PARETO LINEAR PROGRAMMING

1. PARETO LINEAR PROGRAMMING

2. 3.4 Pareto Linear Programming • The Problem: P-opt Cx s.t Ax ≤ b x ≥ 0 where C is a kxn matrix so that Cx = (c(1)x, c(2)x, ..., c(k)x) where c(j) = jth row of C.

3. Example z1 z2 • P- max {3x1 + 2x2, 2x1 – x2} • S.t. • x1 ≤ 2 • 3x1 + x2 ≤ 9 • x1, x2 ≥ 0 Here z = (z1, z2).

4. We refer to the Pareto solutions also as efficient points, or non-dominated points. • We shall focus on opt = max. • For simplicity we shall assume that the set of feasible solution, namely X := {x in Rn: Ax ≤ b, x ≥ 0} is bounded. • Let Z:= {Cx: x in X} observing that Z is a subset of Rk, and usually k <<<<<n.

5. GRAPHICAL INSIGHT

6. We can solve for the efficient points graphically if we only have 2 variables and 2 objective functions.This also gives us a good picture of what is going on, and ideas for generalizing. • P- max {3x1 + 2x2, 2x1 – x2} • S.t. • x1 ≤ 2 • 3x1 + x2 ≤ 9 • x1, x2 ≥ 0

7. So more generally:Geometry ZRk X Rn Cx x  X Ax ≤ b x ≥ 0

8. Find X and Z for the problem below. • P- max {3x1 + 2x2, 2x1 – x2} • S.t. • x1 ≤ 2 • 3x1 + x2 ≤ 9 • x1, x2 ≥ 0

9. How do we generate all the Pareto Solutions? • Idea: • Generate the non-dominated extreme points of Z • Use them to generate the other efficient points of Z. • Rationale: • Other efficient points of Z are convex combinations of the efficient extreme points of Z

10. k=2 (i.e 2-objectives) z2 z1

11. k=2 z2 Z z1

12. k=2 Efficient points (Pareto solutions) z2 Z z1

13. k=2 Efficient points z2 Every efficient point can be expressed as a convex combination of the efficient extreme points. So we first aim to generate the efficient extreme points. Efficient extreme points Z z1

14. An extremely important theorem • Idea: • Theorem 4.6.1 • An extreme point of Z must be produced by an extreme point of X, ie if z* is an extreme point of Z then z*=Cx* where x* is an extreme point of X • Proof: • By contradiction. Assume that z* is an extreme point of Z but x* in X for which z*=Cx*, is not an extreme point of X.

15. Geometry z* ZRk X Rn x* Cx x  X Ax ≤ b x ≥ 0

16. Since x* is not an extreme point of X we have x* = x’ + (1-)x” , 0 <  < 1 for some points x’ and x” in X. Thus, z* = Cx* = C(x’ + (1-)x”) = Cx’ + (1-)Cx” = z’ + (1-)z” , (z’=Cx’, z”=Cx”) This, however, contradicts the assertion that z* is an extreme point of Z. (This is true only when z’ and z” are different and neither of z’ and z” is an extreme point)

17. Comment z* ZRk X Rn x* Cx x  X Ax ≤ b x ≥ 0 If z* extreme then x* extreme. BUT not necessarily vice versa.

18. THEORY LEADING TO MORE GENERAL SOLUTIONS

19. One way to generate the efficient extreme points of Z is by deploying the following well known results: • Theorem 4.6.2: • If z* is an efficient point of Z then there must be a  in Rk such that z* is an optimal solution to the problem: • max { z: z in Z} (****) i.e. max {1 z1+ 2 z2+…+ k zk : z in Z} • Furthermore,  > 0 (all components are strictly positive)

20. Theorem 4.6.3: If > 0 then any optimal solution to max {1 z1+ 2 z2+…+ k zk : z in Z} (****) is an efficient point of Z.

21. Proof of Theorem 4.6.3: • Assume that  > 0 and let z* be any optimal solution to (****). Contrary to the Theorem, assume that z* is not an efficient point of Z. This means that there exists some z’ in Z such that • z’j ≥ z*j , for all j=1,2,...,k • and • z’p > z*p , for some 1 ≤ p ≤ k. • (Remember that we mean Pareto-max here) • Since  >0, this implies that z’ > z*, contradicting the assertion that z* is an optimal solution to (****).

