COLUMN GENERATION IN LINEAR PROGRAMMING

COLUMN GENERATION IN LINEAR PROGRAMMING EXAMPLE: THE CUTTING STOCK PROBLEM A certain material (e.g. lumber) is stocked in lengths of 9, 4, and 6 feet, with respective costs of $5, $9, and $. An order for 3 pieces of length 4 feet, pieces of length 5 feet, and 4 pieces of length 7 feet is to be cut from the stock lengths at minimum cost. NOTATION L, L,... L p are standard lengths kept in stock C, C,..., C p are respective costs for these b i is the required number of pieces of length l i, i=,,,m A j is an m-vector defining the j th cutting pattern, with = the number of pieces of length l i cut from one piece of standard length material. (e.g., and 4 5 7 are valid cutting patterns using the 4 stock) The problem may now be stated as

PROGRAM M Minimize ( ) st, i=,,...,m, x j and integer where x j = number of times pattern j is used, k(j) = standard length index used by pattern j. In realistic problems, there may be an astronomical number of possible patterns (although we can be sure that at most m patterns need to be used in the continuous solution to the problem). Rather than solve the complete problem with its huge number of constraints we try to generate only those patterns (columns) that might be used in the optimal solution. HOW DO WE DO THIS? Given a simplex multiplier vector π = [π π π m ], the pattern [a a... a m ] T which uses stock length L k may be entered into the basis as long as = = [ + + + ] >, and + + +

This leads to SUBPROBLEM k: Find nonnegative integers a, a,, a m to Maximize st Note that if the maximum value of the objective for the subproblem is negative, then for any feasible pattern corresponding to standard length L k the objective is negative. In other words, no column can be generated from the k th subproblem for which the reduced cost is positive with the current vector of simplex multipliers. On the other hand, if the maximum value is positive, then the optimal a i values yield a column for which the reduced cost is positive, and which can therefore be added on to the original problem and made basic to improve the original objective. In general, the complete LP with all possible columns (Program M) is called the MASTER PROBLEM. An LP with some subset of all the possible columns is referred to as a PARTIAL or RESTRICTED MASTER PROBLEM. The column generation algorithm may now be summarized as follows: 3

STEP : Start with an initial set of columns that will yield a feasible region for the resulting restricted master problem, and construct the latter. STEP : Solve this LP and also obtain the vector of simplex multipliers (π). STEP : Use the simplex multipliers to solve p subproblems (one for each standard length L k, k=,,,p). If the maximum value for the k th subproblem is positive, use the optimal solution to the subproblem to generate a column for the master. Add such columns on to the restricted master and return to Step. If all the subproblem minima are nonpositive, then there are no more useful columns left to generate. The current solution to the restricted master must be the optimal solution to the original master problem. STOP. It is important to note that the general column generation problem will solve some subproblem to find a column, but the exact form of the subproblems will depend on that of the original master problem. In some cases (such as the cutting stock problem) may be a set of subproblems, while in others there may be only a single one. In the case of the cutting stock problem the subproblems are constrained; in other instances they may be unconstrained, and sometimes they 4

may even possess a closed form solution. In general, the subproblems are much easier to solve than the restricted master problem. Certainly, it is better to solve a sequence of restricted masters and subproblems, than to try and generate the entire master problem and solve it! Back to the cutting stock problem... Consider the subproblem we got earlier: Find nonnegative integers a, a,, a m to Maximize st Since C k is a constant, as opposed to maximizing, we may just maximize. Thus the subproblem reduces to the following equivalent: Find nonnegative integers a, a,, a m to Maximize st The particular subproblem that results for the cutting stock problem, is commonly referred to as the KNAPSACK PROBLEM, and is a fairly well known one. WARNING: Keep in mind that C must be subtracted from the optimal objective value of the k th knapsack problem in order to obtain the reduced cost!! 5

