Exam 2 Study Guide
PROBLEMS
1.
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?
2.
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.
3.
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
4.
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
3
4
5
1
3
2
1
7
8
9
10
1
0
0
2
5.
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
6.
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.
7.
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.
8.
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.
9.
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?
10.
A company has 12 items in its inventory. Using the data given below classify the items into A, B, and C
classes.
SKU
D120
E111
C140
E151
Annual usage (units)
6850
371
1292
62
Unit $ value
1.20
8.60
13.18
91.80
B180
B120
E149
A180
E110
A155
F120
B150
12667
9625
7010
5100
258
862
1940
967
3.20
10.18
1.27
0.88
62.25
18.10
0.38
2.20
11.
Herbert Adams sells bicycles. One particular model is highly popular with annual sales of 2,000 units per
year. Annual holding cost is $200 per unit and the ordering cost is $40. The store is open 250 days a
year.
a.
What is the economic order quantity?
b.
What is the average number of orders per year?
c.
What is the average time between orders in days?
d.
What is the annual total cost?
12.
Montegut Manufacturing produces a product for which the annual demand is 10,000. Production
averages 100 per day, while demand is 40 per day. Holding costs are $1.50 per unit per year; set-up
costs $200.00. If they wish to produce this product in economic batches, what size batch should be
used? What is the length of time in days to producing one lot? What is the maximum inventory level?
What is the time between orders in days? How many order cycles are there per year? Determine the
total annual inventory cost?
13.
The annual demand, ordering cost, and the inventory carrying cost rate for a certain item are D = 600
units, S = $10/order and holding cost is 30% of item price. Price is established by the following quantity
discount schedule. What should the order quantity be in order to minimize the total annual cost?
Quantity
Unit price
14.
1 to 49
$5.00
50 to 249
$4.50
250 and up
$4.10
A warehouse store sells laser printer cartridges in bulk. The company places restocking orders 1000
boxes at a time. The annual demand is 8000 boxes. The demand during lead time is given below. The
average demand during lead time is 60 boxes. Assume holding cost of $50 per box per year and a stock
out cost of $40 per box.
Demand during lead time
Probability
40
0.1
50
0.2
60
0.2
70
0.2
80
0.2
90
0.1
Determine the least cost safety stock and the corresponding ROP.
15.
The Winfield Distributing Company has maintained an 80% service level policy for inventory of string
trimmers. Mean demand during the reorder period is 130 trimmers, and the standard deviation is 80
trimmers. What is the value of ROP and SS?
16.
The new office supply discounter, Paper Clips, Etc. (PCE), sells a certain type of ergonomically correct
office chair which costs $300. The annual holding cost rate is 40%, annual demand is 600, and the order
cost is $20 per order. The store is open 300 days per year and PCE has decided to establish a customer
service level of 90%.
a.
Suppose that the lead time is a constant 4 days and the demand is variable with a standard
deviation of 2.4 chairs per day. What is the safety stock and reorder point?
b.
Suppose that the lead time is a variable with an average of 4 days and standard deviation of 3
days. Further suppose that the demand is constant. What is the safety stock and reorder point?
c.
Suppose that the lead time is a variable with an average of 4 days and standard deviation of 3
days. Further suppose that the demand is also variable with a standard deviation of 2.4 chairs per
day. What is the safety stock and reorder point?
17.
An oyster bar buys fresh oysters for $3 per pound and sells them for $10 per pound. Unsold oyster at
the end of the day is sold to a grocery store for $1.20 per pound. Determine the pounds of oysters that
must be ordered each day if the daily demand follows normal distribution with mean of 150 pounds and
standard deviation of 12 pounds.
Answers:
1.
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.
2.
100.0%
90.0%
80.0%
70.0%
60.0%
50.0%
40.0%
30.0%
20.0%
10.0%
0.0%
3.
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.
4.
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)
5. 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.
6.
(a)
7.
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.
8.
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
9.
N=
n=
c=
Pd =
nPd =
Pa =
AOQ =
AOQ =
5000
200
5
3%
6
0.446
<-- from Poisson table
.03(.446)(5000-200)/5000
0.0128448 i.e. = 1.28%
10.
No. SKU
1 B120
2 B180
3 C140
4 E110
5 A155
6 E149
7 D120
8 E151
9 A180
10 E111
11 B150
12 F120
Annual
usage
(units) Unit $ value
9625
10.18
12667
3.20
1292
13.18
258
62.25
862
18.10
7010
1.27
6850
1.20
62
91.80
5100
0.88
371
8.60
967
2.20
1940
0.38
Annual Dollar
volume
97,982.50
40,534.40
17,028.56
16,060.50
15,602.20
8,902.70
8,220.00
5,691.60
4,488.00
3,190.60
2,127.40
737.20
220565.66
11.
