SECTION III Service Systems
8.5 Time Savings of Human Operators
(9)
Accordingly, it is useful to recognize repeated cash flow patterns over time and use the relationship above to simplify calculations.
8-12 Occupational Ergonomics: Design and Management of Work Systems
Some Alternative Forms of Analysis with This Example
An alternative but analogous form of analysis is to use continuously compounded interest at the equivalent MARR or 11.333% in this example. The present worth of existing operations without the tool or device can thus be computed as follows:
With the tool or device, the present worth calculation is:
The resulting difference of $261 is the net present worth of savings over the principal. Although, this second form of analysis gives the same result, accounting for the actual time differences within the year shows that ignoring within-year time effects can sometimes mislead the decision maker.
An associated form of analysis is annual worth analysis. This annual worth amount can be computed directly from the present worth using the annual-worth-to-present-worth factor as follows:
The value of C that satisfies the above equation is the annual worth, or $276. One could compute the equivalent annual worth without and with the tool or device as follows:
where the two respective C values are Cw/o device = $4,922 and Cw device = $4,646. The difference between these two annual worths is the net annual worth of the cost savings which is $262 per year.
Another form of analysis which is quite popular but sometimes very misleading is the internal rate of return basis for comparing alternative projects. In such cases, the internal rate of return over an entire year is found by setting the present worth of the savings, without specifying the interest rate, against the investment principal. For the present example, the calculations could be:
200 (1 + i) = $4,699 or i = 22.495
This result shows that this investment earned 2,250% of the supplied principal. Here again, this analysis ignores within-year time differences. While the internal-rate-of-return method of analysis has difficulties, it does point out those actions that should be avoided. If the internal rate of return falls below MARR, that action should not be undertaken. Also, with two alternative actions that essentially do the same thing but one costs more and provides a better return, one can test the more expensive alternative to see if the added return earns at least the MARR on the added investment. Only when it does can the more expensive alternative be considered economical.
P = $24, 000
[
1 e− −0 11333 0 19606( )]
=0 11333 4 654
. .
. $ , .
P = $24, 000
[
1 e− −0 11333 0 17645( )]
+ =0 11333 200 4 393
. .
. . $ , .
P = C 1 e
j jk C 1 e
[
− −]
=[
− −. ( )]
=. .
11333 1
0 11333 261
C e
C e
w o device
w device
1
0 11333 4 654
1
0 11333 4 392
11333 1
11333 1
[
−]
=[
−]
=− ( )
− ( )
.
.
. ,
. ,
(10) In Equation (10), α and β are constant parameters of the model. In effect, this model predicts performance on the next product unit as the current performance time multiplied by the constant α plus a constant β. Since learning is improvement, α is defined as a fraction (0 < α < 1). The values of yn+1 decrease progressively with each new unit produced but at a decreasing rate which approaches the asymptote y*:
(11) The asymptote associated with this learning curve provides an advantage in using this model of learning over many others.4 A more effective prediction equation for this discrete model is as follows:
(12) Equation (12) describes the performance time on the nth production unit based on performance on the first production unit and the asymptote y*. Another useful feature of this learning curve model is the differences between sequential performance times, calculated as follows:
(13) If performance times are generally decreasing, the average ∆yn is negative. Also, as Equation (13) denotes,
∆yn is a linear function of yn. With a short sequence of performance times, then one can compute the forward differences ∆∆∆∆yn as a function of yn. For instance, suppose that the following performance times were recorded on the first five production units:
3The discrete exponential model shown by Pegels (1969) and by Buck, Tanchoco, and Sweet (1976) is based on the first-order difference equation with constant coefficients which Goldberg (1961) describes in detail. Hutchings and Towill (1975) describe a similar but not identical model.
