SF2812 Applied Linear Optimization. Preparatory questions. Lecture 6: Stochastic programming. Anders Forsgren

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1 SF2812 Applied Linear Optimization Lecture 6: Stochastic programming. Anders Forsgren SF2812 Applied Linear Optimization, KTH 1 / 18 Lecture 6, 2016/2017 Preparatory questions 1. Try to solve theory question Study the GAMS file farm.gms. SF2812 Applied Linear Optimization, KTH 2 / 18 Lecture 6, 2016/2017

2 Stochastic programming Consider a linear program in the form: min c T x subject to Ax = b, Tx = h, x 0. Assume that T and h are stochastic. How can this be handled? We assume that the x-variables must be determined before the outcome of T and h are known. The constraint Tx = h can then normally not be fulfilled. SF2812 Applied Linear Optimization, KTH 3 / 18 Lecture 6, 2016/2017 Two-stage model with recourse Assume that the stochastics of T and h may be represented by S scenarios, where scenario s occurs with probability p s, in which T and h take the values T s and h s respectively. We call the x-variables stage-one variables. In a second stage one may compensate for deviations h s T s x by nonnegative second-stage variables y s, s = 1,..., S, and a recourse matrix W according to Wy s = h s T s x, s = 1,..., S. The cost of this compensation is d T s y s. Stage 1 (x) Randomness Stage 2 (y s, s = 1,..., S) SF2812 Applied Linear Optimization, KTH 4 / 18 Lecture 6, 2016/2017

3 Simple recourse One example is simple recourse, where ( ) ( + y ( ) s y s =, W = I I and d s = y s This implies y s+ ys = h s T s x, y s+ 0, ys 0. d s + d s ). SF2812 Applied Linear Optimization, KTH 5 / 18 Lecture 6, 2016/2017 The aggregate optimization problem For scenario s one obtains the optimal recourse cost Q(x, s), for a given x, according to Q(x, s) = min d T s y s subject to Wy s = h s T s x, y s 0. Minimization of the expected cost gives the optimization problem RP = min c T x + subject to Ax = b, x 0. p s Q(x, s) SF2812 Applied Linear Optimization, KTH 6 / 18 Lecture 6, 2016/2017

4 The aggregate optimization problem, cont. The aggregate optimization problem may be written as one problem: RP = min c T x + p s ds T y s subject to Ax = b, T s x + Wy s = h s, s = 1,..., S, x 0, y s 0, s = 1,..., S. RP denotes recourse problem. SF2812 Applied Linear Optimization, KTH 7 / 18 Lecture 6, 2016/2017 The constraint matrix of the aggregate problem The constraint matrix of the aggregate problem becomes A T 1 W T 2 W..... W T S The matrix has a staircase structure. Note that S may be very large. Special techniques may utilize the structure of the matrix. For very large problems one may need to sample the outcomes. SF2812 Applied Linear Optimization, KTH 8 / 18 Lecture 6, 2016/2017

5 A farming example Example from Birge and Louveaux (in the appended material). A farmer has wheat, corn and sugar beets on his 500 ha land. Problem data: Wheat Corn Sugar beet Yield (ton/ha) Plant cost (SEK/ha) Sales price (SEK/ton) ( 6000 ton) 10 (> 6000 ton) Purchase price (SEK/ton) Min. requirement (ton) Available land: 500 ha SF2812 Applied Linear Optimization, KTH 9 / 18 Lecture 6, 2016/2017 A farming example, cont. i = 1 corresponds to wheat, Introduce index i, where i = 2 corresponds to corn, i = 3 corresponds to sugar beet. Variables: x i, ha land with crop i, i = 1, 2, 3. w i, number of tons of crop i sold to normal price, i = 1, 2, 3. v 3, number of tons of sugar beets sold to excess price. y i, purchased number of tons of crop i, i = 1, 2, 3. SF2812 Applied Linear Optimization, KTH 10 / 18 Lecture 6, 2016/2017

