chapter 17 learning curves

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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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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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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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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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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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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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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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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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

R t d E l


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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

R t d E l


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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.

St D E l


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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

St D E l


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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

St D E l


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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

P d ti R t


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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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Using the historical airframe data below for a Navy aircraft

program, estimate the learning curve equations using Unit 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



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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%



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Cumulative Average Theory results...




Equation in Unit Space: CAC = 4359.43 * N ^ -0.21


Average Actual CAC 1896.912






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Production Rate Theory results...




Equation in Unit Space: Cost = 3704.095 * Unit No ^ -0.206 * Qty ^ -0.026


Average Actual Cost 1670.000






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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

chapter 17 learning curves

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