MATH2070 Optimization and Financial Mathematics Week 5-6
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MATH2070 Week 5-6
PART I The Simplex Algorithm
PART 1.1 Standard form
Maximise Z = C1X1 + C2X2 +... + CnXn
Subject to a11X1 + a12X2 +... + a1nXn ≤ b1
a21X1 + a22X2 +... + a2nXn ≤ b2
… …
am1X1 + am2X2 +... + amnXn ≤ bm
with X1, X2, ... , Xn ≥ 0
PART 1.2 The simplex algorithm
Algebraic representation of corner points
a) At the corner point, basic variables are non-zero and non-basic variables are zero.
b) Two corner points are adjacent if they differ in precisely one non-basic variable (or equivalently, in precisely one basic variable)
The simplex algorithm
Aim: move from one FCP to an adjacent FCP with largest potential increase in the objective function
Z.
a) Initialization: start on a feasible corner point (FCP).
i.e by setting decision variables to zero and slack variables to resource constraints b) Iteration steps: move to an adjacent FCP with best potential increase of Z.
1) Which non-basic variable enters the basis?
Largest coefficient rule: increasing the non-basic variable Xl to Xl > 0 with the largest cost coefficient Cl might yield the largest increase of Z
2) Which basic variables leave the basis?
Choose X such that bl is minimal
c) Stopping rule: stop at FCP if its Z is greater than the Z-value of all its adjacent FCPs. Stop when all
modified cost coefficient C ≥ 0.
PART II Non-standard Linear Programming
PART 2.1 Minimising the objective function
To minimise Z = ciXi, define Z = − Z, then minimising Z is equivalent to maximising Z :
i=1
min Z = max Z
Assignment Example:
PART 2.2 Negative resource elements
Suppose that for some constraint a1X1 + a2X2 + a3X3 +... + anXn ≤ b < 0, this is equivalent to − a1X1 − a2X2 − a3X3 − ... − anXn ≥ b > 0
Assignment Example:
PART 2.3 Greater-than or equal-to constraints
For a constraint a1X1 + a2X2 + a3X3 +... + anXn ≥ b ≥ 0, introduce a surplus variable Xn+1 such that
a1X1 + a2X2 + a3X3 +... + anXn − Xn+1 = b
Assignment Example:
PART 2.4 Negative decision variables
If x ≤ 0, introduce a variable x = − x , then x ≥ 0.
Assignment Example:
PART 2.5 Unrestricted decision variables
If x is unrestricted in sign (x ∈ R), introduce two new variables x ≥ 0 and x ≥ 0, then
k k k
Assignment Example:
PART 2.6 Absolute value in objective functions
If max Z = − |f(x)| + g(y) or min Z = |f(x)| + g(y), introduce U = |x|, then U ≥ x and U ≥ − x.
Assignment Example:
PART 2.7 Finding an initial FCP
a) Artificial variables: added into each constraint without a slack variable.
b) Decision variables and surplus variables: non-basic.
Slack and artificial variables: basic.
c) A feasible solution of the new LP problem is a feasible solution of the original LP problem if and only if the artificial variables are zero.
PART III Duality
PART 3.1 Primal problem
PART 3.2 Dual Problem
PART 3.3 Fundamental duality theorem
Theorem 1 (weak duality theorem)
If and are feasible solutions of the primal and dual problems respectively, then
If and are feasible primal and dual solutions such that , then and are primal and dual optimal solutions with V* = Z*.
Theorem 2
If both the primal and dual LP problems are feasible, then both have optimal solutions and the optimal values of the objective functions of the primal and dual are equal.
PART IV Assignment questions
PART V Nonlinear optimization without constraints
PART 5.1 Taylor’s theorem
Taylor’s theorem for one variable
Taylor’s theorem for multivariate functions
PART 5.2 Hessian matrix
Two-dimensional example
Multi-index notation
For α = (α1, α2, α3,..., αn), αi ≥ 0:
a) α! = α 1 ! α2 !... αn !
b) |α| = α 1 + α2 +... + αn
c) xα = x1(α)1 x2(α)2 x xn(α)n
d)
Multinomial theorem
For any x = (x1, x2, x3,..., xn) ∈ R and any positive integer k:
(x1 + x2 + x3 +... + xn)k = ∑ xα
|α|=k
Characterization of extrema
Critical point:
a) x0 is a local minimum if Q(ℎ, ℎ) > 0, ∀ℎ ≠ 0
b) x0 is a local minimum if Q(ℎ, ℎ) < 0, ∀ℎ ≠ 0
c) x0 is a saddle point if Q(ℎ, ℎ) > 0 for some ℎ and Q(ℎ, ℎ) < 0 for some other ℎ
d) No conclusion can be drawn if
- Q(ℎ, ℎ) ≥ 0 for all ℎ and Q(ℎ, ℎ) = 0 for at least one ℎ ≠ 0
- Q(ℎ, ℎ) ≤ 0 for all ℎ and Q(ℎ, ℎ) = 0 for at least one ℎ ≠ 0
If f is convex, any local minimizer is a global minimizer of f. If in addition f is differentiable, then any critical point a global minimizer of f.
- The function f is a convex function if its domain D is a convex set, and if for any two points x and y in D, we have:
f(λx + (1 − λ)y) ≤ λf(x) + (1 − λ)f(y), ∀λ ∈ [0, 1]
Lagrange’s method of completing the square of Q
Example 1: Given
Let H = H(x0):
a) H is positive definite if and only if all of its eigenvalues are positive;
b) H is negative definite if and only if all of its eigenvalues are negative;
c) H is indefinite If and only If it has both positive and negative eigenvalues;
d) H is positive semidefinite if and only if all of its eigenvalues are non-negative and at least one eigenvalue is zero;
e) H is negative semidefinite if and only if all of its eigenvalues are non-positive and at least one eigenvalue is zero.
Example 2: Let Q = 7x1(2) + 4x1x2 + 4x2(2) − 4x2x3 + 7x3(2) − 8x1x3 . Rewrite the quadratic form Q as , where B is a symmetric matrix and x = (x1, x2, x3). Is B positive definite? Check your answer by completing the square in the quadratic form.
Example 3: Consider the function f(x, y) = 6x2 + 8y3 + 3(2x + y − 1)2 . Show analytically that this function has two critical points. By considering the Hessian matrix at each of these critical points, show that at one the function attains a local minimum, with f(x, y) = , and that at the other the function has a saddle point.
2023-01-31