Column generation algorithms

Author: Lorena Garcia Fernandez (lgf572)

Introduction

Column Generation techniques have the scope of solving large linear optimization problems by generating only the variables that will have an influence on the objective function. This is important for big problems with many variables where the formulation with these techniques would simplify the problem formulation, since not all the possibilities need to be explicitly listed.

Theory, methodology and algorithmic discussions

Theory

The way this method work is as follows; first, the original problem that is being solved needs to be split into two problems: the master problem and the sub-problem.

• The master problem is the original column-wise (i.e: one column at a time) formulation of the problem with only a subset of variables being considered.

• The sub-problem is a new problem created to identify a new promising variable. The objective function of the sub-problem is the reduced cost of the new variable with respect to the current dual variables, and the constraints require that the variable obeys the naturally occurring constraints. The subproblem is also referred to as the RMP or “restricted master problem”. From this we can infer that this method will be a good fit for problems whose constraint set admit a natural breakdown (i.e: decomposition) into sub-systems representing a well understood combinatorial structure.

To execute that decomposition from the original problem into Master and subproblems there are different techniques. The theory behind this method relies on the Dantzig-Wolfe decomposition.

In summary, when the master problem is solved, we are able to obtain dual prices for each of the constraints in the master problem. This information is then utilized in the objective function of the subproblem. The subproblem is solved. If the objective value of the subproblem is negative, a variable with negative reduced cost has been identified. This variable is then added to the master problem, and the master problem is re-solved. Re-solving the master problem will generate a new set of dual values, and the process is repeated until no negative reduced cost variables are identified. The subproblem returns a solution with non-negative reduced cost, we can conclude that the solution to the master problem is optimal.

Methodology

Column generation schematics

Consider the problem in the form:

(IP) $z=max\left\{\sum _{k=1}^{K}c^{k}x^{k}:\sum _{k=1}^{K}A^{k}x^{k}=b,x^{k}\epsilon X^{k}\;\;\;for\;\;\;k=1,...,K\right\}$

Where $X^{k}=\left\{x^{k}\epsilon Z_{+}^{n_{k}}:D^{k}x^{k}\leq d^{_{k}}\right\}$ for $k=1,...,K$. Assuming that each set $X^{k}$ contains a large but finite set of points $\left\{x^{k,t}\right\}_{t=1}^{T_{k}}$, we have that $X^{k}=$:

$\left\{x^{k}\epsilon R^{n_{k}}:x^{k}=\sum _{t=1}^{T_{k}}\lambda _{k,t}x^{k,t},\sum _{t=1}^{T_{k}}\lambda _{k,t}=1,\lambda _{k,t}\epsilon \left\{0,1\right\}for\;\;k=1,...,K\right\}$

Note that, by assuming that each of the sets $X^{k}=$ is bounded for $k=1,...,K$ the approach essentially involves solving an equivalent problem of the form:

$max\left\{\sum _{k=1}^{K}\gamma ^{k}\lambda ^{k}:\sum _{k=1}^{K}B^{k}\lambda ^{k}=\beta ,\lambda ^{k}\geq 0\;\;integer\;\;for\;\;k=1,...,K\right\}$

where each matrix $B^{k}$ has a very large number of columns, one for each of the feasible points in $X^{k}$, and each vector $\lambda ^{k}$ contains the corresponding variables.

Now, substituting for $x^{k}=$ leads to an equivalent IP Master Problem (IPM):

(IPM) ${\begin{matrix}z=max\sum _{k=1}^{K}\sum _{t=1}^{T_{k}}\left(c^{k}x^{k,t}\right)\lambda _{k,t}\\\sum _{k=1}^{K}\sum _{t=1}^{T_{k}}\left(A^{k}x^{k,t}\right)\lambda _{k,t}=b\\\sum _{t=1}^{T_{k}}\lambda _{k,t}=1\;\;for\;\;k=1,...,K\\\lambda _{k,t}\epsilon \left\{0,1\right\}\;\;for\;\;t=1,...,T_{k}\;\;and\;\;k=1,...,K.\end{matrix}}$

To solve the Master Linear Program, we use a column generation algorithm. This is in order to solve the linear programming relaxation of the Integer Programming Master Problem, called the Linear Programming Master Problem (LPM):

