To determine the initial output time for the first unit produced, we need to follow several steps based on the learning curve formula [tex]\( Y_x = aX^b \)[/tex]:
1. Given Parameters:
- Total number of iterations ([tex]\(X\)[/tex]) = 20
- Steady state time ([tex]\(Y_x\)[/tex]) = 2.5 hours
- Learning rate = 75% or 0.75
2. Calculate the Slope of the Learning Curve ([tex]\(b\)[/tex]):
The slope ([tex]\(b\)[/tex]) of the learning curve can be calculated using:
[tex]\( \log (\text{learning rate}) / \log (2) \)[/tex]
Given:
- Learning rate = 0.75, hence [tex]\(\log(learning rate) = \log(0.75) \)[/tex]
- [tex]\(\log(2)\)[/tex] is a constant.
Therefore, [tex]\( b \)[/tex] is calculated as:
[tex]\[
b = \frac{\log(0.75)}{\log(2)} = -0.4150374992788438
\][/tex]
3. Determine the Steady State Unit ([tex]\(X\)[/tex]):
Already given is [tex]\( X = 20 \)[/tex].
4. Use the Learning Curve Formula to Find Initial Output Time ([tex]\(a\)[/tex]):
Rearrange the formula [tex]\( Y_x = aX^b \)[/tex] to solve for [tex]\(a\)[/tex]:
[tex]\[
a = \frac{Y_x}{X^b}
\][/tex]
Plugging in the values:
- [tex]\( Y_x = 2.5 \)[/tex] hours
- [tex]\( X = 20 \)[/tex]
- [tex]\( b = -0.4150374992788438 \)[/tex]
Calculate:
[tex]\[
a = \frac{2.5}{20^{-0.4150374992788438}}
\][/tex]
Simplifying the denominator:
[tex]\[
20^{-0.4150374992788438} = \text{approximately} \, 0.288482
\][/tex]
Now, calculate [tex]\( a \)[/tex]:
[tex]\[
a = \frac{2.5}{0.288482} = 8.667946947883458 \text{ hours}
\][/tex]
5. Interpret the Results:
- The slope of the learning curve [tex]\( b \)[/tex] is [tex]\(-0.4150374992788438\)[/tex].
- The number of iterations to reach steady state [tex]\( X \)[/tex] is [tex]\( 20 \)[/tex].
- The initial output time for the first unit produced [tex]\( a \)[/tex] is [tex]\( 8.667946947883458 \)[/tex] hours.
Therefore, the initial output time for the first unit produced is approximately [tex]\( 8.67 \)[/tex] hours.
