Chapter 2- Solutions
1.
a.
b.
Anniversary = 300 / 8 = 37.5 meals/worker; Wedding = 240 / 6 = 40 meals/worker.
Possible reasons are differences in the menu, number of courses, time of day,
facilities, and worker skills/experience.
2.
1
4
96
Labor Productivity per
Worker
24 yards
2
3
72
24
3
4
92
23
4
2
50
25
5
3
69
23
6
2
52
26
Week
Crew Size
Yards Installed
Notes:
Labor Productivity per Worker = Yards Installed / Crew Size
We can determine the Average Labor Productivity per Worker for each crew size as follows:
Crew Size of 2: (25 + 26) / 2 = 25.5
Crew Size of 3: (24 + 23) / 2 = 23.5
Crew Size of 4: (24 + 23) / 2 = 23.5
A crew size of 2 seems to work best with an Average Labor Productivity per Worker = 25.5
yards installed per worker.
3.
Week
Output
Number
of
Workers
1
30,000
6
450
2,880
4,320
2,700
9,900
3.03
2
33,600
7
470
3,360
5,040
2,820
11,220
2.99
3
32,200
7
460
3,360
5,040
2,760
11,160
2.89
4
35,400
8
480
3,840
5,760
2,880
12,480
2.84
Material
(lbs.)
Labor
Cost
Overhead
Cost
Material
Cost
Total
Cost
Notes:
Labor Cost = Number of Workers x 40 hours x $12/hour
Overhead Cost = Labor Cost x 1.50
Material Cost = Material (lbs.) x $6/lb.
Total Cost = Labor Cost + Overhead Cost + Material Cost
Multifactor Productivity (MFP) = Output / Total Cost (rounded to two decimals)
Multifactor productivity dropped steadily from a high of 3.03 to a low of 2.84.
4.
a.
Prior to Buying New Equipment:
Labor Productivity = Carts per Worker per Hour = 80 / 5 = 16 Carts per Worker per
Hour.
MFP
b.
After Buying New Equipment:
Labor Productivity = Carts per Worker per Hour = (80 + 4) / (5 – 1) = 84 / 4 =
21 Carts per Worker per Hour.
Prior to Buying New Equipment:
Multifactor Productivity = Carts per Dollar (Labor + Equipment)
Labor = 5 workers x $10/hour = $50/hour
Equipment = Machine Cost = $40/hour
Multifactor Productivity = 80 carts / ($50 + $40) = 0.89 Carts per Dollar (rounded to
two decimals)
After Buying New Equipment:
Multifactor Productivity = Carts per Dollar (Labor + Equipment)
Labor = 4 workers x $10/hour = $40/hour
Equipment = Machine Cost = $40/hour + $10/hour = $50/hour
Multifactor Productivity = 84 carts / ($40 + $50) = 0.93 Carts per Dollar (rounded to
two decimals)
c.
𝐿𝑎𝑏𝑜𝑟 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 𝐺𝑟𝑜𝑤𝑡ℎ =
=
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦−𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦
𝑥100
𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦
21 − 16
5
𝑥100 =
𝑥100 = 31.25% (𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑡𝑜 𝑡𝑤𝑜 𝑑𝑒𝑐𝑖𝑚𝑎𝑙𝑠)
16
16
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 − 𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦
𝑥100
𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦
0.93 − 0.89
0.04
=
𝑥100 =
𝑥100
0.89
0.89
= 4.49% (𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑡𝑜 𝑡𝑤𝑜 𝑑𝑒𝑐𝑖𝑚𝑎𝑙𝑠)
∗ 𝐸𝑥𝑐𝑒𝑙 𝑚𝑎𝑦 𝑠ℎ𝑜𝑤 𝑎 𝑟𝑜𝑢𝑛𝑑𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 5%
𝑀𝐹𝑃 𝐺𝑟𝑜𝑤𝑡ℎ =
CHAPTER 03
FORECASTING
SOLUTIONS
1.
a. Plotting each data set reveals that blueberry muffin orders are stable, varying around an
average. Therefore, the naïve forecast is the last value, 33. The demand for cinnamon
buns has a trend. The last change was from 31 to 33 (33 – 31 = 2). Using the last value
and adding the last trend change, the forecast is 33 + 2 = 35. Demand for cupcakes has an
apparent seasonal variation with peaks every five days. Day 1 = 45, Day 6 = 48, and Day
11 = 47. Since the peaks occur every five days, the next peak would be at Day 16. We
could predict that demand will be the same as it was the last season—here this value
would equal 47.
b. The use of sales data instead of demand implies that sales adequately reflect demand. We
are assuming that no stockouts occurred because demand equals sales if there are no
shortages.
2.
Given:
Month
Sales (000
units)
Feb.
Mar.
Apr.
May
Jun.
Jul.
Aug.
19
18
15
20
18
22
20
a.
Sales
20
0
b.
20.
F
M
A
M
Month
J
J
A
S
1)
Using the naïve approach, the forecast for the next month (September) will equal
2)
A five-month moving average is shown below:
MA5 
15  20  18  22  20
 19.00 (round to two decimals)
5
3) A weighted using average using 0.60 for August, 0.30 for July, and 0.10 for June is shown
below:
0.10(18) + 0.30(22) + 0.60(20) = 20.40 (round to two decimals)
4) Exponential smoothing, with alpha = 0.20 and an initial forecast for March of 19 are
shown below (round to two decimals):
Month
Forecast
April
18.80
F(old) + .20[Actual – F(old)]
=
= 19
+ .20[ 18 – 19
]
May
18.04
= 18.80
+ .20[ 15
– 18.80 ]
June
18.43
= 18.04
+ .20[ 20
– 18.04 ]
July
18.34
= 18.43
+ .20[ 18
– 18.43 ]
August
19.07
= 18.34
+ .20[ 22
– 18.34 ]
September 19.26
= 19.07
+ .20[ 20
– 19.07 ]
5) A linear trend forecast is shown below (round b & a to two decimals):
t
1
Y
19
t*Y
19
t2
1
2
18
36
4
3
15
45
9
4
20
80
16
5
18
90
25
6
22
132
36
7
20
140
49
28
132
542
140
b
n  tY   t  Y 7(542)  28(132)

