Chapter 6.EA, Problem 1EA

In this application, we set up a mathematical model for determining the total costs in setting up a training program, such as a hospital might use. Then we use calculus to find the time interval between training programs that produces the minimum total cost. The model assumes that the demand for trainees is constant and that the fixed cost of training a batch of trainees is known. Also, it is assumed that people who are trained, but for whom no job is readily available, will be paid a fixed amount per month while waiting for a job to open up.

The model uses the following variables.

D = demand for trainees per month

N = number of trainees per batch

$C_1$ = Fixed cost of training a batch of trainees

$C_2$ = marginal cost of training per trainee per month

$C_3$ = salary paid monthly to a trainee who has not yet been given a job after training

m = time interval in months between successive batches of trainees

t = length of training program in months

$Z\left(m\right)$ = total monthly cost of program

The total cost of training a batch of trainees is given by $C_1+Nt{C}_2$. However, $N=mD$, so that the total cost per batch is $C_1+mDt{C}_2$.

After training, personnel are given jobs at the rate of D per month. Thus, $N-D$ of the trainees will not get a job the first month, $N-2D$ will not get a job the second month, and so on. The $N-D$ trainees who do not get a job the first month produce total costs of $\left(N-D\right)C_3$, and so on. Since $N=mD$, the costs during the first month can be written as.

$\left(N-D\right)C_3=\left(mD-D\right)C_3=\left(m-1\right)D{C}_3$

While the costs during the second month are $\left(m-2\right)D{C}_3$, and so on. The total cost for keeping the trainees without a job is thus

$\left(m-1\right)D{C}_3+\left(m-2\right)D{C}_3+\left(m-3\right)D{C}_3+\cdots +2D{C}_3+D{C}_3$

Which can be factored to give

$D{C}_3\left[\left(m-1\right)+\left(m-2\right)+\left(m-3\right)+\cdots +2+1\right]$

The expression in bracket is the sum of the terms of an arithmetic sequence, discussed in most algebra texts. Using formulas for arithmetic sequences, the expression in brackets can be shown to equal $\frac{m\left(m-1\right)}{2}$, so that we have

$D{C}_3\left[\frac{m\left(m-1\right)}{2}\right](1)$

As the total cost for keeping jobless trainees.

The total cost per batch is the sum of the training cost per batch. $C_1+mDt{C}_2$, and the cost of keeping trainees without a proper job, given by equation (1), since we assume that a batch of trainees is trained every m months, the total cost per month, $Z\left(m\right)$, is given by

$\begin{array}{rllll}Z\left(m\right)& & =& & \frac{C_1+mDt{C}_2}{m}+\frac{D{C}_3\left[\frac{m\left(m-1\right)}{2}\right]}{m}\\ & & =& & \frac{C_1}{m}+Dt{C}_2+D{C}_3\left(\frac{m-1}{2}\right)\end{array}$

Source: P.I., Goyal and S.K. Goyal.

Find $Z^{\prime }\left(m\right)$

To determine

To find:

The expression $Z^{\prime }\left(m\right)$ for the given function $Z\left(m\right)=\frac{C_1}{m}+Dt{C}_2+D{C}_3\left(\frac{m-1}{2}\right)$.

Answer to Problem 1EA

Solution:

The expression $Z^{\prime }\left(m\right)$ for the given function is $Z^{\prime }\left(m\right)=-\frac{C_1}{m^2}+\frac{D{C}_3}{2}$.

Explanation of Solution

Given:

The given function is:

$Z\left(m\right)=\frac{C_1}{m}+Dt{C}_2+D{C}_3\left(\frac{m-1}{2}\right)$.

Approach:

Differentiate the given function with respect to m.

Calculation:

Consider the given function,

$Z\left(m\right)=\frac{C_1}{m}+Dt{C}_2+D{C}_3\left(\frac{m-1}{2}\right)$

Differentiate with respect to m:

$Z^{\prime }\left(m\right)=-\frac{C_1}{m^2}+\frac{D{C}_3}{2}$