The L-Shaped Method. Operations Research. Anthony Papavasiliou 1 / 38

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1 1 / 38 The L-Shaped Method Operations Research Anthony Papavasiliou

2 Contents 2 / 38 1 The L-Shaped Method 2 Example: Capacity Expansion Planning 3 Examples with Optimality Cuts [ 5.1a of BL] 4 Examples with Feasibility Cuts [ 5.1b of BL]

3 Table of Contents 3 / 38 1 The L-Shaped Method 2 Example: Capacity Expansion Planning 3 Examples with Optimality Cuts [ 5.1a of BL] 4 Examples with Feasibility Cuts [ 5.1b of BL]

4 4 / 38 Extensive Form Stochastic linear program in extensive form: (EF ) : min c T x + E ω [min q(ω) T y(ω)] Ax = b T (ω)x + W (ω)y(ω) = h(ω) x 0, y(ω) 0 First-stage decisions: x R n 1 second-stage decisions: y(ω) R n 2 First-stage parameters: c R n 1, b R m 1, A R m 1 n 1 Second-stage parameters: q(ω) R n 2, h(ω) R m 2, T (ω) R m 2 n 1 and W (ω) R m 2 n 2

5 Why L-Shaped? 5 / 38 Multi-stage decision making process Idea: ignore constraints of future stages

6 6 / 38 Value Function Second-stage value function: (S ω ) : Q ω (x) = min q T y ω y W ω y = h ω T ω x y 0. Interpretation: cost of best possible reaction to x and ω Expected value function: N V (x) = p ω Q ω (x). ω=1 Interpretation: cost of best possible reaction to x before knowing ω

7 7 / 38 Complete Recourse Define K 1 = {x : Ax = b, x 0} K 2 (ω) = {x : y, T ω x + W ω y = h ω, y 0} K 2 = dom V Interpretation of K 1, K 2 (ω), K 2? Relative complete recourse: obeying first-stage constraints ensures feasible second-stage decisions exist: pos W = R m 2 Complete recourse: feasible second-stage decision exists, regardless of first-stage decision and realization of uncertainty: K 2 = R n 1

8 Properties of Value Functions 8 / 38 Dual of (S ω ): (D ω ) : max π T (h ω T ω x) π π T W ω q T ω Denote π ω0 as dual optimal multipliers of (S ω ) given x 0 : 1 V (x) and Q ω (x) are piecewise linear convex functions of x 2 π T ω0 (h ω T ω x) is a supporting hyperplane of Q ω (x) at x 0 3 N ω=1 p ωπ T ω0 (h ω T ω x) is a supporting hyperplane of V (x) at x 0 We recall a previous result for the proof

9 9 / 38 Proof: 1 D ω has finite number of dual optimal multipliers 2 Strong duality and π ω0 Q ω (h ω T ω x 0 ) 3 Follows from previous bullet and V (x) = N ω=1 p ωq ω (x)

10 10 / 38 (S) : min y N p ω qω T y ω ω=1 Wy = h Tx y 0 where y T = [y1 T,..., y N T ] h T = [h1 T,..., ht N ] T = diag T ω W = diag W ω Is there a relationship between the feasible regions of the duals of (S) and (S ω )?

11 Supporting Hyperplanes of V 11 / 38 Denote V : the set of extreme vertices of {π : π T W q T } V ω : the set of extreme vertices of {π : π T W ω qω T } Then V = {(p 1 π1 T,..., p NπN T )T : π 1 V 1,..., π N V N }

12 Domain of V 12 / 38 Denote R: the set of extreme rays of {π : π T W q T } R ω : the set of extreme rays of {π : π T W ω q T ω } Then R = {(0,..., σ T ω,..., 0) T : σ ω R ω, ω = 1,..., N}

13 13 / 38 Deterministic Equivalent Program The original problem (EF) can be written as a deterministic equivalent program: min c T x + θ Ax = b σ T (h Tx) 0, σ R θ π T (h Tx), π V x 0

14 Master Problem 14 / 38 Define master problem as (M) : z k = min c T x + θ Ax = b σ T (h Tx) 0, σ R k R (1) θ π T (h Tx), π V k V (2) x 0 Feasibility cuts: equation 1 Optimality cuts: equation 2

Bounds and Exchange of Information Solution of master provides: lower bound z k z candidate solution x k under-estimator θ k V (x k ) Solution of all (S ω ) with input x

15 Bounds and Exchange of Information Solution of master provides: lower bound z k z candidate solution x k under-estimator θ k V (x k ) Solution of all (S ω ) with input x k provides upper bound c T x k + N ω=1 p ωq T ω y ω,k+1 z new vertex π k+1 = (p 1 π T 1,k+1,..., p Nπ T N,k+1 )T or new extreme ray σ k+1 = (0,..., σ T ω,..., 0) T 15 / 38

