Repetition of the same operation generally results in a reduction in the time and effort required to perform subsequent operations, i.e., repetitive tasks become easier to perform with experience.  This reduction tends to be regular and predictive.  The observed characteristic of the improved performance is known as "learning".  This learning can be shown graphically or mathematically on a learning curve.  A learning curve is also known as a "manufacturing process function" , "experience curve", and "dynamic curve".  In simple terms, it means that the costs of production can be lowered with increasing quantity of production or increased learning.

The principle of learning curves can be used effectively in procurement and production. If the learning rate is known, it may be possible to negotiate the cost of producing additional units based on the learning information.  In cost estimating the bidding, pricing, and capital requirements are based partially on the learning curve concept.

Learning is realized at two levels.  The first, and most obvious, is with the learning or experience acquired by direct labor.  As a worker continues to produce a particular unit, it is natural to improve.  The second is in the learning acquired in the management process.  This is accomplished through engineering programs which improve production, encourage high quality production, reduce design complexity, create technology progress, or foster product improvement.

Of the two types of learning, the greatest savings are generally expected to be in the managerial process.  Approximately 15% of the cost reduction is expected from the learning of direct labor while 85% of the cost reduction is expected from management (50% product engineering, 35% industrial engineering).

The application of learning curve principles is diverse.  It can be applied in ship building, aircraft industries, computers, machine tools, building construction, refinery construction, etc.

The concept of learning curves applies best to a very specific type of operation.  The criteria to be met are as follows:

* high cost
* low volume
* discrete item production

Thus, the concept does not apply to operations involving a high volume of units or if the cost (labor cost) is low.  Some general points that apply to learning curves are as follows:

 Amount of time and cost required to produce each unit tends to decrease
  for successive units
 Amount of time to produce each unit decreases at a decreasing rate
 The reduction in time required to produce each unit follows a specific
  estimating model.  That is, the rate of improvement (learning) can be
  predicted by mathematical models

The mathematical formulations for learning curve applications will be given.  There are two types of perspectives.  The first relates to the time (generally in worker hours) required to perform each unit of work and the other relates to the average time per unit for a given number of units.  The model for predicting the time to perform each unit of work will be given first.

 Model to predict the time to complete a particular unit:

  TN = KT  * NS

 where:
  TN  = effort required to complete the nth unit
  N   = unit number
  KT  = constant
   (theoretically KT = T1)
  s  = slope parameter or improvement rate
                         (this is a negative value)
     s =  log ? / log 2
     ? = rate of improvement (generally based on doubled units - log 2 implies doubled units).  If ? = .80, then the second unit is done with 80% of the effort of the first unit.  The 4th unit would require 64% of the effort of the first unit.
 

Example:  The first unit of construction is completed in 10,000 hours.  A learning rate of 80% is expected on doubled units (always assume doubled units unless specifically stated otherwise).  How much time will be required to complete the 8th unit?

  s = log ? / log 2 = log .8 / log 2 = -.3219

  TN = KT * NS
 
  T8 = 10,000 * (8)-.3219

  T8 = 5120 hours
 
 

Example:  If the learning rate is 95%, how much time will be required to complete the 8th unit?
  s = log .95 / log 2 = -.074
  T8 = 10,000 * (8)-.074
  T8 = 8574 hours
 What happens if the learning rate is not known?  It must then be computed from the available information.

The learning rate can be determined if information on two units is known.  Suppose that the time of effort is known for two units.