(LPM) ${\begin{matrix}z^{LPM}=max\sum _{k=1}^{K}\sum _{t=1}^{T_{k}}\left(c^{k}x^{k,t}\right)\lambda _{k,t}\\\sum _{k=1}^{K}\sum _{t=1}^{T_{k}}\left(A^{k}x^{k,t}\right)\lambda _{k,t}=b\\\sum _{t=1}^{T_{k}}\lambda _{k,t}=1\;\;for\;\;k=1,...,K\\\lambda _{k,t}\geq 0\;\;for\;\;t=1,...,T_{k},\;k=1,...,K\end{matrix}}$

Where there is a column ${\begin{pmatrix}c^{k}x\\A^{k}x\\e_{k}\end{pmatrix}}$ for each $x$ ${\textstyle \in }$ ${\textstyle X^{k}}$. On the next steps of this method, we will use $\left\{\pi _{i}\right\}_{i=1}^{m}$ as the dual variables associated with the joint constraints, and $\left\{\mu _{k}\right\}_{k=1}^{K}$ as dual variables for the second set of constraints, known as convexity constraints. The idea is to solve the linear program by the primal simplex algorithm. However, the pricing step of choosing a column to enter the basis must be modified because of the enormous number of columns. Rather than pricing the columns one by one, the problem of finding a column with the largest reduced price is itself a set of K optimization problems.

Initialization: we suppose that subset of columns (at least one for each $k$) is available, providing a feasible Restricted Linear Programming Master Problem:

(RLPM) ${\begin{matrix}z^{LPM}=max{\tilde {c}}{\tilde {\lambda }}\\{\tilde {A}}{\tilde {\lambda }}=b\\{\tilde {\lambda }}\geq 0\end{matrix}}$

where ${\tilde {b}}={\begin{pmatrix}b\\1\\\end{pmatrix}}$, ${\tilde {A}}$ is generated by the available set of columns and ${\tilde {c}}{\tilde {\lambda }}$ are the corresponding costs and variables. Solving the RLPM gives an optimal primal solution ${\tilde {\lambda ^{*}}}$ and an optimal dual solution $\left(\pi ,\mu \right)\epsilon \;R^{m}\times R^{k}$

Primal feasibility: Any feasible solution of RLMP is feasible for LPM. In particular, ${\tilde {\lambda ^{*}}}$ is a feasible solution of LPM, and so ${\tilde {z}}^{LPM}={\tilde {c}}{\tilde {\lambda ^{*}}}=\sum _{i=1}^{m}\pi _{i}b_{i}+\sum _{k=1}^{K}\mu _{k}\leq z^{LPM}$

Optimality check for LPM: We need to check whether $\left(\pi ,\mu \right)$ is dual feasible for LPM. This involves checking for each column, that is for each $k$, and for each $x\;\epsilon \;X^{k}$ whether the reduced price $c^{k}x-\pi A^{k}x-\mu _{k}\leq 0$. Rather than examining each point separately, we treat all points in $X^{k}$ implicitly by solving an optimization subproblem:

$\zeta _{k}=max\left\{\left(c^{k}-\pi A^{k}\right)x-\mu _{k}\;:\;x\;\epsilon \;X^{k}\right\}.$

Stopping criteria: If $\zeta _{k}>0$ for $k=1,...,K$ the solution $\left(\pi ,\mu \right)$ is dual feasible for LPM, and so $z^{LPM}\leq \sum _{i=1}^{m}\pi _{i}b_{i}+\sum _{k=1}^{K}\mu _{k}$. As the value of the primal feasible solution ${\tilde {\lambda }}$ equals that of this upper bound, ${\tilde {\lambda }}$ is optimal for LPM.

Generating a new column: If $\zeta _{k}>0$ for some $k$, the column corresponding to the optimal solution ${\tilde {x}}^{k}$ of the subproblem has positive reduced price. Introducing the column ${\begin{pmatrix}c^{k}x\\A^{k}x\\e_{k}\end{pmatrix}}$ leads to a Restricted Linear Programming Master Problem that can be easily reoptimized (e.g., by the primal simplex algorithm)

Numerical example: The Cutting Stock problem

Suppose we want to solve a numerical example of the cutting stock problem, specifically a one-dimensional cutting stock problem.

Problem Overview

A company produces steel bars with diameter $45$ millimeters and length $33$ meters. The company also takes care of cutting the bars for their different customers, who each require different lengths. At the moment, the following demand forecast is expected and must be satisfied:

Pieces needed	Piece length(m)	Type of item
144	6	1
105	13.5	2
72	15	3
30	16.5	4
24	22.5	5

The objective is to establish what is the minimum number of steel bars that should be used to satisfy the total demand.

