weeks.
Activity
A
B
C
D
E
F
G
H
I
a.
b.
c.
d.
e.
Immediate
Predecessor(s)
A
A, B
B
C
D, E
E
F, G
Normal
Time
7
8
9
8
9
10
5
10
5
Crash
time
4
5
7
8
8
8
5
8
4
Normal
cost
20000
50000
80000
30000
10000
90000
25000
32000
28000
Crash
cost
38000
74000
110000
30000
12000
124000
25000
40000
35000
Draw an AON network.
Identify all the unique paths from start to finish and determine the critical path, normal
project completion time, and normal project cost.
Compute MAC, Cost of crashing/week.
Which activity would you crash first and by how many weeks?
Determine the project time and cost after crashing the activity selected in (d).
5.
Consider the following CPM Solver model.
a) Determine the successor activities in cells I2 to I10.
b) Determine the Excel formulas for the following cells: F2, G2, C15, C18, D18, D21, C25, G19, G16,
G15, H15, B27, B28, and B29.
c) What is the Solver Target cell for minimizing the project completion time?
d) What is the Solver changing cell range?
e) What are the Solver constraints?
6.
What is the forecast for May based on a 3-period MA and a weighted 3-period moving average
applied to the following past demand data? Let the weights be, 3, 3, and 4 (last weight is for
most recent data). Compute MAD, MSE and MAPE for both cases and compare.
Nov.
37
7.
Dec.
36
Jan.
40
Feb.
42
Mar.
47
April
43
Sales of music stands at the local music store over the past ten days are shown in the table below.
Forecast demand using exponential smoothing with an  of .6 (initial forecast = 16).
a) Compute the forecast for period 6 and the MAD.
b) Compute the tracking signal for periods 1 to 5. What do you recommend for this forecasting process?
t
Demand
1
13
2
21
3
28
4
37
5
25
8.
Weekly sales of copy paper at Cubicle Suppliers are in the table below. Forecast week 8 with a trend
projection model.
Week
Sales (cases)
1
17
2
22
3
27
4
32
5
35
6
37
7
41
9.
The quarterly sales for specific educational software over the past three years are given in the following
table. Compute the four seasonal indices and find forecast for Year 4 if the annual demand for year 4 is
estimated to be 10% more than that of year 3.
Quarter 1
Quarter 2
Quarter 3
Quarter 4
10.
YEAR 1
1690
940
2625
2500
YEAR 2
1800
900
2900
2360
YEAR 3
1850
1100
2930
2615
Arnold Tofu owns and operates a chain of 12 vegetable protein "hamburger" restaurants in northern
Louisiana. Sales figures and profits for the stores are in the table below. Sales are given in millions of
dollars; profits are in hundred thousand dollars.
Store
1
2
3
4
5
6
7
8
9
10
11
12
a.
b.
Sales
7
2
6
4
14
15
16
12
14
20
15
7
Profits
15
10
13
15
25
27
24
20
27
44
34
17
Calculate a regression line for the data. What is your forecast of profit for a store with sales of
$24 million? $30 million?
Calculate the standard error of the estimate.
c.
11.
Determine the value of the coefficient of correlation between sales and profit and the value of
the coefficient of determination.
A restaurant manager tracks complaints from the diner satisfaction cards that are turned in at each table.
The data collected from the past week’s diners appear in the following table.
Complaint
Food taste
Food temperature
Order mistake
Slow service
Table/utensils dirty
Too expensive
Frequency
80
9
2
16
47
4
Prepare a Pareto chart. To cover 80% of problems which complaints must be address first?
12.
A list of issues that led to incorrect formulations in Tuncey Bayrak’s jam manufacturing unit in
New England is provided below:
Incorrect measurement
Variability in scale accuracy
Antiquated scales
Equipment in disrepair
Lack of clear instructions
Technician calculation off
Damaged raw material
Jars mislabeled
Operator misreads display
Temperature controls off
Inadequate cleanup
Incorrect weights
Incorrect maintenance
Priority miscommunication
Inadequate flow controls
Inadequate instructions
Create a fish-bone diagram and categorize these issues using the “four Ms” method.