SOLUTION OF SUBPROBLEMS A column (cutting pattern) might be added to the (restricted master) problem for each standard length L k, which yields a positive reduced cost; or alternatively, only the column corresponding to the subproblem yielding the "most positive of all positive" reduced costs might be added. A heuristic procedure may be used to solve the knapsack subproblem (since we only need a column with positive reduced cost). When heuristics fail to produce a column with positive reduced cost, we may resort to an optimal approach (such as dynamic programming) to solve the subproblem. If DP is used, all the knapsack subproblems can be simultaneously solved with little additional effort over that required to solve each one individually. HEURISTIC KNAPSACK FILLING Maximize st a i and integer Order the requested lengths according to the ratios of return to length, π i /l i : 6

Fill the knapsack with as much as possible of the length l i (with maximum return per unit length) where (. ) is the floor function (rounding down). Fill what remains empty (L k l i a i ) with as much as possible of the next most "valuable" length (l i ) etc., until the knapsack is full, or the empty space is too small for any length. Let us use the column generation algorithm to solve our cutting stock problem. The data for this problem may be summarized as follows: STOCK REQUIREMENTS k 3 i 3 L k 9 4 6 l i 4 5 7 C k 5 9 b i 3 3 4 To initiate our column generation algorithm, we shall (arbitrarily) propose the following three patterns cut from the 9' stock: =, =, = 4 5 7 7

Our initial LP is Minimize 5x + 5x + 5x 3 st x -S = 3 x -S = x 3 -S 3 = 4 x, S i for i=,,3, with solution x * =5, x * =, x * 3 =4, S * = S * = S * 3 = and π * =5/, π * =5, π * 3 =5, Z = COST = 375 Let us now try to generate a cutting pattern from the longest (6') stock by seeking a heuristic solution to the knapsack problem: Maximize (5/)a + 5 a + 5a 3 st 4a + 5a + 7a 3 6, a i and integer Ordering the ratios of return to length:,.., 5 5 5 7 5/ 4 We fill the knapsack with as much as possible with l a = 6/5 = 3 units This leaves us with L 3 -l a = 6-3*5 = unit of space that cannot accommodate either of lengths l 3 and l 8

So the column pattern generated is = 3 with cost C 4 = Checking the reduced cost for this candidate column, = = 5 5 3 = 5 > Since the reduced cost is positive, this column when added to the basis will improve the objective. Adding A 4 to the restricted master we get a new LP: Minimize 5x + 5x + 5x 3 + x 4 st x -S = 3 x + 3x 4 -S = x 3 -S 3 = 4 all x, S i After entering x 4 into the basis this LP has the solution x * =5, x * * 3 =4, x 4 =/3 x * = S * = S * = S * 3 = and π * =5/, π * =/3, π * 3 =5, Z = 34.67 ======================= Let us return to the subproblem and try to generate another pattern from the 6' stock by solving Maximize (5/)a + (/3)a + 5a 3 st 4a + 5a + 7a 3 6, a i and integer 9

Ordering the ratios of return to length:,.., 5 7 /3 5 5/ 4 So we fill the knapsack with as much as possible of l 3 which in this case is a 3 = 6/7 = units. This leaves us with L 3 -l 3 a 3 = units of space, so that a =a =. So the candidate column pattern generated is = Checking the reduced cost for this column, = with cost C 5 = = = 5 = Adding this column to the basis will not improve the objective! So let us try to solve the subproblem for the 4' stock. For this only the RHS of the subproblem on the previous page is altered (from 6 to 4). Solving this problem, we get the same pattern as before (except that the knapsack is completely filled this time), at a cost of $9 so that the reduced cost is given by = = 5 9 = > Adding this column will therefore improve the objective value of the restricted master, i.e., so let us do so.

The new restricted master problem is thus given by Minimize 5x + 5x + 5x 3 + x 4 + 9x 5 st x -S = 3 x + 3x 4 -S = x 3 +x 5 -S 3 = 4 all x, S i This LP (after entering x 5 into the basis) has the solution x * =5, x 4 * =/3, x 5 * =, x 3 * = x * = S * = S * = S 3 * = and π * =5/, π * =/3, π 3 * =9/, Z = 3.67 Let us return to the subproblem again... For the 4' stock we have Maximize (5/)a + (/3)a + (9/)a 3 st 4a + 5a + 7a 3 4, a i and integer Ordering the ratios of return to length:,.., /3 5 9/ 7 5/ 4 So we fill the knapsack with as much as possible of l which in this case is a = 4/5 = units. This leaves us with 4 units of space, so that a 3 = and a =.