 0.50
n  t 2  ( t ) 2
7(140)  (28) 2
a
 Y  b  t 132  0.50(28)

 16.86
n
7
For Sept., t = 8, and Yt = 16.86 + 0.50(8) = 20.86 = 20,860
c. The linear trend approach seems to be the least appropriate because the data appear to vary
around an average of about 19 [18.86] and because the slope is close to zero (0.50).
d. Sales are reflective of demand (i.e., no stockouts or backorders occurred).
3.
a. Exponential smoothing forecast for September with alpha = 0.10:
88 + 0.10(89.6 – 88) = 88.16 (round to two decimals)
b. Exponential smoothing forecast for October with alpha = 0.10:
88.16 + 0.10(92 – 88.16) = 88.54 (round to two decimals)
4.
Given:
Week
Requests
1
20
2
22
3
18
4
21
5
22
a. Naïve approach forecast for Week 6 = Demand in Week 5 = 22
b. Four-period moving average forecast for Week 6:
22  18  21  22
 20 .75 (round to two decimals)
4
c. Exponential smoothing with alpha = 0.30 and a Week 2 Forecast = 20 (round to two
decimals):
F3 = 20 + 0.30(22 – 20) = 20.60
F4 = 20.60 + 0.30(18 – 20.6) = 19.82
F5 = 19.82 + 0.30(21 – 19.82) = 20.17
F6 = 20.17 + 0.30(22 – 20.17) = 20.72
5.
a. Annual sales are increasing by 15,000 bottles per year (the slope of the line)
b. Forecast for Year 6:
t = 6, Yt = 80 + 15(6) = 170, which is 170,000 bottles.
9.
a.
t
1
Y
200
t*Y
200
t2
1
2
214
428
4
3
211
633
9
4
228
912
16
5
235
1,175
25
6
232
1,332
36
7
248
1,736
49
8
250
2,000
64
9
253
2,277
81
10
267
2,670
100
11
281
3,091
121
12
275
3,300
144
13
280
3,640
169
14
288
4,032
196
15
310
4,650
225
120
3,772
32,136
1,240
Round b & a to two decimals:
b
n  tY   t  Y 15(32,136)  120(3,772)

 7.00
n  t 2  ( t ) 2
15(1,240)  (120) 2
a
 Y  b  t 3,772  7.00(120)