A possible model for the problem, proposed by Gilmore and Gomory in the 1960ies is the one below:

Sets

$K=\left\{1,2,3,4,5\right\}$: set of item types;

${\textstyle S}$: set of patterns (i.e., possible ways) that can be adopted to cut a given bar into portions of the need lengths.

Parameters

${\textstyle M}$: bar length (before the cutting process)

${\textstyle L_{k}}$: length of item ${\textstyle k}$ ${\textstyle \in }$ ${\textstyle K}$;

${\textstyle R_{s}}$ : number of pieces of type ${\textstyle k}$ ${\textstyle \in }$ ${\textstyle K}$ required;

${\textstyle N_{k,s}}$ : number of pieces of type ${\textstyle k}$ ${\textstyle \in }$ ${\textstyle K}$ in pattern ${\textstyle s}$ ${\textstyle \in }$ ${\textstyle S}$

Decision variables

${\textstyle Y_{s}}$ : number of bars that should be portioned using pattern ${\textstyle s}$ ${\textstyle \in }$ ${\textstyle S}$

Model

${\begin{matrix}\min(y)\sum _{s=1}^{S}y_{s}\\\ s.t.\sum _{k}N_{ks}y_{s}\geq J_{k}\forall k\in K\\y_{s}\in \mathrm {Z} _{+}\forall s\in S\end{matrix}}$

Solving the problem

The model assumes the availability of the set ${\textstyle K}$ and the parameters ${\textstyle N_{k,s}}$ . To generate this data, you would have to list all possible cutting patterns. However, the number of possible cutting patterns is a big number. This is why a direct implementation of the model above is not partical in real-world problems. In this case is when it makes sense to solve the continuous relaxation of the above model. This is because, in reality, the demand figures are so high that the number of bars to cut is also a large number, and therefore a good solution can be determined by rounding up to the next integer each variable $y_{s}$found by solving the continuous relaxation. In addition to that, the solution of the relaxed problem will become the starting point for the application of an exact solution method (for instance, the Branch-and Bound).

Key take-away: In the next steps of this example we will analyze how to solve the continuous relaxation of the model.

As a starting point, we need any feasible solution. Such a solution can be constructed as follows:

1. We consider any single-item cutting patterns, i.e., $\|K\|$ configurations, each containing ${\textstyle {\textstyle N_{k,s}}=\llcorner {\frac {W}{L_{k}}}\lrcorner }$ pieces of type $k$;
2. Set ${\textstyle {\textstyle y_{k}}=\llcorner {\frac {R_{s}}{N_{k,s}}}\lrcorner }$ for pattern $k$ (where pattern $k$ is the pattern containing only pieces of type $k$).

This solution could also be arrived to by applying the simplex method to the model (without integrality constraints), considering only the decision variables that correspond to the above single-item patterns:

${\begin{aligned}{\text{min}}&~~y_{1}+y_{2}+y_{3}+y_{4}+y_{5}\\{\text{s.t}}&~~15y_{1}\geq 144\\\ &~~6y_{2}\geq 105\\\ &~~6y_{3}\geq 72\\\ &~~6y_{4}\geq 30\\\ &~~3y_{5}\geq 24\\\ &~~y_{1},y_{2},y_{3},y_{4},y_{5}\geq 0\\\end{aligned}}$

In fact, if we solve this problem (for example, use CPLEX solver in GAMS) the solution is as below:

Y1	28.8
Y2	52.5
Y3	24
Y4	15
Y5	24

Next, a new possible pattern (number $6$) will be considered. This pattern contains only one piece of item type number $5$. So the question is if the new solution would remain optimal if this new pattern was allowed. Duality helps answer ths question. At every iteration of the simplex method, the outcome is a feasible basic solution (corresponding to some basis $B$) for the primal problem and a dual solution (the multipliers $u^{t}=c^{t}BB^{-1}$) that satisfy the complementary slackness conditions. (Note: the dual solution will be feasible only when the last iteration is reached)

The inclusion of new pattern Failed to parse (syntax error): {\displaystyle "6"} corresponds to including a new variable in the primal problem, with objective cost $1$ (as each time pattern $6$ is chosen, one bar is cut) and corresponding to the following column in the constraint matrix:

$D_{6}={\begin{bmatrix}\ 1\\\ 0\\\ 0\\\ 0\\\ 1\\\end{bmatrix}}$

These variables create a new dual constraint. We then have to check if this new constraint is violated by the current dual solution  (or in other words, if the reduced cost of the new variable with respect to basis $B$ is negative)

The new dual constraint is:$1\times u_{1}+0\times u_{2}+0\times u_{3}+0\times u_{4}+1\times u_{5}\leq 1$

The solution for the dual problem can be computed in different software packages, or by hand. The example below shows the solution obtained with GAMS for this example:

(Note the solution for the dual problem would be: $u=c_{T}^{B}B^{-1}$)

Dual variable	Variable value
D1	0.067
D2	0.167
D3	0.167
D4	0.167
D5	0.333

Since $0.2+1=1.2>1$, the new constraint is violated.

