Question

In the planning of the monthly production for the next four months, in each month a company must operate either a normal shift or an extended shift (but not both) if it produces. It may choose not to produce in a month. A normal shift costs $100,000 per month and can produce up to 5,000 units per month. An extended shift costs $140,000 per month and can produce up to 7,500 units per month.

The cost of holding inventory is estimated to be $2 per unit per month (based on the average inventory held during each month) and the initial inventory is 3,000 units (i.e., inventory at the beginning of Month 1). The inventory at the end of month 4 should be at least 2,000 units. The demand for the company's product in each of the next four months is estimated to be as shown below:

Month	1	2	3	4
Demand	6000	6500	7500	7000

Production constraints are such that if the company produces anything in a particular month it must produce at least 2,000 units. The company wants a production plan for the next four months to meet its demands. Formulate an integer programming model to solve the problem at minimum cost.

Answer 1

Let,
xt = 1 if we operate a normal shift in month t (t=1,2,...,4)

   = 0 otherwise

yt = 1 if we operate an extended shift in month t (t=1,2,...,4)

   = 0 otherwise

Pt (>= 2000) be the amount produced in month t (t=1,2,...,4)

It be the closing inventory (amount of stock left) at the end

   of month t (t=1,2,...,4)

Constraints:

We can operate either normal or extended shift but not both

xt + yt <= 1    t=1,2,...,4

Constraint fro production limit

Pt <= 5000xt + 7500yt    t=1,2,...,4

No stockouts constraint

It >= 0    t=1,2,...,4

We know,

Closing stock = opening stock + production - demand

I0 = 3000. Assuming Dt = demand for the month t (t=1,2,...,4),

Assuming

• opening stock in period t = closing stock in period t-1
• production in period t is available to meet demand in period t

we get

It = It-1 + Pt - Dt t=1,2,...,4

inventory level at the end of month 4 should be at least 2000 units which gives

I4 >= 2000

Objective funtion:

Minimize (100000xt + 140000yt + 2It)