 195.47
n
15
Forecasts for periods 16 through 19 using Linear Trend are (round to two decimals):
Y16 = 195.47 + (7.00)(16) = 307.47
Y17 = 195.47 + (7.00)(17) = 314.47
Y18 = 195.47 + (7.00)(18) = 321.47
Y19 = 195.47 + (7.00)(19) = 328.47
b. Round values to two decimals.
Initial Trend =
228  200
 9.33
3
Period
5
Actual
235
St-1 + Tt-1 = TAFt
228.00 + 9.33 = 237.33
TAFt + .3(At – TAFt) = St
237.33 + .3(235 – 237.33) = 236.63
Tt–1 + .2 (TAFt – TAFt–1 – Tt–1) = Tt
9.33
6
232
236.63 + 9.33 = 245.96
245.96 + .3(232 – 245.96) = 241.77
9.33 + .2(245.96 – 237.33 – 9.33) = 9.19
7
248
241.77 + 9.19 = 250.96
250.96 + .3(248 – 250.96) = 250.07
9.19 + .2(250.96 – 245.96 – 9.19) = 8.35
8
250
250.07 + 8.35 = 258.42
258.42 + .3(250 – 258.42) = 255.89
8.35 + .2(258.42 – 250.96 – 8.35) = 8.17
9
253
255.89 + 8.17 = 264.06
264.06 + .3(253 – 264.06) = 260.74
8.17 + .2(264.06 – 258.42 – 8.17) = 7.66
10
267
260.74 + 7.66 = 268.40
268.40 + .3(267 – 268.40) = 267.98
7.66 + .2(268.40 – 264.06 – 7.66) = 7.00
11
281
267.98 + 7.00 = 274.98
274.98 + .3(281 – 274.98) = 276.79
7.00 + .2(274.98 – 268.40 – 7.00) = 6.92
12
275
276.79 + 6.92 = 283.71
283.71 + .3(275 – 283.71) = 281.10
6.92 + .2(283.71 – 274.98 – 6.92) = 7.28
13
280
281.10 + 7.28 = 288.38
288.38 + .3(280 – 288.38) = 285.87
7.28 + .2(288.38 – 283.71 – 7.28) = 6.76
14
288
285.87 + 6.76 = 292.63
292.63 + .3(288 – 292.63) = 291.24
6.76 + .2(292.63 – 288.38 – 6.76) = 6.26
15
310
291.24 + 6.26 = 297.50
297.50 + .3(310 – 297.50) = 301.25
6.26 + .2(297.50 – 292.63 – 6.26) = 5.98
16
301.25 + 5.98 = 307.23
10.
The initial estimate of trend is based on the net change of 30 for the three periods from 1 to 4,
for an average of +10 units. Use  = .5 and  = .4. Round values to two decimals.
Initial trend = (240 – 210)/3 = 10.00
Period
1
Actual
210
Model
2
224
Development
3
229
4
240
5
255
St + Tt = TAFt
TAFt + .5(At – TAFt) = St
Tt–1 + .4 (TAFt – TAFt–1 – Tt–1) = Tt
240.00 + 10.00 = 250.00 250.00 + .5(255 – 250.00) = 252.50 10.00
6
265
252.50 + 10.00 = 262.50 262.50 + .5(265 – 262.50) = 263.75 10.00 + .4(262.50 – 250.00 – 10.00) = 11.00
7
272
263.75 + 11.00 = 274.75 274.75 + .5(272 – 274.75) = 272.38 11.00 + .4(274.75 – 262.50 – 11.00) = 11.50
8
285
272.38 + 11.50 = 284.88 284.88 + .5(285 – 284.88) = 284.94 11.50 + .4(284.88 – 274.75 – 11.50) = 10.95
9
294
284.94 + 10.95 = 295.89 295.89 + .5(294 – 295.89) = 294.95 10.95 + .4(295.89 – 284.88 – 10.95) = 10.97
Actual
Model Test
Next
Forecast
10
294.95 + 10.97 = 305.92
11.
Yt = 70 + 5t
t = 0 (June of last year)
t = 1 (July of last year)
t = 7 (January of this year)
t = 8 (February of this year)
t = 9 (March of this year)
t = 19 (January of next year)
t = 20 (February of next year)
t = 21 (March of next year)
YJan. = 70 + (5)(19) = 165
YFeb. = 70 + (5)(20) = 170
YMar. = 70+ (5)(21) = 175
Forecast = Trend * Seasonal Relative (round to two decimals):
Month
January
12.
Trend * Seasonal Relative
165 * 1.10 = 181.50
February
170 * 1.02 = 173.40
March
175 * 0.95 = 166.25
The current quarter is Quarter 1 = t = 4. Quarter 1 from one year ago = t = 0.
Quarter 1 next year = t = 8.
Quarter
Value of t
Trend component, Ft
Quarter relative
Forecast
Next Year,
Q1
8
116.00
x 1.1 =
127.60
Next Year,
Q2
9
143.50
x 1.0 =
143.50
Next
Year, Q 3
10
175.00
x 0.6 =
105.00
Next
Year, Q 4
11
210.50
x 1.3 =
273.65
Trend component calculations: 𝐹𝑡 = 40 − 6.5𝑡 + 2𝑡 2 (round to two decimals):
𝐹8 = 40 − 6.5(8) + 2(82 ) = 116.00
𝐹9 = 40 − 6.5(9) + 2(92 ) = 143.50
𝐹10 = 40 − 6.5(10) + 2(102 ) = 175.00
𝐹11 = 40 − 6.5(11) + 2(112 ) = 210.50
𝐹12 = 40 − 6.5(12) + 2(122 ) = 250.00
Two
Years, Q 1
12
250.00
x 1.1 =
275.00
13.
Given:
Quarter
1
2
3
4
Year 1
2
6
2
5
Year 2
3
10
6
9
Year 3
7
18
8
15
Year 4
4
14
8
11
SA method (round season averages to three decimals and seasonal relatives to two decimals):
Quarter
1
2
3
4
Year 1
2
6
2
5
Year 2
3
10
6
9
Year 3
7
18
8
15
Year 4
4
14
8
11
Season
Average
4.000
12.000
6.000
10.000
8.000
Overall
Average
Sum of Seasonal Relatives = 0.50 + 1.50 + 0.75 + 1.25 = 4.00
Seasonal Relative
0.50 =(4.000/8.000)
1.50 =(12.000/8.000)
0.75 =(6.000/8.000)
1.25 =(10.000/8.000)
22.
a. Compute MSE & MAD for each forecast method (round to two decimals). Round % to
two
decimals.
Period
Demand
F1
1
770
771
-1
2
789
785
4
(e/Demand)
F2
x 100 (%)
1 1
0.13% 769
4 16
0.51% 787
1
1
(e/Demand)
x 100 (%)
1
0.13%
2
2
4
0.25%
3
794
790
4
4 16
2
2
4
0.25%
4
780
784
-4
4 16
-18
18
324
2.31%
5
768
770
-2
2
0.26% 774
0.52% 770
-6
6
36
0.78%
6
772
768
4
2
2
4
0.26%
7
760
761
-1
0.13% 759
0.52% 775
1
1
1
0.13%
8
775
771
4
4 16
0
0
0
0.00%
9
786
784
2
2
4
0.25% 788
0.25% 788
-2
2
4
0.25%
10
790
788
2
2
4
2
2
4
0.25%
Sum
12
-16
36
382
e e2
e
4
4 16
1
1
28 94
0.50% 792
0.51% 798
3.58%
e
e
e2
MAD F1: 28/10 = 2.80
MAD F2: 36/10 = 3.60
MSE F1: 94/(10-1) = 10.44
MSE F2: 382/(10-1) = 42.44
F1 has both lower MAD and lower MSE so it seems better.
b. Compute MAPE for each forecast method (round to two decimals).
MAPE F1: 3.58%/10 = 0.36%
MAPE F2: 4.61%/10 = 0.46%
4.61%
CHAPTER 05
STRATEGIC CAPACITY PLANNING FOR PRODUCTS AND
SERVICES
Solutions
1.
a.
Utilizatio n 
Efficiency 
b. Utilization 
Efficiency 
Actual output
7
 x100%  70.00%
Design capacity 10
Actual output
7
 x100%  87.50%
Effective capacity 8
Actual output
4
 x100%  66.67%
Design capacity 6
Actual output
4
 x100%  80.00%
Effective capacity 5
c. This is not necessarily true. If the design capacity is relatively high, the utilization could
be low even though the efficiency is high.
2.
Given:
Effective capacity = .50 (Design Capacity)
Actual output = .80 (Effective capacity)
Actual output desired = 8 jobs per week
By substitution:
Actual output = (.80) x [(.50) (Design capacity)]
Actual output = (.40) (Design capacity)
Re-arranging terms:
Design Capacity 
Actual output
.40
By substitution (Actual output desired = 8 jobs from above):
Design capacity 
8
 20 jobs
.40
3.
Given:
FC = $9,200/month
v=
$ .70/unit
R = $ .90/unit
a.
QBEP 
FC
$9,200

