Crawford Learning Curve

it would help if you could provide information about what a crawford learning curve is, this being an excel forum not a crawford learning curve forum :wink:

cross post

http://www.excelforum.com/excel-charting/660191-crawford-learning-curve.html

Sorry about the cross post I will re-read the rules.

This is what I used to help me better understand the Crawford Learning Curve.

The equation used in Crawford's model is as follows:
Y = aK<SUP>b
</SUP>where: Y = the incremental unit time (or cost) of the lot midpoint unit.
K = the algebraic midpoint of a specific production batch or lot.

X (i.e., the cumulative number of units produced) can be used in the equation instead of K, to find the unit cost of any particular unit, but determining the unit cost of the last unit produced is not useful in determining the cost of a batch of units. The unit cost of each unit in the batch would have to be determined separately. This is obviously not a practical way to solve for the cost of a batch that may involve hundreds, or even thousands of units. A practical approach involves calculating the midpoint of the lot. The unit cost of the midpoint unit is the average unit cost for the lot. Thus, the cost of the lot is found by calculating the cost of the midpoint unit and then multiplying by the number of units in the lot.
Since the relationships are non linear, the algebraic midpoint requires solving the following equation:
K = [L(1+b)/(N2<SUP>1+b</SUP> - N1<SUP>1+b</SUP>)]<SUP>-1/b</SUP>
where:
K = the algebraic midpoint of the lot.
L = the number of units in the lot.
b = log of learning rate / log of 2
N1 = the first unit in the lot minus 1/2.
N2 = the last unit in the lot plus 1/2.
Once Y<SUB>c</SUB> is determined for the algebraic midpoint of a lot, then the cost of the entire lot is found by multiplying Y<SUB>c</SUB> by the number of units in the lot as indicated above.
An example of an 80 percent learning curve based on Crawford's unit time (or cost) model.

TABLE 2
<TABLE width="100%" border=1><TBODY><TR><TD vAlign=top align=middle width="20%" bgColor=#c0c0c0>1
Cumulative Output
X

</TD><TD vAlign=top align=middle width="20%" bgColor=#c0c0c0>2</B>
Incremental Unit Labor Hours
Yh

</TD><TD vAlign=top align=middle width="20%" bgColor=#c0c0c0>3</B>
Cumulative Total Labor Hours
Kh (Yh)

</TD><TD vAlign=top align=middle width="20%" bgColor=#c0c0c0>4</B>
Incremental Unit Labor Cost
Yc

</TD><TD vAlign=top align=middle width="20%" bgColor=#c0c0c0>5</B>
Cumulative Total Labor Cost
Kc(Yc)

</TD></TR><TR><TD vAlign=top align=middle width="20%" bgColor=#bbddff>1 2</B>
4
8


</TD><TD vAlign=top align=middle width="20%" bgColor=#bbddff>100 80</B>
64
51.2


</TD><TD vAlign=top align=middle width="20%" bgColor=#bbddff>100 180</B>
314.2
534.6


</TD><TD vAlign=top align=middle width="20%" bgColor=#bbddff>$2,000 1,600</B>
6,284
1,024


</TD><TD vAlign=top align=middle width="20%" bgColor=#bbddff>$2,000 3,600</B>
6,284
10,692


</TD></TR></TBODY></TABLE></B>
Notice from Table 2 that the unit labor hours (column 2) and unit labor cost (column 4) decrease by 20% each time the cumulative output is doubled. However, the cumulative total labor hours (column 3) and cumulative total labor cost (column 5) increase by a variable rate. This means that columns 3 and 5 are much more difficult to develop. To illustrate the use of the algebraic midpoint equation and Crawford's approach, assume that the firm in the example above has produced 2 units and wants to determine the cost of producing 4 additional units. One way to find the answer is to calculate the unit cost for each unit 3 through 6 and then sum those values. That works reasonably well for a lot of 4 units, but would not be a practical way to determine the cost of 40, 400, or 4,000 additional units. The midpoint of the lot is:
K = [L(1+b)/(N2<SUP>1+b</SUP> - N1<SUP>1+b</SUP>)]<SUP>-1/b</SUP> = [4(.678)/(6.5<SUP>.678</SUP> - 2.5<SUP>.678</SUP>)]<SUP>1/.322
</SUP>= [2.712/(3.55758 - 1.86124)]<SUP>3.10559</SUP> = 4.29385

The cost of the mid point unit is:
Y<SUB>c</SUB> = $2,000(4.29385)<SUP>-.322</SUP> = $1,250.99
and the total cost for the lot of 4 = 4(1,250.99) = $5,005

An alternative is to use the equation for hours as follows:
Y<SUP>h</SUP> = 100(4.29385)<SUP>-.322</SUP> = 62.5494 hours
Then the total cost for the lot of 4 is 4(62.5494)($20) = $5,004.

This explaination can be found at http://maaw.info/LearningCurveSummary.htm.

I posted a solution at excelforum.