Chapter 3.12, Problem 4P

Explanation of Solution

Linear Programming (LP) to maximize the profit:

• Assume the following,
  • Let “${\text{WS}}_{\text{m}}$” be the bushels of wheat sold during the month “m”.
  • Let “${\text{WB}}_{\text{m}}$” be the bushels of wheat purchased during the month “m”.
  • Let “${\text{WI}}_{\text{m}}$” be the bushels of wheat in inventory at the end of month “m”.
• The Linear programming (LP) to maximize the profit earned over the next “10” months is as follows,

  $\begin{array}{l}{\text{Max 3WS}}_{\text{1}}{\text{+ 6WS}}_{\text{2}}{\text{+7WS}}_{\text{3}}{\text{+WS}}_{\text{4}}{\text{+4WS}}_{\text{5}}{\text{+5WS}}_{\text{6}}{\text{+5WS}}_{\text{7}}{\text{+WS}}_{\text{8}}{\text{+3WS}}_{\text{9}}{\text{+2WS}}_{\text{10}}\hfill \\ -{\text{8WB}}_{\text{1}}-{\text{8WB}}_{\text{2}}-{\text{2WB}}_{\text{3}}-{\text{3WB}}_{\text{4}}-{\text{4WB}}_{\text{5}}-{\text{3WB}}_{\text{6}}-{\text{3WB}}_{\text{7}}-{\text{2WB}}_{\text{8}}-{\text{5WB}}_{\text{9}}-{\text{5WB}}_{\text{10}}\hfill \end{array}$

$\begin{array}{l}\text{S}\text{.T}\text{.}\\ {\text{WI}}_{\text{0}}\text{=6000}\\ {\text{WS}}_1-{\text{WB}}_{\text{1}}-{\text{WI}}_0+{\text{WI}}_{\text{1}}=0\\ {\text{WS}}_2-{\text{WB}}_2-{\text{WI}}_1+{\text{WI}}_2=0\\ {\text{WS}}_3-{\text{WB}}_3-{\text{WI}}_2+{\text{WI}}_3=0\\ {\text{WS}}_4-{\text{WB}}_4-{\text{WI}}_3+{\text{WI}}_4=0\\ {\text{WS}}_5-{\text{WB}}_5-{\text{WI}}_4+{\text{WI}}_5=0\end{array}$