Problems:
1.
BEP = ($12/60) / $30 = .0067
a. (1) 4 percent defective: .0067 < .04  100% inspection is justified.
(2) 1 percent defective: .0067 < .01  100% inspection is justified. However, if the percent
defective is unstable, sampling inspection may be used.
b. When percent defective is around .67%..
Instructor’s Manual, Chapter 10 Supplement
1
2.
n = 20, N = 4,000
c= 1
1.00
.90
.80
.70
Pac .60
[Use binomial table because n / N = 20/4000 = .005 ≤ .1; use x = c in the table]
p
a. Pac = Pr (x ≤ 1 | n = 20, p)
b. AOQ = P * Pac
.05
.7358
.037
.10
.3917
.039
.15
.1756
.026
a. OC
.20
.0692
.014
Curve
.25
.0243
.006
.30
.0076
.002
.50
.05
.40
.04
.30
.03
AOQ
.02
.20
.10
0
AOQ Curve
AOQL  .039
b.
.01
0
.05 .10 .15 .20 .25 .30
Proportion defective
.05 .10 .15 .20 .25 .30
Proportion defective
a.
1.00
3.
N = 8,000
n = 15 [use binomial because n / N < .1]
.80
c=0
a.
p
Pac
.60
.05
.4633
.10
.2059
Pac
.15
.0874
.40
.20
.0352
.25
.0134
OC curve
.20
0
2
.05
.10
.15
Proportion defective
.20
.25
Operations Management 4/C/E
3. (continued)
b. N = 8,000
n = 150 [use Poisson because n/N < .1 and nP < 5 (see below)]
c = 0
p
.001
.002
.003
.004
.005
.006
.008
.010
.012
np
.15
.30
.45
.60
.75
.90
1.20
1.50
1.80
c. P × Pac = AOQ
Pac = Pr (x = 0 | µ = np)
.861
.741
.638
.549
.472
.407
.301
.223
.165
.0009
.0015
.0019
.0022
.0024
.0024
.0024
.0022
.0020
c.
b.
1.00
AOQ
Curve
.003
AOQL  .0024
.80
OC Curve
AOQ .002
.60
Pac
.001
.40
0
.20
.002 .004
.006
.008
Proportion defective
.010
.012
.05
.06
0
.002 .004
.006
.008
Proportion defective
4.
a.
.010
.012
N = 3000
n = 100 [use Poisson because n/N < .1 and nP < 5 (see below)]
a.
c=0 c=1 c=2
Pac
Pac
Pac
1.00
p
np
.001
.1 .905
.995 1.000
.002
.2 .819
.982
.999
.80
.003
.3 .741
.963
.996
.004
.4 .670
.938
.992
.60
C=2
.005
.5 .607
.910
.986
C
=
1
Pac
.006
.6 .549
.878
.977
C=0
.007
.7 .497
.844
.966
.40
.008
.8 .449
.809
.953
.010 1.0 .368
.736
.920
.20
.015 1.5 .223
.558
.809
.020 2.0 .135
.406
.677
.00
.030 3.0 .050
.199
.423
.01
.02
.03
.040 4.0 .018
.092
.238
Proportion defective
.050 5.0 .007
.040
.125
Instructor’s Manual, Chapter 10 Supplement
.04
3
4. (continued)
[Binomial table; n/N < .1]
n=5
n = 20
Pac
Pac
b.
c=2
p
.05
.10
.15
.20
.25
.30
.35
.40
.45
.50
.55
.60
.65
.70
b.
Pr(x ≤ 2 | n, p)
.9988
.9914
.9734
.9421
.8965
.8369
.7648
.6826
.5931
.5000
.4069
.3174
.2352
.1631
Pr(x ≤ 2 | n, p)
.9245
.6769
.4049
.2061
.0913
.0355
.0121
[Poisson table; n/N < .1 and nP < 5]
n = 120
Pac
p
μ = np
Pr(x ≤ 2 | μ = np)
.005
.010
.015.
.020
.025
.030
.035
.040
.045
.6
1.2
1.8
2.4
3.0
3.6
4.2
4.8
5.4
.977
.880
.731
.570
.423
.303
.210
.143
.095
1.00
.80
n=5
Pac
n = 20
.60
n = 120
.40
.20
.00
5.