16 16 / 38 The L-Shaped Algorithm Step 0: Set k = 0, V 0 = R 0 = Step 1: Solve (M) If (M) is feasible, store x k If (M) is infeasible, exit: infeasible Step 2: For ω = 1,..., N, solve (S ω ) with x k as input If (S ω ) is infeasible, let S k+1 = S k {σ k+1 }, where σ k+1 is an extreme ray of (S ω ), let k = k + 1 and return to step 1 If (S ω ) is feasible, store π ω,k+1 Step 3: Let V k+1 = V k {(p 1 π 1,k+1,..., p N π N,k+1 )} If V k = V k+1 then terminate with (x k, y k+1 ) as the optimal solution. Else, let k = k + 1 and return to step 1

17 Table of Contents 17 / 38 1 The L-Shaped Method 2 Example: Capacity Expansion Planning 3 Examples with Optimality Cuts [ 5.1a of BL] 4 Examples with Feasibility Cuts [ 5.1b of BL]

18 Example: Capacity Expansion Planning min x,y 0 s.t. n m (I i x i + E ξ C i T j y ij (ω)) i=1 j=1 n y ij (ω) = D j (ω), j = 1,..., m i=1 m y ij (ω) x i, i = 1,... n 1 j=1 I i, C i : fixed/variable cost of technology i D j (ω), T j : height/width of load block j y ij (ω): capacity of i allocated to j x i : capacity of i Note: D j is not dependent on ω 18 / 38

19 Problem Data Two possible realizations of load duration curve: Reference scenario: 10% 10x wind scenario: 90% Duration (hours) Level (MW) Level (MW) Reference scenario 10x wind scenario Base load Medium load Peak load / 38

20 Slave Problem 20 / 38 (S ω ) : min y 0 (λ j (ω)) : (ρ i (ω)) : n i=1 j=1 m C i T j y ij n y ij = D j (ω), j = 1,..., m i=1 m y ij x i, i = 1,... n 1 j=1 where x has been fixed from the master problem

21 Sequence of Investment Decisions Iteration Coal (MW) Gas (MW) Nuclear (MW) Oil (MW) / 38

22 22 / 38 Observations Investment candidate in each iteration necessarily different from all past iterations Greedy behavior: low capital cost in early iterations

23 Table of Contents 23 / 38 1 The L-Shaped Method 2 Example: Capacity Expansion Planning 3 Examples with Optimality Cuts [ 5.1a of BL] 4 Examples with Feasibility Cuts [ 5.1b of BL]

24 24 / 38 Example 1 z = min 100x x 2 + E ξ (q 1 y 1 + q 2 y 2 ) s.t. x 1 + x y y 2 60x 1 8y 1 + 5y 2 80x 2 y 1 d 1, y 2 d 2 x 1 40, x 2 20, y 1, y 2 0 ξ = (d 1, d 2, q 1, q 2 ) = { (500, 100, 24, 28), p 1 = 0.4 (300, 300, 28, 32), p 2 = 0.6

25 Iteration 1 Step 1. min{100x x 2 x 1 + x 2 120, x 1 40, x 2 20} x 1 = (40, 20) T, θ 1 = Step 3. For ξ = ξ 1 solve min{ 24y 1 28y 2 6y y , 8y 1 + 5y y 1 500, 0 y 2 100} w 1 = 6100, y T = (137.5, 100), π T 1 = (0, 3, 0, 13) For ξ = ξ 2 solve min{ 28y 1 32y 2 6y y , 8y 1 + 5y y 1 300, 0 y 2 300} w 2 = 8384, y T = (80, 192), π2 T = ( 2.32, 1.76, 0, 0) 25 / 38

26 Iteration 1: Optimality Cut h 1 = (0, 0, 500, 100) T, h 2 = (0, 0, 300, 300) T T,1 = ( 60, 0, 0, 0) T, T,2 = (0, 80, 0, 0) T e 1 = 0.4 π1 T h π2 T h 2 = 0.4 ( 1300)+0.6 (0) = 520 E 1 = 0.4 π1 T T πt 2 T = 0.4(0, 240) + 0.6(139.2, 140.8) = (83.52, ) w 1 = 520 (83.52, ) x 1 = w 1 = > θ 1 =, therefore add the cut 83.52x x 2 + θ / 38

27 Iteration 2 27 / 38 Step 1. Solve master min{100x x 2 + θ x 1 + x 2 120, x 1 40, x 2 20, 83.52x x 2 + θ 520} z = , x 2 = (40, 80) T, θ 2 = Step 3. Add the cut 211.2x 1 + θ 1584

28 Iteration 3 28 / 38 Step 1. Solve master. z = , x 3 = (66.828, ) T, θ 3 = Step 3. Add the cut 115.2x x 2 + θ 2104

29 29 / 38 Iteration 4 Step 1. Solve master. z = 889.5, x 4 = (40, 33.75) T, θ 4 = 9952 Step 3. There are multiple solutions for ξ = ξ 2. Select one, add the cut x x 2 + θ 0

30 30 / 38 Iteration 5 Step 1. Solve master min{100x x 2 + θ x 1 + x 2 120, x 1 40, x 2 20, 83.52x x 2 + θ 520, 211.2x 1 + θ x x 2 + θ 2104, x x 2 + θ 0} z = , x 5 = (46.667, 36.25) T, θ 5 = Step 3. w 5 = 520 (83.52, ) x 5 = = θ 5, stop. x = (46.667, 36.25) T is the optimal solution.