 46,000 units
R  v $.90  $.70
b. Profit = R x Q – (FC + v x Q)
1. P61,000 = $.90(61,000)  [$9,200 + $.70(61,000)] = $3,000
2. P87,000 = $.90(87,000)  [$9,200 + $.70(87,000)] = $8,200
Specified profit  FC
$16,000  9,200

 126,000 units
R v
$.90 / unit  $.70 / unit
Total Revenue
$23,000
d. Total Revenue = R x Q, so Q =

 25,555.56 units
R
$.90 / unit
c.
Q
e.
$100,000
TR = $90,000 @ Q = 100,000 units
TC = $79,200 @ Q = 100,000 units
TR
TC
Cost
$50,000
$9,200
0
Volume
(units)
100,000
1.
1.
Given:
a.
FC
A: $40,000
R
$15/unit
v
$10/unit
B: $30,000
$15/unit
$11/unit
QBEP 
FC
R v
Q BEP,A 
$40,000
 8,000 units
$15 / unit  $10 / unit
QBEP,B 
$30,000
 7,500 units
$15 / unit  $11/ unit
b. Profit = Q(R – v) – FC
[A’s Profit]
[B’s Profit]
Q($15 – $10) – $40,000 = Q($15 – $11) – $30,000
$5Q - $40,000 = $4Q - $30,000
$5Q - $4Q = - $30,000 + $40,000
Q = 10,000 units
c. PA = 12,000($15 – $10) – $40,000 = $20,000 [A is higher]
PB = 12,000($15 – $11) – $30,000 = $18,000
5.
Given:
Demand = 30,000 = Q
FC = $25,000
v = $.37/pen
a. Given: R = $1.00/pen
QBEP 
FC
$25,000

 39,682.54 units
R  v $1.00  $.37
b. Given: Demand = 30,000 units. Specified profit = $15,000
Specified profit  FC $15,000  $25,000