This means that the current primal solution (in which the new variable is $y_{6}=0$) may not be optimal anymore (although it is still feasible). The fact that the dual constraint is violated means the associated primal variable has negative reduced cost:

the norm of$c_{6}=c_{6}-u^{T}D_{6}=1-0.4=0.6$

To help us solve the problem, the next step is to let $y_{6}$ enter the basis. To do so, we modify the problem by inserting the new variable as below:

${\begin{aligned}{\text{min}}&~~y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}\\{\text{s.t}}&~~15y_{1}+y_{6}\geq 144\\\ &~~6y_{2}\geq 105\\\ &~~6y_{3}\geq 72\\\ &~~6y_{4}\geq 30\\\ &~~3y_{5}+y_{6}\geq 24\\\ &~~y_{1},y_{2},y_{3},y_{4},y_{5},y_{6}\geq 0\\\end{aligned}}$

If this problem is solved with the simplex method, the optimal solution is found, but restricted only to patterns $1$ to $6$. If a new pattern is available, a decision should be made whether this new pattern should be used or not by proceeding as above. However, the problem is how to find a pattern (i.e., a variable; i.e, a column of the matrix) whose reduced cost is negative (i.e., which will mean it is convenient to include it in the formulation). At this point one can notice that number of possible patterns exponentially large,and all the patterns are not even known explicitly. The question then is:

Given a basic optimal solution for the problem in which only some variables are included, how can we find (if any exists) a variable with negative reduced cost (i.e., a constraint violated by the current dual solution)?

This question can be transformed into an optimization problem: in order to see whether a variable with negative reduced cost exists, we can look for the minimum of the reduced costs of all possible variables and check whether this minimum is negative:

${\bar {c}}=1-u^{T}z$

Because every column of the constraint matrix corresponds to a cutting pattern, and every entry of the column says how many pieces of a certain type are in that pattern. In order for $z$ to be a possible column of the constraint matrix, the following condition must be satisfied:

${\textstyle {\begin{matrix}z_{k}\in \mathrm {Z} _{+}\forall k\in K\\\ \sum _{k}L_{k}z_{k}\leq M\end{matrix}}}$

And by so doing, it enables the conversion of the problem of finding a variable with negative reduced cost into the integer linear programming problem below:

${\begin{matrix}\min \ {\bar {c}}=1-sum_{k=1}^{K}u_{k}\times z_{k}\\\ s.t.\sum _{k}L_{k}z_{k}\leq M\\z_{k}\in \mathrm {Z} _{+}\forall k\in K\end{matrix}}$

which, in turn, would be equivalent to the below formulation (we just write the objective in maximization form and ignore the additive constant 1):

${\begin{matrix}\max \sum _{k=1}^{K}u_{k}\times z_{k}\\\ s.t.\sum _{k}L_{k}z_{k}\leq M\\z_{k}\in \mathrm {Z} _{+}\forall k\in K\end{matrix}}$

The coefficients $z_{k}$ of a column with negative reduced cost can be found by solving the above integer "knapsack" problem (which is a traditional type of problem that we find in integer programming).

In our example, if we start from the problem restricted to the five single-item patterns, the above problem reads as:

${\begin{aligned}{\text{min}}&~~0.067z_{1}+0.167z_{2}+0.167z_{3}+0.167z_{4}+z_{5}\\{\text{s.t}}&~~6z_{1}+13.5z_{2}+15z_{3}+16.5z_{4}+22.5z_{5}\leq 33\\\ &~~z_{1},z_{2},z_{3},z_{4},z_{5}\geq 0\\\end{aligned}}$

which has the following optimal solution: $z^{T}=[1\quad 0\quad 0\quad 0\quad 1]$

This matches the pattern we called $D6$, earlier on in this page.

Optimality test

If : ${\textstyle \sum _{k=1}^{K}z_{k}^{*}u_{k}^{*}\leq 1}$

then $y^{*}$ is an optimal solution of the full continuous relaxed problem (that is, including all patterns in ${\textstyle S}$)

If this condition is not true, we go ahead and update the master problem by including in ${\textstyle S^{'}}$ the pattern $\lambda$ defined by $N_{s,\lambda }$ (in practical terms this means that the column $y^{*}$ needs to be included in the constraint matrix) Then,go to Step1.

