# Learning Curve

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A pharmacist in a drugstore must check $$1,000$$ receipts per day. A new pharmacist was hired in the drugstore. In $$1\text{st}$$ week, the pharmacist was able to check $$100$$ receipts per day. Estimate the number of receipts the pharmacist can check for the second week.

### Example 2

Suppose that a news is spread by mass media according to the law described by the learning curve. What should be the initial percentage of the population who know about this news so that it could reach $$50\%$$ of the population in $$1$$ week and $$90\%$$ of the population in $$4$$ weeks?

### Example 1.

A pharmacist in a drugstore must check $$1,000$$ receipts per day. A new pharmacist was hired in the drugstore. In $$1\text{st}$$ week, the pharmacist was able to check $$100$$ receipts per day. Estimate the number of receipts the pharmacist can check for the second week.

Solution.

By setting the initial skill level to zero: $$M = 0,$$ we can describe the learning process by the following law:

$L\left( t \right) = {L_{\max }}\left( {1 - {e^{ - kt}}} \right).$

Determine the parameter $$k$$ knowing the number of receipts completed in $$1$$ week:

$100 = 1000\left( {1 - {e^{ - kt}}} \right),\;\; \Rightarrow 1 - {e^{ - kt}} = 0.1,\;\; \Rightarrow e^{ - kt} = 0.9,\;\; \Rightarrow - kt = \ln 0.9,\;\; \Rightarrow k = - \frac{{\ln 0.9}}{t}.$

By substituting $$t = 1\,\text{week},$$ we find:

$k = - \frac{{\ln 0.9}}{1} \approx - 0.105$

Now we can calculate productivity of the new pharmacist for the second week:

$L\left( {t = 2} \right) = 1000\left( {1 - {e^{ - 0.105 \cdot 2}}} \right) = 1000\left( {1 - {e^{ - 0.21}}} \right) \approx 1000\left( {1 - 0.811} \right) = 189\left[ {\frac{\text{receipts}}{\text{day}}} \right].$

### Example 2.

Suppose that a news is spread by mass media according to the law described by the learning curve. What should be the initial percentage of the population who know about this news so that it could reach $$50\%$$ of the population in $$1$$ week and $$90\%$$ of the population in $$4$$ weeks?

Solution.

The news is spread according to the law:

$L\left( t \right) = {L_{\max }} - M{e^{ - kt}}.$

We know two points on this curve: at $$t = 1\,\text{week}$$ and $$t = 4\,\text{weeks}.$$ Therefore, we can write the following two equations:

$\left\{ \begin{array}{l} L\left( {t = 1} \right) = {L_{\max }} - r{L_{\max }}{e^{ - k \cdot 1}}\\ L\left( {t = 4} \right) = {L_{\max }} - r{L_{\max }}{e^{ - k \cdot 4}} \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} 0.5{L_{\max }} = {L_{\max }} - r{L_{\max }}{e^{ - k \cdot 1}}\\ 0.9{L_{\max }} = {L_{\max }} - r{L_{\max }}{e^{ - k \cdot 4}} \end{array} \right.$

Here the parameter $$M$$ is represented as $$r{L_{\max }},$$ where $$r$$ lies in the interval $$0 \le r \le 1,$$ and the time $$t$$ is expressed in weeks.

Dividing by $${L_{\max }}$$ gives the system of equations with two unknowns: $$r$$ and $$k:$$

$\left\{ \begin{array}{l} 0.5 = 1 - r{e^{ - k}}\\ 0.9 = 1 - r{e^{ - 4k}} \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} r{e^{ - k}} = 0.5\\ r{e^{ - 4k}} = 0.1 \end{array} \right..$

By taking logarithms in each equation, we obtain:

$\left\{ \begin{array}{l} \ln r - k = \ln 0.5\\ \ln r - 4k = \ln 0.1 \end{array} \right..$

Eliminate the parameter $$k$$ from the system by multiplying the first equation by $$-4$$ and then adding both the equations:

$\left. {\left\{ \begin{array}{l} - 4\ln r + 4k = - 4\ln 0.5\\ \ln r - 4k = \ln 0.1 \end{array} \right.} \right| + ,\;\; \Rightarrow - 3\ln r = \ln 0.1 - 4\ln 0.5,\;\; \Rightarrow \ln r = \frac{1}{3}\left( {4\ln 0.5 - \ln 0.1} \right).$

Easy calculations give the following answer:

$\ln r = \frac{1}{3}\left( {4\ln 0.5 - \ln 0.1} \right) \approx \frac{1}{3}\left[ {4 \cdot \left( { - 0.693} \right) - \left( { - 2.302} \right)} \right] = \frac{1}{3}\left( { - 0.469} \right) = - 0.156$

It follows from here that the parameter $$r$$ is equal to

$r = {e^{ - 0.156}} \approx 0.855$

Thus, in the given case the initial level of penetration of the news must be equal to

$L\left( {t = 0} \right) = {L_{\max }} - r{L_{\max }} = {L_{\max }} - 0.855{L_{\max }} = 0.145{L_{\max }} \approx 15\% {L_{\max }}.$