Maximin and Maximal Solutions for Linear Programming Problems with - - PowerPoint PPT Presentation

Maximin and Maximal Solutions for Linear Programming Problems with Possibilistic Uncertainty

Erik Quaeghebeur, Nathan Huntley, Keivan Shariatmadar, Gert de Cooman

Ghent University, SYSTeMS Research Group, Belgium

Linear programming problems under uncertainty

maximize cTx subject to ax ≤ b, x ≥ 0 Variables x: optimization vector. Parameters c: objective function coefficient vector, a: constraint coefficient matrix, b: constraint coefficient vector. Independence of components of A and B is assumed. Give meaning by reformulating as a decision problem with utility functions Gx := cTxIAx≤B + LIAxB = L + (cTx − L)IAx≤B L: penalty value; L < cTx for ‘feasible’ x.

Linear programming problems under uncertainty

maximize cTx subject to Ax ≤ B, x ≥ 0 with given uncertainty model for (A, B) Variables x: optimization vector. Parameters c: objective function coefficient vector, A: constraint coefficient matrix with uncertain components, B: constraint coefficient vector with uncertain components. Independence of components of A and B is assumed. Give meaning by reformulating as a decision problem with utility functions Gx := cTxIAx≤B + LIAxB = L + (cTx − L)IAx≤B L: penalty value; L < cTx for ‘feasible’ x.

Linear programming problems under uncertainty

maximize cTx subject to Ax ≤ B, x ≥ 0 with given uncertainty model for (A, B) Variables x: optimization vector. Parameters c: objective function coefficient vector, A: constraint coefficient matrix with uncertain components, B: constraint coefficient vector with uncertain components. Independence of components of A and B is assumed. Give meaning by reformulating as a decision problem with utility functions Gx := cTxIAx≤B + LIAxB = L + (cTx − L)IAx≤B L: penalty value; L < cTx for ‘feasible’ x.

Running example

maximize 2x1 + 3x2 subject to 1x1 + 3x2 ≤ 2, 1x1 + 1x2 ≤ B2, −3x1 − 3x2 ≤ −1, x1 ≥ 0, x2 ≥ 0 ≡ maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 maximize cTx := 2x1 + 3x2 subject to x ⊳ 1 x1 x2 (1

2, 1 2) 2 3

1

1 3 1 3

Running example

maximize 2x1 + 3x2 subject to 1x1 + 3x2 ≤ 2, 1x1 + 1x2 ≤ B2, −3x1 − 3x2 ≤ −1, x1 ≥ 0, x2 ≥ 0 ≡ maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 maximize cTx := 2x1 + 3x2 subject to x ⊳ 1 x1 x2 (1

2, 1 2) 2 3

1

1 3 1 3

Running example

maximize 2x1 + 3x2 subject to 1x1 + 3x2 ≤ 2, 1x1 + 1x2 ≤ B2, −3x1 − 3x2 ≤ −1, x1 ≥ 0, x2 ≥ 0 ≡ maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 maximize cTx := 2x1 + 3x2 subject to x ⊳ 1 x1 x2 (1

2, 1 2) 2 3

1

1 3 1 3

Penalty value choice L := 0 in the running example.

Probabilistic case (probability mass function)

Maximizing expected utility P(Gx) = L + (cTx − L)P(Ax ≤ B) maximize cTx subject to Ax ≤ B, x ≥ 0 with given p → maximize P(Gx) subject to x ≥ 0

Probabilistic case (probability mass function)

Maximizing expected utility P(Gx) = L + (cTx − L)P(Ax ≤ B) maximize cTx subject to Ax ≤ B, x ≥ 0 with given p → maximize P(Gx) subject to x ≥ 0 Running example maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 with pB2 =

1/5 3/5 2 3 1 4 3

↓ maximize P(B2 ≥ b)

maximize

cTx subject to x ⊳ b

• subject to

b ∈ {2/3, 1, 4/3} x1 x2

2 3 2 3

1

4 3 1 3 1 3

(1

2, 1 2)

(1, 1

3)

Optimality criteria for lower & upper previsions

Generalizations of maximizing expected utility for P & P: Maximinity those x ≥ 0 are optimal that maximize lower expected utility; P(Gx) = L + (cTx − L)P(Ax ≤ B). Maximality those x ≥ 0 are optimal that are undominated by all other vectors z ≥ 0 in the sense that P(Gx−Gz) = P

(cTx−L)IAx≤B−(cTz−L)IAz≤B ≥ 0.

Optimality criteria for lower & upper previsions

Generalizations of maximizing expected utility for P & P: Maximinity those x ≥ 0 are optimal that maximize lower expected utility; P(Gx) = L + (cTx − L)P(Ax ≤ B). Maximality those x ≥ 0 are optimal that are undominated by all other vectors z ≥ 0 in the sense that P(Gx−Gz) = P

(cTx−L)IAx≤B−(cTz−L)IAz≤B ≥ 0.

