SECTION III Service Systems
6.3 Method Application: Sample Calculations for Evaluating and Tracking Costs and Benefits
of Proposals, Projects, and Programs
A series of examples is presented to illustrate the application of the equations presented in the previous section. The reader is reminded that the primary purpose for the chapter is to present a model of cost categories, and that specific equations are only suggested methods for obtaining line item values for the Program Record and Ergonomic Project Worksheets. Companies may have established methods for estimating many of these line items. Additionally, the references that are cited throughout the chapter may provide some alternative calculation methods that are better suited for a particular company.
Example 1: A. Ergonomics Program — Ergonomics Team Meetings
Company XYZ’s ergonomics team meets once a week for 1 hour. The team consists of three hourly operators (at $10/hr), one engineer, the plant nurse, the plant manager, the human resources manager, and one first line supervisor. Accounting, the plant manager, and the HR manager have decided that a charge of $25/hr per nondirect team member is appropriate. The cost of each meeting would then be calculated as follows, using Equation 1:
3Note that this cost could be workers’ compensation, benefits, and taxes paid directly by the company, or this cost could be a portion of the insurance premium the company pays out that can be fairly attributed to the employee’s absence.
C W T
Meeting cost = C C
hr 1 hr meeting meetings month 3 operators hr hr meeting 4 meetings month 5 members
month
LP h h
LP-direct labor LP-indirect labor
= ×
+
=
(
× × ×)
+× × ×
( )
=
$ $
$
10 4
25 1
620
6-14 Occupational Ergonomics: Design and Management of Work Systems
Example 2: B. Turnover and Training — On-the-job training
Operation E has an annual turnover rate of three employees. The hourly wage cost to the company is $16/hr (includes benefits, taxes); employees work 8 h per day. It takes about 90 days for an employee to become fully productive on this job. Over the 90 days, a new employee typically performs at about 70% of the efficiency of a seasoned employee. The annual cost of this turnover would be calculated as follows per Equation 2:
Therefore, the annual cost to replace three employees would be $10,368 ( = $3456 × 3).
If turnover were reduced to one employee per year, the savings attributed to the ergonomics program would be calculated as follows:
Example 3. B. Turnover and Training — Overtime to compensate for lost production
Management estimates that Operation K runs ten Saturdays each year to compensate for reduced pro- duction associated with training new employees. Hourly wage is $18/hr; overtime is time and a half. Two operators are required. Use Equation 3 to calculate overtime costs.
Assume employees were able to learn the job more quickly, as a result of changes in training methods or production methods, for example. If new employees were able to learn the job more quickly, which resulted in a reduction of overtime to only four Saturdays per year, the following recurring savings could be attributed to the ergonomics program:
Example 4: B. Turnover and Training — Costs due to an open position Part a: Lost production due to an open position. Operation G loses 3 days worth of production for each turnover due to a temporary vacancy. This production is not made up through overtime. The turnover rate is 4 employees/year. The hourly wage cost to the company is $11/hr (includes benefits, taxes);
employees work 8 h per day. Lost production could be calculated using Equation 4.
C W eff T
Annual cost replaced employee = $16 hr 8 hr day days employee year
LC= d× −
( )
× d(
×)
× −( )
×= 1
1 7 90 3456
.
$
Savings = Cost Cost
year
after change− before change
= −$3456− −$10 368, =$6912
C W T
Cost = $9 hr 8 hr day 10 days 2 employees year
OT = Eh× h
× × ×
=$1440
Savings = Cost Cost
year
after change− before change
= −$576− −$1440=$864
Part b. Overtime to compensate for lost production. If, instead, management opts to utilize overtime to make up for the lost production, costs and savings would be calculated as follows, based on the same turnover rate reduction from 4 per year to 2, an hourly wage cost of $11/hr, and 3 days of overtime per turnover. Using Equation 3,
Example 5: C. Absenteeism
Department S has an average of 6 lost time back injuries per year. The company is self-insured, and so pays all medical and compensation to sick or injured employees. Average lost time for a back injury in this department is 10 days; average medical costs are $5000. Absent workers are not replaced. Compen- sation costs are ⅔ the usual $9/hr labor cost.
Savings = Cost Cost
year
after change− before change
= −$528− −$1056=$528
C W T
Savings = Cost Cost
Hourly excess overtime rate hours days number of turnovers Hourly excess overtime rate hours days number of turnovers
hr 8 hr 3 days 2 turnovers hr 8 hr 3 days 4 turnovers
year
OT Eh h
after change before change
after change
before change
= ×
−
=
(
× × ×)
−× × ×
( )
=
(
× × ×)
−× × ×
( )
= − − − =
$ . $ .
$ $ $
5 50 5 50
264 528 264
Annual costs back injury = Cost Cost Cost hr 8 hr day 10 days
hr 8 hr day 10 days
back injury
compensation+ medical+ lost production
= × × +
+
× ×
= + + =
$ $ $
$ $ $ $
6 5000 9
480 5000 720 6200
6-16 Occupational Ergonomics: Design and Management of Work Systems
Therefore, total cost for six back injuries in a year would be $37,200 ( = $6200 × 6). A proposed change in methods is expected to reduce the number of back injuries to 3 per year. The expected savings would be:
Example 6: D. Productivity
Production was 80 units per 8 hours for Operation F (10 units/hour, or 0.1 hour/unit). A 13% reduction in per unit production time was realized following methods improvements. Direct labor costs are $12/hr.