13.
Cartons of Plaster of Paris are supposed to weigh exactly 32 oz. Inspectors want to develop process
control charts. They take ten samples of six boxes and weigh them. Based on the following data, compute
the lower and upper control limits and determine whether the process is in control.
Sample
1
2
3
4
5
14.
Mean
33.8
34.4
34.5
34.1
34.2
Range
1
0.3
0.5
0.7
0.2
Sample
6
7
8
9
10
Mean
34.3
33.9
34.0
33.8
34.0
Range
0.4
0.5
0.8
0.3
0.3
McDaniel Shipyards wants to develop control charts to assess the quality of its steel plate. They take ten
sheets of 1" steel plate and compute the number of cosmetic flaws on each roll. Each sheet is 20' by 100'.
Based on the following data, develop limits for the control chart and determine whether the process is in
control.
Sheet
Number of flaws
Sheet
Number of flaws
1
6
6
2
2
1
7
1
3
4
5
3
2
1
8
9
10
0
0
2
15.
Rancho No Tengo Orchards wants to establish control limits for its mangos before they are sent to the
retailers. They randomly take six containers (assume it is enough) of one hundred mangos in an attribute
testing plan and find some mangos with blemishes. What should be the limits on the control chart? Is the
process in control?
Container
Number of mangos with blemishes
1
5
2
3
3
1
4
3
5
4
6
2
16.
A woodworker is concerned about the quality of the finished appearance of her work. In sampling units
of a split-willow hand-woven basket, she has found the following number of finish defects in ten units
sampled: 4, 0, 3, 0, 1, 0, 1, 1, 0, 2.
a. Calculate the average number of defects per basket
b. If 3-sigma control limits are used, calculate the lower control limit, centerline, and upper control
limit.
17.
The specifications for a plastic liner for concrete highway projects call for a thickness of 6.0 mm ± 0.1
mm. The standard deviation of the process is estimated to be 0.02 mm. What are the upper and lower
specification limits for this product? The process is known to operate at a mean thickness of 6.04 mm.
Determine the values of Cpk and Cp for this process. Is the process capable? Explain.
18.
A medical supplies company buys its supplies in bulk and redistributes them to doctor’s offices and
clinics. The receive thermometers in lots of 500 from the vendor. They are considering a sampling plan
of n = 50 and c = 1.
a. Develop a OC curve for this sampling plan. (Use Poisson Tables)
b. Determine the producer’s risk if the AQL is 2%.
c. Determine the consumer’s risk if the LTPD is 14%.
d. Develop a curve for AOQ and determine the value of AOQL.
19.
An acceptance sampling plan has lots of 5000 units, a sample size of 200 and c is 5. Suppose that the
incoming lots have percentage defective of 3%. What is AOQ?
Answers:
1.
The energy modifications did not generate the expected savings; labor and capital productivity
decreased.
Given data
Production
Labor
Capital =
Energy =
Last year
4000
350
15000
3000
Now
4100
375
18000
2600
Labor productivity (Units/hr) =
11.4286
10.9333
Capital productivity (units/$) =
Cost of capital =
Energy productivity (Units/KWH) =
0.2667
3000
1.3333
0.2278
3600
1.5769
4200
15000
1200
8400
0.4762
4500
18000
1040
9140
0.4486
0.4762
8610
530
Labor cost = Hours x $12 =
