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8/3/2019 Chapter 17 Learning Curves
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Learning Curves
Chapter 17
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Learning Curve Analysis
Developed as a tool to estimate the recurring costs in aproduction process
recurring costs: those costs incurred on each unit ofproduction
Dominant factor in learning theory is direct labor based on the common observation that as a task is
accomplished several times, it can be be completed inshorter periods of time
each time you perform a task, you become better at
it and accomplish the task faster than the previoustime
Other possible factors
management learning, production improvements such as
tooling, engineering
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Learning Theory
$0.00
$20.00
$40.00
$60.00
$80.00
$100.00
$120.00
1 3 5 7 9 11 13 15
Qty
UnitC
ost
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Log-Log Plot of Linear Data
$10.00
$100.00
1 10 100
Qty
UnitCost
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Linear Plot of Log Data
$0.00
$0.50
$1.00
$1.50
$2.00
$2.50
0 0.5 1 1.5
Log Qty
Log
UnitCo
st
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Learning Theory
Two variations:
Cumulative Average Theory
Unit Theory
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Cumulative Average Theory
If there is learning in the production process, thecumulative average cost of some doubled unit equals thecumulative average cost of the undoubled unit times theslope of the learning curve
Described by T. P. Wright in 1936
based on examination of WW I aircraft production costs
Aircraft companies and DoD were interested in the regularand predictable nature of the reduction in production coststhat Wright observed
implied that a fixed amount of labor and facilities wouldproduce greater and greater quantities in successiveperiods
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Unit Theory
If there is learning in the production process, the cost ofsome doubled unitequals the cost of the undoubled unittimes the slope of the learning curve
Credited to J. R. Crawford in 1947
led a study of WWII airframe production commissionedby USAF to validate learning curve theory
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Basic Concept of Unit Theory
As the quantity of units produced doubles, the cost1
toproduce a unit is decreased by a constant percentage
For an 80% learning curve, there is a 20% decrease incost each time that the number of units produceddoubles
the cost of unit 2 is 80% of the cost of unit 1
the cost of unit 4 is 80% of the cost of unit 2
the cost of unit 8 is 80% of the cost of unit 4, etc.
1 The Cost of a unit can be expressed in dollars, labor hours, or other units of measurement.
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80% Unit Learning Curve
100
66.9254.98
44.638
80
$0.00
$20.00
$40.00
$60.00
$80.00
$100.00
$120.00
0 2 4 6 8 10 12 14 16
Qty
UnitCost
$0.00
$0.50
$1.00
$1.50
$2.00
$2.50
0 0.5 1 1.5
Log Qty
Log
UnitCost
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Unit Theory
Defined by the equation Yx = Axb
where
Yx = the cost of unit x (dependent variable)
A = the theoretical cost of unit 1 (a.k.a. T1)
x = the unit number (independent variable)
b = a constant representing the slope (slope = 2b)
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Learning Parameter
In practice, -0.5 < b < -0.05 corresponds roughly with learning curves between 70%
and 96%
learning parameter largely determined by the type ofindustry and the degree of automation
for b = 0, the equation simplifies to Y = A which meansany unit costs the same as the first unit. In this case,the learning curve is a horizontal line and there is nolearning.
referred to as a 100% learning curve
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Learning Curve Slope versusthe Learning Parameter
As the number of units doubles, the unit cost is reduced bya constant percentage which is referred to as the slope ofthe learning curve
Cost of unit 2n = (Cost of unit n) x (Slope of learning curve)
Taking the natural log of both sides:
ln (slope) = b x ln (2)
b = ln(slope)/ln(2)For a typical 80% learning curve:
ln (.8) = b x ln (2)
b = ln(.8)/ln(2)
Slope of learning curveCost of unit n
Cost of unit n
A n
A n
b
b b 2 2
2( )
( )
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Slope and 1st Unit Cost
To use a learning curve for a cost estimate, a slope and 1stunit cost are required
slope may be derived from analogous productionsituations, industry averages, historical slopes for thesame production site, or historical data from previousproduction quantities
1st unit costs may be derived from engineeringestimates, CERs, or historical data from previousproduction quantities
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Slope and 1st Unit Costfrom Historical Data
When historical production data is available, slope and 1stunit cost can be calculated by using the learning curveequation
Yx = Axb
take the natural log of both sides:
ln (Yx) = ln (A) + b ln (x)
rewrite as Y = A + b X and solve for A and b usingsimple linear regression
A = eA
no transformation for b required
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Example
Given the following historical data, find the Unit learningcurve equation which describes this productionenvironment. Use this equation to predict the cost (inhours) of the 150th unit and find the slope of the curve.