31 31 / 38 Example 2 z = min E ξ (y 1 + y 2 ) s.t. 0 x 10 y 1 y 2 = ξ x y 1, y p 1 = 1/3 ξ = 2 p 2 = 1/3 4 p 3 = 1/3 K 2 = R

32 L-Shaped Method in Example 2 32 / 38 Iteration 1, Step 1: x 1 = 0 Iteration 1, Step 3: x 1 not optimal, add cut: θ 7/3 x Iteration 2, Step 1: x 2 = 10 Iteration 2, Step 3: x 2 not optimal, add cut: θ x 7/3 Iteration 3, Step 1: x 3 = 7/3 Iteration 3, Step 3: x 3 not optimal, add cut: θ (x + 1)/3 Iteration 4, Step 1: x 4 = 1.5 Iteration 4, Step 3: x 4 not optimal, add cut: θ (5 x)/3 Iteration 3, Step 1: x 5 = 2 Iteration 3, Step 3: x 5 is optimal

33 33 / 38 V (x) 1 x 1 1 x 3 4 V (x 1 ) = 7/3 and V (x) = 7/3 x around x 1 (7 x)/3 is the optimality cut at x 1

34 Table of Contents 34 / 38 1 The L-Shaped Method 2 Example: Capacity Expansion Planning 3 Examples with Optimality Cuts [ 5.1a of BL] 4 Examples with Feasibility Cuts [ 5.1b of BL]

35 Feasibility Cuts 35 / 38 Consider the following problem: (F) : min w = e T v + + e T v s.t. Wy + Iv + Iv = h k T k x v y 0, v + 0, v 0 with dual multipliers σ v. Define D r+1 = (σ v ) T T k d r+1 = (σ v ) T h k Step 2 of L-shaped method: For k = 1,..., K solve (F). If w = 0 for all k, go to Step 3. Else, add D r+1 x d r+1, set r = r + 1 and go to Step 1.

36 36 / 38 Example min 3x 1 + 2x 2 E ξ (15y y 2 ) s.t. 3y 1 + 2y 2 x 1, 2y 1 + 5y 2 x 2 0.8ξ 1 y 1 ξ 1, 0.8ξ 2 y 2 ξ 2 x, y 0 ξ = (4, 4), p 1 = 0.25 (4, 8), p 2 = 0.25 (6, 4), p 3 = 0.25 (6, 8), p 4 = 0.25

37 Generating a Feasibility Cut For x 1 = (0, 0) T, ξ = (6, 8) T, solve min v + v +,v 1 + v 1 + v v 2 + v v 3 +,y v v 4 + v v 5 + v v 6 s.t. v + 1 v 1 + 3y 1 + 2y 2 0, v + 2 v 2 + 2y 1 + 5y 2 0 v + 3 v 3 + y 1 4.8, v + 4 v 4 + y v + 5 v 5 + y 1 6, v + 6 v 6 + y 2 8 We get w = 11.2, σ 1 = ( 3/11, 1/11, 1, 1, 0, 0) h = (0, 0, 4.8, 6.4, 6, 8) T, T,1 = ( 1, 0, 0, 0, 0, 0) T, T,2 = (0, 1, 0, 0, 0, 0) T D 1 = ( 3/11, 1/11, 1, 1, 0, 0) T = (3/11, 1/11), d 1 = ( 3/11, 1/11, 1, 1, 0, 0) h = /11x 1 + 1/11x / 38

38 38 / 38 Induced Constraints Going by the book: Iteration 2 master problem: x 2 = (41.067, 0) T Iteration 2 feasibility cut: x Iteration 3 master problem: x 3 = (33.6, 22.4) T Iteration 3 feasibility cut: x Iteration 4 master problem: x 4 = (27.2, 41.6) T is feasible Induced Constraints: Observe that for ξ = (6, 8) T, y 1 4.8, y This implies x , x , which should be added directly to the master Personal experience: feasibility cuts are impractical

The L-Shaped Method. Operations Research. Anthony Papavasiliou 1 / 44

The L-Shaped Method. Operations Research. Anthony Papavasiliou 1 / 44 1 / 44 The L-Shaped Method Operations Research Anthony Papavasiliou Contents 2 / 44 1 The L-Shaped Method [ 5.1 of BL] 2 Optimality Cuts [ 5.1a of BL] 3 Feasibility Cuts [ 5.1b of BL] 4 Proof of Convergence