Chapter 3, Problem 54RP

Explanation of Solution

LP to maximize the profit of Priceler:

• Assume the following,
  • Let “${\text{MS}}_{\text{m}}$” be the number of sedans made during the month “m”.
  • Let “${\text{MW}}_{\text{m}}$” be the number of wagons made during the month “m”.
  • Let “${\text{S}}_{\text{m}}$” be the number of sedans that are sold during the month “m”.
  • Let “${\text{WS}}_{\text{m}}$” be the number of wagons that are sold during the month “m”.
  • Let “${\text{SI}}_{\text{m}}$” be the number of sedans in the inventory at the end of month “m”.
  • Let “${\text{WI}}_{\text{m}}$” be the number of wagons in the inventory at the end of month “m”.
• The Linear Programming (LP) to maximize the profit of Priceler during the next “3” months is as follows,

$\begin{array}{c}\text{Max}{\text{z=8000(S}}_{\text{1}}{\text{+S}}_2+{\text{S}}_3){\text{+9000(WS}}_1{\text{+WS}}_2+{\text{WS}}_3)\\ -6000(M{S}_1+M{S}_2+M{S}_3)-7500(M{W}_1+M{W}_2+M{W}_3)\\ -150(S{I}_1+S{I}_2+S{I}_3)-200(W{I}_1+W{I}_2+W{I}_3)\end{array}$

• The above function subject to the following conditions,
  • “${\text{MS}}_{\text{1}}{\text{+MW}}_{\text{1}}\le 1500$” denotes that the maximum number of vehicles made during the month “1” must be “1500”.
  • “${\text{MS}}_2{\text{+MW}}_2\le 1500$” denotes that the maximum number of vehicles made during the month “2” must be “1500”.
  • “${\text{MS}}_3{\text{+MW}}_3\le 1500$” denotes that the maximum number of vehicles made during the month “3” must be “1500”.
  • “${\text{MS}}_{\text{1}}\ge \frac{{\text{2(MS}}_{\text{1}}+{\text{MW}}_{\text{1}})}{3}$” represents that at least “$\frac{2}{3}$” of all vehicles made during the month “1” should be sedans