Unit # Hours Ln (X) Ln (Y)
(X) (Y) X' Y'5 60 1.60944 4.09434
12 45 2.48491 3.80666
35 32 3.55535 3.46574
125 21 4.82831 3.04452
b = -0.32546 =SLOPE($E$3:$E$6,$D$3:$D$6)
A' = 4.618 =INTERCEPT($E$3:$E$6,$D$3:$D$6)
A = 101.298 =EXP(4.618)
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Example
Or, using the Regression Add-In in Excel...Regression Statistics
Multiple R 1.0000
R Square 0.9999
Adjusted R Square 0.9999
Standard Error 0.0042
Observations 4
ANOVA
df SS MS F Significance F
Regression 1 0.614 0.614 35482.785 2.818E-05
Residual 2 0.000 0.000
Total 3 0.614
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 4.618 0.006 799.404 0.000 4.593 4.643LN(Unit No.) -0.325 0.002 -188.369 0.000 -0.333 -0.318
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The equation which describes this data can be written:Yx = Ax
b
A = e4.618 = 101.30
b = -0.32546
Yx = 101.30(x)-0.32546
Solving for the cost (hours) of the 150th unit
Y150 = 101.30(150)-0.32546
Y150 = 19.83 hours
Slope of this learning curve = 2b
slope = 2-0.32546 = .7980 = 79.8%
Example
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Estimating Lot Costs
After finding the learning curve which best models theproduction situation, the estimator must now use thislearning curve to estimate the cost of future units.
Rarely are we asked to estimate the cost of just oneunit. Rather, we usually need to estimate lot costs.
This is calculated using a cumulative total cost equation:
where CTN = the cumulative total cost of N units
CTN may be approximated using the following equation:
N
x
bbbb
NxANAAACT
1
)()2()1(
1
)( 1
b
NACT
b
N
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Estimating Lot Costs
Compute the cost (in hours) of the first 150 units from theprevious example:
To compute the total cost of a specific lot with first unit #Fand last unit # L:
approximated by:
hrsb
ACT
b
4.44171325.0
)150)(3.101(
1
)150( 1325.01
150
1
11
,
F
x
b
L
x
b
LFxxACT
1
)1(
1
11
,
b
FA
b
ALCT
bb
LF
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Estimating Lot Costs
Compute the cost (in hours) of the lot containing units 26through 75 from the previous example:
hrsCT
b
L
F
A
b
FA
b
ALCT
bb
LF
8.14481325.0
)126)(3.101(
1325.0
)75)(3.101(325.0
75
26
3.101
1
)1(
1
1325.01325.0
75,26
11
,
= -
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Fitting a Curve Using Lot Data
Cost data is generally reported for production lots (i.e., lotcost and units per lot), not individual units
Lot data must be adjusted since learning curvecalculations require a unit number and its associatedunit cost
Unit number and unit cost for a lot are represented byalgebraic lot midpoint (LMP) and average unit cost (AUC)
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Fitting a Curve Using Lot Data
The Algebraic Lot Midpoint (LMP) is defined as thetheoretical unit whose cost is equal to the average unit costfor that lot on the learning curve.
Calculation of the exact LMP is an iterative process. Iflearning curve software is unavailable, solve byapproximation:
For the first lot (the lot starting at unit 1):
If lot size < 10, then LMP = Lot Size/2
If lot size 10, then LMP = Lot Size/3 For all subsequent lots:
lot.ainnumberunitlasttheL
andlot,ainnumberunitfirsttheF
where
4
2
FLLF
LMP
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Fitting a Curve Using Lot Data
The LMP then becomes the independent variable (X) whichcan be transformed logarithmically and used in our simplelinear regression equations to find the learning curve forour production situation.
The dependent variable (Y) to be used is the AUC which canbe found by:
The dependent variable (Y) must also be transformed
logarithmically before we use it in the regression equations.