4
.10
.20
.30
.40
fraction defective
.50
.60
.70
Replacement is $6.25/unit
Inspection cost is $1/unit
a. BEP = $1 / $6.25 = .16 = proportion defective
b. n = 15, AQL = .02 and Pac at .02 = .95.
n/N < .1 and np = 15(.02) = .3 < 5  can use Poisson. Try various c values: 0, 1, …
Pr (x = 0 | µ = .3) = .741, Pr (x ≤ 1| µ = .3) = .963, …
The closest to .95 is .963  c = 1
c. If P = 1%, AQL = 2%, n = 15, np = 15(.01) =.15
(1)
Accept the lot.
(2)
Pac = Pr (x = 0 | µ = np = .15) = .861
(3)
Pre = 1 - Pac = 1 - .861 = .139
d. If P = .03, AQL = 2%, n = 15, np = 15(.03) = .45
(1)
Reject the lot.
(2)
Pac = Pr (x = 0 | µ = np = .45) = .638
(3)
Pre = 1 - Pac = 1 - .638 = .362
Operations Management 4/C/E
6.
7.
2 defectives in the sample
a. Reject because there are 2 defectives in the sample and c = 1. Type I error is possible.
b. Accept, because 2 < 3 = c. Type II error is possible.
c. AOQ? if c = 1 and n = 15.
Assume a large lot size  Binomial
AOQ = P × Pac= P × Pr (x ≤ 1 | n = 15, P)
(1) .05(.8290) = .0415
(2) .10(.5490) = .0549
(3) .15(.3186) = .0478
(4) .20(.1671) = .0334
N = 432, AQL = .65%, Inspection level II, normal type inspection
Instructor’s Manual, Chapter 10 Supplement
5
7 (Cont’d).
Therefore, for single sampling, use n = 80 and c = 1
6
Operations Management 4/C/E
8.
Plan 1, C&B&A (using Inspection Level II)
Thus, MIL-STD-105E  n = 13 and c = 0. The manufacturer uses n = 8 and c = 0 for C&B and
c = 1 for A. The manufacturer should increase its sample size from 8 to 13 and use c = 0 for A’s
as well. Note: For A’s, Inspection level I (or even a special inspection level) may be more
appropriate than level II, but in this case it results in the sample sampling plan.
Plan 2, C&B&A
& Plan 3, C&B&A:
Thus, MIL-STD-105E  n = 2 and c = 0. The manufacturer uses n = 5 and c = 1 for Plan 2 and
n = 3 and c = 0 for Plan 3. The manufacturer should reduce its sample size to n = 2 and use c = 0.
Note: For A’s, Inspection level I (or even a special inspection level) may be more appropriate
than level II, but in this case it results in the sample sampling plan.
9.
n = 100
c=2
a. Pac if P = .01? Can assume Poisson.
Pac = P (X ≤ 2 | µ = np = 100(.01) = 1) = .92
b. Protection of buyer = Pac if P = .05? Can assume Poisson.
Pac = P (X ≤ 2 | µ = np = 100(.05) = 5) = .125
c. For p = .01, AOQ = Pac (P) = .92(.01) = .0092 or .92%
d. For p = .05, AOQ = Pac (P) = .125(.05) = .00625 or .625%
e.
(1.25c + 1) = (.4/100)[1.25(2) + 1] = .014 or 1.4%
Instructor’s Manual, Chapter 10 Supplement
7
10.
LTPD = 5%
N = 1000
= 2.2%
100% inspection of rejected lots
Sampling plan method?
The data fit Dodge-Romig LTPD method.
.44
LTPD × N = .05(1000) = 50
From figure DR1, the intersection of x = .44 and y = 50 is associated with c = 7.
From figure DR2, the intersection of x = 50 and curve c = 7 is y = 11.
11 = n (LTPD)  n = 11/.05 = 220
Therefore, the sampling plan is n = 220 and c = 7.
From figure DR3, the intersection of x = .44 and y = 50 is between curves 15 and 20, perhaps 17.
17 = Imin (LTPD)  Imin = 17/.05 = 340 = Minimum expected total items inspected.
11.
