Production Planning Revisited

The National Steel Corporation (NSC) produces a special-purpose steel that is used in the aircraft and aerospace industries. The marketing department of NSC has received orders for 2400, 2200, 2700 and 2500 tons of steel during each of the next four months. NSC can meet these demands by producing the steel, by drawing from its inventory or by a combination of both.

The production costs per ton of steel during each of the next four months are projected to be $7400, $7500, $7600 and $7800. Because of these inflationary costs, it might be advantageous for NSC to produce more steel than it needs in a given month and store the excess, although production capacity can never exceed 4000 tons in any month. All production takes place at the beginning of the month and immediately thereafter the demand is met. The remaining steel is then stored in inventory at a holding cost of $120/ton for each month that it remains there.

If the inventory at the beginning of the first month is 1000 tons of steel and the inventory level at the end of the fourth month should be at least 1500 tons. Formulate a mathematical program for NSC that will minimize the total production cost over the next four months while satisfying the demand.

Solution:
 
 

Decision variables

Let Pi be the tons of steel produced in month i.

Let Ii be the inventory at the end of month i. Note that I0 is the initial inventory (i.e. at the start of Month 1).

Objective Function:

We wish to minimize the production cost which is given by 7400 P1 + 7500 P2 + 7600 P3 + 7800 P4 and the holding cost which is given by 120 (I1 + I2 + I3).

The objective function is thus 7400 P1 + 7500 P2 + 7600 P3 + 7800 P4 + 120 (I1 + I2 + I3).

Constraints:

First, we have to meet the demand. Thus, we have the following constraints.

P1 + I0 >= 2400

P2 + I1 >= 2200

P3 + I2 >= 2700

P4 + I3 >= 2500

Next, we consider the inventory levels at the end of each month.

I0 = 1000

I1 = P1 + I0 - 2400

I2 = P2 + I1 - 2200

I3 = P3 + I2 - 2700

I4 = P4 + I3 - 2500

I4 >= 1500

Since at most 4000 tons of steel can be produced each month, we have the following constraints.

Pi <= 4000 for i = 1, 2, 3, 4

Finally, we have to ensure that the Pi and Ii variables are all non-negative.

Pi >= 0 for i = 1,2,3,4

Ii >= 0 for i = 1,2,3,4
 
 

Example 6

An optimal solution for Example 6 is shown below.

The cost of this production plan is $78,332,000.

Now suppose that if the production level is increased or decreased from one month to the next, then NSC incurs a cost for implementing these changes. Specifically, for each ton of increased or decreased production over the previous month, the cost is $50. The production of the first month, however, is exempt from this cost. Thus, the solution shown above would incur an extra cost (4000 - 2300) ($50) =$85,000 for increasing the production from 2300 to 4000 tons month 1 to month two.

How does this change the formulation?

Solution

We add the following expression to the objective function $50 (|P2 - P1 | + |P3 - P2| + |P4 - P3|).
 
 

Example 8: LP Formulation with Absolute Values

Formulate the problem in Example 7 as a linear program.

In a linear program (LP) we are not allowed to use the absolute value function.

Solution:

Let Y2, Y3 and Y4 be the increases in production in months 2, 3 and 4, respectively.

Let Z2, Z3 and Z4 be the decreases in production in months 2, 3 and 4, respectively.

Note that be definition, the Y and Z variable must be nonnegative.

Suppose we add the term $50 (Y2 + Z2) to the objective function and the constraints

Y2 >= P2 - P1 and

Z2 >= P1 - P2.

If P1 = P2, then the LP can set Y2 = Z2 = 0 and there is no additional charge to the production cost from the objective function.

Suppose the production is increased in month 2. For example, consider the solution from Example 6 where P2 = 4000 and P1 = 2300. In this case, the LP must make Y2 >= 4000 - 2300 = 1700 and it must make

Z2 >= 2300 - 4000 = -1700. Since the objective function wants to minimize $50 (Y2 + Z2), the LP will try to make Y2 and Z2 as small as possible. Thus, it will make Y2 = 1700 and Z2 = 0 (z2 must be >= 0).

On the other, hand if the solution decreases production, say P2 = 2300 and P1 = 4000, then we will get Y2 = 0 and z2 = 1700.

Thus by combining the constraints on the Y?s and Z?s and with the objective function, we can represent the absolute value function, which is not linear, in and LP.
 
 

Putting it all together, the LP is

minimize 7400 P1 + 7500 P2 + 7600 P3 + 7800 P4 + 120 I1 + 120 I2 + 120 I3 +

50 Y2 + 50 Z2 + 50 Y3 + 50 Z3 + 50 Y4 + 50 Z4

subject to

P1 + I0 >= 2400

P2 + I1 >= 2200

P3 + I2 >= 2700

P4 + I3 >= 2500

I0 = 1000

P1 + I0 - I1= 2400

P2 + I1 - I2= 2200

P3 + I2 - I3 = 2700

P4 + I3 - I4 = 2500

I4 >= 1500

P1 <= 4000

P2 <= 4000

P3 <= 4000

P4 <= 4000

Y2 >= P2 - P1

Y3 >= P3 - P2

Y4 >= P4 - P3

Z2 >= P1 - P2

Z3 >= P2 - P3

Z4 >= P3 - P4

Now, it is optimal to produce 2575 tons in each month and the total cost is $78,520,500.