SizeLot
CostLotTotalAUC
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Example
Given the following historical production data on a tankturret assembly, find the Unit Learning Curve equationwhich best models this production environment andestimate the cost (in man-hours) for the seventh productionlot of 75 assemblies which are to be purchased in the nextfiscal year.
Lot # Lot Size Cost (man-hours)1 15 36,7502 10 19,0003 60 90,000
4 30 39,0005 50 60,0006 50 in process, no data available
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Solution
The Unit Learning Curve equation:
Yx = 3533.22x-0.217
Lot # Lot Size Cost Cum Qnty LMP AUC ln (LMP) ln (AUC)
1 15 36,750 15 5.00 2450 1.609 7.8042 10 19,000 25 20.25 1900 3.008 7.5503 60 90,000 85 51.26 1500 3.937 7.3134 30 39,000 115 99.97 1300 4.605 7.1705 50 60,000 165 139.42 1200 4.938 7.0906 50 ? 215
b = -0.217A' = 8.17 A = 3533.22
slope = 2b = 2-0.217 = .8604 = 86.04%
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Solution
To estimate the cost (in hours) of the 7th production lot:
the units included in the 7th lot are 216 - 290
hrsCT
b
L
F
A
b
FA
b
ALCT
bb
LF
7.866,791217.0
)1216)(22.3533(
1217.0
)290)(22.3533(
217.0
290
216
22.3533
1
)1(
1
1217.01217.0
290,216
11
,
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Cumulative Average Theory
As the cumulative quantityof units produced doubles, theaverage costofall units produced to dateis decreased by aconstant percentage
For an 80% learning curve, there is a 20% decrease inaverage cost each time that the cumulative quantity
produced doubles
the average cost of 2 units is 80% of the cost of 1unit
the average cost of 4 units is 80% of the average cost
of 2 units the average cost of 8 units is 80% of the average cost
of 4 units, etc.
C l ti A
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Cumulative AverageLearning Theory
Defined by the equation YN
= ANb
where
YN = the average cost of N units
A = the theoretical cost of unit 1
N = the cumulative number of units produced
b = a constant representing the slope (slope = 2b) Used in situations where the initial production of an item is
expected to have large variations in cost due to:
use of soft or prototype tooling
inadequate supplier base established early design changes
short lead times
This theory is preferred in these situations because theeffect of averaging the production costs smoothes out
initial cost variations.
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80% Cumulative Average Curve
100
6451.2
40.96
80
$0.00
$20.00
$40.00
$60.00
$80.00
$100.00
$120.00
0 2 4 6 8 10 12 14 16
Cum Qty
Cum Avg Cost
$0.00
$0.50
$1.00
$1.50
$2.00
$2.50
0 0.5 1 1.5
Log Cum Qty
LogCum Avg Cost
L i C Sl
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Learning Curve Slope versusthe Learning Parameter
As the number of units doubles, the average unit cost isreduced by a constant percentage which is referred to asthe slope of the learning curve
Average Cost of units 1 thru 2n
= Slope of learning curve x Average Cost of units 1 thru n
Taking the natural log of both sides:
ln (slope) = b x ln (2)
b=ln(slope)/ln(2)
Slope of LCAvg Cost of units thru n
Avg Cost of units thru n
A n
A n
b
b
b 1 2
1
22
( )
( )
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Slope and 1st Unit Cost
To use a learning curve for a cost estimate, a slope and 1stunit cost are required
slope may be derived from analogous productionsituations, industry averages, historical slopes for thesame production site, or historical data from previous
production quantities
1st unit costs may be derived from engineeringestimates, CERs, or historical data from previousproduction quantities
Sl d 1 t U it C t
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Slope and 1st Unit Costfrom Historical Data
When historical production data is available, slope and 1stunit cost can be calculated by using the learning curveequation
YN = ANb
take the natural log of both sides:
ln (YN) = ln (A) + b ln (N)
rewrite as Y = A + b N and solve for A and b usingsimple linear regression
A = eA
no transform for b required
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The equation which best fits this data can be written:
Yx = 12.278(N)-0.33354
Slope of this learning curve = 2b
slope = 2-0.33354
= .7936
Lot Cum Cum Cum Avg Ln (N) Ln (Y)
Lot Qnty Cost ($M) Qnty (N) Cost Cost (Y) N' Y'22 98 22 98 4.455 3.09104 1.4939336 80 58 178 3.069 4.06044 1.1213440 80 98 258 2.633 4.58497 0.9679940 78 138 336 2.435 4.92725 0.88986
b = -0.33354 =SLOPE($G$3:$G$6,$F$3:$F$6)A' = 2.5078 =INTERCEPT($G$3:$G$6,$F$3:$F$6)A = 12.27789 =EXP(2.5078)
Example
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Example
Or, using the Regression Add-In in Excel...