22. Understanding Theorem 4.6.2 requires some fundamental results!!! • We will omit slides 17 - 40 (not examinable). • From slide 41 on is examinable.

23. Review(618-261) • Convex set: • If y’ and y” are elements of Y then the entire line segment connecting these points is also in Y. • Line segment: • Line segment connecting y’ and y” is the set of all the convex combinations of y’ and y”. • Convex combinations: • y =y’ + (1-)y” , 0 ≤  ≤ 1.

25. X2 Extreme Points Convex comb. of X1 and X2 X1 X2 X1 X2 • A point y in Y is an extreme point of Y if it cannot be expressed as a convex combination of two other points in Y. Not convex comb. of X1 and X2 X1

26. 2 extreme points 4 extreme points (the corners) Entire boundary is extreme

27. Properties of Convex Sets • If S is a convex set, then for any  in R+, S := {s: s in S} is a convex set. • If S and T are convex sets, then so is S + T : = {s+t: s in S, t in T} • The intersection of any collection of convex sets is convex.

28. S S A origin S+T S B T

29. facts about hyperplanes • Fact 1: • Let  be a non-zero element of Rn and let b be a real number. Then the set H := {x in Rn: x = b} is a hyperplane in Rn. • Example: • In R3, the set of all points satisfying 3x1 + 3x2 + x3 =5 is a hyperplane (set b=5 and  = (3,3,1) )

30. Fact 2: • Let H be a hyperplane in Rn. Then there is a non-zero vector  in Rn and a constant b such that • H = S:= {x in Rn: x=b}.

31. bottom line • A hyperplane in Rn is the set of solutions to a single linear equation.

32. Half Spaces • Given a hyperplane, say H := {x in Rn : x = b} we shall consider the two closed half spaces it generates: H+ := {x in Rn: x ≥ b} H– := {x in Rn: x ≤ b}

33. H+ H– H := {x in Rn: x=b}

34. Main Results(for our purposes) • Theorem 4.6.4Separating Hyperplanes • Given a convex set S and a point x exterior to its closure*, there is a hyperplane containing x that contains S in one of its half spaces. • Closure of S: the smallest closed set containing S. • Closed set: A set with the property that any point that is arbitrarily close to it is a member of the set.

35. x H S

36. x S H x S

37. Theorem 4.6.5 Supporting Hyperplane • Let S be a convex set and let x be a boundary point of S. Then there is a hyperplane containing x and containing S in one of its closed half spaces. x S Supporting hyperplane at x

38. Theorem 4.6.6 • Let S be a convex set, H a supporting hyperplane of S and I the intersection of H and S. • Then every extreme point of I is an extreme point of S.

39. I One common extreme point S H Face two extreme points in common I S Facet H

40. And so ...... • For every extreme point in Z there is a supporting hyperplane • Each extreme point of Z is an optimal solution to max {z : z in Z} for some non zero  in Rk.

41. And so ...... • For each efficient extreme point of Z there is a strictly positive  in Rk such that the point is an optimal solution to max {z : z in Z}. • Each extreme point of Z is an optimal solution to max {z : z in Z} for some non zero  in Rk. • And also 1 +  + k = 1 (See supplementary notes for details).

42. USING THE THEORY

43. Bottom line • We can generate the efficient extreme points associated with P-opt Cx Ax ≤ b x ≥ 0 by solving maxCx Ax ≤ b x≥ 0 for all  > 0 .

44. Example

45. k=2 • Thus we need two multipliers 1 and 2. • The objective function of the parametric linear programming problem will therefore be of the form: z() := 1c(1)x + 2c(2)x • But since the parameters are positive, we can divide say by 1, so obtain an equivalent objective function of the form z() := c(1)x + c(2)x ,  =2 / 1 The parametric problem is thus:

46. Note that  varies from 0 to infinity but cannot equal 0.

47. Efficient points z2 Efficient extreme points Z z1

48. z2 Z z1

49. z1 + z2 = Constant z2 Z z1

50. z1 + z2 = Constant z2 Z z1