R v
R  $.37
$40,000
Q
R  $.37
Q
By substitution:
30,000 
8.
$40,000
R  $.37
30,000 x (R - $.37) = $40,000
30,000R - $11,100 = $40,000
30,000R = $40,000 + $11,100
30,000R = $51,100
R = $51,100/30,000
R = $1.71 [rounded up]
Given:
Source
Internal 1
FC
$200,000
v
$17
Internal 2
240,000
14
Vendor A
20 up to 30,000 units
Vendor B
22 for 1 to 1,000; 18 each if larger amount
Vendor C
21 for 1 to 1,000; 19 each for additional units
a.
TC for 10,000 units
Int. 1: 200,000 + 17(10,000) = $370,000
TC for 20,000 units
200,000
+ 17(20,000) = $540,000
Int. 2: 240,000 + 14(10,000) = $380,000
240,000
+ 14(20,000) = $520,000
Vend A:
20(10,000) = $200,000
20(20,000) = $400,000
Vend B:
18(10,000) = $180,000
18(20,000) = $360,000
Vend C: 21,000 + 19(9,000) = $192,000
21,000
+ 19(19,000) = $382,000
At 10,000 units, total cost is lowest when using Vendor B. At 20,000 units, total cost is
lowest when using Vendor B.
b.
Given:
Cost functions for each alternative:
Internal 1:
$200,000 + $17Q
Internal 2:
$240,000 + $14Q
Vendor A:
$20Q (Q ≤ 30,000)
Vendor B:
$22Q (Q ≤ 1,000)
Vendor C:
$21Q (Q ≤ 1,000)
$18Q for all units when Q > 1,000
$21Q + $19(Q - 1,000) when Q > 1,000
First, we analyze the range of 1 - 1,000 units:
Vendor A exhibits lower total cost over this range than do Vendor B and Vendor C;
therefore, we can eliminate Vendors B & C from consideration for this range.
Next, we could graph the costs functions of the remaining three options for the range of 1
– 1,000 units:
Internal 1:
$200,000 + $17Q
Internal 2:
$240,000 + $14Q
Vendor A:
$20Q
As shown in the Excel chart below, Vendor A provides the lowest total cost over this
entire range. If the manager is going to purchase between 1 to 1,000 units, Vendor A is
preferred.
300 000
250 000
Int. 2
Int. 1
200 000
$ 150 000
100 000
50 000
Vend A
0
0
1000
Units
Second, we analyze the range of 1,001 units or more to determine the total costs if we
purchase > 1,000 units:
Total Cost Functions (when purchasing 1,001 units or more):
Internal 1: $200,000 + $17Q
Internal 2: $240,000 + $14Q
Vendor A: $20Q (≤ 30,000 units)
Vendor B: $18Q
Vendor C: $21Q + $19(Q – 1,000)
Looking at the cost functions above, we can see that Vendor B dominates Vendor A and
Vendor C. Therefore, we can eliminate Vendor A and Vendor C from further consideration.
We must compare the total costs of Internal 1, Internal 2, and Vendor B when purchasing
more than 1,000 units.
We can plot these costs functions on a graph as shown in the Excel chart below:
1 600 000
Int. 1
1 400 000
1 200 000
1 000 000
$ 800 000
600 000
Int. 2
400 000
Vend B
200 000
0
0
10000
20000
30000
Units
40000
50000
60000
70000
We can see in the chart above that Vendor B has the lowest total cost until its total cost function
intersects the total cost function of Internal 2.
Our next step is to determine the indifference point between Vendor B and Internal 2.
Set the two cost functions equal and solve for Q:
$18Q = $240,000 + $14Q
$18Q - $14Q = $240,000
$4Q = $240,000
Q = $240,000/$40
Q = 60,000 units
Therefore, Vendor B has lower total cost in the range of 1,001 units – 59,999 units.
Internal 2 has lower total cost > 60,000 units.
Summary:
Purchase Quantity:
1 – 1,000 units
Prefer Vendor A
1,001 – 59,999 units
Prefer Vendor B
60,000 units
Indifferent between Vendor B & Internal 2
> 60,000 units
Prefer Internal 2
Note: Internal 1 and Vendor C are never best.
9. Given: Actual output will be 225 per day per cell. 240 working days/year. Projected annual
demand = 150,000 within 2 years.
Annual capacity per cell = 225 units/day x 240 days/year = 54,000
Cells :
150,000
 2.78, round up to 3 cells
54,000
10. Given: Our objective is to select one type of machine to purchase. We are given the data below:
Machine Type
1
2
Product
001
002
003
Annual
Demand
(units)
12,000
10,000
18,000
Purchasing
Cost/Machine
$10,000
$14,000
Process Time per
Unit on Type 1
(min.)
4
9
5
Process Time per
Unit on Type 2
(min.)
6
9
3
a. Number of machines of each type needed if the machines will operate 60 minutes per hour, 8
hours per day, 250 days per year.
Using Machine Type 1:
Each Machine Type 1 is available 250 x 8 x 60 = 120,000 minutes per year
Processing Requirements using Machine Type 1:
Product 001: 12,000 x 4 min. = 48,000 min.
Product 002: 10,000 x 9 min. = 90,000 min.
Product 003: 18,000 x 5 min. = 90,000 min.
Total = 228,000 min.
Number of Machine Type 1 Needed = processing time needed / processing time capacity per
unit = 228,000 / 120,000 = 1.9 = 2 machines (round up)
Capacity = 2 x 120,000 minutes = 240,000 minutes
Capacity cushion = 240,000 – 228,000 = 12,000 minutes
Using Machine Type 2:
Each Machine Type 2 is available 250 x 8 x 60 = 120,000 minutes per year
Processing Requirements using Machine Type 2:
Product 001: 12,000 x 6 min. = 72,000 min.
Product 002: 10,000 x 9 min. = 90,000 min.
Product 003: 18,000 x 3 min. = 54,000 min.
Total = 216,000 min.
Number of Machine Type 2 Needed = processing time needed / processing time capacity per
unit = 216,000 / 120,000 = 1.8 = 2 machines (round up)
Capacity = 2 x 120,000 minutes = 240,000 minutes
Capacity cushion = 240,000 – 216,000 = 24,000 minutes
b. If we faced high uncertainty of annual demand, we would select the type of machine with the
higher capacity cushion (Machine Type 2). If we faced low uncertainty of annual demand, we
would select the type of machine with the lower capacity cushion (Machine Type 1).
c. Given: Operating costs = $6/hour for Type 1 & $5/hour for Type 2
Purchase Cost for Machine Type 1 = 2 machines x $10,000/machine = $20,000
Total Operating Time for Machine Type 1 = 228,000 minutes = 3,800 hours
Total Operating Cost = 3,800 hours x $6/hour = $22,800
Total Cost = $20,000 + $22,800 = $42,800
Purchase Cost for Machine Type 2 = 2 machines x $14,000/machine = $28,000
Total Operating Time for Machine Type 2 = 216,000 minutes = 3,600 hours
Total Operating Cost = 3,600 hours x $5/hour = $18,000
Total Cost = $28,000 + $18,000 = $46,000
Conclusion: Machine Type 1 would minimize total cost.
11. a. Given: 10 hrs. or 600 min. of operating time per day per machine.
250 days x 600 min. = 150,000 min. per year operating time per machine.
Machine purchase costs: A = $40,000; B = $30,000; C = $80,000.
Total processing time by machine
Product
1
A
48,000
B
64,000
C
32,000
2
48,000
48,000
36,000
3
30,000
36,000
24,000
4
60,000
60,000
30,000
Total
186,000
208,000
122,000
186,000
 1.24  2 machines
150,000
208,000
NB 
 1.38  2 machines
150,000
122,000
NC 
 .81  1 machine
150,000
NA 
Options:
Buy two A machines at a total purchase cost of 2 x $40,000 = $80,000.
Buy 2 B machines at a total purchase cost of 2 x $30,000 = $60,000.
Buy 1 C machine at a total purchase cost of $80,000.
Conclusion: We should buy 2 of the B machines at a total cost of $60,000.
b. Given: Operating Costs: A = $10/hour/machine; B = $11/hour/machine; C =
$12/hour/machine.
Total cost for each type of machine:
A (2): 186,000 min. / 60 min./hour = 3,100.00 hrs. x $10 = $31,000 + $80,000 = $111,000
B (2): 208,000 min. / 60 min./hour = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133
C(1): 122,000 min. / 60 min./hour = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400
Conclusion: Buy 2 Bs—these have the lowest total cost.
CH 5S-Problems Solutions
2.
Given: P(Low Demand) = .3 and P(High Demand) = .7.
a. Determine the best expected profit of the alternatives from Problem 1
Expected Profit:
Do nothing =
.3($50) + .7($60) = $15 + $42 = $57
Expand =
.3($20) + .7($80) = $6 + $56 = $62
Subcontract =
.3($40) + .7($70) = $12 + $49 = $61
Conclusion: Expand is the best alternative because it has the highest expected value.
b. Decision Tree Analysis to Select an Alternative:
.3
$57
Do Nothing
$62
Expand
$61
Subcontr.
.7
.3
.7
.3
.7
$50
$60
$20
$80
$40
$70
Expected Value Calculations:
Do nothing =
.3($50) + .7($60) = $15 + $42 = $57
Expand =
.3($20) + .7($80) = $6 + $56 = $62
Subcontract =
.3($40) + .7($70) = $12 + $49 = $61
Conclusion: Expand is the best alternative because it has the highest expected value
($62).
c. Expected Value of Perfect Information:
Expected value of perfect information (EVPI) = Expected payoff under certainty –
Expected payoff under risk
Find the best payoff under each state of nature:
Low Demand: Best Payoff = $50
High Demand: Best Payoff = $80
Expected payoff under certainty (apply the probabilities of each state of nature) =
= (Prob. of Low Demand x Best Payoff) + (Prob. of High Demand x Best Payoff)
= .3($50) + .7($80) = $15 + $56 = $71
Expected payoff under risk = Expected value of the alternative selected = $62
4.
EVPI = $71 - $62 = $9
1) Draw the tree diagram:
$400,000 (1)
Demand Low (.4)
Maintain
$50,000 (2)
Build Small
Demand High (.6)
2
Expand
1
$450,000 (3)
Build Large
Demand Low (.4)
$-10,000 (4)
Demand High (.6)
$800,000 (5)
2) Analyze decisions from right to left (i.e., work backwards from the end of the tree
towards the root). For instance, begin with decision 2 and choose expansion because
it has a higher present value ($450,000 vs. $50,000). Draw a double slash through the
Maintain alternative.
3) Determine the product of the chance probabilities and their respective payoffs for the
remaining branches.
Build Small
Demand Low:
Demand High:
.4($400,000) = $160,000
.6($450,000) = $270,000
Build Large
Demand Low:
Demand High:
.4(-$10,000) = -$4,000
.6($800,000) = $480,000
4) Determine the expected value of each initial alternative.
Build Small = $160,000 + $270,000 = $430,000
Build Large = -$4,000 + $480,000 = $476,000
Conclusion: Because the expected value of building a large plant is highest, select
the large plant alternative. Draw a double slash through the Build Small alternative.
b. Expected payoff under certainty:
.4(400,000) + .6(800,000) = $640,000
-Expected payoff under risk:
-476,000
Expected value of perfect information:
$164,000
7.
Given: Probability that the motel’s application will be approved = .35.
The probability that the motel will be rejected = 1.00 - .35 = .65.
Alternative
a. Renew
Expected Value
(.35)500,000 + (.65)4,000,000 = $2,775,000
Relocate
(.35)5,000,000 + (.65)100,000 = $1,815,000
Conclusion:
Renew lease.
b.
Approve (.35)
$500,000
E.V.
$2,775,000
Renew
Reject (.65)
$4,000,000
Approve (.35)
Relocate
$5,000,000
Reject (.65)
$1,815,000
$100,000
Conclusion: Renew lease.
c. Expected value of perfect information (EVPI) = Expected payoff under certainty –
Expected payoff under risk
Find the best payoff under each state of nature:
Motel Approved: Best Payoff = $5,000,000
Motel Rejected: Best Payoff = $4,000,000
Expected payoff under certainty (apply the probabilities of each state of nature) =
= (Prob. of Motel Approved x Best Payoff) + (Prob. of Motel Rejected x Best Payoff)
= .35($5,000,000) + .65($4,000,000) = $1,750,000 + $2,600,000 = $4,350,000
Expected payoff under risk = Expected value of the alternative selected = $2,775,000
EVPI = $4,350,000 - $2,775,000 = $1,575,000Conclusion: Yes, the manager should sign
the lease for $24,000 because this cost is less than the EVPI of $1,575,000.
Chapter 13- Solutions
3.
Given:
D = 1,215 bags per year
S = $10
H = $75
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
a. Determine the EOQ:
Q0 
2DS
2(1,215)10

