1. Foundations for LP Sean Eom

2. In this course, LP is the foundation of all other quantitative models. Study of LP and related techniques (Integer LP and Goal Programming) requires 2/3 of our time in the semester. It is critical to understand some building blocks of LP conceps.

3. Foundational Concepts/tools Drawing lines on a two-dimensional space Identifying the feasible area Identifying extreme points and their values Understanding the properties of the objective function line

4. Drawing lines on a two-dimensional space An LP model consists of the following: Max 3x 1 + 3x 2 s.t.(subject to) 2 x 1 + 4x 2 = = 0

5. 2 x 1 + 4x 2 =< 12 LP constraints must be drawn on a 2D space. To do so, at minimum two sets of values that are on the constraint line ( 2 x 1 + 4x 2 = 12) must be found. For example, x 1 = 4, x 2 = 1 or x 1 = 2, x 2 = 2 x 1 = 6, x 2 = 0 or x 1 = 0, x 2 = 3

6. 2x1+4x2=<12

7. Variations of linear constraints Draw a graph for x 1 =>.5 ( x 1 + x 2 ) You need to rearrange the constraint. x 1 =>.5 ( x 1 + x 2 ) x 1 -.5 ( x 1 + x 2 ) =>0 x 1 -.5 x 1 -.5 x 2 ) =>0.5 x 1 -.5 x 2 ) =>0

8. 6 x 1 + 4x 2 =< 24 At minimum two sets of values that are on the constraint line (6x 1 + 4x 2 = 24) must be found. For example, x 1 = 2, x 2 = 3 or x 1 = 4, x 2 = 0 x 1 = 0, x 2 = 6 It is easier for you to find values with ( x 1 = 0, x 2 =? Or x 1 = ?, x 2 = 0 )

9. 6 x 1 + 4x 2 =< 24

10. All LP constraints on one space

11. The Objective Function (Max 3x 1 + 3x 2 ) The difference between the objective function and constraint lines is the absence of the right-hand side value. Arbitrary value may be added before drawing the line. You can use any numbers, but add the smallest number that can be dividable by both coefficients of x 1 and x 2 In this case, 3.

12. 3x 1 + 3x 2 = 3

13. All lines on one space

14. About the object function line (3x1 + 3x2 = 3)

15. The objective function line indicates the $ amount of profits/costs “3x1 + 3x2 = 3” can be interpreted this way: Any combination of production of the two products along the objective function line produces same profit, $3. Therefore, it is often called “ISO- profit/cost line. Iso=same.

16. 3x1 + 3x2 = 6

17. 3x1 + 3x2 = 60

18. Interpretation of the objective function line 3x1 + 3x2 = 6 3x1 + 3x2 = 60 The larger right hand side value indicates the larger profit ($)

19. Figure 1

20. Figure 2

21. Figure 3

22. Feasible area & Extreme points in figure 3 The green colored area is called the feasible area. There are 4 extreme points in the feasible area. Extreme points are “corners” (vertices) of the feasible area. The optimal solution = one of the extreme points

23. Computing the optimal answer In figure 3, the optimal point is the intersection of an extreme point and the objective function line. What is the values of the optimal point (x1=? and x2=?)? How do you compute them?

24. Computing the optimal answer Identify the two constraints that intersect. 2 x 1 + 4x 2 =< 12 6 x 1 + 4x 2 =< 24 Ignore in-equality signs 2 x 1 + 4x 2 = 12 (1) 6 x 1 + 4x 2 = 24 (2)

25. Computing the optimal answer Eliminate either x1 or x2. In this case, equations 1 and 2 have same objective function coefficient of x2, let’s eliminate x2 first. 2 x 1 + 4x 2 = 12 (1) - ( 6 x 1 + 4x 2 = 24) (2) -4 x 1 = -12 x 1 = 3 Use x 1 = 3 and substitute this value to find x 2 using either one of the two equation. 2*3 + 4x 2 = 12 6 + 4x 2 = 12 4x 2 = 12-6=6 x 2 = 1.5

26. The final answer is x 1 = 3 x 2 = 1.5 Z= 3x1 + 3x2 = 3*3 + 3*1.5 =9 + 4.5 =13.5