4The powerform learning curve, noted by Snoddy (1926) and later by Wright (1936), which is much better known in the U.S., does not have a natural asymptote and so an artificial asymptote at n = 1000 is often used. Hax and Majluf (1982) used this same model for describing improvements over time by whole industries.
n
1 2 3 4 5
yn
time (hours) = 10.000 9.920 9.842 9.765 9.689
∆ yn = –0.0800 –0.078 –0.077 –0.0755 —
yn+1=αyn+β
y* =1 β
α
−
yn=αn
[
y1−y *]
+y *∆yn=yn+1−yn=αyn+ −β yn=
(
α−1)
yn+β8-14 Occupational Ergonomics: Design and Management of Work Systems
When ∆ yn is plotted as a linear function of yn, the following equation results:
∆ yn = 0.1055 – 0.01853 yn
This result is obtained by using a simple linear regression routine on a calculator where the coefficient of determination is r = 0.99. Based on Equation (13), α = 0.98 and β = 0.12. By substituting these values in Equation (11), one finds the asymptote y* is 6.0 hours per unit and the production time on each production unit can be described with the following equation:
With this discrete exponential model of learning, the total production time required to produce a total of n units is as follows:
(14) In the example above, the total production time can be described with the following equation:
In Equation (14) the last term on the right-hand side is the asymptotic values times the number of items produced. In essence, that value is the steady-state value and the first term in that equation shows the transition time required by the learning process. For the example above, the transition and total production hours for selected values of n are shown in Table 8.4. The last row of this table reports the transition production time (added time for learning) as a function of the total time. As the production run gets longer, a smaller and smaller fraction of production time is required for learning.
An economic question with production is the economic production run length. Stated otherwise,
“How many items should the company make each time it starts up production of this item?” Typically, a company builds up a component that is used in several products over the course of a year. If it makes the entire yearly production in one single run, the company takes advantage of the learning process and minimizes the number of production setups needed per year. But with a single annual production run, many items must be placed in inventory until needed and the inventory costs are often substantial.
Consider as an example the product unit made by the company which needs 80 lots per year uniformly over the year. This usage rate is about 1.6 units per week or 6.67 per month, averaging 0.04 per production hour. Now let us consider several production strategies in separate sections below.
Produce Lots as Needed
At one extreme, the company could make a unit of the product during the first 10 hours and then cease production. After every 15 production hours, the company could start producing another unit and then
TABLE 8.4 Some Basic Data on the Learning Curve in the Numerical Example n
10 20 30 40 50 60 70
Total Hours Yn = 96.59 186.48 270.90 350.86 427.16 500.49 571.38 Trans. Hours 36.59 66.48 90.90 110.86 127.17 140.49 151.38
% Trans. 37.9 35.9 33.2 31.6 29.8 28.1 26.5
yn=0 98 10 00. n
[
. −6 00.]
+6 00.Yn y y * y * n
n
= − 1
−
[
−]
+1 1
α α
Yn n
= − n
−
[
−]
+1 0 98
1 0 98. 10 0 6 0 6 0
. . . .
if it started immediately. However, only the initial run of one lot starts immediately, the remaining 79 lots occur every 15 production hours. Accordingly, the total PW of production is:
However, before each production run can occur, a setup crew comes in and gets everything ready.
Setup costs are estimated at 15 operator hours times $12/hr or $180 each, and each is done off shift just prior to production. So the setup cost component present worth is:
The only other cost component to be considered is the inventory cost (also known as storage or holding cost). In the particular case where only one lot is produced at a time and production is timed to finish just as the lot is issued, there are no inventory and zero inventory costs. The other two components sum to $22,687.68 each year.
Make the Entire Year’s Production at One Time
At the other extreme, the company could make the entire year’s requirements in a single production run.
If the learning curve continued as described, the required production time yielded by Equation (14) would be computed as follows:
For a company which works single shifts all year, this production strategy requires 16.65 weeks, 83.24 shifts, or 0.32 years. Actual production costs are $12/hour or $24,000 per year at a uniform rate so that present worth of the production operation is a step function of $24,000/year over the 0.32 years as:
Total PW = $119.97 1+ e.11333 e e
25
− 2000 − −
+ +…+
⎡
⎢⎢ ⎤
⎥⎥
.11333 50 .