 18 bags
H
75
b. Determine the average inventory:
Q/2 = 18/2 = 9 bags
c. Determine the number of orders per year:
D
1,215 bags

 67.5 orders
Q 18 bags / order
d. Determine the total cost of ordering and carrying flour:
TC = Carrying cost + Ordering cost
𝑄
𝐷
18
1,215
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 = ( ) 75 + (
) 10 = $675 + $675 = $1,350
2
𝑄
2
18
e. Assuming that holding cost per bag increases by $9/bag/year, what would happen to total
cost?
New H = $75 + $ 9 = $84.
Q0 
2(1,215)(10)
 17 bags
84
𝑄
𝐷
17
1,215
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 = ( ) 84 + (
) 10 = $714 + $714.71 = $1,428.71
2
𝑄
2
17
Increase in cost = $1,428.71 – $1,350 = $78.71 per year
4.
Given:
D = 40/day x 260 days/yr. = 10,400 boxes
S = $60. H = $30.
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
a. Determine the EOQ:
Q0 
2DS

H
2(10,400)60
 203.96  204 boxes
30
b. Determine total cost:
TC = Carrying cost + Ordering cost
𝑄
𝐷
204
10,400
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 = (
) 30 + (
) 60 = $3,060 + $3,058.82 = $6,118.82
2
𝑄
2
204
c. Yes, annual ordering and carrying costs always are equal at the EOQ (except when
rounding).
d. Determine the total cost for Q = 200 and compare to current total cost:
𝑄
𝐷
200
10,400
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 = (
) 30 + (
) 60 = $3,000 + $3,120 = $6,120
2
𝑄
2
200
$6,120 – $6,118.82 = $1.18 higher per year for Q = 200 (this should be acceptable).
5.
Given:
D = 750 pots/mo. x 12 mo./yr. = 9,000 pots/yr.
C = $2. H = (.30)($2) = $.60/unit/year
S = $20
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
a. Determine the additional annual cost for using Q = 1,500:
Step 1:
Determine total cost for Q = 1,500.
𝑄
𝐷
1,500
9,000
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 = (
) . 60 + (
) 20 = $450 + $120 = $570
2
𝑄
2
1,500
Step 2:
Determine EOQ.
Q0 
2DS
2(9,000)20