D = 2000, No. of days = 250, H = $200, S = $40
a.
EOQ =
Dollar %
44.4%
18.4%
7.7%
7.3%
7.1%
4.0%
3.7%
2.6%
2.0%
1.4%
1.0%
0.3%
100%
Cum. %
for $
44.4%
62.8%
70.5%
77.8%
84.9%
88.9%
92.6%
95.2%
97.3%
98.7%
99.7%
100.0%
2(2000)40
 28
200
b.
N = D/Q = 2000/28 = 71.4
c.
d = D/No. of days per year = 2000/250 = 8, T = Q/d = 28/8 = 3.5 days
d.
Annual total cost = (D/Q)S + (Q/2)H = (2000/28)40 + (28/2)200 = $5,657
12.
D = 10,000, H = $1.50, S = $200, p = 100/day, d = 40/day
EPQ = √
2𝐷𝑆
𝑑
𝑝
𝐻(1− )
2(10000)200
=√
1.50(1−
40
)
100
= 2108
Production time = Q/p = 2108/100 = 21.08 days
Imax = (Q/p)(p - d) = (2108/100)(100 - 40) = 1264.80
Cum. % for
no. of items
8.3%
16.7%
25.0%
33.3%
41.7%
50.0%
58.3%
66.7%
75.0%
83.3%
91.7%
100.0%
Class
A
A
B
B
B
B
C
C
C
C
C
C
Average number of orders per year = D/Q = 10000/2108 = 4.74
Time between orders = Q/d = 2108/40 = 52.7 days
Annual holding cost = (Imax/2) x H = (1264.80/2) x 1.50 = $948.60
Annual setup cost = (D/Q) x S = (10000/2108) x 200 = $948
Total cost = 948.60 + 948 = $1,896.60
13.
D = 600
S = 10
Q
Price Holding cost
1 - 49
5.00
1.50
50 - 249
250 &
above
4.50
1.35
4.10
1.23
Q
1 – 49
50 - 249
>= 250
EOQ =
Holding cost = 30%
Formula Q
Candidate Q
Formula Q > upper limit -89 not a candidate
Formula Q is within range, =
94 Candidate Q = Formula Q
Formula Q < lower limit,
99 Candidate Q = lower limit
Price Candidate Q Ordering cost
5.00
4.50
94
63.83
4.10
250
24.00
250 @ P = $4.10
Holding cost
63.45
153.75
14. Number of orders per year = 8000/1000 = 8, H = $50, Cs = $40
Safety
ROP
stock
Carrying cost
Expected stock out
60
0
0
(10x.2 + 20x.2 + 30x.1) = 9
70
10
10 x $50 = $500 (10x.2 + 20x.1) = 4
80
20
20 x $50 = $1000 (10x.1) = 1
90
30
30 x $50 = $1500 0
94
250
Item cost
Total cost
2700
2460
2827.28
2637.75
Stock out cost/year
9 x 8 x 40 = $2,880
4 x 8 x 40 = $1,280
1 x 8 x 40 = $320
$0
Least cost safety stock = 20, ROP = 80
15.
Given dL = 130, dLT = 80, and for 80% service level, Z = 0.84
ROP = 130 + 0.84 x 80 = 197.2, or round up to 198 for at least 80% service level
16.
d = D/No. of days per year = 600/300 = 2 per day, Z for 90% service level = 1.285
a.
Given: L = 4 days Constant, d = 2.4 per day, therefore dLT = 2.4 √4 = 4.8
Safety stock = Z dLT = 1.285 x 4.8 = 6.2 or 7 (round up for at least 90% service level)
Total cost
$2,880
$1,780
$1320
$1500
ROP = dL + SS = (2 chairs/day * 4) + 7 = 15
b.
Given: L = 4 days with L = 3 and demand is constant, dLT = 2 (3) = 6
Safety stock = Z dLT = 1.285 x 6 = 7.7 or 8 (round up for at least 90% service level)
ROP = dL + SS = (2 chairs/day * 4) + 8 = 16
c.
Given: L = 4 days with L = 3 , and d = 2.4 per day,
therefore dLT = √4(2.4)2 + 22 32 = 7.684
Safety stock = Z dLT = 1.285 x 7.684 = 9.9 or 10 (round up for at least 90% service level)
ROP = dL + SS = (2 chairs/day * 4) + 10 = 18
17.
Cs = Lost profit = Selling price per unit – Cost per unit = 10 – 3 = $7
Co = Cost/unit – salvage value/unit = 3 – 1.20 = $1.80
Optimum service level = 7/(7 + 1.80) = 0.795 = 79.5%
From normal table, for 79.5% service level, Z = 0.83
Stock =  + Z  = 150 + 0.825 (12) = 159.9 or 160