17 - 1 Learning Curves Chapter 17. 17 - 2 Learning Curve Analysis Developed as a tool to estimate...

17 - 1

Learning Curves

Chapter 17

17 - 2

Learning Curve Analysis

• Developed as a tool to estimate the recurring costs in a production process– recurring costs: those costs incurred on each unit

of production

• 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 in shorter periods of time» “each time you perform a task, you become

better at it and accomplish the task faster than the previous time”

• Other possible factors– management learning, production improvements

such as tooling, engineering

17 - 3

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

Un

it C

ost

17 - 4

Log-Log Plot of Linear Data

$10.00

$100.00

1 10 100Qty

Un

it C

ost

17 - 5

Linear Plot of Log Data

$0.00

$0.50

$1.00

$1.50

$2.00

$2.50

0 0.5 1 1.5Log Qty

Log

Un

it C

ost

17 - 6

Learning Theory

• Two variations:

– Cumulative Average Theory

– Unit Theory

17 - 7

Cumulative Average Theory

“If there is learning in the production process, the cumulative average cost of some doubled unit equals the cumulative average cost of the undoubled unit times the slope 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 regular and predictable nature of the reduction in production costs that Wright observed– implied that a fixed amount of labor and facilities

would produce greater and greater quantities in successive periods

17 - 8

Unit Theory

“If there is learning in the production process, the cost of some doubled unit equals the cost of the undoubled unit times the slope of the learning curve”

• Credited to J. R. Crawford in 1947– led a study of WWII airframe production

commissioned by USAF to validate learning curve theory

17 - 9

Basic Concept of Unit Theory

• As the quantity of units produced doubles, the cost1 to produce a unit is decreased by a constant percentage

– For an 80% learning curve, there is a 20% decrease in cost each time that the number of units produced doubles

» 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.

17 - 10

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

Un

it C

ost

$0.00

$0.50

$1.00

$1.50

$2.00

$2.50

0 0.5 1 1.5

Log Qty

Log

Un

it C

ost

17 - 11

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)

17 - 12

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

of industry and the degree of automation – for b = 0, the equation simplifies to Y = A which

means any unit costs the same as the first unit. In this case, the learning curve is a horizontal line and there is no learning.» referred to as a 100% learning curve

17 - 13

Learning Curve Slope versus the Learning Parameter

“As the number of units doubles, the unit cost is reduced by a constant percentage which is referred to as the slope of the 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)

17 - 14

Slope and 1st Unit Cost

• To use a learning curve for a cost estimate, a slope and 1st unit cost are required

– slope may be derived from analogous production situations, industry averages, historical slopes for the same production site, or historical data from previous production quantities

– 1st unit costs may be derived from engineering estimates, CERs, or historical data from previous production quantities

17 - 15

Slope and 1st Unit Costfrom Historical Data

• When historical production data is available, slope and 1st unit cost can be calculated by using the learning curve equation

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 using simple linear regression» A = eA’

» no transformation for b required

17 - 16

Example

• Given the following historical data, find the Unit learning curve equation which describes this production environment. Use this equation to predict the cost (in hours) 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 4.0943412 45 2.48491 3.8066635 32 3.55535 3.46574125 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)

17 - 17

Example

• Or, using the Regression Add-In in Excel...

Regression StatisticsMultiple R 1.0000R Square 0.9999Adjusted R Square 0.9999Standard Error 0.0042Observations 4

ANOVAdf SS MS F Significance F

Regression 1 0.614 0.614 35482.785 2.818E-05Residual 2 0.000 0.000Total 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

17 - 18

The equation which describes this data can be written:Yx = Axb

A = e4.618 = 101.30b = -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

17 - 19

Estimating Lot Costs

• After finding the learning curve which best models the production situation, the estimator must now use this learning curve to estimate the cost of future units.– Rarely are we asked to estimate the cost of just one

unit. 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

bbbbN xANAAACT

1

)()2()1(

1

)( 1

b

NACT

b

N

17 - 20


• Compute the cost (in hours) of the first 150 units from the previous example:

• To compute the total cost of a specific lot with first unit # F and last unit # L:

– approximated by:

hrsb

ACT

b

4.44171325.0

)150)(3.101(

1

)150( 1325.01

150

1

11,

F

x

bL

x

bLF xxACT

1

)1(

1

11

,

b

FA

b

ALCT

bb

LF

17 - 21


• Compute the cost (in hours) of the lot containing units 26 through 75 from the previous example:

hrsCT

b

L

F

Ab

FA

b

ALCT

bb

LF

8.14481325.0

)126)(3.101(

1325.0

)75)(3.101(

325.0

75

26

3.1011

)1(

1

1325.01325.0

75,26

11

,

= -

17 - 22

Fitting a Curve Using Lot Data

• Cost data is generally reported for production lots (i.e., lot cost and units per lot), not individual units

– Lot data must be adjusted since learning curve calculations require a unit number and its associated unit cost

– Unit number and unit cost for a lot are represented by algebraic lot midpoint (LMP) and average unit cost (AUC)

17 - 23


• The Algebraic Lot Midpoint (LMP) is defined as the theoretical unit whose cost is equal to the average unit cost for that lot on the learning curve.

• Calculation of the exact LMP is an iterative process. If learning curve software is unavailable, solve by approximation:

• 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. ain number unit last the L

and lot, ain number unit first the F

where4

2

FLLFLMP

17 - 24


• The LMP then becomes the independent variable (X) which can be transformed logarithmically and used in our simple linear regression equations to find the learning curve for our production situation.

• The dependent variable (Y) to be used is the AUC which can be found by:

• The dependent variable (Y) must also be transformed logarithmically before we use it in the regression equations.

SizeLot

CostLotTotalAUC

17 - 25

Example

• Given the following historical production data on a tank turret assembly, find the Unit Learning Curve equation which best models this production environment and estimate the cost (in man-hours) for the seventh production lot of 75 assemblies which 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 39,0005 50 60,0006 50 in process, no data available

17 - 26

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%

17 - 27

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

Ab

FA

b

ALCT

bb

LF

7.866,791217.0

)1216)(22.3533(

1217.0

)290)(22.3533(

217.0

290

216

22.35331

)1(

1

1217.01217.0

290,216

11

,

17 - 28

Cumulative Average Theory

• As the cumulative quantity of units produced doubles, the average cost of all units produced to date is decreased by a constant percentage

– For an 80% learning curve, there is a 20% decrease in average cost each time that the cumulative quantity produced doubles

» the average cost of 2 units is 80% of the cost of 1 unit

» 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.

17 - 29

Cumulative Average Learning Theory

• Defined by the equation YN = AN b

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 the effect of averaging the production costs “smoothes out” initial cost variations.

17 - 30

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 16Cum Qty

Cum Avg Cost

$0.00

$0.50

$1.00

$1.50

$2.00

$2.50

0 0.5 1 1.5Log Cum Qty

Log Cum Avg Cost

17 - 31

Learning Curve Slope versus the Learning Parameter

“As the number of units doubles, the average unit cost is reduced by a constant percentage which is referred to as the 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)

17 - 32

Slope and 1st Unit Cost

• To use a learning curve for a cost estimate, a slope and 1st unit cost are required

– slope may be derived from analogous production situations, industry averages, historical slopes for the same production site, or historical data from previous production quantities

– 1st unit costs may be derived from engineering estimates, CERs, or historical data from previous production quantities

17 - 33

Slope and 1st Unit Costfrom Historical Data

• When historical production data is available, slope and 1st unit cost can be calculated by using the learning curve equation

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 using simple linear regression» A = eA’

» no transform for b required

17 - 34

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

17 - 35

Example

• Or, using the Regression Add-In in Excel...