   Ti = KT * Nis  Tj = KT * Njs

  By dividing:  Ti / Tj = (KT * Nis) / (KT * Njs)

   or Ti / Tj = Nis / Njs
 

  Take the log:  log (Ti / Tj) = s * log  (Ni / Nj)

   s = (log Ti - log Tj) / (log Ni - log Nj)

  Then solve for k:
   Tj = KT * Njs

   log Tj = log KT + (s * log Nj)

  Note the form:  Y = A + BX

   KT = intercept  s = slope

Example:  Suppose the 5th unit was completed in 200 hours and that the 10th unit was completed in 150 hours.  Find the time required to compete the 20th and the 30th units.

   s = (log Ti - log Tj)  /  (log Ni - log Nj)

   s = (log 200 - log 150)  /  (log 5 - log 10)

   s = .1249  /  (-.301)  = -.415

   log Tj =  log KT +  (s  *  log Nj)

   log 150 = log KT +  ((-.415) * log 10)

   log KT =  2.5911

   KT =  390
   Learning rate:  s = log ? / log 2

   -.415 = log ? / log 2

   log ? = -.415 * log 2 = -.1249

   ? = .75 or 75 percent
   (Note: this is observable by inspection)

  For unit 20: T20 = KT * Ns

  T20 = 390 * (20)-.415

  T20 = 112.5 hours

  (Note: this is observable by inspection)
 

For unit 30: T30 = KT * Ns
 
  T30 = 390  *  (30)-.415

  T30 =  95.08 Hours
 

If several data points (more that two) are known, a more accurate and perhaps more realistic learning curve can be developed as a predictive tool.  There are essentially two methods of obtaining learning information from the data.  The first is simply to plot the data points on log-log paper.  A "best fit" straight line should then be drawn "through" those points.  The slope of the line will be the learning rate (it will be negative).  The second is to use the least squares fit method.
 
 
 
 
 
 
 
 
 
 
 

 Least squares fit method:

 M = number of data points

Data:
N T log N log T (log N)2 (log N) * (log T)
10
30
100
150
300 510
210
190
125
71  1.0
 1.4771
 2.0
 2.1761
 2.4771
 9.1303  2.7076
 2.3222
 2.2788
 2.0969
 1.8513
 11.2568  1.0
 2.1818
 4.0
 4.7354
 6.136
 18.0532  2.7076
 3.4301
 4.5576
 4.5631
 4.5859
 19.8443
 

Once s and KT are determined, other units can be easily estimated.
 
 

 CUMULATIVE AVERAGE TIMES

Although the foregoing discussion of predicting the amount of effort required to produce a given unit is helpful, it is not as useful as a predictor of the average unit cost of producing a given number of items.  Since estimating is generally based on the overall average cost per unit (rather than the discrete cost associated with each individual unit), a predictor based on unit averages would be more meaningful.

The cumulative average time for completing a given number of units can be related directly to the individual times as determined in the earlier section.  This cumulative average time (hereinafter called CAT) can be shown algebraically as follows:

  CATN  =  (T1 + T2 + T3 + T4 + ......  TN) / N

  T1-N = N * CATN

 Another method: (This is approximate)

 CATN = KT * (Ns+1 -1) / ( (s + 1) * (N - 1))

 Example:   given: KT = 1000   ? = 80%   or  s = -.3219

 Find:  the CAT value for 100 units or find CAT100

 CAT100 = 1000 * (100-.3219+1 -1) / ((-.3219 + 1) * (100 - 1))

  = 21,706 / 67.129  = 323

 Another method:  (More approximate than the above method)
    Reasonably close if N > 20

 CATN  = KT * Ns / (1 + s)

 Example:  The same problem as above

 CAT100 = 1000 * 100-3219 / (1 - .3219) = 335

Although these approximations may have their place, one never has a clear idea of how "approximate" the answer really is.  Thus, a more accurate and more reliable procedure is more desirable.  The format of the formulation that will be used is identical to that already discussed.  The formula is simply stated as follows:

   CATN = Kc * Ns

Regardless of the apparent similarity between this equation and the one used earlier, there is no apparent direct relationship between the values of Kc and KT or between the values of s that are used in the formulas.  The mechanics of solving the problems is, of course, the same.  That is, if the learning rate for individual units is known, it is not easy to determine what the learning rate is for cumulative units.  Using CAT values does however give one greater flexibility.  The following formulas should make that clear:

  T1-N = N * CATN

  TN = (N * CATN) - ( (N-1) * CATN-1)

So, individual unit times to complete a unit are more easily determined from CAT values than it is to determine CAT values from information about individual units (T values). At any rate, the use of CAT values gives more accurate information.  The formula to remember, as before is:

  CATN = Kc * Ns

The mechanics of solving problems with this equation are identical to the method used to solve for T values.