So the candidate column pattern generated is = Checking the reduced cost for this column, = with cost C 6 = 9 = = 9 =.7 > and adding this column to the restricted master will result in improvement. Adding this on to the restricted master problem we get a new LP, which yields the optimum solution x * =, x 5 * =, x 6 * = x 4 * = x 3 * = x * = S * = S * = S 3 * = and π * =5/, π * =39/, π 3 * =9/, Z = 3 Try again with the 4' stock to see if any further improvement is possible... Maximize (5/)a + (39/)a + (9/)a 3 st 4a + 5a + 7a 3 4, a i and integer Ordering the ratios of return to length:,.., 39/ 5 9/ 7 5/ 4 This leads to the pattern. BUT THIS WAS PATTERN 6 Let us try using the 9' stock... 4 changes to 9 in the RHS above)

a = 9/5 =, a = 4/7 =, a 3 = 4/4 = This candidate pattern costs $5, and has reduced cost = = 5 =.75 > and adding this column to the restricted master will result in improvement. Adding this on to the restricted master problem we get a new LP, which yields the optimum solution x * =5, x 5 * =, x 7 * = x 6 * = x 4 * = x 3 * = x * = S * = S * = S 3 * = and π * =5/, π * =5/, π 3 * =9/, Z = 35 To generate further columns, we must solve knapsack problems which in general, are given by Maximize (5/)a + (5/)a + (9/)a 3 st 4a + 5a + 7a 3 L k, a i and integer Ordering the ratios of return to length:,.., 9/ 7 5/ 4 5/ 5 Using L =9' we get the pattern with = = 5 = -.5 NO GOOD 3

Using L =4' we get the pattern with = = Using L 3 = 5' we get the pattern with = = 9 = NO GOOD = - NO GOOD Our heuristic algorithm has failed to generate any new useful patterns. Could this be the optimal solution to the problem?! At this point we need to resort to solving the subproblems for each pattern optimally (using Dynamic Programming) to verify that we get the true maximum values for each. It may be verified that when we do this we do indeed find that the maximum reduced costs for each length are nonpositive, which means that there are no more useful patterns to generate and no further improvement can be made to the solution yielded by the current basis. The current basis is OPTIMAL ---- STOP. : 4 lengths from the 9 stock; pattern used on 5 pieces of stock ($5) 5: 7 lengths from the 4 stock; pattern used on pieces of stock ($8) 7: 4 + 5 length from the 9 stock; pattern used on pieces of stock ($) 4

THE DANTZIG-WOLFE DECOMPOSITION PRINCIPLE Another instance of the column generation idea... Consider the LP: Minimize cx st Ax b (m rows) Bx b (m rows) x Let X = {x: Bx b, x }, and, i=,,...,k be the extreme points of X, j=,,...,l be the extreme rays of X Recall that by the Resolution Theorem, x X if, and only if, = + for some λ, µ, satisfying =, λ i, µ j for all i and j. We may therefore rewrite the LP as Minimize cx st Ax b x X 5

This in turn may be rewritten as + + = ; ( ) + ( ) ( ) + ( ) = ; Now let us define =, =, =, = 6

Note that and (and therefore,,, and ) are all constants; the variables are λ i and µ j Also note that x,, n c n A m n b m and,, m In summary, we used a change of variables to transform Minimize cx st Ax b, Bx b, x into ( ) + ( ) ( ) + ( ) = ; This (transformed) problem is referred to as the DANTZIG-WOLFE COMPLETE MASTER PROBLEM. 7

Note that this LP has (m +) constraints and (k+l) decision variables, and that = = (scalars) = = (column vectors) In general, the number of extreme points and extreme rays could be astronomical and quite impossible to individually and exhaustively evaluate. This implies that the number of variables in the complete master problem (=k+l) is unmanageably large. Fortunately, we know a priori that at most (m +) of these columns could be basic at the optimum! The Dantzig-Wolfe procedure does not compute and store all (k+l) columns of the complete master problem; instead, at each iteration it generates only a very small subset of the more "attractive" columns. The master problem containing only a subset of the all (k+l) columns is referred to as a "partial" master problem. Suppose now that the Revised Simplex Method is being used in the master problem. 8