SUMMARY OUTPUT

Regression StatisticsMultiple R 0.9952R Square 0.9903Adjusted R Square 0.9855Standard Error 0.0323Observations 4

ANOVA

df SS MS F Significance FRegression 1 0.214 0.214 204.912 0.0048Residual 2 0.002 0.001Total 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 2.931N' -0.334 0.023 -14.315 0.005 -0.434 -0.233

17 - 36


• The learning curve which best fits the production data must be used to estimate the cost of future units– usually must estimate the cost of units grouped into

a production lot» lot cost equations can be derived from the basic

equation YN = AN b which gives the average cost of N units

– the total cost of N units can be computed by multiplying the 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)AN

b

17 - 37


• To compute the total cost of a specific lot with first unit # F and last unit # L:

111, )1(

bbFLLF FLACTCTCT

17 - 38

Example

• Given the following historical production data on a tank turret assembly, find the Cum Avg Learning Curve equation which best models this production and estimate the cost (in man-hours) for the seventh production lot of 75 assemblies which 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 39,0005 50 60,0006 50 in process, no data available

17 - 39

SolutionLot LN of LN of

Lot # Size Cost Cum Qnty Cum Cost Cum Avg Cost Cum Qnty Cum Avg Cost

1 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

17 - 40

Solution

• To estimate the cost of the 7th production lot:– the units included in the 7th lot are 216 - 290

hrsCT

b

F

L

A

FLACT bb

645,80)215(29043.4359

2105.0

216

290

43.4359

)1(

12105.12105.0290,216

11290,216

17 - 41

Unit Versus Cum Avg

1000

1500

2000

2500

0 25 50 75 100 125 150 175 200

Qty

Cost

Cum Avg LCUnit LC

17 - 42

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

17 - 43

Unit versus Cum Avg

• Since the Cum Avg curve is based on the average cost of a production quantity rather than the cost of a particular unit, it is less responsive to cost trends than the unit cost curve.

– A sizable change in the cost of any unit or lot of units is required before a change is reflected in the Cum Avg curve.» 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 lie below it. » Government negotiator prefers to use Unit cost

curve since it is lower and more responsive to recent trends

17 - 44

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, or development/production process may provide justification for choosing one theory over another

– Industry Standards» certain industries gravitate toward one theory

versus another

– Historical Experience» some defense contractors have a history of using

one theory versus another because it has been shown to best model that contractor’s production process

17 - 45

Learning Curve Selection

– Expected Production Environment» certain production environments favor one

theory over another• Cum Avg: contractor is starting production

with prototype tooling, has an inadequate supplier base established, expects early design changes, or is subject to short lead-times

• Unit curve: contractor is well prepared to begin production in terms of tooling, suppliers, lead-times, etc.

– Statistical Measures» best fit, highest r2

17 - 46

Production Breaks

• Production breaks can occur in a program due to funding delays 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 future production items?

• The first question can be answered by using the “Anderlohr Method” for estimating the learning lost.

• The second question can then be answered by using the “Retrograde Method” to “reset” the learning curve.

17 - 47

Anderlohr Method

• To assess the impact on cost of a production break, it is first necessary to quantify how much learning was achieved prior to the break, and then quantify how much of that learning was lost due to the break.

• George Anderlohr, a Defense Contract Administration Services (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.

17 - 48

Anderlohr Method

• Each production situation must be examined and a weight assigned to each category. An example weighting scheme for a helicopter production line might be:

Category Weight

Personnel Learning 30%

Supervisory Learning 20%

Continuity of Production 20%

Tooling 15%

Methods 15%

Total 100%

17 - 49

Anderlohr Method

• To find the percentage of learning lost (Learning Lost Factor or LLF) we must find the learning lost in each category, and then calculate a weighted average based upon the previous weights.

• Example: A contractor who assembles helicopters experiences a six-month break in production due to the delayed issuance of a follow-on production contract. The resident Defense Plant Representative Office (DPRO) conducted a survey of the contractor and provided the following information.– During the break in production, the contractor

transferred many of his resources to commercial and other defense programs. As a result the following can be expected when production resumes on the helicopter program:

17 - 50

Anderlohr Method Example

• 75% of the production personnel are expected to return to this program, the remaining 25% will be new hires or transfers from other programs.

