Learning Curve Solutions

Stephen R. Lawrence (c) 1996

Question 1. The first-unit cost for both Big Blue and Little Red are the same at K(1)= $1,000. However Little Red learns more quickly with R=85% versus Big Blue's learning rate of R=93%. After producing 10 units, the unit cost for each company can be calculated for both companies.

For Big Blue:

• b = log R / log 2 = log 0.93 / log 2 = -0.1047
• K(n) = K(m)( n/m)^b = 1,000 (10/1)^ (-0.1047) = $785.78

For Little Red:

• b = log R / log 2 = log 0.85 / log 2 = -0.2345
• K(n) = K(m)( n/m)^b = 1,000 (10/1)^(-0.2345) = $582.82

The unit costs for other production quantities can be similarly calculated. Summarizing:

         Big Blue  Little Red 
      1 $1,000.00  $1,000.00 
     10    785.78     582.82 
    100    617.45     339.68 
  1,000    485.19     197.97 
 10,000    381.25     115.38 

Clearly, the unit production costs of Little Red are declining much more rapidly than those for Big Blue. Unless Big Blue can somehow change the game it is playing, it will very quickly be at a severe competitive disadvantage relative to Little Red!

Question 2. This ski equipment firm has an historical 90% learning curve. To determine when (and if) production costs for the new process will ever fall to those of the current process, we can use trial and error with LEARN.XLS to determine when unit production costs are equal for the two processes. For the current process, n=5,000 and K1(5,000)=150. For the new process, n=1 and K2(1)=250. By trying different quantitities of addtional production, we find that unit production costs for the two processes will be equal after only 28 more units are produced! That is:

• K1(5,028) = K2(29) = 149.86

So very quickly, unit costs for the new process will fall to those of the old process. After a year, an additional 2,000 pairs of skis will be produced and unit production costs will be K1(7,000) = $142.52 and K2(2,000) = $78.74. Clearly, the new process will provide great savings over time, and should be adopted, all else equal.

Question 3. Summarizing the data for this problem:

             Good Water    WaterPure   
    K(1)        $150          $100     
     n           100         10,000    
    K(n)       $52.50          ??      
     R           ??           0.95     

3a) WaterPure: We want K(n) for n = 10,000. Using LEARN.XLS, enter in the learning rate, K(1) , and n for Waterpure. Examining the the solution for n=10,000, you should find that K(10,000) = $50.48.

This result can also be directly determined. First calculate parameter b:

• b= log R / log 2, with R = 0.95.
• b = log 0.95 / log 2 = - 0.0223 / 0.3010 = - 0.0742.
• Plug this value for b in the equation for K(n) with n=10,000 :
• K(n) = K(10,000) = K(1) * n * b = 100 ( 10,000) -0.0742 = 100 ( 0.5048) = 50.48
• K(10,000) = $50.48

3b) GoodWater: For GoodWater we want the learning rate. Plug in K(1) , n, and K(n) for GoodWater into LEARN.XLS and solve. You should find that GoodWater has a learning rate of R = 85.4%.

To calculate the current learning rate for GoodWater, first calculate the value for b, then use the equation b= log [K(n) - log K(1)] / log n to calculate the value for R.

• b = [log K(n) - log K(1)] / log n = [log 52.50 - log 150 ] / log 100 = -0.4559 / 2 = -0.228

Using the equation R = 10 ( b log 2)

• R = 10 ( - 0.152 * 0.3010 ) = 0.854
• R = 85.4%

3c) Using the learning rates for GoodWater and WaterPure determined above, and target production figures of 2,500 and 12,500, with LEARN.XLS, you should find that K(2,500) = $25.19 for GoodWater and K(12,500) = $49.75 for WaterPure.

To calculate these figures directly, use the values for b obtained above and the given data for K(1)

• GoodWater (n = 2,500): K(2,500) = 150 ( 2,500) - 0.228 = 150 ( 0.1680) = $25.19
• WaterPure (n = 12,500): K(12,500) = 100 (12,500) -0.074 = 100 (0.4966) = $49.75

3d) Inform the company president that within a year, GoodWater will be earning almost $15.00 on every unit they make, even when pricing them at $40.00 a unit. In contrast, at the same time, WaterPure would lose $9.75 per unit if it tried to match GoodWater's price! Some possible options for WaterPure: increase the learning rate of the production process by adopting continuous improvement methods; or, increase the production volume so that even with a lower learning rate, WaterPure will be able to catch up with GoodWater's production costs (very difficult). In any case, WaterPure is facing a serious competitive challenge and had better make some changes!