PRICING OUT A NONBASIC COLUMN OR } } ( ) Let π = [π' π"] be a vector of simplex multipliers note that π' is a vector of m elements and π'' is a singleton (the multiplier corr. to the convexity constraint). Then [π' π"] =π' + π"; [π' π"] =π' so that the reduced cost for the column (z j - c j ) is π' + π" - = π' + π" - = [π' ] + π" ( extr. pts.) OR π' + π" - = π' - = [π' ] ( extr. dirs.) Thus, for a minimization (maximization) problem if the reduced cost is > (<), the column is attractive and should be entered into the basis. 9

SELECTING THE ENTERING COLUMN To select the column to be entered into the basis, we choose the one that has the most negative (most positive) reduced cost for a MAX (for a MIN) problem. We therefore solve the following SUBPROBLEM (assuming MIN) Maximize [π' ] st x X = {x: Bx b, x }, NOTE THE SIMILARITY WITH THE COLUMN GENERATION PROCEDURE THAT WE USED FOR THE CUTTING STOCK PROBLEM... The subproblem is an LP whose solution, if bounded is an extreme point of the set X, i.e., for some s between and k. If this LP is unbounded, then it is unbounded along an extreme ray of X, i.e., for some s between and l. Therefore, a column for the master problem with maximum reduced cost can be found by solving the subproblem, (without explicitly computing the rest of the nonbasic columns). Note that if the subproblem has an optimal solution which is negative, then the reduced cost for all columns must also be negative, and we are therefore at the optimum solution to the master problem.

To summarize the procedure: MASTER Select an optimal linear combination of "proposals" and submitted by the SUBPROBLEM; determine shadow prices (simplex multipliers) π' for scarce resources b shadow prices π' proposal or SUBPROBLEM Assume that the scarce resources (b ) are unlimited but that a premium price π' must be paid for them; find the optimal plan or to be submitted to the MASTER problem for consideration. The process can be interpreted as being PRICE DIRECTED.

SEPARATION OF THE D-W SUBPROBLEM Suppose that the matrix B in our LP Minimize cx st Ax b, Bx b, x is "BLOCK-DIAGONAL" that is, our LP has the structure MINIMIZE C J x J + C J x J +... + C Jq x Jq ST A J x J + A J x J +... + A Jq x Jq b x J (b ) I x J... (b ) I : x Jq (b ) Iq x where INDEX SETS J J... J q I rows I rows I 3 rows B = I q rows

SUBPROBLEM Maximize (π'a J -C J ) x J + (π'a J -C J ) x J +... + (π'a Jq -C Jq )x Jq st x J (b ) I x J... (b ) I : x Jq (b ) Iq x J, x J,... x Jq may now be separated into q different subproblems: Max (π'a J -C J ) x J Max (π'a J -C J ) x J Max (π'a Jq -C Jq ) x Jq st x J (b ) I st x J (b ) I... st x Jq (b ) Iq x J x J x Jq because, in the subproblem, there is no interaction among the subset of variables x J, x J,... x Jq. After each subproblem is (separately) solved, the q solutions,,,,, could be submitted as a single "combined proposal" to the MASTER. Q. What if one of the subproblems is UNBOUNDED??? A. Find the direction along which it is unbounded, and generate a column 3

AN EXAMPLE: Minimize -x - x - x 3 st x + x + x 3 } Ax b -x + x -x + x 8 Bx b x 3 3 x, x, x 3 c = [- - -]; A = [ ] (NOTE: A is a matrix!) b =[] B = b = 8 3 The set X = {x: Bx b, x } x 3 x 7 6 5 4 3 (,,3) (,,) (,,3) (,,) 3 4 5 6 7 8 9 (4,6,3) (4,6,) x 4

