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

2 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 [ 5.1c of BL] 5 Example: Capacity Expansion Planning

3 Table of Contents 3 / 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 [ 5.1c of BL] 5 Example: Capacity Expansion Planning

4 4 / 44 Extensive Form 2-Stage Stochastic Linear Program K (EF ) : min c T x + p k qk T y k s.t. Ax = b k=1 T k x + Wy k = h k, k = 1,..., K x 0, y k 0, k = 1,..., K K realizations of random vector ξ, with probabilities p k, k = 1,..., K Randomness: q k, h k, T k y k : second-stage decision given realization k

5 Block Structure 5 / 44 Multi-stage decision making process A T 1 W T 2 W. T K W Idea: ignore constraints of future stages

6 6 / 44 Master Problem We know that K V (x) = {min p k qk T y k Wy k = h k T k x, y k 0} k=1 is a piecewise linear function of x Define master problem as (M) : min z = c T x + θ (1) s.t. Ax = b D l x d l, l = 1,..., r (2) E l x + θ e l, l = 1,..., s (3) x 0, θ R Feasibility cuts: equation 2 Optimality cuts: equation 3

7 Optimality Cuts 7 / 44 Consider a trial first-stage decision x v Let πk v be simplex multipliers of second-stage problem: min w = qk T y s.t. Wy = h k T k x v y 0 e s+1 E s+1 x supports V (x) at x v, where E s+1 = e s+1 = K p k (πk v )T T k (4) k=1 K p k (πk v )T h k (5) k=1

8 8 / 44 V (x) e s+1 E s+1 x x v x

9 9 / 44 The L-Shaped Algorithm Step 0. Set r = s = v = 0 Step 1. Solve master problem (M). Let (x v, θ v ) be an optimal solution. If s = 0 (no optimality cuts), remove θ from (M) and set θ 0 = Step 2. If x / K 2, add feasibility cut (equation 2) and return to Step 1. Otherwise, go to Step 3. K 2 = {x y : Wy = h k T k x, y 0, k = 1,..., K } Step 3. Compute E s+1, e s+1. Let w v = e s+1 E s+1 x v. If θ v w v, stop with x v an optimal solution. Otherwise, set s = s + 1, add optimality cut to equation 3 and return to Step 1.

10 Table of Contents 10 / 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 [ 5.1c of BL] 5 Example: Capacity Expansion Planning

11 11 / 44 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

12 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) 12 / 44

13 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 / 44

14 Iteration 2 14 / 44 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

15 Iteration 3 15 / 44 Step 1. Solve master. z = , x 3 = (66.828, ) T, θ 3 = Step 3. Add the cut 115.2x x 2 + θ 2104

16 16 / 44 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

17 17 / 44 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.

18 18 / 44 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

19 L-Shaped Method in Example 2 19 / 44 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

20 20 / 44 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

Geometric Interpretation of Optimality Cuts Q(x, ξ) = min y {q(ω) T y W (ω)y = h(ω) T (ω)x, y 0} yields dual optimal multipliers π T = q B (ω) T B(ω) 1 From linear programming duality Q(x, ξ) = q B

21 Geometric Interpretation of Optimality Cuts Q(x, ξ) = min y {q(ω) T y W (ω)y = h(ω) T (ω)x, y 0} yields dual optimal multipliers π T = q B (ω) T B(ω) 1 From linear programming duality Q(x, ξ) = q B (ω) T B(ω) 1 (h(ω) T (ω)x) B(ω) is optimal basis as long as y B = B(ω) 1 (h(ω) T (ω)x) 0, y N = 0 and q B (ω) T B(ω) 1 W q(ω) T Taking expectation, V (x) = E ξ {q B (ω) T B(ω) 1 (h(ω) T (ω)x)} in the set ξ Ξ {x B(ω) 1 (h(ω) T (ω)x) 0} This is exactly the expression for optimality cuts at x v 21 / 44

22 Table of Contents 22 / 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 [ 5.1c of BL] 5 Example: Capacity Expansion Planning

23 Feasibility Cuts 23 / 44 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.

24 24 / 44 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

25 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 / 44

26 26 / 44 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

27 27 / 44 Complete Recourse and Relative Complete Recourse pos W = {z z = Wy, y 0} K 1 = {x Ax = b}, K 2 = {x y : Wy = h k T k x, k = 1,..., K } Feasibility cuts are not necessary when we have: Complete recourse: pos W = R m 2 Relative complete recourse: K 2 K 1

28 Table of Contents 28 / 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 [ 5.1c of BL] 5 Example: Capacity Expansion Planning

