Question

There are two possible causes for a breakdown of a machine. To check the first possibility would cost C1 dollars, and, if that were the cause of the breakdown, the trouble could be repaired at a cost of R1 dollars. Similarly, there are costs C2 and R2 associated with the second possibility. Let p and 1 − p denote, respectively, the probabilities that the breakdown is caused by the first and second possibilities. Under what conditions on p, Ci, Ri, i = 1, 2, should we check the first possible cause of breakdown and then the second, as opposed to reversing the checking order, so as to minimize the expected cost involved in returning the machine to working order?

Short answer

The expected amount of money that is needed to repair is minimized when,

$E\left(X_1\right)=E\left(X_2\right)$and it happen if and only if

$C_1=p/{\left(1-p\right)}^{}{C}_2$.

Step-by-step solution

Step 1:Given Information

We have two options to check the causes. we can check the first cause, and then the first one was not the problem, check and repair according to the second cause.

similarly,we can check the second cause and it was not the problem repair and solve aacording to the first cause

Step 2:Explanation

$E\left(X_1\right)=C_1+p.R_1+\left(1-p\right)\left(C_2+R_2\right)$

$E\left(X_2\right)=C_2+\left(1-p\right)R_2+p\left(C_1+R_1\right)$

$E\left(X_1\right)<E\left(X_2\right)$

$\iff C_1+p·R_1+(1-p)\left(C_2+R_2\right)<C_2+(1-p)R_2+p\left(C_1+R_1\right)$

$\iff C_1+(1-p)C_2<C_2+p{C}_1$

$\iff C_1<\frac{p}{1-p}{C}_2$

Step 3:Final Answer

The expected amount of money that is needed to repair is minimized when

$E\left(X_1\right)=E\left(X_2\right)$and it happens if and only if

$C_1=p/(1-p)C_2$.