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.9952
R Square 0.9903
Adjusted R Square 0.9855
Standard Error 0.0323
Observations 4
ANOVA
df SS MS F Significance F
Regression 1 0.214 0.214 204.912 0.0048
Residual 2 0.002 0.001
Total 3 0.216
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 2.508 0.098 25.485 0.002 2.084 2.931
N' -0.334 0.023 -14.315 0.005 -0.434 -0.233
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Estimating Lot Costs
The learning curve which best fits the production data mustbe used to estimate the cost of future units
usually must estimate the cost of units grouped into aproduction lot
lot cost equations can be derived from the basic
equation YN= ANb
which gives the average cost of Nunits
the total cost of N units can be computed by multiplyingthe average cost of N units by the number of units N
where CTN = the cumulative total cost of N units
the cost of unit N is approximated by (1 + b)ANb
CT AN N ANNb b
( )
1
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Estimating Lot Costs
To compute the total cost of a specific lot with first unit #Fand last unit # L:
111, )1(
bb
FLLF FLACTCTCT
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Example
Given the following historical production data on a tankturret assembly, find the Cum Avg Learning Curve equationwhich best models this production and estimate the cost (inman-hours) for the seventh production lot of 75 assemblieswhich are to be purchased in the next fiscal year.
Lot # Lot Size Cost (man-hours)1 15 36,7502 10 19,0003 60 90,0004 30 39,000
5 50 60,0006 50 in process, no data available
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Solution
Lot LN of LN of
Lot # Size Cost Cum Qnty Cum Cost Cum Avg Cost Cum Qnty Cum Avg Cost1 15 36,750 15 36750 2450.00 2.70805 7.80384
2 10 19,000 25 55750 2230.00 3.21888 7.70976
3 60 90,000 85 145750 1714.71 4.44265 7.44700
4 30 39,000 115 184750 1606.52 4.74493 7.38183
5 50 60,000 165 244750 1483.33 5.10595 7.30205
6 50 ? 215
b = -0.21048
A' = 8.3801 A = 4359.43
slope = 2b = 2-0.21048 = .8643 = 86.43%
The Cum Avg Learning Curve equation:
YN = 4359.43N-0.21048
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Solution
To estimate the cost of the 7th production lot:
the units included in the 7th lot are 216 - 290
hrsCTb
F
L
A
FLACTbb
645,80)215(29043.4359
2105.0
216
290
43.4359
)1(
12105.12105.0290,216
11
290,216
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Unit Versus Cum Avg
1000
1500
2000
2500
0 25 50 75 100 125 150 175 200
Qty
Cost
Cum Avg LCUnit LC
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Unit Versus Cum Avg
3
3.1
3.2
3.3
3.4
3.5
0 0.5 1 1.5 2 2.5
Log Qty
Log Cost
Cum Avg LC
Unit LC
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Unit versus Cum Avg
Since the Cum Avg curve is based on the average cost of aproduction quantity rather than the cost of a particularunit, it is less responsive to cost trends than the unit costcurve.
A sizable change in the cost of any unit or lot of units is
required before a change is reflected in the Cum Avgcurve.
Cum Avg curve is smoother and always has a higher r2
Since cost generally decreases, the Unit cost curve will
roughly parallel the Cum Avg curve but will always liebelow it.
