Chapter 4.14, Problem 3P

Explanation of Solution

Formulation of LP to minimize Steelco’s costs:

From the information given, observe that during the next 3 months, steeco faces has three demands for steel.

Consider the following terms:

$i_x$=Tons of steel in inventory at the end of month “t”.

$p_x$=Tons of steel produced in month “t”.

$w_x$=Workers employed during month “t”.

$z$=Cost of satisfy demand.

Now, the one can write the following equations,

$\begin{array}{l}{i}_x=i_l\text{'}-i_l\text{'}\text{'}\\ x_x=x_l\text{'}-x_l\text{'}\text{'}\end{array}$

Here,

$i_l\text{'}$=Tons of steel left in inventory at the end of month “t”.

$i_l\text{'}\text{'}$=Tons of steel demanded after all inventory used up in month “t”.

$x_l$=Change of number of workers from month “(l-1)” to month “t”.

$x_l\text{'}$=Workers hired at the start of month “t”.

$x_l\text{'}\text{'}$=Workers fired at the start of the month “t”.

Objective function:

The objective function as follows,

Minimize

$\begin{array}{c}z=300(p_1+p_2+p_3)+100(i_1{}^x+i_2{}^x+i_3{}^{-1})\\ +70(i_1\text{'}\text{'}+i_2\text{'}\text{'}+i_3\text{'}\text{'})+5000(w_1+w_2+w_3)+\\ +3000(x_1\text{'}\text{'}+x_2\text{'}\text{'}+x_3\text{'}\text{'})+4000(x_1\text{'}+x_2\text{'}+x_3\text{'})\end{array}$

Subject to the constraints,

$\begin{array}{c}{i}_1=i_1\text{'}-i_1\text{'}\text{'}[invent\mathrm{o</}\end{array}$