For this example we find that the optimality test is met as $0.4<1$ so we have have found an optimal solution of the relaxed continuous problem (if this was not the case we would have had to go back to Step 1 as described in the methodology section of this page)

$\sum _{k=1}^{K}z_{k}^{*}u_{k}^{*}=4\leq 1$

Algorithm discussion

The column generation subproblem is the critical part of the method is generating the new columns. It is not reasonable to compute the reduced costs of all variables $y_{s}$ for $s=1,...,S$, otherwise this procedure would reduce to the simplex method. In fact, n$n$ can be very large (as in the cutting-stock problem) or, for some reason, it might not be possible or convenient to enumerate all decision variables. It is then necessary to study a specific column generation algorithm for each problem; only if such an algorithm exists (and is efficient), the method can be fully developed. In the one-dimensional cutting stock problem, we transformed the column generation subproblem into an easily solvable integer linear programming problem. In other cases, the computational effort required to solve the subproblem may be so high as to make the full procedure unpractical.

Applications

As previously mentioned, column generation techniques are most relevant when the problem that we are trying to solve has a high ratio of number of variables with respect to the number of constraints. As such some common applications are:

• Bandwith packing
• Bus driver scheduling
• Generally, column generation algorithms are used for large delivery networks, often in combination with other methods, helping to implement real-time solutions for on-demand logistics.We discuss a supply chain scheduling application below.

Bandwidth packing

The objective of this problem is to allocate bandwidth in a telecommunications network to maximize total revenue. The routing of a set of traffic demands between different users is to be decided, taking into account the capacity of the network arcs and the fact that the traffic between each pair of users cannot be split The problem can be formulated as an integer programming problem and the linear programming relaxation solved using column generation and the simplex algorithm. A branch and bound procedure which branches upon a particular path is used in this particular paper that looks into bandwidth routing, to solve the IP. The column generation algorithm greatly reduces the complexity of this problem.

Bus driver scheduling

Bus driver scheduling aims to find the minimum number of bus drivers to cover a published timetable of a bus company. When scheduling bus drivers, contractual working rules must be enforced, thus complicating the problem. In this research, we develop a column generation algorithm that decomposes this complicated problem into a master problem and a series of pricing subproblems. The master problem selects optimal duties from a set of known feasible duties, and the pricing subproblem augments the feasible duty set to improve the solution obtained in the master problem. The proposed algorithm is empirically applied to the realistic problems of several bus companies. The numerical results show that the proposed column generation algorithm can solve real‐world problems and obtain bus driver schedules that are better than those developed and used by the bus companies. Copyright © 2016 John Wiley & Sons, Ltd.

Supply Chain scheduling problem

A typical application is where we consider the problem of scheduling a set of shipments between different nodes of a supply chain network. Each shipment has a fixed departure time, as well as an origin and a destination node, which, combined, determine the duration of the associated trip. The aim is to schedule as many shipments as possible, while also minimizing the number of vehicles utilized for this purpose. This problem can be formulated by an integer programming model and an associated branch and price solution algorithm. The optimal solution to the LP relaxation of the problem can be obtained through column generation, solving the linear program a huge number of variables, without explicitly considering all of them. In the context of this application, the proposed methodology utilizes a master problem that schedules the maximum possible number of shipments using only a small set of vehicle-routes, and a column generation (colgen) sub-problem that generates cost-effective vehicle-routes which are fed into the master problem. After finding the optimal solution to the LP relaxation of the problem, the algorithm branches on the fractional decision variables (vehicle-routes), in order to reach the optimal integer solution.

Conclusions

Column generation is a way of beginning with a small, manageable parts of a problem (specifically, a few of the variables), solving that part, analyzing that partial solution to discover the next part of the problem (specifically, one or more variables) to add to the model, and then resolving the extended model. Column generation repeats the algorithm steps until it achieves an optimal solution to the entire problem.

More formally, column generation is a way of solving a linear programming problem that adds columns (corresponding to constrained variables) during the pricing phase of the simplex method of solving the problem. Generating a column in the primal simplex formulation of a linear programming problem corresponds to adding a constraint in its dual formulation.

Column generation provides an advantage to the simplex method as the solvers (when computing the solution with software) will not need to access all the variables of the problem simultaneously. In fact, a solver could begin work with only the basis (a particular subset of the constrained variables) and then use reduced cost to decide which other variables to access as needed.