 774.60  775 pots
H
.60
Step 3:
Determine total cost for Q = 775.
𝑄
𝐷
775
9,000
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 = (
) . 60 + (
) 20 = $232.50 + $232.26 = $464.76
2
𝑄
2
775
Step 4:
Determine annual savings from using the EOQ.
$570 – $464.76 = $105.24.
b. The benefit of using the EOQ is that about one half of the storage space would be needed.
9.
Given:
p = 5,000 hotdogs/day
u = 250 hotdogs/day
Factory operates 300 days per year
D = 250 * 300 = 75,000 hotdogs per year
S = $66
H = $0.45 per hotdog per year
Note: Round Qp to an integer value, but round any other values to a maximum of two
decimals.
a. Find the optimal run size:
Qp 
2 DS
H
p

pu
2(75,000)66
5,000
 4,812.27  4,812 hotdogs
0.45
5,000  250
b. Number of runs per year:
D / Qp = 75,000 / 4,812 = 15.59 runs per year
c. Days to produce the optimal run quantity:
Qp / p = 4,812 / 5,000 = 0.96 days
10.
Given:
A chemical firm produces 100-pound bags. Demand for the product = 20 tons per day. The
capacity = 50 tons per day. Setup cost = $100, and storage and handling costs = $5 per ton a
year. The firm operates 200 days a year. Note: 1 ton = 2,000 pounds.
p = 50 tons per day * 2,000 pounds per ton = 100,000 pounds per day = 100,000 pounds per
day / 100 pounds per bag = 1,000 bags per day
u = 20 tons per day * 2,000 pounds per ton= 40,000 pounds per day = 40,000 pounds per day
/ 100 pounds per bag = 400 bags per day
D = 400 bags per day * 200 days per year = 80,000 bags per year
S = $100
H = $5 per ton per year = $5 per ton per year / 20 bags per ton = $0.25 per bag per year
1.
Note: Round Qp to an integer value, but round any other values to a maximum of two
decimals.
a.
Qp 
b.
I max 
2 DS
H
Qp
p
p

pu
( p  u) 
Average Inventory =
c. Run length =
Qp
d. Runs per year:
p

2(80,000)100
1,000
 10,327.97  10,328 bags
0.25
1,000  400
10,328
(1,000  400)  6,196.8 bags
1,000
I max 6,196.8

 3,098.4 bags
2
2
10,328
 10.33 days
1,000
D 80,000

 7.75 runs per year
Q 10,328
e. S = $25:
Qp 
I max 
2 DS
H
Qp
p
p

pu
( p  u) 
2(80,000)25
1,000
 5,163.98  5,164 bags
0.25
1,000  400
5,164
(1,000  400)  3,098.4 bags
1,000
𝐼𝑚𝑎𝑥
𝐷
3,098.4
80,000
)𝐻 + ( )𝑆 = (
) 0.25 + (
) 25 =
2
𝑄
2
5,164
387.30 + 387.30 = $774.60
𝑇𝐶(𝑆 = $25) = (
𝐼𝑚𝑎𝑥
𝐷
6,196.8
80,000
𝑇𝐶(𝑆 = $100) = (
)𝐻 + ( )𝑆 = (
) 0.25 + (
) 100 =
2
𝑄
2
10,328
774.60 + 774.59 = $1,549.19
Savings when S = $25 = $1,549.19 – $774.60 = $774.59 per year.
12.
Given:
p = 800 units per day
u = 300 units per day
Q = 2,000 units per batch
Company operates 250 days a year
a. Number of batches of heating elements per year:
D 75,000

 37.5 batches per year
Q
2,000
b. Amount of inventory after 2 days of production:
The number of units produced in 2 days = (2 days)(800 units/day) = 1600 units
The number of units used in 2 days = (2 days) (300 units per day) = 600 units
Inventory build up after the first 2 days of production = 1,600 – 600 = 1,000 units
Current inventory of the heating unit = 0 units
Total inventory after the first 2 days of production = Beginning Inventory + Inventory
Buildup after 2 Days of Production = 0 + 1,000 = 1,000 units.
c. Average Inventory:
I max 
Q
2,000
( p  u) 
(800  300)  1,250 units
p
800
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝐼𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 =
𝐼𝑚𝑎𝑥 1,250
=
= 625 units
2
2
d. The other component requires 4 days (including setup). Setup time for the heating
element =
0.5 days. Is there enough time to run the other component between batches of
heating elements?
How much time is available to run the other component? The other component must be
finished during the pure consumption time for the heating element. The end of the pure
consumption time is when inventory of the heating element falls to 0 units. If the other
component takes longer than the pure consumption time, we will run out of inventory of
the heating element.
𝑄 2,000
𝐶𝑦𝑐𝑙𝑒 𝑇𝑖𝑚𝑒 = =
= 6.67 𝑑𝑎𝑦𝑠
𝑢
300
This is the time between starting production runs of the heating element.
𝐶𝑦𝑐𝑙𝑒 𝑇𝑖𝑚𝑒 = 𝑅𝑢𝑛 𝑇𝑖𝑚𝑒 + 𝑃𝑢𝑟𝑒 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑇𝑖𝑚𝑒
𝑄
2,000
+ 𝑆𝑒𝑡𝑢𝑝 𝑇𝑖𝑚𝑒 =
+ .5 = 2.5 + .5 = 3 𝑑𝑎𝑦𝑠
𝑝
800
Plugging in values and solving for Pure Consumption Time:
𝐶𝑦𝑐𝑙𝑒 𝑇𝑖𝑚𝑒 = 𝑅𝑢𝑛 𝑇𝑖𝑚𝑒 + 𝑃𝑢𝑟𝑒 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑇𝑖𝑚𝑒
6.67 𝑑𝑎𝑦𝑠 = 3 𝑑𝑎𝑦𝑠 + 𝑃𝑢𝑟𝑒 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑇𝑖𝑚𝑒
𝑃𝑢𝑟𝑒 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑇𝑖𝑚𝑒 = 6.67 − 3 = 3.67 𝑑𝑎𝑦𝑠
Conclusion: There will not be enough time to run the other component because the other
component requires 4 days, which is .33 (4 – 3.67) days too many.
𝑅𝑢𝑛 𝑇𝑖𝑚𝑒 =
13.
Given:
D = 18,000 boxes/year
S = $96
H = $.60/box/year
Price Schedule:
Number of Boxes
Price per Box (P)
1,000-1,999
$1.25
2,000-4,999
$1.20
5,000-9,999
$1.15
10,000+
$1.10
a. Determine the optimal order quantity (round to an integer value):
Step 1:
Compute the common minimum point.
Q
2DS
2(18,000)96