N = 1000
Inspection cost/unit = $.76
Repair cost/unit = $15.20
Suggested plan: n = 75 and c = 2
= 3.4%
a. BEP = .76/15.20 = .05 ≈ .034  sampling inspection is justified
b. If LTPD = .04, Pac?
n/N < .1, np = 75(.04) = 3 < 5  use Poisson
Pac = P(X ≤ 2 | µ = np = 3) = .38  too large for LTPD
 the sampling plan is not discriminating enough
c.
 N  n
AOQ  Pac  p 

 N 
= .38(.04)(1000-75)/1000 = .014 or 1.4%.
12.
8
a.
AQL = 1%
α = .05
LTPD = 5%
β = .10
n and c?
From Larson’s nomograph, n = 80 and c = 1.
Operations Management 4/C/E
13.
b.
Reject the lot because 2 > 1 = c.
N = 10,000
AQL = 1%
LTPD = 10%
n = 100 and c = 2
α and β?
This requires using Larson’s nomograph in reverse. Find the point n=100 and c=2. Connect it to
.01 on the left vertical line. Extend it to the right vertical line. This gives the value of 1- α = .98
 α = .02. Similarly, connect the point n=100 and c=2 to .1 on the left vertical line. Extend it
to the right vertical line. This gives the value of β = .001.
14.
Because www.sqconline.com/mil-std-105.html is requiring subscription after a few uses, we will here use
the MIL-STD-105E tables from e.g., http://www.dianyuan.com/bbs/u/39/1142140688.pdf:
Instructor’s Manual, Chapter 10 Supplement
9
N = 500, AQL = 1%.
10
Operations Management 4/C/E
a.
From Table I, the sample size code letter for lot size of 500 units under general inspection level II
is H.
From Table IIIA (double sampling plan- normal inspection), for code letter H, n1 = 32 and n2 =
32. Under AQL = 1% column, we find c1 = 0, r1 = 2, c2 = 1, and r2 = 2.
In words, take a random sample of 32 units; if the number of defectives = 0 accept the lot; if it is
≥ 2, reject the lot; if it is = 1, take a 2nd ransom sample of 32. It the total number of defectives is
still 1, accept the lot; if it is 2 or more, reject the lot.
b.
N = 500, AQL = 1.5%. The sample size code letter is still H.
From Table IIA (single sampling plan- normal inspection), for code letter H, n = 50. Under AQL
= 1.5% column, we find c = 2 and r = 3.
c.
p = .1
Pac by the manufacturer’s plan?
n/N = 32/500 < .1 and np = 32(.1) = 3.2 < 5 for the first sample. These are slightly violated for
the cumulative sample. However, we will still use Poisson distribution to find the probability.
Let X = no. of defectives in the first sample, and Y = no. of defectives in the second sample.
Pac = P(X=0 | µ=3.2) + P(X=1 | µ=3.2)P(Y=0 | µ=3.2) = .041 + .171(.041) = .048
d.
p = .1
Pac by the customer’s plan?
n/N = 50/500 < .1 and np = 50(.1) = 5. Therefore, we can use Poisson.
Pac = P(X ≤ 2 | µ=5) = .125
e.
p = .1
Pac by the manufacturer and customer’s plan?
We can assume independence of the two plans
Pac = .048(.125) = .006 or .6%.
15.
AOQL = 4%
N = 125
= .01
.25
× N = .01(125) = 1.25
From figure DR4, x = .25 and y = 1.25 intersect in c = 0 area
From table DR1, for c = 0, we pick up x = 1.00 and y = .3679
n = yN / (AOQL × N + y) = .3679(125) / (.04 × 125 + .3679) = 8.6, round to 9
Thus, the sampling plan is n = 9 and c = 0.
AOQL occurs at Pm = x / n = 1.00 / 9 = .111 or 11.1%
Instructor’s Manual, Chapter 10 Supplement
11
16.
N = 4(12)(12) = 576
a.
AQL = 2.5%, Normal type, Inspection level II. We will use the Tables given for problem 14.
From Table I, the sample size code letter for lot size of 576 units under general inspection
level II is J.
From Table IIA (single sampling plan- normal inspection), for code letter J, n = 80.
Under AQL = 2% column, we find c = 5 and r = 6.
b.
12
3 < 5  accept the lot.
Operations Management 4/C/E