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Tricks for “Linearizing” Some Non-Linear Functions

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

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Tricks for “Linearizing” Some Non-Linear Functions

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

  2. 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.

  3. 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.

  4. 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.

  5. LP Formulation

  6. Optimal Solution for NSC: cost = $78, 332,000

  7. 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.

  8. 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

  9. 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.

  10. 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)

  11. 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 (60 - 30) + (40 - 20) = 50 miles to the fire.

  12. 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

  13. Example 2: Feasible Solution

  14. Example 2: Better Solution

  15. b h d h ¯ h L D i i t t t t t e e e m a x m u m s a n c e o e r e o u s e . D i m n d d l b D D M i i ¸ t t t s s a n c e o e u r y . . D D V i ¸ t t s a n c e o e r g e n n e s l D D B i i ¸ t t t s a n c e o r s o Example 2: Min Max Formulation

  16. b h d h ¯ h L D i i t t t t t e e e m a x m u m s a n c e o e r e o u s e . d b h d f h ¯ h L i t t t t e x a n y e e c o o r n a e s o e r e o u s e . D i m n j j j j D 1 0 2 0 ¸ t ¡ + ¡ s x y . . j j j j D 6 0 2 0 ¸ ¡ + ¡ x y j j j j D 4 0 3 0 ¸ ¡ + ¡ x y Example 2: Min Max Formulation with Non-Linear Constraints

  17. b h d h ¯ h L D i i t t t t t e e e m a x m u m s a n c e o e r e o u s e . d b h d f h ¯ h L i t t t t e x a n y e e c o o r n a e s o e r e o u s e . D i m n ( ) ( ) D D 1 0 2 0 ¸ ¸ t ¡ + ¡ ¡ s x y . . ( ) ( ) D D 1 0 2 0 ¸ ¸ ¡ ¡ ¡ ¡ x y ( ) ( ) D D 6 0 2 0 ¸ ¸ ¡ + ¡ ¡ x y ( ) ( ) D D 6 0 2 0 ¸ ¸ ¡ ¡ ¡ ¡ x y ( ) ( ) D D 4 0 3 0 ¸ ¸ ¡ + ¡ ¡ x y ( ) ( ) D D 4 0 3 0 ¸ ¸ ¡ ¡ ¡ ¡ x y Example 2: Min Max Formulation with Linear Constraints

  18. D i m n D 3 0 ¸ t + ¡ s x y . . D 3 0 ¸ ¡ ¡ + x y D 1 0 ¸ ¡ + x y D 1 0 ¸ ¡ + ¡ x y D 8 0 ¸ + ¡ x y D 8 0 ¸ ¡ ¡ + x y D 4 0 ¸ ¡ ¡ x y D 4 0 ¸ ¡ + + x y D 7 0 ¸ + ¡ x y D 7 0 ¸ ¡ ¡ + x y D 1 0 ¸ ¡ ¡ x y D 1 0 ¸ ¡ + + x y Example 2: Min Max Formulation with Linear Constraints

  19. Example 3: Optimal Solution

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