Government negotiator prefers to use Unit cost curvesince it is lower and more responsive to recent trends
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Learning Curve Selection
Type of learning curve to use is an important decision.Factors to consider include:
Analogous Systems
systems which are similar in form, function, ordevelopment/production process may provide
justification for choosing one theory over another
Industry Standards
certain industries gravitate toward one theory versusanother
Historical Experience
some defense contractors have a history of using onetheory versus another because it has been shown tobest model that contractors production process
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Learning Curve Selection
Expected Production Environment
certain production environments favor one theoryover another
Cum Avg: contractor is starting production withprototype tooling, has an inadequate supplier baseestablished, expects early design changes, or issubject to short lead-times
Unit curve: contractor is well prepared to beginproduction in terms of tooling, suppliers, lead-times, etc.
Statistical Measures best fit, highest r2
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Production Breaks
Production breaks can occur in a program due to fundingdelays or technical problems.
How much of the learning achieved has been lost (forgotten)due to the break in production?
How will this lost learning impact the costs of futureproduction items?
The first question can be answered by using the AnderlohrMethod for estimating the learning lost.
The second question can then be answered by using theRetrograde Method to reset the learning curve.
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Anderlohr Method
To assess the impact on cost of a production break, it isfirst necessary to quantify how much learning was achievedprior to the break, and then quantify how much of thatlearning was lost due to the break.
George Anderlohr, a Defense Contract AdministrationServices (DCAS) employee in the 1960s, divided the
learning lost due to a production break into five categories:
Personnel learning;
Supervisory learning;
Continuity of productivity;
Methods; Special tooling.
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Anderlohr Method
Each production situation must be examined and a weightassigned to each category. An example weighting schemefor a helicopter production line might be:
Category Weigh
Personnel Learning 30%
Supervisory Learning 20%
Continuity of Production 20%
Tooling 15%
Methods 15%
Total 100%
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Anderlohr Method
To find the percentage of learning lost (Learning LostFactor or LLF) we must find the learning lost in eachcategory, and then calculate a weighted average based uponthe previous weights.
Example: A contractor who assembles helicoptersexperiences a six-month break in production due to thedelayed issuance of a follow-on production contract. Theresident Defense Plant Representative Office (DPRO)conducted a survey of the contractor and provided thefollowing information.
During the break in production, the contractortransferred many of his resources to commercial andother defense programs. As a result the following can beexpected when production resumes on the helicopterprogram:
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Anderlohr Method Example
75% of the production personnel are expected to return to this
program, the remaining 25% will be new hires or transfers fromother programs.
90% of the supervisors are expected to return to this program, theremainder will be recent promotes and transfers.
During the production break, two of the four assembly lines were
torn down and converted to other uses, these assembly lines willhave to be reassembled for the follow-on contract.
An inventory of tools revealed that 5% of the tooling will have tobe replaced due to loss, wear and breakage.
Also during the break, the contractor upgraded some capitalequipment on the assembly lines requiring modifications to 7% of
the shop instructions
Finally, it is estimated that the assembly line workers will lose35% of their skill and dexterity, and the supervisors will lose 10%of the skills needed for this program during the production break.
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Anderlohr Method Example
Personnel 75% returned X 65% skill retained 48.75% learning retained
51.25% learning lost
Supervisory 90% returned X 90% skill retained 81% learning retained
19% learning lost
Continuity of Production 2 of 4 assembly lines torn down 50% learning lost
Tooling 5% lost, worn or broken 5% learning lost
Methods 7% of shop instructions need modification 7% learning lost
Calculation of Learning Lost Factor (LLF)
Category Weight
Percent
Lost
Weighted
LossPersonnel Learning 30% 51.25% 15.4%
Supervisory Learning 20% 19% 3.8%
Continuity of Production 20% 50% 10.0%
Tooling 15% 5% 0.8%
Methods 15% 7% 1.1%
Total Learning Lost Factor (LLF) 31.0%
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Retrograde Method
Once the Learning Lost Factor (LLF) has been estimated, weuse the LLF to estimate the impact of the cost on futureproduction using the Retrograde Method.
Hrs
Qty
HrsLost
Units of Retrograde
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Retrograde Method
The theory is that because you lose hours of learning, theLLF should be applied to the hours of learning that youachieved prior to the break.
The result of the Anderlohr Method gives you the number ofhours of learning lost.
These hours can then be added to the cost of the first unitafter the break on the original curve to yield an estimate ofthe cost (in hours) of that unit due to the break inproduction.