29 29 / 44 Proof of Convergence Key observation: when we solve the master problem (M) subject to x K 1 K 2 either θ v V (x v ) or θ v < V (x v ) The first case is our termination criterion and implies optimality. The second case can occur finitely many times. K 1 = {x Ax = b, x 0}, K 2 = {x y : Wy = h k T k x, y 0, k = 1,..., K }

30 30 / 44 θ v V (x v ) Is Equivalent to the Termination Criterion Q(x, ξ(ω)) = min y {q(ω) T y Wy = h(ω) T (ω)x, y 0} From linear programming duality, Q(x v, ξ k ) = (π v k )T (h k T k x v ) Taking expectation, V (x) = E ξ Q(x v, ξ) = K k=1 p k (π v k )T (h k T k x v ) Our termination criterion is θ v w v, where K K w v = e s+1 E s+1 x v = ( p k (πk v )T h k ) ( p k (πk v )T T k )x v k=1 k=1

31 31 / 44 θ v V (x v ) Implies Optimality Fact: Q(x, ξ) is convex in x, and π v k is a subgradient at x v : Q(x, ξ k ) (π v k )T h k (π v k )T T k x Taking expectation: V (x) K k=1 p k (π v k )T (h k T k )x Full problem can be written as min c T x + θ s.t. V (x) θ x K 1 K 2 Therefore master (M) is a relaxation and gives lower bound If there were x K 1 K 2 with c T x + V ( x) < c T x v + V (x v ) then x v would not be optimal for (M) since (M) would be at most c T x + V ( x)

32 θ v < V (x v ) Occurs Finitely Many Times 32 / 44 We have seen that each optimality cut is equivalent to θ V (x l ) + V (x l )(x x l ), l = 1,..., s Every time θ v < V (x v ) occurs, the set of new multipliers (πk v ), k = 1,..., K must be different from those generated previously Optimal multipliers πk v correspond to optimal bases of min y {qk T y Wy = h k T k x, y 0} There are finitely many bases, therefore finitely many combinations of optimal multipliers (πk v ), k = 1,..., K

33 33 / 44 Finite Termination of Feasibility Cuts Key observations: 1 Only a finite number of feasibility cuts can be generated 2 The feasibility cuts of equation 2 are valid Valid feasibility cuts: remove current candidate x v from consideration do not remove any other candidates x K 2 from consideration

34 Finite Number of Possible Cuts 34 / 44 Recall feasibility cuts: D l x d l, l = 1,..., r + 1, where D r+1 = (σ v ) T T k d r+1 = (σ v ) T h k and σ v the dual multipliers of (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 σ v corresponds to a basis B of W : σ v = c T B B 1 W has a finite number of bases

35 35 / 44 Valid Feasibility Cuts: x v Is Cut Off If w > 0 then by linear programming duality (σ v ) T (h k T k x v ) > 0, therefore D r+1 x d r+1 removes x v from consideration

36 Valid Feasibility Cuts: No Other x K 2 Is Cut Off Consider the following subset of R m 2: {z z = h k T k x, x K 2 } This set is contained in pos W = {z z = Wy, y 0} by definition of K 2 The half-space defined by (σ v ) T z 0 contains pos W To see this, note that the reduced cost of y must be 0: c T N c T B B 1 N = (σ v ) T W 0 where c T N = 0, (σv ) T = c T B B 1, N = W Any element of pos W can be expressed as z = Wy, y 0, hence from the previous inequality (σ v ) T z 0 Therefore, (σ v ) T (h Tx) 0 will be a valid feasibility cut 36 / 44

37 Feasibility Cuts 37 / 44 h k T k x v R m 2 σ r+1 pos W

38 Table of Contents 38 / 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 [ 5.1c of BL] 5 Example: Capacity Expansion Planning

39 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 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 39 / 44

40 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 / 44

41 Slave Problem 41 / 44 (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 j=1 where x has been fixed from the master problem

42 Sequence of Investment Decisions Iteration Coal (MW) Gas (MW) Nuclear (MW) Oil (MW) / 44

43 43 / 44 Observations Investment candidate in each iteration necessarily different from all past iterations Greedy behavior: low capital cost in early iterations

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

The L-Shaped Method. Operations Research. Anthony Papavasiliou 1 / 38 1 / 38 The L-Shaped Method Operations Research Anthony Papavasiliou Contents 2 / 38 1 The L-Shaped Method 2 Example: Capacity Expansion Planning 3 Examples with Optimality Cuts [ 5.1a of BL] 4 Examples