132
Problem 22 – Table 7. Maintenance labor hours (Low Skill Technician)
Year
Configuration
1
2
3
4
5
6
7
8
9
10
Total
Configuration
“A”
0
0
37
74
150
225
225
225
130
93
1,159
Configuration
“B”
0
0
29
59
117
117
117
117
102
74
912
Maintenance labor hours = (Maintenance Actions)
()M
(Number of Personnel)
Table 8 includes both the low–skilled and high–skilled technicians for corrective
maintenance (split evenly). For instance, the corrective maintenance labor hours for
Unit “A” of Configuration “A” for Year 3 is (18 Maintenance Actions) (5 hrs.) (2
Technicians) = 180 MLH. This is from MMH = (Maintenance Actions)
(Mct
or
)Mpt
(Number of People).
Problem 22 – Table 8. Maintenance Labor Hours at Intermediate Level
Life–Cycle Year
Item
1
2
3
4
5
6
7
8
9
10
Total
Configuration “A”
Corrective
Maintenance
Unit “A”
0
0
180
370
730
1,100
1,100
1,100
640
460
5,680
Unit “B”
0
0
232
464
936
1,400
1,400
1,400
816
584
7,232
Unit “C”
0
0
28
56
116
176
176
176
104
72
904
Total
0
0
440
890
1,782
2,676
2,676
2,676
1,560
1,11
6
13,816
Preventive
Maintenance
Unit “A”
0
0
320
640
1,280
1,920
1,920
1,920
1,120
800
9,920
Configuration “B”
Corrective
Maintenance
133
Unit “A”
0
0
180
370
730
1,100
1,100
1,100
640
460
5,680
Unit “B”
0
0
150
290
580
880
880
880
510
370
4,540
Unit “C”
0
0
36
72
138
210
210
210
120
90
1,086
Total
0
0
366
732
1,448
2,190
2,190
2,190
1,270
920
11,306
Preventive
Maintenance
Unit “B”
0
0
240
480
960
1,440
1,440
1,440
840
600
7,440
Problem 22 – Table 9. System XYZ Operator Labor Hours
Year
Item
1
2
3
4
5
6
7
8
9
10
Total
System XYZ
0
0
146
292
584
876
876
876
511
365
7,300
The next step is to determine operator and maintenance personnel costs by applying
the above MLH and Maintenance Action values, and the individual cost factors stated
in the problem. The results are given in Table 10.
For corrective maintenance at the intermediate level, one–half of the maintenance
labor hours are at $20 per hour and one–half are at $30 per hour.
Maintenance facility costs are based on the total corrective and preventive
maintenance labor hours at the intermediate facility multiplied by the given burden
rate (for each configuration).
Maintenance data costs are based on the number of corrective and preventive
maintenance actions multiplied by the dollar rate per maintenance action.
134
Problem 22 – Table 10. Personnel Costs ($)
Life–Cycle Year
Item
1
2
3
4
5
6
7
8
9
10
Total Cost
($)
Configuration “A”
1. Operator Cost
0
0
5,840
11,680
23,360
35,040
35,040
35,040
20,440
14,600
181,040
2. Maintenance Cost
Organization
0
0
740
1,480
3,000
4,500
4,500
4,500
2,600
1,860
23,180
Intermediate
a. Corrective Maint.
0
0
11,00
22,250
44,550
66,900
66,900
66,900
39,000
27,900
345,400
b. Preventive Maint.
0
0
9,600
19,200
38,400
57,600
57,600
57,600
33,600
24,000
297,600
Total
0
0
27,180
54,610
109,310
164,040
164,040
164,040
95,640
68,360
847,220
Configuration “B”
1. Operator Cost
0
0
5,840
11,680
23,360
35,040
35,040
35,040
20,440
14,600
181,040
2. Maintenance Cost
Organization
0
0
580
1,180
2,340
4,540
3,540
3,540
2,040
1,480
18,240
Intermediate
a. Corrective Maint.
0
0
9,150
18,300
36,200
54,750
54,750
54,750
31,750
23,000
282,650
b. Preventive Maint.
0
0
7,200
14,400
28,800
43,200
43,200
43,200
25,200
18,000
223,200
Total
0
0
22,770
45,560
90,700
136,530
136,530
136,530
79,430
57,080
705,130
Personnel costs are entered for each configuration in Table 11.
Problem 22 – Table 11. Summary of Operations and Maintenance Cost ($)
Life–Cycle Year
Item
1
2
3
4
5
6
7
8
9
10
Total Cost
($)
Configuration “A”
1. Personnel Cost
—
—
27,180
54,610
109,310
164,040
164,040
164,040
95,640
68,360
847,220
2. Material Cost
a. Spare Units
—
42,000
21,000
42,000
21,000
—
—
—
—
—
126,000
b. Component Spares
—
—
15,500
31,250
62,750
94,250
94,250
94,250
55,000
39,250
486,500
3. Maintenance Facilities
—
—
760
1,530
3,062
4,596
4,596
4,596
2,680
1,916
23,736
4. Maintenance Data
—
—
1,850
3,725
7,475
11,225
11,225
11,225
6,550
4,675
57,950
Total
—
42,000
66,290
133,115
203,597
274,111
274,111
274,111
159,870
114,201
1,541,406
Configuration “B”
1. Personnel Cost
—
—
22,770
45,560
90,700
136,530
136,530
136,530
79,430
57,080
705,130
2. Material Cost
a. Spare Units
—
46,000
23,000
46,000
23,000
—
—
—
—
—
138,000
b. Component Spares
—
—
11,750
23,500
46,500
70,250
70,250
70,250
40,750
29,500
362,750
3. Maintenance Facilities
—
—
606
1,212
1,408
3,630
3,630
3,630
2,110
1,510
18,746
4. Maintenance Data
—
—
1,475
2,950
5,850
8,825
8,825
8,825
5,125
3,700
45,575
Total
—
46,000
59,601
119,222
168,458
219,235
219,235
219,235
127,415
91,800
1,270,201
136
Spare part costs are related to Unit spares (one set per intermediate level maintenance shop and one set at the depot), and
Component spares which are a function of individual maintenance actions. Spares cost are summarized in Table 12.
Problem 22 – Table 12. Spares Cost ($)
Life–Cycle Year
Item
1
2
3
4
5
6
7
8
9
10
Total Cost
($)
Configuration “A”
Spare Units
—
42,000
21,000
42,000
21,000
—
—
—
—
—
126,000
Component Spares
Corrective Maintenance
—
—
13,500
27,250
54,750
82,250
82,250
82,250
48,000
34,250
424,500
Preventive Maintenance
—
—
2,000
4,000
8,000
12,000
12,000
12,000
7,000
5,000
62,000
Total
—
42,000
36,500
72,250
83,750
94,250
94,250
94,250
55,000
39,250
612,500
Configuration “B”
Spare Units
—
46,000
23,000
46,000
23,000
—
—
—
—
—
138,000
Component Spares
Corrective Maintenance
—
—
9,750
19,500
38,500
58,250
58,250
58,250
33,750
24,500
300,750
Preventive Maintenance
—
—
2,000
4,000
8,000
12,000
12,000
12,000
7,000
5,000
62,000
Total
—
46,000
34,750
69,500
69,500
70,250
70,250
70,250
40,750
29,500
500,750
138
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(b) Problem 22 – Figure 1. System XYZ Configuration “B” Cost Profile
by Benjamin S. Blanchard and Wolter J. Fabrycky. © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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