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# Agenda - PowerPoint PPT Presentation

Agenda. Duality Geometric Picture Piecewise linear functions. Dual Problem. Original: max profit from running plant s.t. capacity not exceeded variables are production quantities Dual: min cost to buy all capacity s.t. willing to sell capacity instead of produce

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Presentation Transcript

• Duality

• Geometric Picture

• Piecewise linear functions

Original:

max profit from running plant

s.t. capacity not exceeded

variables are production quantities

Dual:

min cost to buy all capacity

s.t. willing to sell capacity instead of produce

variables are prices

Original:

max \$840 profit * S cars + …

s.t. 3hr * S + 2hr * F + 1hr * L <= 120hr engine shop capacity

1hr * S + 2hr * F + 3hr * L <= 80hr body shop capacity

variables S, F, L are production quantities

Dual:

min price E * 120 hr engine shop capacity + …

s.t. 3hr * E + 1hr * B + 2hr * SF >= \$840 (standard car profit)

2hr * E + 2hr * B + 3hr * FF >= \$1120 (fancy car profit)

variables E, B, SF, FF, FL are prices

• constraint becomes dual variable

• constraint bound goes into dual objective

• shadow price = optimal dual variable

• variable becomes dual constraint

• objective coefficient is dual constraint bound

• optimal value = dual shadow price

• max problem becomes min problem

• solutions the same

• unbounded problem becomes infeasible

maxx pTx

s.t. Ax <= c

x >= 0

equivalent to

miny cTy

s.t. ATy >= p

y >= 0

• Customer demand d

• Generator i has cost ci and capacity bi

• Production xi on generator i

• Goal: meet demand with little cost

minx cTx

s.t. x1+x2+…+xn >= d

xi <= bi for i=1,..,n

x >= 0

Original:

minx cTx

s.t. x1+x2+…+xn >= d

xi <= bi for i=1,…,n

x >= 0

Dual:

maxp,y dp - bTy

s.t. p - yi <= ci for i=1,…,n

p >= 0, y >= 0

Dual

maxp,y dp - bTy

s.t. p - yi <= ci for i=1,…,n

p >= 0, y >= 0

p = market price for power

yi = profit rate at generator i

constraint: yi >= p - ci

Goal: max net revenue

(after paying out-sourced generators their profit)

• min f(x) = - max -f(x)

• g(x) <= b same as -g(x) >= -b

• x <= 5 same as -x >= -5

maxx pTx

s.t. Ax ? c

x ? 0

miny cTy

s.t. ATy ? p

y ? 0

• for max problem

<= constraint becomes variable >= 0

>= constraint becomes variable <= 0

= constraint becomes variable without bound

• for min problem the opposite

minx c1(x1) + c2x2

s.t. x1+x2 >= d

x >= 0

minx,z z + c2x2

s.t. x1+x2 >= d

x >= 0

z >= s1 x1

z >= s2 x1 + t

c1(x1)