The set X has 6 extreme points and extreme rays, 3,, 3, 4 6, 4 6 3, Using the Resolution Theorem, and following the derivations from before, we could rewrite the above LP in an equivalent form, where the decision variables are the weights for the 6 extreme points (λ i ) and for the extreme directions (µ j ) as + + = ; where = = with c = [- - -]; = = with A = [ ] (Again note that A is a matrix and and are column vectors) Thus =, = 3, = 4, = 7, = 6, = 9 = [], = [3], = [], = [5], = [], = [3] = 4, = 9 = [3], = [] 5

This yields the COMPLETE MASTER PROBLEM below s RHS 3 4 7 6 9 4 st 3 5 3 3 Ext. pts. Ext. Dir. Slack We shall start NOT with the complete master problem, but with the PARTIAL master consisting of the columns for slack variables s and weight for the extreme point = [ ] T This is given by s ; opt. sol. has λ =, s=, (A B ) - =, = [ ] Thus '=[]; "= FIRST DANTZIG-WOLFE SUBPROBLEM: To find a new column which maximizes the reduced cost, i.e., Maximize [π' ] st x X 6

Max x + x + x 3 st -x + x -x + x 8 x 3 3 x, x, x 3 SEPARATION Max x + x st -x + x -x + x 8 x, x Max x 3 st x 3 3 x 3 Optimum x * 3 =3 8 3 4 4 4 3 6 6 4 UNBOUNDED SOLUTION! As S increases by unit, x increases by units and x by unit 7

Thus is an EXTREME DIRECTION. In the x -x -x 3 space, this extreme direction is = (Note that we could also have obtained the extreme direction = tableau!) at the first Since [π' ] = [][ ] - [- - -] = 4, the maximum reduced cost is positive, and a column may be generated from and added to the master problem: = [ ] = = 3 with cost = [ ] = = 4. Master Problem with the new column added is s µ RHS 4 3 8

"Update" the new column: (P B ) - 3 = 3 = 3 New Basis is [µ λ ] with (P B ) - = 3 = /3 Simplex multiplier vector is π = [-4 ] /3 =[-4/3 ]= [ ' "] New basic solution is l = /3 = 4 SECOND DANTZIG-WOLFE SUBPROBLEM: Maximize [π' ] ; st x X [π' ] = [4][ ] - [- - -] = [-/3 /3 -/3] Max -(/3)x + (/3)x (/3)x 3 st -x + x -x + x 8 x 3 3 x, x, x 3 Max -(/3)x + (/3)x st -x + x -x + x 8 x, x Max (-/3)x 3 st x 3 3 x 3 Optimum x * 3 =, Z= Optimum x * =4, x * =6; Z=-8/3 9

Thus the extreme point = 4 6 is generated, and since [π' ] +π"= {[-4/3] [ ] - [- - -]} 4 6 + = 8/3, the maximum reduced cost is positive, and a column may be generated from and added to the master problem: = 4 [ ] 6 = = with cost = [ ] 4 6 = = 6 Master Problem with the new column added is s µ RHS 4 6 3 "Update" the new column: (P B ) - 3 = /3 = /3 Current BFS has l = 4 From the usual ratio test, Min {4 (/3), ) =, so that the second basic variable leaves the basis 3

New Basis is [µ λ ] with (P B ) - = 3 = /3 /3 Simplex multiplier vector is π = [-4-6] /3 /3 =[-4/3-8/3]= [ ' "] New basic solution is l = /3 /3 = /3 SECOND DANTZIG-WOLFE SUBPROBLEM: Maximize [π' ] ; st x X [π' ] = [/3] [ ] - [- - -] = [-/3 /3 -/3] Since the subproblem is unchanged from the Second D-W subproblem, we get the same optimum solution which generates the same extreme point = 4 6. This in turn generates the same column with cost = -6. The maximum reduced cost is [π' ] -π"= {[-4/3] [ ] - [- - -]} 4 6 + (-8/3) = Since the maximum reduced cost is nonpositive, reduced costs corresponding to all of the other columns in the master problem must also be nonpositive. Thus no further improvement is possible and the OPTIMUM SOLUTION is given by = + () 4 6 = 6/3 /3, with cost cx * = -56/3 NOTE THAT x * IS NOT AN EXTREME POINT OF X! 3