2000 113331975
2000
Total PW = $119.97 e e
−
−
−
−
⎡
⎣
⎢⎢
⎢
⎤
⎦
⎥⎥
⎥
=
.
. $ , .
11333 25 200080 11333 25
2000
1 1
9 073 71
Total PW = $180.00 e e
−
−
−
−
⎡
⎣
⎢⎢
⎢
⎤
⎦
⎥⎥
⎥
=
.
. $ , .
11333 25 200080 1133325
2000
1 1
13 613 97
Y80= − h
−
[
−]
+( )
=1 0 98
1 0 98 10 0 6 0 6 0 80 640 27
. 80
. . . . .
PW e
production= ⎛ −
⎝⎜
⎞
⎠⎟=
− ( )
$ . $ , .
. .
24000 1
11333 7 542 41
11333 32
8-16 Occupational Ergonomics: Design and Management of Work Systems
During these 640.27 production hours, the company issues some components from production, and inventories the production surplus. The number of issues during production can be described as follows:
If 80 lots are produced during production but during the same duration 25.6 are dispensed, the net inventory at the end of production is 54.4 lots. Inventory increases from the start of production. Although the rate of lot production is not constant, due to the learning curve, an average rate would be 80/640.27 or about 0.125 per production hour. Lots are issued out of inventory at the rate of 0.04 per hour so that inventory builds during production at approximately 0.085 per hour. Accordingly, the inventory cost component during the production of 80 lots is approximated by a ramp which increases at the rate of 0.085 per production hour or 170 lots per year times $1000 per lot per year. The present worth of this inventory buildup is:
There is also the single setup cost of $180 at time zero. The inventory cost from the end of production over the rest of the year decreases uniformly, and so the inventory cost can be described as a down-ramp (see Table 8.2). The step portion of this down-ramp has an initial inventory level of 54.4 lots at a $1000 per year per lot is:
At the end of the production (0.32 years) inventory is at the level of 54.02 lots. The present worth for the remaining inventory can be found by computing the present worth of the maximum inventory continuing over the remainder of the year minus the present worth of a ramp that describes the issuing rate from the inventory. The first part of that computation is a step of 54.4 times $1000, or:
The other part is the ramp or:
TABLE 8.5 The Effect of the Production Run Size on Cumulative Storage (Inventory) Requirements no. made at a time
10 20 30 40 50 60 70 80
Manuf. hours 96.59 186.48 270.90 350.86 427.16 500.49 571.38 640.27
Issued during manuf. 3.86 3.60 3.38 3.56 3.05 2.93 2.84 2.76
Stored during manuf. 6.14 6.40 6.62 6.44 6.95 7.07 7.16 7.24
Cumul. storage 6.14 12.54 19.16 25.60 32.55 39.62 46.78 54.02
640 27. production hours 0.04 lots per hour = 25.6 lots×
PW e e
inv= ⎛ −
⎝⎜
⎞
⎠⎟−⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎢
⎤
⎦⎥
⎥=
− ( ) − ( )
$ ,
.
.
. $ , .
. . . .
170 000 1
113333
0 32
11333 8 496 4
11333 32 2
113333 32
PW e
step= ⎡ −
⎣⎢ ⎤
⎦⎥ =
− ( )
54400 1
11333 35 603
11333 68
. .
. $ , .
PW e e
ramp= ⎛ −
⎝⎜
⎞
⎠⎟−⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎢
⎤
⎦⎥
⎥=
− ( ) − ( )
$ ,
.
.
. $ , .
. . . .