Using Equation 5,
Annual savings would be based on the number of units produced.
Example 7. E. Rework and Scrapped Product — Rework — Direct labor costs/savings
Records show that on average 1 unit in 50 requires rework in Department C. The production rate is 800 units/day; labor costs are $10/hr. On average rework time is 1 min/unit. Using Equation 6,
The annual rework cost for a production rate of 800/day, running 250 days per year would be $667 ( = $0.0033 × 800 × 250). Note that if a production worker is pulled off production work to perform rework tasks, lost production should also be accounted for.
Example 8: E. Rework and Scrapped Product — Scrapped Product
Records show that on average 1 unit in 400 is scrapped in Department C. The production rate is 800 units/day; labor costs are $10/hr; direct material costs are $1.20/unit. Salvageable materials are valued at
$.30/unit
Savings = Costafter change−Costbefore change
= −$18 600, − −$37 200, =$18 600,
S W T
Savings per unit = $12 hr 13% 0.1hr unit unit
U= h× eff × RU
× ×
=
∆
$ .0 156
CR W R T
Rework direct labor cost = labor cost rework rate time requirement
= $10 hr 1 unit 50 production units 1 min unit 1 hr 60 min production unit
dl = h× pu× ru
× ×
× × ×
=$ .0 0033
Scrapped product cost unit = direct labor + direct materials salvage value
hr 8 hr day units unit unit
unit
(
−)
=
[ (
×) ( )
+]
−=
$ $ . $ .
$ .
10 800 1 20 0 30
1 00
Example 10: Capital Expenditure
Scenario. BVB Company is evaluating the purchase of a lift table for a particular location in its plant.
The rate of occurrence of back injury at that location averages one in three years. The average cost of a back injury is $5200, per the company’s insurance carrier and the human resources director (Category C. Absenteeism). The lift table will cost $3200 (Category I. One-Time Costs), and is expected to last for 5 years, with proper maintenance (expected maintenance costs of $500/year, charged under Category G.
Overhead). The lift table is expected to improve productivity by 5% (current production rate is 40 units per hour, 7 hours per day, 250 days per year; wages for employees at that location are calculated at $15/hr;
appears under Category D. Productivity). To be conservative, these are the only benefits that are used to evaluate the purchase. If the company has established an MARR of 10%, should the lift table be purchased, based on economic projections?
Solution. Economic decision analysis always involves comparison of alternatives, even if there is appar- ently only one proposal, because Do Nothing is always considered an alternative.
First consider the Do Nothing option. If back injury rate is one in three years, with an average cost of $5200, the annual cost of a back injury can be assessed as $5200/3, or $1733/year. Use a cash flow diagram to depict this alternative (refer to Figure 6.3a).
Next, evaluate each of the applicable line item categories for the lift table alternative. The initial cost is $3200, but maintenance costs of $500/year are also anticipated. Productivity is anticipated to improve by 5%. Using Equation 5,
The costs and benefits associated with the lift table alternative are presented in Figure 6.3b. Figure 6.3c combines the two, and is the cash flow diagram that will be analyzed. Each of the four methods of analysis will be demonstrated. A project life of 5 years will be used, since that is the life of the table.
PB = C S
Project cost, C = $12 belt 10 belts = $120
Savings unit = $15 hr employee 10 employees 1% 0.33 hr unit = $0.495 unit Annual savings, S = $0.495 unit 24 units day 250 days yr = $2970
Payback period, PB = $120 $2970 = 0.04 years, or about1 2 month.
×
× × ×
× ×
S W T
Savings per unit = $15 hr 5% 1 hr 40 units unit
Annual savings = $0.0188 unit 280 units day 250 days yr = $1316 yr
U= h× eff × RU
× ×
( )
=× ×
∆
$ .0 0188
6-18 Occupational Ergonomics: Design and Management of Work Systems
Setting i = 10%, n = 5 years, results in an NPW = $6463. Since NPW is greater than $0, the project should be undertaken.
Setting i = 10%, n = 5 years, results in an NAW = $1705. Since the NAW is greater than $0, the project should be undertaken.
B/C analysis can be performed with PW or AW calculations. Using AW, i = 10% and n = 5 years:
Therefore, B/C = 3049/1344 = 2.3. Since this exceeds 1, the project should be undertaken.
FIGURE 6.3 Cash flow diagrams for economic assessment of a) Do Nothing alternative, b) lift table purchase, and c) comparison of lift table purchase to Do Nothing alternative.
NPW = PW PW PW PW
i i 1+ i
i i 1+ i
i i 1+ i
savings in back injury costs productivity savings lift table cost maintenance costs n
n
n n
n n
+ − −
=
[ (
+)
−]
( )
+(
+)
−[ ]
( )
− −(
+)
−[ ]
1733 1 1
( )
1316 1 1
3200 500 1 1
NAW = AW AW AW AW
i 1+ i i
savings in back injury costs productivity savings lift table cost maintenance costs n
n
+ − −
= + −
( )
(
+)
−[ ]
−1733 1316 3200
1 1 500
AWB = AW AW
AWC = AW AW
savings in back injury costs productivity savings
lift table cost maintenance costs
+ = + =
+ = + =
1733 1316 3049 844 500 1344