## Related questions with answers

A company must meet (on time) the following demands: quarter 1—30 units; quarter 2—20 units; quarter 3—40 units. Each quarter, up to 27 units can be produced with regular-time labor, at a cost of $40 per unit. During each quarter, an unlimited number of units can be produced with overtime labor, at a cost of$60 per unit. Of all units produced, 20% arc unsuitable and cannot be used to meet demand. Also, at the end of each quarter, 10% of all units on hand spoil and cannot be used to meet any future demands. After each quarter’s demand is satisfied and spoilage is accounted for, a cost of $ 15 per unit is assessed against the quarter’s ending inventory. Formulate an LP that can be used to minimize the total cost of meeting the next three quarters’ demands. Assume that 20 usable units are available at the beginning of quarter 1.

Solution

VerifiedLet us $q_t$ denote demand for $q_t$ units in quarter $t.$ We are given $q_1=30,q_2=20$ and $q_3=40.$ Up to $27$ units can be produced during regular time labour in the quarter, while unlimited amount of units can be produced in overtime labour. In the regular time labour cost per unit is $40\$$ and in the overtime labour it is$60\$.$ Immediately we see that in optimum case it is impossible that regular time labour is not fulfilled completely while there is overtime labour. However, we'll be more precise. Notice that $20\%$ of units are unsuitable for meeting the demand and at the end of each quarter $10\%$ of all units on hand spoil. As usual, a cost for units in inventory at the end of the quarter is assesed, which is in this case $15\$$ per unit. First let us introduce decision variables$i_t,$for quarter$t,$which denotes number of units in the inventory after removing the unsuitable units and meeting the demand. It is given that at the beggining of quarter$1$there are$20$usable units. Also let us$x_t$and$y_t$be number of units made in the regular time labour and number of units made in the overtime labour (in quarter$t$ofcourse). Thus we have:$ $i_t=0.9(i_{t-1}+0.8x_t+0.8y_t-d_t)$ $Let us$z$be the objective function. Thus we have:$ $\min z=40(x_1+x_2+x_3)+60(y_1+y_2+y_3)+15(i_1+i_2+i_3)$ $Obviously all variables are nonnegative. The other constraints are:$ $i_1=0.9(0.8x_1+0.8y_1-10),i_2=0.9(i_1+0.8x_2+0.8y_2-20)$ $i_3=0.9(i_2+0.8x_3+0.8y_3-40),x_1,x_2,x_3\leq27.$$ Thus we are done formulating an LP that minimizes the cost of meeting the demands.

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