6 A farming example, cont. Problem formulation: min 150x x x y 1 170w y 2 150w 2 36w 3 10v 3 subject to x 1 + x 2 + x 3 500, 2.5x 1 + y 1 w 1 200, 3x 2 + y 2 w 2 240, 20x 3 w 3 v 3 0, w , x 1, x 2, x 3, y 1, y 2, w 1, w 2, w 3, v 3 0. SF2812 Applied Linear Optimization, KTH 11 / 18 Lecture 6, 2016/2017 A farming example, cont. Consider a simple stochastic model with correlated outcome for the three crops, either high (+20%), medium (as above) or low ( 20%). Modelled as s = 1, 2, 3. Each scenario is assigned probability 1/3. Utilization of land must be determined in beforehand: x i, i = 1, 2, 3. Sales and purchases may be determined when outcome is known: w is, i = 1, 2, 3, s = 1, 2, 3, v 3s, s = 1, 2, 3, y js, j = 1, 2, s = 1, 2, 3. SF2812 Applied Linear Optimization, KTH 12 / 18 Lecture 6, 2016/2017

7 A farming example, cont. Problem formulation: min 150x x x (238y 1s 170w 1s +210y 2s 150w 2s 36w 3s 10v 3s ) subject to x 1 + x 2 + x 3 500, (1 + α s )2.5x 1 + y 1s w 1s 200, s = 1, 2, 3, (1 + α s )3x 2 + y 2s w 2s 240, s = 1, 2, 3, (1 + α s )20x 3 w 3s v 3s 0, s = 1, 2, 3, w 3s 6000, s = 1, 2, 3, x 1, x 2, x 3, y 1s, y 2s, w 1s, w 2s, w 3s, v 3s 0, s = 1, 2, 3, where α 1 = 0.2, α 2 = 0 and α 3 = 0.2. SF2812 Applied Linear Optimization, KTH 13 / 18 Lecture 6, 2016/2017 Complete information If one (hypothetically) could wait for the outcome we obtain: WS = S p s min c T x s + ds T y s subject to Ax s = b, T s x s + Wy s = h s, x s 0, y s 0. WS reads wait and see. Randomness Stage 2 (x s, y s, s = 1,..., S) SF2812 Applied Linear Optimization, KTH 14 / 18 Lecture 6, 2016/2017

8 Expected value of perfect information The expected value of perfect information, EVPI, is given by the difference between RP and WS, i.e., EVPI = RP WS. Gives a measure of how much it would be worth to have perfect information. RP is given by: Stage 1 (x) Randomness Stage 2 (y s, s = 1,..., S) WS is given by: Randomness Stage 2 (x s, y s, s = 1,..., S) Farming example: RP = SEK , WS = SEK , EVPI = SEK SF2812 Applied Linear Optimization, KTH 15 / 18 Lecture 6, 2016/2017 The mean-value problem A primitive way is to take mean values, which gives where d = EV = min c T x + d T y p s d s, T = subject to Ax = b, T x + Wy = h, x 0, y 0, p s T s and h = EV means expected value. p s h s. Let x denote the optimal x-solution of this problem. SF2812 Applied Linear Optimization, KTH 16 / 18 Lecture 6, 2016/2017

9 Expected value of EV solution Recourse for x fixed to x gives the optimization problem EEV = min c T x + Equivalently: p s ds T y s subject to A x = b, T s x + Wy s = h s, s = 1,..., S, y s 0, s = 1,..., S. EEV = c T x + S p s min d T s y s subject to Wy s = h s T s x, y s 0. EEV means expected result of using EV solution. SF2812 Applied Linear Optimization, KTH 17 / 18 Lecture 6, 2016/2017 Expected value of stochastic solution The expected value of the stochastic solution, VSS, is given by the difference between EEV and RP, i.e., VSS = EEV RP. VSS means value of stochastic solution. Gives a measure of how much it is worth to solve the stochastic optimization problem. EEV is given by: Compute x, ȳ Randomness Decision (y s, s = 1,..., S) RP is given by: Stage 1 (x) Randomness Stage 2 (y s, s = 1,..., S) Farming example: EEV = SEK , RP = SEK , VSS = SEK SF2812 Applied Linear Optimization, KTH 18 / 18 Lecture 6, 2016/2017