Optimality criteria for lower & upper previsions

Generalizations of maximizing expected utility for P & P: Maximinity those x ≥ 0 are optimal that maximize lower expected utility; P(Gx) = L + (cTx − L)P(Ax ≤ B). Maximality those x ≥ 0 are optimal that are undominated by all other vectors z ≥ 0 in the sense that P(Gx−Gz) = P

(cTx−L)IAx≤B−(cTz−L)IAz≤B ≥ 0.

Dominance x ≥ 0 is undominated by z ≥ 0 in pointwise comparison of utility functions if Gz = Gx

• r

max(Gx − Gz) > 0,

• r, equivalently, cTx ≥ max(Ax≤B)=(Az≤B) cTz,

cTx > max(Ax≤B)⊂(Az≤B) cTz.

Maximin solutions in the interval case

maximize cTx subject to Ax ≤ B, x ≥ 0 with a ≤ A ≤ a, b ≤ B ≤ b → maximize cTx subject to ax ≤ b, x ≥ 0

Maximin solutions in the interval case

maximize cTx subject to Ax ≤ B, x ≥ 0 with a ≤ A ≤ a, b ≤ B ≤ b → maximize cTx subject to ax ≤ b, x ≥ 0 Running example maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 with B2 ∈ [2/3, 4/3] ↓ maximize cTx subject to x ⊳ 2/3 x1 x2 (1, 1

3) 2 3 2 3 4 3 1 3 1 3

Maximal solutions in the interval case

maximize cTx subject to Ax ≤ B, x ≥ 0 with a ≤ A ≤ a, b ≤ B ≤ b → find all x subject to ax ≤ b, x ≥ 0, cTx ≥ maxaz≤b cTz, dominance

Maximal solutions in the interval case

maximize cTx subject to Ax ≤ B, x ≥ 0 with a ≤ A ≤ a, b ≤ B ≤ b → find all x subject to ax ≤ b, x ≥ 0, cTx ≥ maxaz≤b cTz, dominance Running example maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 with B2 ∈ [2/3, 4/3] ↓ find all x subject to x ⊳ 4/3 cTx ≥ maxz⊳2/3 cTz, cTx ≥ max1z1+1z2≤1x1+1x2 cTz (dominance) x1 x2 (1, 1

3) 2 3 2 3 4 3

1

1 3

Maximin solutions in the possibilistic case

maximize cTx subject to Ax ≤ B, x ≥ 0 with given π ↓ maximize L + (1 − t)

maximize

cTx − L subject to atx ≤ bt, x ≥ 0

• subject to

0 ≤ t < 1

Maximin solutions in the possibilistic case

Running example maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 with πB2 =

1/5

1

2 3 1 4 3

↓ maximize (1 − t)

maximize

cTx subject to x ⊳ b2t

• subject to

t ∈ {0, 1/5} x1 x2 (1

2, 1 2)

(1, 1

3) 2 3 2 3

1

4 3 1 3 1 3

Maximal solutions in the possibilistic case

◮ No analytical reduction to a standard optimization problem known. ◮ Numerical approach:

◮ Make a grid in the solution set of the corresponding interval case. ◮ Compare grid points and remove the dominated ones. ◮ This is computationally expensive.

Maximal solutions in the possibilistic case

◮ No analytical reduction to a standard optimization problem known. ◮ Numerical approach:

◮ Make a grid in the solution set of the corresponding interval case. ◮ Compare grid points and remove the dominated ones. ◮ This is computationally expensive.

Running example (numerical) x1 x2 (1, 1

3) 2 3 2 3 4 3

1

1 3

Maximal solutions in the possibilistic case

Running example (analytical) maximize cTx := 2x1 + 3x2 subject to x ⊳ B2 with πB2 =

9/10

1

2 3 1 4 3

↓ find all x either subject to x ⊳ 1, x ⊳ 2/3 but not cTx < max1z1+1z2≤1x1+1x2 cTz

• r subject to

x ⊳ 4/3, x ⊳ 1, cTx ≥ max1z1+1z2≤1x1+1x2 cTz (dominance) but not cTx < 10/9 maxz⊳1 cTz (cf. green-filled dot) x1 x2 (1, 1

3) 2 3 2 3

1

1 3 1 3 10 9 25 18

(5

9, 5 9)

Conclusions

◮ Problem is very hard in general.

(Even without dominance.)

◮ But some specific cases can be tackled, as shown by our results. ◮ Extension to problems with uncertainty in the goal function . . . ◮ Using different utility functions . . .