0. Tricks for “Linearizing” Some Non-Linear Functions
Updated 3 February 2005

1. Example 1: The Absolute Value Function
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.
NSC currently has an inventory of 1000 tons of steel.

2. NSC Problem Continued
The production costs per ton of steel during each of the next four months are projected to be $7400, $7500, $7600 and $7800.
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.
The inventory level at the end of the fourth month must be at least 1500 tons.

3. NSC Formulation Decision Variables
Let Pi be the tons of steel produced in month i
Let Ii be the tons of steel in inventory at the end of month i.

4. LP Formulation

5. Optimal Solution for NSC: cost = $78, 332,000
Variable
Value P1
2300 P2
4000 P3 P4 I1
900 I2
2700 I3 I0
1000 I4
1500

6. Increase/Decrease Penalty
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 (except for month 1).
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 2.

7. New Objective Function
The new objective function is
To make the objective function linear define
Yi = increase in production from month i-1 to month i
Zi = decrease in production from month i-1 to month i

8. Additional Constraints
Yi  0 for i = 2, 3, 4
Zi  0 for i = 2, 3, 4
Yi  Pi - Pi-1 for i = 2, 3, 4
Zi  Pi-1 - Pi for i = 2, 3, 4
Alternatively we can write Yi - Zi = Pi - Pi-1 for i = 2, 3, 4
Examples
If P1 = P2, then Y2 = 0, and Z2 = 0
If P1 = 2300 and P2 = 4000 then Y2 = 1700, and Z2 = 0
If P1 = 4000 and P2 = 2300 then Y2 = 0, and Z2 = 1700
Now, it is optimal to produce 2575 tons in each month
and the total cost is $78,520,500.

9. Example 2: Min Max Functions
Addison county is trying to determine where to place the county fire station.
The locations of the county's three major towns are as follows (each town's location is given in terms of (x,y) coordinates where x = miles north of the center of the county and y = miles east of the county center)
Middlebury: (10, 20)
Vergennes: (60, 20)
Bristol: (40, 30)

10. Example 2 Continued
The county wants to build a fire station in a location (to be specified in terms of (x,y) coordinates as above) that minimizes the largest distance that a fire engine must travel to respond to a fire.
Since most roads run in either an east-west or north-south direction, we assume that a fire engine must always be traveling in a north-south or east-west direction.
Example: if the fire station is at (30,40) and a fire occurred at Vergennes, the fire engine would have to travel ( ) + ( ) = 50 miles to the fire.

11. Example 2 Continued
Formulate a linear program to determine where the fire station should be located.
Define all variables and briefly justify each constraint.
Hint: |a| + |b|  c if and only if
-c  a + b  c, and
-c  a - b  c

12. Example 2: Feasible Solution

13. Example 2: Better Solution

14. Example 2: Min Max FormulationD b h m a x i u d s n c o r . M l y V g B

15. Example 2: Min Max Formulation with Non-Linear ConstraintsD b h m a x i u d s n c o r . y f j 1 + 2 6 4 3

16. Example 2: Min Max Formulation with Linear ConstraintsD b h m a x i u d s n c o r . y f ( 1 ) + 2 6 4 3

17. Example 2: Min Max Formulation with Linear ConstraintsD s . t x + y 3 1 8 4 7

18. Example 3: Optimal Solution