  ***************************************
   What if Work is Interrupted?
  ***************************************

If the flow of work is interrupted for some reason (inclement weather, labor strike, extended holiday season, plant shutdown, job reassignment, etc.) the learning that has occurred will also be effected.  Eventually, the learning will regress to that level which had existed earlier when the first unit was produced.  This learning can be formulated as follows:

  F = 1 - ( 1 / log (D + 10) )

  F = the forgetting function
  D = the number of time units of delay
                 (in the same units as N, usually days)

This function is then used in the following equation:

 CATAD = CATBD + (F * (1 - CATBD) )

where:  CATAD = CAT after the delay (fraction of CAT1)

 CATBD = CAT before the delay (fraction of CAT1)
 
 
 
 
 
 
 

Example:  Given:  Kc = 1000 ? = .90 s = -.152
   interruption of 5 days after 100 units

 Find:  The CAT for units 101 to 105 or find CAT101-105
 
  F = 1 - (1 / log (D + 10)) = 1 - (1 / log 15)  = .1497

  CATBD = 100-.152 = .4966

  CATAD = CATBD + F (1 - CATBD)
   = .4966 + (.1497 (1 - .4966) = .57196

  .57196 = Ns

  N = 39.47

Learning has regressed as a result of the delay as if only 39.47 units had been produced. So after the delay the rate of production will be the same as it was for units 39.47 through 44.47.  That is, after the work resumes the learning rate will resume at the same pace as it had before.  The delay simply shifted the production rate "back up" the learning curve.  So the production for units 101 through 105 will be calculated as follows:

     CAT44.47 = 1000 * (44.47)-.152 = 561.60

     CAT39.47 = 1000 * (39.47)-.152 = 571.96

     CAT39.47-44.47  = (44.47 * (561.68) - 39.47 * (571.96)) / 5

     = 480.53
 
 
 
 
 
 

   LEARNING CURVE PROBLEMS

1. The third unit of an aircraft assembly was completed in 2700 hours.  The 15th unit was built in 2100 hours.  What is the learning rate (based on doubled units)? How long would it take to build the 100th unit?
  (Answer: ? = 89.7%, T100 = 1561.6 hours)

2. A tower (the first one) was installed in 900 hours.  The improvement rate for doubled units is expected to be 90%.  What will be the time required to complete the 5th tower?  The 30th tower?
  (Answer: T5= 704.7 hours, T30 = 536.7 hours)

3. A contractor was in the business of building storage tanks.  The learning rate was 0.80.  He broke even on the 12th unit which brought the average of all the units to 1500 hours per unit with a labor cost of $18,000 per unit.  What was his profit margin on a contract for 30 such units?  Assume that this was the first contract and that he was paid the same price per storage tank.  Give the profit margin in dollars and in percent.
  (Answer: Profit = $137, 944.80 or 34.3%)

4. The CAT value for 5 sluice gate installations was 160 hours (learning was 0.90 for doubled averages).  Find the average time to install 25 sluice gates.  What is the time required to install the 20th sluice gate?
  (Answer: CAT25 = 125.28 hours, T20 = 110.33 hours)

5. A tower was erected in 7000 hours.  The company feels that the learning rate per tower will be 0.80 on doubled units.  How many towers will have to be erected before the time per tower is reduced to 3000 hours?  How much time will be consumed on the 50th tower.
  (Answer: N = 14, T50 = 1986.8 hours)