Finally, we can then back up the curve (retrograde) to thepoint where production costs were equal to our newestimate.
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Retrograde Example
Continuing with our previous example
Assume 10 helicopters were produced prior to the sixmonth production break.
The first helicopter required 10,000 man-hours to completeand the learning slope is estimated at 88%. Using the LLFfrom the previous example, estimate the cost of the nextten units which are to be produced in the next fiscal year.
Hrs
201 10
10,000
Qty
d l
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Retrograde Example
Step 1 - Find the amount of learning achieved to date.
hrsAL
hrsAXY
YAL
YYAL
b
460,3540,6000,10..
6540)10(000,10
000,10..
..
)2ln(
)88.0ln(
10
10
101
d l
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Retrograde Example
Step 2 - Estimate the number of hours of learning lost.
In this case we achieved 3,460 hours of learning, but welost 31% of that, or 1,073 hours, due to the break inproduction.
hrs
LLFAL
073,1%)31()460,3(LostLearning
)(.).(LostLearning
R d E l
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Retrograde Example
Step 3 - Estimate the cost of the first unit after the break.
The cost, in hours, of unit 11 is estimated by adding thecost of unit 11 on the original curve to the hours oflearning lost found in the previous step.
hrsY
hrsAXY
YY
b
499,7073,1426,6426,6)11(000,10
LostLearning
*
11
)2ln(
)88ln(.
11
11
*
11
R t d E l
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Retrograde Example
Step 4 - Find the unit on the original curve which is
approximately the same as the estimated cost, in hours, ofthe unit after the break.
This can be done using actual data, but since the actualdata contains some random error, it is best to use the unitcost equation to solve for X. In this case, X = 5.
576.4000,10
499,7 )88ln(.)2ln(
1
b
b
b
AYX
AYX
AXY
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Retrograde Example
Step 5 - Find the number of units of retrograde.
The number of units of retrograde is how many units youneed to back up the curve to reach the unit found in step 4.In this example, since the estimated cost of unit 11 isapproximately the same as unit 5 on the original curve you
need to back up 6 units to estimate the cost of unit 11 andall subsequent units.
6511Retrograde
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Retrograde Example
Step 6 - Estimate lot costs after the break.
This can be done by applying the retrograde number to ourstandard lot cost equation.
To estimate the cost, in hours, of units 11 through 20, wesubtract the units of retrograde from the units in questionand instead solve for the cost of units 5 through 14.
Therefore, our estimate of cost for the next 10 helicoptersis 66,753 man-hours.
hrsxxTC
xxATC
x
b
x
b
F
x
b
L
x
b
LF
753,66000,10
146205611
4
1
14
1
14,5
1
11
,
St D F ti
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Step-Down Functions
A Step-Down Function is a method of estimating the
theoretical first unit production cost based upon prototype(development) cost data.
It has been found, in general, that the unit cost of aprototype is more expensive than the first unit cost of acorresponding production model.
The ratio of production first unit cost to prototype averageunit cost is known as a Step-Down factor.
An estimate for the Step-Down factor for a given weaponsystem can be found by examining historical similarweapon systems and developing a cost estimatingrelationship, with prototype average unit cost as theindependent variable.
Once an appropriate CER is developed, it can be used withactual or estimated prototype costs to estimate the firstunit production cost.
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Step-Down Example
We desire to estimate the first unit production cost for a
new missile radar system (APGX-99). The slope of thesystem is expected to be a 95% unit curve. The estimatedaverage prototype cost is expected to be $3.5M for 8prototype radars.
The following historical data on similar radar systems has
been collected:
Radar
System
Production
Cost (Unit 150)
No. of
Prototypes
Prototype
AUC
APG-63 0.985M 13 7.47MAPG-66 0.414M 12 2.78M
Patriot 6.500M 4 65.20M
Phoenix 1.752M 11 13.25M
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Step-Down Example
Because the data gives production costs for unit 150, we
can develop our CER based on unit 150 and then back upthe curve to the first unit.
Using Simple Linear Regression, we find the relationshipbetween development average cost (DAC) and the unit 150production cost (PR150) to be:
150,625$
5.30957.02902.0
0957.02902.0
150
150
150
PR
PR
DACPR
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Step-Down Example
We can now estimate our first unit cost using our unit cost
equation and the (assumed) 95% slope as follows:
This result gives an estimate for the first unit productioncost of $905,766. We have stepped down from adevelopment cost estimate to a production cost estimate.