Capital $ =
Energy $ = $0.40 x Energy =
Total input $ =
Multifactor productivity (Units/$) =
Target productivity =
Target input = 4100/0.4762 =
Reduction in input needed = 9140 – 8610 =
#2
Number serviced
Dollar value/unit
Production in $
Labor hours = 3 workers x 8 hrs. =
Labor productivity = 1260/24 =
Multifactor productivity
Labor cost = 3x8x$12 =
Material =
Overhead =
Total input cost =
Productivity = 1260/538 =
5% improvement in MF productivity =
Target productivity after 5% improvement =
Input for improved productivity =
Reduction in input needed =
LM
10
50
500
24
52.50
$
$
$
$
288
200
50
538
2.3420
0.1171
2.4591
512.38
25.62
Change
-0.4952
Change %
-4.33%
-0.0389
-14.58%
0.2436
18.27%
-0.0276
-5.8%
R
W
2
3
200
120
400
360
per day
per hour of labor
1260
<-- Total $
= 288 + 200 + 50
per $ input
<-- Output/Productivity = 1260/2.4591
<-- 538 – 512.38
3. (a)
B
D
E
Start
A
I
G
C
F
H
J
Fin
ish
Task
A
B
C
D
E
F
G
H
I
J
Task
Start
A
B
C
D
E
F
G
H
I
J
Finish
a
4
2
8
1
6
2
2
6
4
1
m
7
8
12
2
8
3
2
8
8
2
t
ES
7
9
12
2
10
3
2
8
8
2
0
7
7
16
19
19
22
22
29
37
39
B
10
20
16
3
22
4
2
10
12
3
EF
0
7
16
19
18
29
22
24
30
37
39
t
7
9
12
2
10
3
2
8
8
2
LS
0
8
7
17
19
24
27
31
29
37
39
Variance
1.0000
64/36
256/36
64/36
4/36
LF
0
7
17
19
19
29
27
29
39
37
39
Slack
0
1
0
1
0
5
5
9
0
0
Critical
Critical
Critical
Critical
Critical
Critical path = A-C-E-I-J, Project completion time TE = 39
Variance for project completion time = 2p = 1 + 388/36 = 11.7778; p =3.4319
b. For P(T <=42), Z = (42 – 39)/3.4319 = 0.87, Table area = 0.80785, Probability = 0.80785
c. For P(T <=35), Z = (35 – 39)/3.4319 = -1.17, Table area = .879; Probability = 1 - .879 = 0.121
d. Z for 99% confidence = 2.325, T = 39 + 2.325(3.4319) = 46.98
4.
C
A
F
I
Finish
Start
G
B
D
H
E
Normal
Time
Crash
time
Normal
cost
Crash
cost
MCA
Crashing
cost/week
A
7
4
20000
38000
3
6000
B
8
5
50000
74000
3
8000
C
9
7
80000
110000
2
15000
D
8
8
30000
30000
0
E
9
8
10000
12000
1
2000
F
10
8
90000
124000
2
17000
G
5
5
25000
25000
0
H
10
8
32000
40000
2
4000
I
5
4
28000
35000
1
7000
Sum =
365,000
Activity
Paths
A-C-F-I
A-D-G-I
B-D-G-I
B-E-G-I
B-E-H
Path time
31
25
26
27
27
Predecessor(s)
A
A, B
B
C
D, E
E
F, G
Critical path
Normal project time = 31 weeks
Normal project cost = 365,000
Activity to crash = A – among the critical activities (A, C, F, I) the crashing cost/week for A is the smallest.
Weeks to crash = Minimum{MTR of A, Project time – time of second longest path}
i.e. = Minimum{3, 31-27} = 3
Project time after crashing A 3 weeks = 31 – 3 = 28 weeks
Project cost after crashing A = 365,000 + 3 x 6,000 = 383,000
5.
(a)
Activity
Successors(s)
A
C, D
B
D, E
C
F
D
G
E
G, H
F
I
G
I
H
Finish
I
Finish
(b)
F2
G2
C15
D18
E18
D21
C25
G19
F19
G16
G15
H15
B27
B2-C2
(E2-D2)/F2
B2-B15
Max(E15,E16)
D18+C18
Max(E18,E19)
Max(E22,E23)
Min(F21,F22)
G19-C19
Min(F18,F19)
Min(F17,F18)
F15-D15 or G15-E15
Sum(D2:D10)
B28
B29
Sumproduct(BG15:B23,G2:G10)
B27+B28
(c)
(d)
(e)
Solver Target cell for minimizing the project completion time = C25
Changing cell range = B15:B23
What are the Solver constraints?
B15:B23 <= F2:F10
B15:B23 = Integer (Optional)
6.
Month
Demand
(At)
Nov.
Dec.
Jan.
Feb.
Mar.
April
37
36
40
42
47
43
3-MA
Forecast
|Et|
|Et|/At
Et2
Weight
Weighted
3-MA
|Et|
|Et|/At
3
3
4
37.67
4.33
0.1031
18.7489
37.90
4.1
0.0976
39.33
7.67
0.1632
58.8289
39.60
7.4
0.1574
43.00
0
0.0000
0
43.40
0.4
0.0093
Forecast =
MAD =
4
44.00
MAPE =
8.88%
MSE
=25.8593
Forecast
=
MAD
= 3.97
43.90
7.
Period
1
2
3
4
5
Demand
13
21
28
37
25
Ft
Et
|Et|
16.00
-3.00
3.00
14.20
6.80
6.80
18.28
9.72
9.72
24.11
12.89
12.89
31.84
-6.84
6.84
F6 =
27.74
MAD =
7.85
Period
1
2
3
4
5
Demand
13
21
28
37
25
Ft
Et
16.00
-3.00
3.00
14.20
6.80
6.80
3.80
18.28
9.72
9.72
13.52
24.11
12.89
12.89
26.41
31.84
-6.84
6.84
19.57
Week
1
2
3
4
5
6
7
Sales
17
22
27
32
35
37
41
Tracking signal
|Et|
CFEt
CAEt
MADt
TS
-3.00
3.00
3
-1
9.80
4.9
0.78
19.52
6.51
2.08
32.41
8.1
3.26
39.25
7.85
2.49
8.
28
b
211
 XY  n X Y
 X  nX
2
a  Y  bX
2
b
XY
X2
17
1
n=
7
X2 =
140
44
4
X =
28
954
81
9
Y =
211
XY = 
b=
128
16
=
4.0000
175
25
=
30.14
222
36
287
954
49
140
a=
954  7(4)(30.14)
 3.9286
140  7(4) 2
a  30.14  3.9286(4)  14.4286
Regression equation: Ŷ = 14.4286 + 3.9286t
F8 = 14.4286 + 3.9286(8) =
45.8571
9.
Quarter
Year 1
Demand
Year 2
Year 3
Average
Index
3.9286
14.4286
MAPE =
8.81%
1690
940
2625
2500
1
2
3
4
1800
900
2900
2360
1850
1100
2930
2615
Overall average =
1780.00
0.8823
980.00
0.4857
2818.33
1.3969
2491.67
2017.50
1.2350
Year 3 sum =
8495
Annual demand for year 4 = 1.1 x 8495 = 9345
Demand/season = 2336
Forecast for year 4
Quarter
1
2
3
4
Average demand
2336
2336
2336
2336
Seasonal Index
0.8823
0.4857
1.3969
1.2350
Forecast
2061
1135
3263
2885
10.
Store
1
2
3
4
5
6
7
8
9
10
11
12
Sum =
X
24
30
(b)√
(c)
Sales (X)
7
2
6
4
14
15
16
12
14
20
15
7
132
Profits (Y)
15
10
13
15
25
27
24
20
27
44
34
17
271
Y
43.2923
52.8503
Estimated
profit
$ 4,329,230
$ 5,285,030
7159−5.0601(271)−1.593(3529)
12−2
Y2
225
100
169
225
625
729
576
400
729
1936
1156
289
7159
= 4.0738
12(3529)−(132)(271)
√[12(1796)−1322 ][12(7159− 2712 ]
r2 = 0.8403
X2
49
4
36
16
196
225
256
144
196
400
225
49
1796
XY
105
20
78
60
350
405
384
240
378
880
510
119
3529
=
6576
7173.8258
=0.9167
n=
X =
Y =
=
=
12
1796
X2 =
132
XY =  3529
271
7159
Y2 =
11
b = 1.5930
22.5833
a = 5.0601
Ft = 5.0601 + 1.593 X
11.
Complaint
Food taste
Table/utensils dirty
Slow service
Food temperature
Too expensive
Order mistake
Frequency
80
47
16
9
4
2
158
%
Cum %
50.6%
29.7%
10.1%
5.7%
2.5%
1.3%
100.0%
50.6%
80.4%
90.5%
96.2%
98.7%
100.0%
Frequency
Pareto Chart: Complaints
90
80
70
60
50
40
30
20
10
0
To cover 80% of complaints, Food Taste and dirty utensils must be addressed first.
12.
100.0%
90.0%
80.0%
70.0%
60.0%
50.0%
40.0%
30.0%
20.0%
10.0%
0.0%
13.
Sample
1
2
3
4
5
6
7
8
9
10
R
1.0
0.3
0.5
0.7
0.2
0.4
0.5
0.8
0.3
0.3
33.8
34.4
34.5
34.1
34.2
34.3
33.9
34.0
33.8
34.0
𝑋̿ = 34.1
n=6
0.483
A2 =
A2
=
0.2415
LCL = 𝑋̿ - A2 =
UCL = 𝑋̿ + A2 =
33.86
34.34
D2 =
0
D3 =
2.004
LCLR =
0
UCLR =
1.002
= 0.5
The process is not in control, since the 𝑥̅ values for samples 1, 2, 3, and 9 fall outside the control
limits. Although all the sample ranges fall within 0 and 1.002, the assignable causes should be
investigated and eliminated.
14.
Use c-chart
𝑐̅ = total defects/number of sheets = 1.8
UCLc = 1.8 + 3 1.8 = 1.8 + 4.02 = 5.825
LCLc = 1.8 - 3 1.8 = 1.8-4.02 = converts to zero
Sheet number 1
has too many flaws; investigate the cause.
0.03 (1−0.03)
15. LCLp = 0. 03 −
3√
100
0.03 (1−0.03)
LCLp = 0. 03 + 3√
100
= 0.03 - (3 * 0.017) = -0.02; can’t be zero, so, round to 0
= 0.03 + (3 * 0.017) = 0.081
Limits are LCL = 0 and UCL = 0.081. All six points are in control; there is no pattern or trend in
the data.
= 1.2; (b) LCLc = 1.2 – 3 √1.2 = -2.0862, or zero
UCLc = 1.2 + 3 √1.2 = 4.49.
16.
(a)
17.
LSL = 5.9 mm, USL = 6.1 mm.
Cpk = min{(6.1-6.04)/(3*0.02), (6.04 - 5.9)/(3*0.02) = min{1.00, 2.33} = 1.
Cp = (6.1 – 5.9)/(6*.02) = 1.67
Since Cpk is < 1.333 the process is not capable. Since Cp = 1.67, the process variability is small enough to be
within the desired specification range. Therefore, the process needs to be centered to achieve a Cpk of at least
1.33.
18.
Pd
0.00
nPd
0.00
Pa
1.000
0.01
0.02
0.03
0.04
0.05
0.50
1.00
1.50
2.00
2.50
0.910
0.736
0.558
0.406
0.288
0.06
0.07
0.08
0.09
0.10
3.00
3.50
4.00
4.50
5.00
0.199
0.137
0.092
0.061
0.040
For AQL = 2%, Pa= 0.736
i.e., Producer’s risk = 1 – 0.736 = 0.264
For LTPD = 14%, nPd = 50 x 0.14 = 7.0,
Pa from Poisson table = 0.007
i.e. Consumer’s risk = 0.007
OC-Curve for the sampling plan, n=50, C = 1
1.0
0.9
P(Accepting lot)
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.00
0.02
0.04
0.06
0.08
0.10
0.12
Pd (% of defectives in the lot)
Pd
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
Pa
1.000
0.910
0.736
0.558
0.406
0.288
0.199
0.137
0.092
0.061
0.040
AOQL = 0.01507
AOQ
0.00000
0.00819
0.01325
0.01507
0.01462
0.01294
0.01075
0.00860
0.00662
0.00494
0.00360
Percent defective accepted
(d)
AOQ
0.02000
0.01500
0.01000
0.00500
0.00000
0.00
0.05
0.10
Pd
0.15
19.
N=
n=
c=
Pd =
5000
200
5
3%
nPd =
Pa =
AOQ =
AOQ =
6
0.446
<-- from Poisson table
.03(.446)(5000-200)/5000
0.0128448 i.e. = 1.28%