 2,400 boxes
H
.60
This quantity is feasible in the range 2000-4,999.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower
unit costs.
𝑄
𝐷
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 + 𝑃𝐷
2
𝑄
TC2,400 =
2,400
18,000
(.60) 
($96)  $1.20(18,000 )  $23,040
2
2,400
TC5,000 =
5,000
18,000
(.60) 
($96)  $1.15(18,000)  $22,545.60
2
5,000
TC10,000 =
10,000
18,000
(.60) 
($96)  $1.10(18,000 )  $22,972 .80
2
10,000
Conclusion: Optimal order quantity = 5,000 boxes.
b. Determine number of orders per year:
D 18,000

 3.6 orders per year (round to a maximum of two decimals)
Q 5,000
14.
Given:
D = 25 stones/day * 200 days/year = 5,000 stones/year
S = $48
Price Schedule:
Number of Stones Price per Stone (P)
1-399
$10
400-599
$9
600+
$8
a. H = $2. Determine the optimal order quantity:
Step 1:
Compute the common minimum point.
Q
2DS
2(5,000)48

 489.90  490 stones
H
2
This quantity is feasible in the range 400-599.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower
unit costs.
𝑄
𝐷
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 + 𝑃𝐷
2
𝑄
490
5,000
𝑇𝐶490 = (
)2 + (
) 48 + 9(5,000) = $45,979.80
2
490
600
5,000
𝑇𝐶600 = (
)2 + (
) 48 + 8(5,000) = $41,000
2
600
Conclusion: Optimal order quantity = 600 stones.
b. H = 30% of unit cost
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until
you find a feasible minimum point.
Minimum point P = $8:
2 DS

H
2(5,000)48
 447.21  447 Not feasible
.30(8)
Minimum point P = $9:
2 DS

H
2(5,000)48
 421.64  422 Feasible
.30(9)
Step 2:
Compare the total cost at Q = 422 to Q = 600.
𝑄
𝐷
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 + 𝑃𝐷
2
𝑄
422
5,000
𝑇𝐶422 = (
) (.30 ∗ 9) + (
) 48 + 9(5,000) = $46,138.42
2
422
600
5,000
𝑇𝐶600 = (
) (.30 ∗ 8) + (
) 48 + 8(5,000) = $41,120
2
600
Conclusion: Optimal order quantity = 600 stones.
c. Lead time = 6 working days. Determine ROP:
ROP = 25 stones/day * 6 days = 150 stones.
15.
Given:
D = 4,900
S = $50
H = 40% of purchase cost
Price Schedule:
Range
Price per Unit (P)
1-999
$5.00
1,000-3,999
$4.95
4,000-5,999
$4.90
6,000+
$4.85
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you
find a feasible minimum point.
Minimum point P = $4.85:
2 DS

H
2(4,900)50
 502.57  503 Not feasible
.40(4.85)
Minimum point P = $4.90:
2 DS

H
2(4,900)50
 500 Not feasible
.40(4.90)
Minimum point P = $4.95:
2 DS
2(4,900)50

 497.47  497 Not feasible
H
.40(4.95)
Minimum point P = $5.00:
2 DS
2(4,900)50

 494.97  495 Feasible
H
.40(5.00)
Step 2:
Compare the total cost at Q = 495 to Q = 1,000, 4,000, & 6,000.
𝑄
𝐷
𝑇𝐶 = ( ) 𝐻 + ( ) 𝑆 + 𝑃𝐷
2
𝑄
495
4,900
𝑇𝐶495 = (
) (.40 ∗ 5.00) + (
) 50 + 5.00(4,900) = $25,489.95
2
495
1,000
4,900
𝑇𝐶1,000 = (
) (.40 ∗ 4.95) + (
) 50 + 4.95(4,900) = $25,490
2
1,000
𝑇𝐶4,000 = (
4,000
4,900
) (.40 ∗ 4.90) + (
) 50 + 4.90(4,900) = $27,991.25
2
4,000
𝑇𝐶6,000 = (
6,000
4,900
) (.40 ∗ 4.85) + (
) 50 + 4.85(4,900) = $29,625.83
2
6,000
Conclusion: Optimal order quantity = 495 units. Note: The total cost for 1,000 units is only
$.05 different.