Chapter 3, Problem 34RP

Explanation of Solution

 Formulation of Linear Programming (LP) to minimize the cost of meeting air conditioner demands for the next three months:

 Let consider,

• “x_i” be the number of production of air conditioners during month “i”:
• “y_i” be the number of production of air conditioners during month “i”:
• “I_i” be the inventory at the end of month “i”:

 The objective is to minimize the cost of meeting the three months’ demand.

 $z=\left[\begin{array}{l}\left(\text{cost of production of AC in LA}\right)\left(\text{number of production of AC in LA}\right)+\\ \left(\text{cost of production of AC in NY}\right)\left(\text{number of production of AC in NY}\right)+\\ \left(\text{cost to hold an AC}\right)\left(\text{AC in inventory for i month}\right)\end{array}\right]$

 $z=400\left(x_1+x_2+x_3\right)+350\left(y_1+y_2+y_3\right)+100\left(I_1+I_2+I_3\right)$

 Therefore, the objective function is,

 Minimize

 $z=400\left(x_1+x_2+x_3\right)+350\left(y_1+y_2+y_3\right)+100\left(I_1+I_2+I_3\right)$

 Constraint 1:

 During month 1, 300 air conditioners are in demand and at the beginning of month 1 Airco has 200 air conditioner.

 $\left(\text{inventory at the end of month 1}\right)=\left[\begin{array}{l}\left(\text{inventory of AC at the beginning of month 1}\right)+\\ \left(\text{production during month 1}\right)-\\ \left(\text{number of AC required during month 1}\right)\end{array}\right]$

 $\begin{array}{l}{I}_1=200+x_1+y_1-300\\ I_1-200-x_1-y_1+300=0\end{array}$

 Constraint 2:

 During month 2, 400 air conditioners are in demand.

 $\left(\text{inventory at the end of month 2}\right)=\left[\begin{array}{l}\left(\text{inventory of AC at the beginning of month 2}\right)+\\ \left(\text{production during month 2}\right)-\\ \left(\text{number of AC required during month 2}\right)\end{array}\right]$

 $\begin{array}{l}{I}_2=I_1+x_2+y_2-400\\ I_2-I_1-x_2-y_2+400=0\end{array}$

 Constraint 3:

 During month 3, 500 air conditioners are in demand