80 000 1 11333
0 68
11333 17 572
11333 68 2
11333 68
where the difference is $18,031. However, that final inventory cost starts at 0.32 years so that the present worth is:
The total inventory present worth using a single production run is calculated as the sum of the first phase during production and the second phase after production or:
Two other cost components are the production work present worth of $7,742 and the single setup of
$180 for a grand total of:
Alternative Strategies between These Extremes
There are many alternative production strategies, starting with making 2 items in each run and restarting production every 40 hours of operation time. Based on the learning curve above, the initial production would require 19.92 hours. During the 20.08 hours before restarting production (i.e., 40 hours less 19.92 hours), forgetting would be expected to be very small and so the next learning curve is estimated to start at 9.92 hours for the first unit. Continuing this strategy of 2 lots at a time would result in the annual 2000 h divided by 40-hour cycles between successive production starts for a total of 50 cycles a year.
Other strategies of 3, 4, 5, …, 10 lots of production in each cycle are shown in Figure 8.1. Longer production runs are illustrated along with longer gaps between subsequential runs. Table 8.6 provides additional data on these alternative strategies. It is assumed in this example that the first cycle is the initial production run, so the second cycle production is affected by forgetting after the first cycle, but all subsequent cycles have production times similar to those shown in the second cycle. These cycle times were based on the first cycle’s initial time of 10 hours and on the fact that the second and subsequent cycles started production 10 hours before the next lot was needed for issue.
In this general situation, an initial cycle follows the learning curve stated above. Following that, a forgetting curve occurs that shows the amount of learning lost before production restarts. This forgetting curve is based on the calculation 150 operating hours (3 months) as the longest time knowledge is retained. It is assumed here that the time gap between ending production on the nth run and restarting production on the next run will increase forgetting (or decrease learned skill). For the time gap g, the
FIGURE 8.1
PWdownramp=$18 031, e.11333 .32( )=$17 389, .
PWinventory=$ ,8 496. $+ 17 389, =$25 885, .
PWtotal=$ ,7 742+25 885 180, + =$33 807,
8-18 Occupational Ergonomics: Design and Management of Work Systems
fraction (1- g/150 h) is multiplied by (y1 – ym) and that value is subtracted from y1. That computed value is repeated in the next learning curve until the actual curve is lower. This computational procedure of forgetting brings the process back to an earlier value on the original learning curve. For example, suppose the company produced 7 lots at a time. The initial production follows the learning curve from 10.0 down to 9.543 h. The amount learned is 10.0 – 9.473 h or a learning effect of 0.5275 h. The time gap between two successive production runs is 107 hours. Thus, the fraction computed is (1 – 107/150) or 28.67% of the 0.5275 h or 0.1512 h retained. This is 10.0 – 0.5275 = 9.4725 h for the restart initial time. Arzi and Shtub (1996) and Dar-El and Vollichman (1996) found that interrupted tasks which are restarted come back to the original learning curves after a short transition period. Their transition (forgetting) period was not precisely described and the one shown above is my conjecture. Table 8.6 shows production durations on the 1st run where the initial learning occurred. On the second and subsequent runs there is some forgetting as reflected by the time required to make the first unit, and the combination of learning and forgetting is reflected by the duration of the 2nd run compared to the 1st run. Note in the 2nd run how the time required to make the first unit decreases down to the case of 6 lots production and thereafter increases. Longer run lengths have greater learning but also longer time gaps between production runs.
Between runs of length 6 and 7, the gap effect overcame the longer learning effect.
Specific production and inventory levels are given in Figure 8.2 for the case of producing 7 lots in a production run. Note that this figure shows the 1st and 2nd production runs at the top. At the bottom of this figure are part of the 10th run, the 11th run and part of the last (12th) run. Runs 2 through 11 are identical in character. Since 11 runs each produce 7 lots, the combined production of 77 lots is 3 short of the required 80 lots per year and the final production run completes those 3 lots and inventory goes to zero on New Year’s eve. As there are only 3 different types of runs, economic calculations for each of the 3 types are computed separately in Table 8.7. Each separate calculation starts with the temporary assumption that the production started at time zero. Within each production run the initial phase of manufacturing and dispensing inventory, while the final phase is only dispensing inventory.