6. The cumulative average time for constructing 30 electrical assemblies is 170 hours.  The rate of learning for doubled averages is about 0.95.  How much time was consumed in building the 50th unit?  What was the total time required for all units after 100 units were completed?  What total hours were needed to assemble units 100 through 150?
  (Answer: T50 = 151.69, T1-100 = 15,550.90 hours,
    T100-150 = 7230 hours)

7. The learning rate of 0.90 on cumulative averages is realized when constructing metal storage tanks.  The first unit took 1400 hours to construct.  After 6 units were constructed, the hunting season opened and the entire work crew took off a total of 5 days.  Ignoring the impact of weekends, determine the time required to complete units 7 through 12.  If wages are $10.00 per hour, what was the cost of the hunting season in terms of lost productivity?  In other words, what is the maximum size of bonus that could be justified by the company in order to keep the workers on the job?
 
 
 

8. A welder and a carpenter decided to get out of the construction industry and build farm trailers instead.  From having built a few trailers on weekends, they estimated that the first trailer will take about $700 of their own labor to build and that an 85% learning rate could be anticipated on the cumulative average time as each trailer was built.  (Note:  They decided that their hourly wages should be no less than they were receiving in the construction trades.)  The material costs for each trailer will be about $500 and the craftsmen do not see any way that this can be reduced.  They estimate that each trailer can be sold for $1000.  In addition to making their wages on labor, they want to make 15% profit on the trailer materials.  How many trailers must be built before this rate of profit can be realized?  What is the labor cost on the 50th trailer?

9. A learning rate of 0.92 on cumulative averages is realized when constructing metal storage tanks.  The first tank took 1200 hours to construct.  After 6 tanks were constructed, the work was stopped for two days and then the work was resumed. Find the time required to complete units 7 through 10.  If wages are $15.00 per hour, what was the cost of the delay from lost productivity?  What was the total cost of constructing 15 tanks?

10. Study the following data:

   Unit  #  Labor Cost/Unit
        1    $1000
        5      900
       25      810

Based on the trend of production costs on these units, develop a formula that will be usable to predict future costs, i.e.  find the value of s.  Find the cost for unit 100.

11. An off-shore drilling operation requires a large number of welders to build a platform.  The contractor is paying welders $15.00 per hour on a 40-hour (8 hrs/day) work week.  The contractor has never had the same welders return to the platform after they have been on the job for one week.  He is now considering a 7-day work week so that he can take advantage of the "experience" gained in the previous five days of work.  Of course he would have to pay triple time for anything beyond 40 hours.  He figures that all other expenses would equalize (5 trips for 35 work days with 7-day work weeks or 7 trips for 35 work days with 5-day work weeks.  He consistently incurs a cost of $2000 per welder on the first day of work week, but does not know what type of learning rate must take place for the "scheme" to work.  Find the learning rate (to the nearest whole percent) at which the new plan becomes feasible.  What (expressed as a percent) is the benefit or loss on the 7-day work week if the learning rate is 0.90?  Clearly indicate if it is a gain or a loss.
 
12. A contractor has kept accurate records on the installation of pressure vessels in a refinery.  He had the following data:

   Unit #  CAT
      3   1530
      4   1450
      7   1370
     10   1325
     12   1300
     15   1275
 A strike delayed further installations for 2 weeks (16 days), before the 16th unit could be installed.  The contractor had a crew size of 11 workers who receive an average wage of $13.00 per hour.  The rental on the equipment and the company overhead is $240 per day.  The contractor noted that he "broke even" after having completed the 8th unit.  Find the theoretical time to complete the first unit.  Find the learning rate (in percent).  When work resumes, how many units must be completed before the crew is working at the same pace it had when it completed the 15th unit?  How long will it take to complete the 16th unit?  What is the cost of the delay on units 16 through 20?  What is the profit margin on 20 units if no delay had occurred?