Introduction column generation

Transcription

1 Introduction column generation Han Hoogeveen Institute of Information and Computing Sciences University Utrecht The Netherlands May 28, 2018

2 Contents Shadow prices Reduced cost Column generation

3 Example: Continuous Knapsack problem We are given n items with integral weight a j ; integral value c j. B is a given integer. Goal: Find a subset of the items with maximum value and total weight B. Assume that it is possible to select an item partially.

4 Linear Programming formulation Use variables x j to model how much of item j is selected (x j = 1: take item j completely). max n c j x j n a j x j B. (1) 0 x j 1 j = 1,..., n. (2)

5 Shadow price Consider constraint i; right-hand side value b. Decrease (increase) b by ɛ Change in outcome value (ɛ) Shadow price of this constraint is (ɛ) ɛ Comparable to the left (right) derivative of the outcome value with respect to b. The shadow price signals the unit price you want to pay for a little more. The shadow price is a by-product of the LP-solution.

6 Shadow price for Continuous Knapsack (1) Optimal solution: Split each item j into a j subitems with size 1 and value c j a j. Renumber the items such that c 1 a 1 c 2 a 2... c n a n Fill the knapsack with the subitems with highest value: first item 1, then item 2, etc. until the knapsack is full.

7 Shadow price for Continuous Knapsack (2) Increase (decrease) B by ɛ. Result: Suppose that item k has been chosen partially. Increase (decrease) the amount of item k that you choose in the knapsack by ɛ. Increase (decrease) in value: ɛ c k Shadow price: c k a k. Alternative situation: item k has been chosen entirely (and none of item k + 1) a k. If you decrease B, then you find a shadow price of c k a k ; If you increase B, then you find a shadow price of c k+1 a k+1.

8 Reduced cost New activity (item) 0; variable x 0. Add x 0 to the problem (with c 0 and a 0 ). Compute net gain (gain minus investment ) of putting x 0 ɛ. Reduced cost x 0 is the net gain per unit. Adding item 0 will improve the LP-solution only if its reduced cost is positive in case of maximization and negative in case of minimization.

9 Reduced cost for Continuous Knapsack (1) First step: add x 0 to the LP-formulation n max c 0 x 0 + c j x j n a 0 x 0 + a j x j B. (3) 0 x j 1 j = 0,..., n. (4)

10 Reduced cost for Continuous Knapsack (2) Put x 0 ɛ and rewrite the LP: n c 0 ɛ + max c j x j n a j x j B a 0 ɛ (5) 0 x j 1 j = 1,..., n. (6)

11 Reduced cost for Continuous Knapsack (3) Estimate the change in cost value compared to the outcome of the LP without item 0. There is an additional term c 0 ɛ in the objective. The righthand-side of the knapsack constraint has increased by a 0 ɛ. Using the concept of shadow price we find that the change in righthand-side will increase the outcome value by a 0 ɛ c k a k The total change in outcome value is therefore equal to c 0 ɛ a 0 ɛ c k a k = ɛ(c 0 a 0 c k a k )

12 Reduced cost for Continuous Knapsack (4) The reduced cost is the change per unit, so divide by ɛ: (c 0 a 0 c k a k ) If this is positive, then adding item 0 will improve the solution. This will be the case if and only if c 0 a 0 > c k a k If there are more constraints, then add up all the changes due to a change in righthand-side (weighted by the corresponding reduced cost).

13 Column generation Take any LP-problem (maximization) with large number of variables. Only a few variables will get positive value. Remainder of the variables is lumber. Remedy Solve the LP-problem for a small subset of the variables The remaining variables are ignored (they get value 0). Is it worthwhile to add a variable? Check reduced cost. Smart implementation: pricing problem Maximize the reduced cost over all possible variables.

14 Example: Beun de Haas s problem Beun de Haas is an independent entrepreneur (ZZZP er). Clients contact him for small jobs; a job must be started and finished on the same day. For each job j (j = 1,..., p) is given: the reward (c j ); the time it takes (a j ); the days on which they can be done (characterized by binary parameters h jt ). Planning period: days 1,..., T. Beun has Q t time on day t (t = 1,..., T ). Goal. Choose and plan the work to earn as much as possible.

15 Standard ILP formulation Use binary variables y jt to signal if job j is done on day t. max p t=1 T c j y jt T y jt 1 j (7) t=1 p a j y jt Q t t (8) Preprocessing: y jt = 0 if h jt = 0, for all j, t. y jt {0, 1} j, t (9) The formulation is correct, but has the disadvantage that the LP-relaxation yields a poor lower bound.

16 Advanced ILP formulation Formulation with day plans. There are N day plans; each day plan refers to a specific day at which it should be executed and it consists of a set of jobs that Beun should execute when following this day plan. Dayplan j is characterized by parameters g ij (i = 1,..., p) and d jt (t = 1,..., T ): gij = 1 if job i is part of this day plan (and 0 otherwise). d jt = 1 if day plan j is meant for day t (and 0 otherwise). The reward of day plan j is equal to C j (the total reward of the included jobs). Use a binary variable x j for each day plan j: x j = 1 if day plan j is chosen.

17 Advanced ILP formulation (2) N max C j x j N g ij x j 1 i (10) N d jt x j 1 t (11) x j {0, 1} j (12) The feasibility of the day plans ensures Availability of chosen jobs. Beun does not have to work too long per day.

