Question

A bus travels between the two cities A and B, which are $100$miles apart. If the bus has a breakdown, the distance from the breakdown to city A has a uniform distribution over $(0,100)$. There is a bus service station in city A, in B, and in the center of the route between A and B. It is suggested that it would be more efficient to have the three stations located $25,50,\mathrm{and}75$miles, respectively, from A. Do you agree? Why?

Short answer

The new scheme is efficient because the expected value for new scheme is less.

Step-by-step solution

Step 1. Given information.

Here, it is given that:

The distance between City A and B $=100miles$

In case of breakdown of the bus, the distance from the breakdown to city A has a uniform distribution over$\left(0,100\right)$.

Step 2. Find the probability density function.

Let, the distance between the place where bus breaks down to city A be $X$and the distance to the nearest service station be Y.

The random variable X follows uniform distribution with $\left(0,100\right)$.

The probability density function of uniform distribution can be defined as:

$f\left(x\right)=\frac{1}{B-A}=\frac{1}{100}$

Step 3. Find the expected time under current scheme.

If three stations are located $0,50\mathrm{and}100$miles, then the towing distance will be given by:

role="math" localid="1646658782747" $Y=g_1\left(x\right)=\left\{\begin{array}{l}X0<X<25\\ \left|X-50\right|25\le X\le 75\\ 100-X75<X<100\end{array}\right.$

The expected time under the current scheme is:

role="math" localid="1646659067639" $$\begin{gathered} E\left(Y\right)={\int }_0^{100}{g}_1\left(x\right)f\left(x\right)dx \\ ={\int }_0^{25}x\left(\frac{1}{100}\right)dx+{\int }_{25}^{75}\left|x-50\right|\left(\frac{1}{100}\right)dx+{\int }_{75}^{100}\left(100-x\right)\left(\frac{1}{100}\right)dx \\ =\frac{1}{100}\left[{\left(\frac{x^2}{2}\right)}_0^{25}+{\left(50x-\frac{x^2}{2}\right)}_{25}^{50}+{\left(\frac{x^2}{2}-50x\right)}_{50}^{75}+{\left(100x-\frac{x^2}{2}\right)}_{75}^{100}\right] \\ =\frac{1}{100}\left[\frac{625}{2}+\frac{625}{2}+\frac{625}{2}+\frac{625}{2}\right] \\ =12.5 \end{gathered}$$

Step 4. Find the expected time under new scheme.

If the service stations are located at $25,50,\text{and}100$miles then the towing distance will be given by:

$Y=g_1\left(x\right)=\left\{\begin{array}{l}\left|X-25\right|0<X<37.5\\ \left|X-50\right|37.5\le X\le 62.5\\ \left|X-75\right|62.5<X<100\end{array}\right.$

The expected time under the new scheme is:

$$\begin{gathered} E\left(Y\right)={\int }_0^{100}{g}_2\left(x\right)f\left(x\right)dx \\ ={\int }_0^{25}\left|25-x\right|\left(\frac{1}{100}\right)dx+{\int }_{25}^{75}\left|x-50\right|\left(\frac{1}{100}\right)dx+{\int }_{75}^{100}\left|x-75\right|\left(\frac{1}{100}\right)dx \\ ={\int }_0^{25}\left(25-x\right)dx+{\int }_{25}^{37.5}\left(x-25\right)dx+{\int }_{37.5}^{50}\left(50-x\right)dx+{\int }_{50}^{62.5}\left(x-50\right)dx+{\int }_{62.5}^{75}\left(75-x\right)dx+{\int }_{75}^{100}\left(x-75\right)dx \\ =\frac{1}{100}\left[{\left(25x-\frac{x^2}{2}\right)}_0^{25}+{\left(\frac{x^2}{2}-25x\right)}_{25}^{37.5}+{\left(50x-\frac{x^2}{2}\right)}_{37.5}^{50}+{\left(\frac{x^2}{2}-50x\right)}_{50}^{62.5}+{\left(75x-\frac{x^2}{2}\right)}_{62.5}^{75}+{\left(\frac{x^2}{2}-75x\right)}_{75}^{100}\right] \\ =\frac{1}{100}\left[\frac{625}{2}+78.125+78.125+78.125+78.125+\frac{625}{2}\right] \\ =9.375 \end{gathered}$$

Step 5. Compare the expected time of both the schemes.

The expected time of new scheme is lesser than the expected time of the current scheme. Therefore, the new scheme is better.