766,905$
)150(150,625$
0740.0)2ln(
)95ln(.
0740.0
150
A
A
AXY
b
b
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Production Rate
A variation of the Unit Learning Curve Model
Adds production rate as a second variable
unit quantity costs should decrease when the rate ofproduction increases as well as when the quantityproduced increases
two independent effects Model: Yx = Ax
bQc
where Yx = the cost of unit x (dependent variable)
A = the theoretical cost of 1st unit
x = the unit number
b = a constant representing the slope (slope = 2b)
Q = rate of production (quantity per period or lot)
c = rate coefficient (rate slope = 2c)
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log Cost
log Rate
log Quantity
Rate Model Advantages
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Rate Model Advantages
It directly models cost reductions which are achieved
through economies of scale quantity discounts received when ordering bulk
quantities
reduced ordering and processing costs
reduced shipping, receiving, and inspection costs
Rate Model Weaknesses
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Rate Model Weaknesses
Appropriate production rate (i.e., annual, quarterly,
monthly) is not always clear
If Q is always increasing, there tends to be a high degree ofcollinearity between the quantity and rate variables
Always estimates decreasing unit costs for increasingproduction rates
when a manufacturers capacity is exceeded, unit costsgenerally increase due to costs of overtime,hiring/training new workers, purchase of new capital,
etc.
When to Consider a Rate Model
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When to Consider a Rate Model
Production involves relatively simple components for which
lot size is a much greater cost driver than cumulativequantity
When production is taking place well down the learningcurve where it flattens out
When there is a major change in production rate
Model Selection Example
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Model Selection Example
Using the historical airframe data below for a Navy aircraft
program, estimate the learning curve equations using Unit theory
Cumulative Average Theory
Rate Theory
FY Quantity
Lot Cost
(CY94$K)
Lot
Midpoint
(LMP)
Lot AUC
(FY94$K)
Cum.
Quantity
Cum. Avg.
Cost
(CY94$K)
89 15 36,750 5 2450 15 2,450
90 10 19,000 20.25 1900 25 2,230
91 60 90,000 51.26 1500 85 1,715
92 30 39,000 99.97 1300 115 1,60793 50 60,000 139.42 1200 165 1,483
Unit Theory Cum Avg Theory Rate Theory
Airframe
Model Selection Example
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Model Selection Example
Unit Theory results...
I. Model Form and Equation
Model Form: Log-Linear Model
Number of Observations: 5
Equation in Unit Space: Cost = 3533.245 * Unit No ^ -0.217
III. Predictive Measures (in Unit Space)
Average Actual Cost 1670.000
Standard Error (SE) 42.817
Coefficient of Variation (CV) 2.60%
Adjusted R-Squared 99.30%
Model Selection Example
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Model Selection Example
Cumulative Average Theory results...
I. Model Form and Equation
Model Form: Log-Linear Model
Number of Observations: 5
Equation in Unit Space: CAC = 4359.43 * N ^ -0.21
III. Predictive Measures (in Unit Space)
Average Actual CAC 1896.912
Standard Error (SE) 13.261
Coefficient of Variation (CV) 0.70%
Adjusted R-Squared 99.90%
Model Selection Example
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Model Selection Example
Production Rate Theory results...
I. Model Form and Equation
Model Form: Log-Linear Model
Number of Observations: 5
Equation in Unit Space: Cost = 3704.095 * Unit No ^ -0.206 * Qty ^ -0.026
III. Predictive Measures (in Unit Space)
Average Actual Cost 1670.000
Standard Error (SE) 31.233
Coefficient of Variation (CV) 1.90%
Adjusted R-Squared 99.70%
Model Selection Example
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Model Selection Example
Which model should we use?
From a purely statistical point of view, we prefer theCumulative Average Theory model, since it gives the beststatistics.
The real answer may depend on other issues.
Model
Standard Error 42.8 13.2 31.2CV 2.6% 0.7% 1.9%Adj R2 99.3% 99.9% 99.7%
Unit Cum. Avg Rate