Chapter 3.5, Problem 36E

a.

To determine

Find the learning curve for this employee.

Answer to Problem 36E

The learning curve for this employee is $N=30\left(1-e^{\frac{t}{20}\ln \frac{11}{30}}\right)$ .

Explanation of Solution

Given:

The given model is $N=30\left(1-e^{kt}\right)$ . New employee produces $19$ units in $20$ days.

Calculation:

  $N=30\left(1-e^{kt}\right)$

Substitute the value of $N=19$ and $t=20$ in the equation $\left(i\right)$ .

  $\begin{array}{l}\Rightarrow 19=30\left(1-e^{k\cdot 20}\right)\\ \Rightarrow \frac{19}{30}=\frac{30}{30}\left(1-e^{20k}\right)\\ \Rightarrow \frac{19}{30}=1-e^{20k}\\ \Rightarrow \frac{19}{30}-1=1-e^{20k}-1\\ \Rightarrow \frac{11}{30}=e^{20k}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{11}{30}=\ln e^{20k}\\ \Rightarrow \ln \frac{11}{30}=20k\\ \Rightarrow \frac{1}{20}\ln \frac{11}{30}=\frac{20}{20}k\\ \Rightarrow k=\frac{1}{20}\ln \frac{11}{30}\end{array}$

Hence the learning curve for this employee is $N=30\left(1-e^{\frac{t}{20}\ln \frac{11}{30}}\right)$ .

b.

To determine

Find the number of days for the givenunits.

Answer to Problem 36E

The employee completes the $25$ units in $36$ days.

Explanation of Solution

Given:

The given model is $N=30\left(1-e^{kt}\right)$ . The given units are $25$ .

Calculation:

  $N=30\left(1-e^{kt}\right)$

Substitute the value of $N=19$ and $t=20$ in the equation $\left(i\right)$ .

  $\begin{array}{l}\Rightarrow 19=30\left(1-e^{k\cdot 20}\right)\\ \Rightarrow \frac{19}{30}=\frac{30}{30}\left(1-e^{20k}\right)\\ \Rightarrow \frac{19}{30}=1-e^{20k}\\ \Rightarrow \frac{19}{30}-1=1-e^{20k}-1\\ \Rightarrow \frac{11}{30}=e^{20k}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{11}{30}=\ln e^{20k}\\ \Rightarrow \ln \frac{11}{30}=20k\\ \Rightarrow \frac{1}{20}\ln \frac{11}{30}=\frac{20}{20}k\\ \Rightarrow k=\frac{1}{20}\ln \frac{11}{30}\end{array}$

Now,

  $N=30\left(1-e^{\frac{t}{20}\ln \frac{11}{30}}\right)$

Substitute the value of $N=25$ in the equation.

  $\begin{array}{l}25=30\left(1-e^{\frac{t}{20}\ln \frac{11}{30}}\right)\\ \Rightarrow \frac{25}{30}=\frac{30}{30}\left(1-e^{\frac{t}{20}\ln \frac{11}{30}}\right)\\ \Rightarrow \frac{5}{6}=1-e^{\frac{t}{20}\ln \frac{11}{30}}\\ \Rightarrow \frac{5}{6}-1=1-e^{\frac{t}{20}\ln \frac{11}{30}}-1\\ \Rightarrow \frac{1}{6}=e^{\frac{t}{20}\ln \frac{11}{30}}\end{array}$

Take natural log on both sides.

  $\begin{array}{l}\Rightarrow \ln \frac{1}{6}=\ln e^{\frac{t}{20}\ln \frac{11}{30}}\\ \Rightarrow \ln \frac{1}{6}=\frac{t}{20}\ln \frac{11}{30}\\ \Rightarrow -1.79=\frac{t}{20}\cdot \left(-1.003\right)\\ \Rightarrow \frac{-1.79}{-1.003}=\frac{t}{20}\cdot \frac{\left(-1.003\right)}{-1.003}\\ \Rightarrow 1.78=\frac{t}{20}\\ \Rightarrow 1.78\cdot 20=\frac{t}{20}\cdot 20\\ \Rightarrow 35.69=t\\ \Rightarrow t\approx 36\end{array}$

Hence the employee completes the $25$ units in $36$ days.