Question

Hallco runs a day shift and a night shift. No matter how many units are produced, the only production cost during a shift is a setup cost. It costs $8,000 to run the day shift and $4,500 to run the night shift. Demand for the next two days is as follows: day 1, 2,000; night 1, 3,000; day 2, 2,000; night 2, 3,000. It costs $1 per unit to hold a unit in inventory for a shift.

Please formulate (do not solve) an IP that can be used by Hallco to determine a production schedule that minimizes the sum of setup and inventory costs. All demand must be met on time.

Answer 1

Let

D1 = demand for day 1 shift

N1 = demand for night 1 shift

D2 = demand for day 2 shift

N2 = demand for night 2 shift

Decision variables:

xD1 = number of units produced on day 1 shift

xN1 = number of units produced on night 1 shift

xD2 = number of units produced on day 2 shift

xN2 = number of units produced on night 2 shift

yD1 = 1 if day 1 shift is run, 0 otherwise

yN1 = 1 if night 1 shift is run, 0 otherwise

yD2 = 1 if day 2 shift is run, 0 otherwise

yN2 = 1 if night 2 shift is run, 0 otherwise

Objective function:

Minimize the sum of setup and inventory costs, which is the sum of setup costs and inventory costs for each shift:

$8,000(yD1 + yD2) + $4,500(yN1 + yN2) + $1[(xD1 + xN1) + (xD2 + xN2)]

Constraints:

Demand must be met on time:

xD1 + xN1 = D1

xD2 + xN2 = D2

Non-negativity constraint:

xD1, xN1, xD2, xN2 >= 0

Production must be less than or equal to shift capacity:

xD1 + xD2 <= 8yD1 + 8yD2

xN1 + xN2 <= 8yN1 + 8yN2

yD1, yN1, yD2, yN2 are binary variables, indicating whether or not a shift is run.