# The cutting stock problem: part 2, solving with column generation

In the previous post, we explored a well-known integer optimization situation in manufacturing, the cutting stock problem. After some details on the decisions, constraints and objectives, we implemented a naive model in JuMP.

One key thing to notice is the explosion of number of variables and constraints and the fact that relaxed solutions (without constraining variables to be integers) are very far from actual feasible solutions.

We will now use an other way of formulating the problem, using a problem decomposition and an associated solution method (column generation).

## Re-stating the cutting stock problem

Remember we used two decisions: $Y_i$ stating if the big roll $i$ is used and $X_{ij}$ expressing the number of cuts $j$ made in the roll $i$. To minimize the number of rolls, it makes sense to put as many small cuts as possible on a big roll. We could therefore identify saturating patterns, that is, a combination of small cuts fitting on a big roll, such that no additional cut can be placed, and then find the smallest combination of the pattern satisfying the demand.

One problem remains: it is impossible to compute, or even to store in memory all patterns, their number is exponentially big with the number of cuts, so we will try to find the best patterns and re-solve the problem, using the fact that not all possible patterns will be necessary.

This is exactly what the Dantzig-Wolfe decomposition does, it splits the problem into a Master Problem MP and a sub-problem SP.

• The Master Problem, provided a set of patterns, will find the best combination satisfying the demand.
• The sub-problem, given an “importance” of each cut provided by the master problem, will find the best cuts to put on a new pattern.

This is an iterative process, we can start with some naive patterns we can think of, compute an initial solution for the master problem, which will be feasible but not optimal, move on to the sub-problem to try to find a new pattern (or column in the optimization jargon, hence the term of column generation).

How do we define the “importance” of a cut $j$? The value of the dual variable associated with this constraint will tell us that. This is not a lecture in duality theory, math-eager readers can check out further documentation on the cutting stock problem and duality in linear optimization.

Moreover, we are going to add one element to our model: excess cuts can be sold at a price $P_j$, so that we can optimize by minimizing the net cost (production cost of the big rolls minus the revenue from excess cuts).

## New formulation

Again, we are going to formulate first possible decisions and then constraints on these decisions for the new version of the problem.

### