ADM 3301 A, B
Assignment #5 - Solution
ASSIGNMENT # 5 - Solutions
Queuing models, Quality Control and SPC
Total points: 20 Marks
Problem # 1 (4 marks)
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The parts department of a large automobile dealership has a counter used exclusively for their
own service mechanics requesting parts. The length of time between requests can be modelled by
an Exponential distribution that has a mean of five minutes. A parts clerk can handle requests at
an average rate of 15 per hour, and this can be modelled by a Poisson distribution. Suppose there
are two parts clerks at the counter.
a) On average, how many mechanics would be at the counter, including those being served?
(1 mark)
b) If a mechanic has to wait, how long would the average wait be? (1 mark)
c) What percentage of time is a clerk idle? (1 mark)
d) If clerks represent a cost of $20 per hour and mechanics represent a cost of $30 per hour,
how many clerks would be optimal in terms of minimizing total cost? (1 mark)
[Multiple servers, Exponential service durations]
Avg. inter-arrival duration = 5 min. à λ = 12 / hr.
µ = 15 / hr.
M = 2 clerks
a.
λ
µ
Th
Ls
12
= .80. , Lq = 0.152
15
λ
= Lq +
µ
= .152 + .80 = .952 mechanics
=
b.
Wa =
Lq
λ
=
0.152
12
=.0126 hr.
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ADM 3301 A, B
c.
Assignment #5 - Solution
12
λ
=
=.40
Mµ
2(15)
ρ=
% idle time = 1 - ρ = 1 - .40 = .60
d.
λ
= Ls
µ
3.2 + .8 = 4
.152 + .8 = .952
.019 + .8 = .819
Clerks Costs
M($20)
$20
40
60
M
1
2
3
Mechanics Costs
$Ls ($30 / hr)
$120
28.56
24.57
Lq +
TC
$140
68.56*
84.57
è 2 clerks.
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Problem # 2 (4 marks)
Essex Corp. of St. Louis is a small manufacturer of parts for Boeing's Aircraft and Missile
Systems. In the late 1990s, Essex was trying to establish a SPC program. Part number 528003-N
was typical. The following samples of two were observed for a critical dimension of the part (in
inches). The lower specification and upper specification values were .4160, .4210 inches,
respectively.
Sample
1
0.419
0.419
2
3
4
5
0.419
0.4185
6
7
8
9
10
0.418
0.418
0.418
0.4179
0.4175
0.418
0.4175
0.4173
0.4183
0.4185
0.4175
0.4179
0.4175
0.4176
0.4184
0.4185
Th
a) Is the process capable? Calculate both Cp and Cpk . (Hint: you may use excel to find
required parameters such as standard deviation of the sample) (2 marks)
USL = .421, LSL = .416
σ = 0.00054 (from Excel)
Process Avg = .4181 (from Excel)
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ADM 3301 A, B
Assignment #5 - Solution
(1 mark)
Cp
= (USL – LSL) / 6 σ
= (.421 – .416) / 6(.00054)
= 1.54
(1 mark)
Cpk
= min { (USL – Avg) / 3 σ , (Avg - LSL) / 3 σ )
= min { (.421 – .4181) / 3(.00054) , (.4181 – .416) / 3(.00054) }
= min { 1.79, 1.30}
= 1.30
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Both Cp and Cpk are greater than 1.00. Therefore, the process is capable.
b) Construct three sigma sample mean and sample range control charts. Is the process in
control? (Hint: notice that control limit formulas are different depending on availability
of standard deviation of the process (population)) (2 marks)
For n = 2, from Shewhart Table, A2 = 1.88, D3 = 0, D4 = 3.27.
=
−
(1 mark) Mean control limits: X ± A2R = .4181 ± 1.88(.0001) = .4181 ± .0002
UCL = .4183, LCL = .4179
−
(1 mark) Range control limits: UCL = D4R = 3.27(.0001) = .0003
−
LCL = D3R = 0(.0001) = 0
The means of samples 1, 2, 5, 7-10 and range of sample 2 fall outside their respective
control limits. Therefore, the process is NOT in control.
Th
Problem # 3 (7.5 marks)
Using four samples of 200 credit card statements each, an auditor found the following number of
erroneous statements in each sample:
Number of errors
Sample
1
4
2
2
3
5
4
9
a) Determine the proportion of defectives in each sample. (0.5 mark)
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ADM 3301 A, B
Assignment #5 - Solution
a.
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b) If the true proportion of defectives for this process is unknown, what is your best estimate
of it? (1 mark)
c) What is your estimate of the mean and standard deviation of the sampling distribution of
the sample proportion of defectives for samples of 200? (1 mark)
d) What control limits would give an alpha risk (Type I error) of .03 for this process? (Hint:
you need use alpha to calculate z for this process) (1 mark)
e) What alpha risk would control limits of .047 and .003 provide? (Hint: first you need to
find z) (1 mark)
f) Using control limits of .047 and .003, is the process in control? (1 mark)
g) Suppose that the long-term proportion of defectives of the process is known to be 2
percent. What are the values of the mean and standard deviation of the sampling
distribution? (1 mark)
h) Construct a p-control chart for the process, assuming a proportion of defectives of 2
percent, and Two Sigma control limits. Is the process in control? (1 mark)
n = 200
Sample 1
4
= .020
200
Sample 2
2
= .010
200
Sample 3
5
= .025
200
Sample 4
9
= .045
200
b. (0.02 + 0.01 + 0.025 + 0.045) / 4 = 0.025
c. mean = .025
Std. dev. =
p (1 − p )
.025(.975)
=
= .011
n
200
d. α = 0.03 à confidence = .97 (two-sided) à z = 2.17
.025 ± 2.17(0.011) = .025 ± .024 = .001 to .049.
Th
e. .025 + z(.011) = .047
Solving, z = 2, leaving .0228 in each tail.
Hence, alpha = 2(.0228) = .0456.
f.
Yes, all sample proportions in part (a) fall within these control limits.
g. mean = .02
Std. dev. =
.02(.98)
= .0099
200
h. .02 ± 2(.0099) = 0.0002 to .0399.
No, the last sample’s proportion is beyond the upper control limit.
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ADM 3301 A, B
Assignment #5 - Solution
Problem # 4 (4.5 mark)
As part of an insurance company's training program, participants learn how to conduct a fast but
effective analysis of clients’ insurability. The goal is to have participants achieve a time less than
45 minutes. There is no minimum time, but the quality of assessment should be acceptable. Test
results for three participants were: Armand, a mean of 38 minutes and a standard deviation of 3
minutes; Jerry, a mean of 37 minutes and a standard deviation of 2.5 minutes; and Melissa, a
mean of 37.5 minutes and a standard deviation of 2.5 minutes.
Which of the participants would you judge to be capable? Explain.
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Let USL = Upper Specification Limit
X = Process mean, σ = Process standard deviation.
USL = 45 minutes,
X Armand = 38 min., σ Armand = 3 min.
X Jerry = 37 min., σ Jerry = 2.5 min.
X Melissa = 37.5 min., σ Melissa = 2.5 min.
(1.5 mark)
For Armand:
USL − X 4.5 − 38
C pk =
=
= .78
3σ
(3)(3)
Because .78 < 1.0, Armand is not capable.
(1.5 mark)
For Jerry:
USL − X 4.5 − 37
C pk =
=
= 1.07
3σ
(3)(2.5)
Th
Because 1.07 > 1.0, Jerry is capable.
(1.5 mark)
For Melissa:
USL − X 4.5 − 37.5
C pk =
=
= 1.0
3σ
(3)(2.5)
Because 1.0 ≥ 1.0, Melissa is capable.
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