These two phases are computed separately in Table 8.7. Note that all of the coefficients are based on an annual basis and so the time units in the calculations are yearly fractions. These units must agree.
Production operations cost of $24,000 per year uniformly over the actual length of the production phase and the present worth calculations are for a step function. Inventory costs during the production and dispensing phase are an up-ramp whereas during the dispensing phase there is a down-ramp, where present worth is computed as a step at the maximum inventory in that run minus an up-ramp at a cost of $80,000 per year. Each type is similarly calculated. However, the 2nd run has 10 replications and setup costs, $180 each, for runs 2 through 12 with 11 replications. The present worth of the final production run is shown under the temporary assumption of a zero time start and then it is corrected for starting TABLE 8.6 Production Run-Size Problem with Learning and Forgetting Curves. The 1st Run is 10 h Short to Produce a Minimum Inventory of 0.4 Lots. The 2nd through the Next-to-the-Last Run, Where the Forgetting Curve Occurs, Completes the Year with Zero Inventory and This Last Run is Either Like the 2nd But with 10 h More Time or the Last Run Produces the Remainder of the 80 Units Not Produced Previously with Fixed Run Sizes.
No. Cycle 1st Run 2nd Run to N-1th. Run Last Run
Run Length Runs Time Duration Gap Duration 1st Unit Gap Duration Gap
1 80.0 25 10.0 5.00 9.925 9.925 15.07 9.925 25.075
2 50.0 50 19.92 20.08 19.65 9.863 30.35 19.65 40.35
3 26.67 75 29.76 35.24 29.23 9.820 45.77 19.64 40.36
4 20.0 100 39.53 50.47 38.72 9.793 61.28 38.72 71.28
5 16.0 125 49.22 65.78 48.18 9.785 76.82 48.18 86.82
6 13.33 150 58.83 81.48 57.63 9.789 92.37 19.58 40.42
7 11.42 175 68.31 96.69 67.38 9.84 107.62 29.46 55.55
8 10.0 200 77.78 112.22 74.52 9.84 125.48 74.52 135.48
9 8.89 225 87.18 127.82 86.44 9.90 138.56 77.77 132.23
10 8.0 250 96.59 143.41 96.27 9.96 153.73 96.27 163.73
0.9575 years instead. Finally, present worth computations of all 3 types of runs, including the initial setup cost, are combined to show the overall present worth for a year.
FIGURE 8.2
TABLE 8.7A Present Worth Computations of the 1st Production Run GENERAL COMPUTATIONS OF THE PRODUCTION-RUN-SIZE PROBLEM 1st Production Run —
Production Labor Cost Present Worth
$12.00/hr × 2000 hrs/year = $24,000/year
Inventory Cost During Production — Up Ramp 0 to 0.0342 y-
@ .06247 lots/hr × 2000hrs/year × $1000/lot = $124,940/year
Inventory Cost During Dispensing Down Ramp from .0342 to .04834 y
@ 0.04 lots/hr × 2000hrs/year × $1000/lot = $80,000/year reaching 4.27 lots inventory
The Total for the 1st Run excluding Setup Cost is:
$818.25 + 72.82 +204.99 – 74.82 = $1,021.05 PW = $ 24, 000 year 1 e
0 11333 818 25
0 11333 0342
⎛ −
⎝⎜
⎞
⎠⎟=
−. (. )
. $ .
PW = $124, 940 ⎛1 e− − e
⎝⎜
⎞
⎠⎟=
−. (. ) −. (. )
.
.
. $ .
1133 0342 2
1133 0342
1133
0342
1133 72 63
PW year e
step= ⎛ − e
⎝⎜
⎞
⎠⎟ =
− ( )
− ( )
$ , . $ .
. .
. .