• 90% of the supervisors are expected to return to this program, the remainder 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 will have to be reassembled for the follow-on contract.

• An inventory of tools revealed that 5% of the tooling will have to be replaced due to loss, wear and breakage.

• Also during the break, the contractor upgraded some capital equipment on the assembly lines requiring modifications to 7% of the shop instructions

• Finally, it is estimated that the assembly line workers will lose 35% of their skill and dexterity, and the supervisors will lose 10% of the skills needed for this program during the production break.

17 - 51

Anderlohr Method ExamplePersonnel 75% returned X 65% skill retained 48.75% learning retained

51.25% learning lost

Supervisory 90% returned X 90% skill retained 81% learning retained19% 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 WeightPercent

LostWeighted

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%

17 - 52

Retrograde Method

• Once the Learning Lost Factor (LLF) has been estimated, we use the LLF to estimate the impact of the cost on future production using the Retrograde Method.

Hrs

Qty

HrsLost

Units of Retrograde

17 - 53

Retrograde Method

• The theory is that because you lose hours of learning, the LLF should be applied to the hours of learning that you achieved prior to the break.

• The result of the Anderlohr Method gives you the number of hours of learning lost.

• These hours can then be added to the cost of the first unit after the break on the original curve to yield an estimate of the cost (in hours) of that unit due to the break in production.

• Finally, we can then back up the curve (retrograde) to the point where production costs were equal to our new estimate.

17 - 54

Retrograde Example

• Continuing with our previous example…• Assume 10 helicopters were produced prior to the six

month production break.• The first helicopter required 10,000 man-hours to

complete and the learning slope is estimated at 88%. Using the LLF from the previous example, estimate the cost of the next ten units which are to be produced in the next fiscal year.Hrs

201 10

10,000

Qty

17 - 55

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

17 - 56

Retrograde Example

• Step 2 - Estimate the number of hours of learning lost.

• In this case we achieved 3,460 hours of learning, but we lost 31% of that, or 1,073 hours, due to the break in production.

hrs

LLFAL

073,1%)31()460,3(Lost Learning

)(.).(Lost Learning

17 - 57

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 the cost of unit 11 on the original curve to the hours of learning lost found in the previous step.

hrsY

hrsAXY

YY

b

499,7073,1426,6

426,6)11(000,10

Lost Learning

*11

)2ln(

)88ln(.

11

11*

11

17 - 58

Retrograde Example

• Step 4 - Find the unit on the original curve which is approximately the same as the estimated cost, in hours, of the unit after the break.

• This can be done using actual data, but since the actual data contains some random error, it is best to use the unit cost equation to solve for X. In this case, X = 5.

576.4000,10499,7 )88ln(.

)2ln(1

b

b

b

AYX

AYX

AXY

17 - 59

Retrograde Example

• Step 5 - Find the number of units of retrograde.

• The number of units of retrograde is how many units you need to back up the curve to reach the unit found in step 4. In this example, since the estimated cost of unit 11 is approximately the same as unit 5 on the original curve you need to back up 6 units to estimate the cost of unit 11 and all subsequent units.

6511Retrograde

17 - 60

Retrograde Example

• Step 6 - Estimate lot costs after the break.• This can be done by applying the retrograde number to

our standard lot cost equation.• To estimate the cost, in hours, of units 11 through 20,

we subtract the units of retrograde from the units in question and instead solve for the cost of units 5 through 14.

• Therefore, our estimate of cost for the next 10 helicopters is 66,753 man-hours.

hrsxxTC

xxATC

x

b

x

b

F

x

bL

x

bLF

753,66000,10

146205611

4

1

14

114,5

1

11,

17 - 61

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 a prototype is more expensive than the first unit cost of a corresponding production model.

• The ratio of production first unit cost to prototype average unit cost is known as a Step-Down factor.

• An estimate for the Step-Down factor for a given weapon system can be found by examining historical similar weapon systems and developing a cost estimating relationship, with prototype average unit cost as the independent variable.