18 What can we do with it? Downside of this formulation: There are very many day plans possible Not all day plans are known We do not want to enumerate them Most of them are useless anyway. Heuristic approach. Select a number of presumably useful day plans. Solve the ILP formulation for this subset

19 How to find this subset? Strategy: Simplify (relax) the problem Solve this relaxation Put the discovered day plans in the subset Solve the original problem Use here the LP-relaxation; solve it using column generation. Underlying idea: the solutions of the relaxation and the integral problem will resemble each other

20 LP relaxation N max C j x j N g ij x j 1 i (13) N d jt x j 1 t (14) x j 0 j (15) Replace x j {0, 1} by 0 x j 1 x j 1 is redundant.

21 Reduced cost Suppose that you have enumerated n (out of the total of N) day plans; the current LP is based on these n day plans. Consider a new day plan 0 (it is not in the current LP). Characterize day plan 0 by paramaters g i and d t g i = 1 if job i is included (and 0 otherwise) d t = 1 if the day plan is to be executed at day t. The value C 0 of day plan 0 is equal to p i=1 c ig i. Compute the reduced cost for day plan 0.

22 Computation of reduced cost of day plan 0 First step: solve the current LP. The shadow prices are π i (i = 1,..., p) and λ t (t = 1,..., T ) (shown after the respective constraints). n max C j x j n g ij x j 1 i = 1,..., p π i (16) n d jt x j 1 t = 1,..., T λ t (17) x j 0 j (18)

23 Derivation reduced cost Add day plan 0 (parameters g i and d t ) with variable x 0. n max C 0 x 0 + C j x j g i x 0 + d t x 0 + n g ij x j 1 i = 1,..., p π i (19) n d jt x j 1 t = 1,..., T λ t (20) x 0, x j 0 j (21)

24 Derivation reduced cost (2) Put x 0 ɛ, rewrite LP: n C 0 ɛ + max C j x j n g ij x j 1 g i ɛ i = 1,..., p π i (22) n d jt x j 1 d t ɛ t = 1,..., T λ t (23) x j 0 j (24)

25 Derivation reduced cost (3) Compare the new LP to the old LP: The objective value has a term C 0 ɛ; The righthand-side values have changed: compute the resulting change in objective value using shadow-prices. Per constraint: the objective value changes by the amount by which the righthand-side has changed times the shadow price. Add up all these terms to find the total change. = The change in objective value amounts to p g i ɛ π i + i=1 T d t ɛ λ t. t=1

26 Derivation reduced cost (4) Add this change because of the righthand-side to C 0 ɛ = ɛ p i=1 g ic i. The total change per ɛ is then equal to [ p ] p T ɛ g i c i g i π i d t λ t. i=1 i=1 The reduced cost is the change per unit = the reduced cost is equal to t=1 p p T g i c i g i π i d t λ t = i=1 i=1 t=1 p T g i (c i π i ) d t λ t. i=1 t=1

27 Pricing problem The reduced cost of this day plan amounts to p g i (c i π i ) i=1 T d t λ t t=1 If p i=1 g i(c i π i ) T t=1 d tλ t > 0, then improvement possible. The goal of the pricing problem is to construct a feasible day plan with maximum reduced cost. If the outcome value of the pricing problem is > 0, then add the corresponding day plan to the current LP and solve it, etc. If the outcome value of the pricing problem is 0 (you can show that it will have value 0 then), then there is no day plan that can improve the current LP: the solution to the current LP is the solution to the LP with all possible day plans. In case of a minimization problem, a new variable is interesting if it has a negative reduced cost; hence, in the pricing problem we then want to find a variable with minimum reduced cost.

28 Solving the pricing problem (1) The pricing problem is easier to solve if the corresponding day is known. Therefore, we solve the pricing problem for each day separately. Assume we look for a dayplan for a given day t 0. Then d t0 = 1 and d t = 0 for all t t 0. If we fill this in, then the reduced cost is equal to p g i (c i π i ) λ t0. i=1 Ignore the constant λ t0 in the maximization.

29 Solving the pricing problem (2) The pricing problem is then equal to: Maximize p i=1 g i(c i π i ) such that (g 1,..., g p ) corresponds to a feasible day plan. The variables are the g i values. Consequence of putting job i in the day plan (g i = 1): Gain c i π i ; Extra work a i in day t 0. Requirements Beun has Q t0 time available on day t 0 Job i can be chosen only if it is available on day t 0.

30 Pricing problem for day t 0 Consider all jobs available on day t 0 Select the most valuable set with total processing time Q t0 Knapsack problem: weight aj value cj π j. Size of knapsack: Qt0. You have to solve the problem for all possible days.

31 Improving the solution heuristically Augment the set of columns (corresponding to day plans) that is used to solve the ILP. After solving pricing problem for day t find second-choice plans by perturbing the solution of the pricing problem: Let j 1,..., j k be the indices of the jobs chosen in the pricing problem. For i = 1,..., k, repeat the following: Remove job ji from the instance of the pricing problem. Solve the pricing problem again for the instance without job ji Store the solution and add job ji again. These additional columns are not added to the current LP, but will be added later to the ILP. The ILP is solved for the set of selected columns (day plans). It is a heuristic, since a column that is necessary in the optimum solution may not be present in the set of selected columns (day plans in this application).

32 Branch-and-price Branch-and-price is a form of branch-and-bound based on column generation. The upper bound is computed by solving for each node in the branching tree the LP-relaxation through column generation. In contrast to the heuristic, it is possible to add new columns (day plans) in the branch-and-bound tree. Branch-and-price will find the optimum solution. Branch-and-price is possible only if you use a branching rule that can be combined with the column generation.