4 270 1
0 11333 204 99
0 11333 04834
1133 03416
PW yr e e
ramp= ⎛ − − e
⎝⎜
⎞
⎠⎟ =
− ( ) − ( )
− ( )
$ ,
.
.
. $ .
. . . .
. .
80 000 1 1133
0483
1133 74 82
1133 0483 2
1133 0483
1133 0342
8-20 Occupational Ergonomics: Design and Management of Work Systems
Table 8.8 summarizes the production, setup, and inventory costs present worth for each of the alter- native production run lengths from 1 at-a-time to 10 at-a-time production. All of the calculations follow similar to those in Table 8.7 covering a single year. These data show the minimum cost strategy of those considered is making run length of 7 lots per run.
There is an economic principle which reappears in the literature of practitioners, but it is not frequently stated. The principle is, “It is rarely very important to be precise in finding the optimal economic alternative, but it is imperative to be near.” The total cost differences between the least cost alternative strategies at producing 7 lots during a production run and those quite similar are quite small. In fact, those differences are likely much smaller than the precision associated with the cost estimates. But differences between the least cost and those alternatives much farther from the minimum cost are very striking. Hence, it is very important to identify the approximate optimum economic choices. There is an ergonomic principle authored by Helson (1949) which is very similar. It is known as his U-Hypothesis and it is: “For most variables of concern in ergonomics, performance, as an inverted function of that variable, is U-shaped. The bottom of the U is nearly flat but the extremes rise almost vertically.” Here again, the point is the same. One location at the bottom of the U is much the same as another, but being TABLE 8.7B Present Worth Calculations for the 2nd Production Run of the Learning and Forgetting Example 2nd Production Run (Tentatively Consider Start at Time 0)
Production Labor Cost Present Worth
$12.00/hr × 2000 hrs/year = $24,000/year
Inventory Cost During Production Up Ramp 0 to 0.0337 years
@ .06389 lots/hr × 2000hrs/year × $1000/lot = $127,780/year
Inventory Cost During Dispensing Down Ramp from .0337 y to .0875 y
@ 0.04 lots/hr × 2000hrs/year × $1000/lot = $80,000/year reaching 4.27 lots inventory
This gives a total cost for the 2nd run, excluding setup costs, of:
806.99 + 72.33 + 170.72 – 114.92 = $935.12
Combined Costs of Production Runs 2 through 11, which are each identical in cost but are 175 hours aor 0.0865 years) apart temporarily let the start of the 2nd run as time zero with 10 repetitions:
Setup Costs of $180.00 Occur at the Beginning of Each of these Runs, plus at the end of the 11th run, and at the beginning of the first run. Excluding the initial setup, the total amount of PW at the beginning of the 2nd run is
These two cost components sum to $8,951.20 + $1,886.20 = $10,837.40 PW = $ 24, 000 year 1 e
0 11333 806 99
0 11333 0337
⎛ −
⎝⎜
⎞
⎠⎟=
−. (. )
. $ .
PW e e
ramp= ⎛ − −
⎝⎜
⎞
⎠⎟=
− ( ) − ( )
$ , .
.
. $ .
. . . .
127 780 1 1133
0337
1133 72 33
1133 0337 2
1133 0337
PW e e
ramp= ⎛ − − e
⎝⎜
⎞
⎠⎟ =
− ( ) − ( )
− ( )
$ , .
.
. $ .
. . . .
. .
80 000 1 1133
0538
1133 114 92
113 0538 2
113 0538
1133 0337
PW = $935.12 1 e e
−
−
⎧⎨
⎪
⎩⎪
⎫⎬
⎪
⎭⎪=
− ( )
− ( )
. .
. . $ , .
1133 0 0865 10 1133 0 0865
1 8 951 20
PW = $180 1 e e
−
−
⎧⎨
⎪
⎩⎪
⎫⎬
⎪
⎭⎪=
− ( )
− ( )
. .
. . $ , .
1133 0 0865 11 1133 0 0865
1 1 886 20