• Once an appropriate CER is developed, it can be used with actual or estimated prototype costs to estimate the first unit production cost.

17 - 62

Step-Down Example

• We desire to estimate the first unit production cost for a new missile radar system (APGX-99). The slope of the system is expected to be a 95% unit curve. The estimated average prototype cost is expected to be $3.5M for 8 prototype 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 2.78MPatriot 6.500M 4 65.20MPhoenix 1.752M 11 13.25M

17 - 63

Step-Down Example

• Because the data gives production costs for unit 150, we can develop our CER based on unit 150 and then back up the curve to the first unit.

• Using Simple Linear Regression, we find the relationship between development average cost (DAC) and the unit 150 production cost (PR150) to be:

150,625$

5.30957.02902.0

0957.02902.0

150

150

150

PR

PR

DACPR

17 - 64

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 production cost of $905,766. We have “stepped down” from a development cost estimate to a production cost estimate.

766,905$

)150(150,625$

0740.0)2ln(

)95ln(.

0740.0

150

A

A

AXY

b

b

17 - 65

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 of production increases as well as when the quantity produced increases”

– two independent effects

• Model: Yx = AxbQc

where Yx = the cost of unit x (dependent variable)

A = the theoretical cost of 1st unitx = the unit numberb = a constant representing the slope (slope

= 2b)Q = rate of production (quantity per period

or lot)c = rate coefficient (rate slope = 2c)

17 - 66

log Costlog Rate

log Quantity

17 - 67

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

17 - 68

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 of collinearity between the quantity and rate variables

• Always estimates decreasing unit costs for increasing production rates– when a manufacturer’s capacity is exceeded, unit

costs generally increase due to costs of overtime, hiring/training new workers, purchase of new capital, etc.

17 - 69

When to Consider a Rate Model

• Production involves relatively simple components for which lot size is a much greater cost driver than cumulative quantity

• When production is taking place well down the learning curve where it flattens out

• When there is a major change in production rate

17 - 70

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 QuantityLot Cost (CY94$K)

Lot Midpoint

(LMP)Lot AUC (FY94$K)

Cum. Quantity

Cum. Avg. Cost

(CY94$K)89 15 36,750 5 2450 15 2,45090 10 19,000 20.25 1900 25 2,23091 60 90,000 51.26 1500 85 1,71592 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

17 - 71


• Unit Theory results...

I. Model Form and EquationModel Form: Log-Linear ModelNumber of Observations: 5Equation in Unit Space: Cost = 3533.245 * Unit No ̂-0.217

III. Predictive Measures (in Unit Space)

Average Actual Cost 1670.000Standard Error (SE) 42.817Coefficient of Variation (CV) 2.60%Adjusted R-Squared 99.30%

17 - 72


• Cumulative Average Theory results...

I. Model Form and EquationModel Form: Log-Linear ModelNumber of Observations: 5Equation in Unit Space: CAC = 4359.43 * N ̂-0.21


Average Actual CAC 1896.912Standard Error (SE) 13.261Coefficient of Variation (CV) 0.70%Adjusted R-Squared 99.90%

17 - 73


• Production Rate Theory results...

I. Model Form and EquationModel Form: Log-Linear ModelNumber of Observations: 5Equation in Unit Space: Cost = 3704.095 * Unit No ̂-0.206 * Qty ̂-0.026


Average Actual Cost 1670.000Standard Error (SE) 31.233Coefficient of Variation (CV) 1.90%Adjusted R-Squared 99.70%

17 - 74


• Which model should we use?

• From a purely statistical point of view, we prefer the Cumulative Average Theory model, since it gives the best statistics.

• The real answer may depend on other issues.

ModelStandard 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

17 - 1 Learning Curves Chapter 17. 17 - 2 Learning Curve Analysis Developed as a tool to estimate...

Documents

Transcript of 17 - 1 Learning Curves Chapter 17. 17 - 2 Learning Curve Analysis Developed as a tool to estimate...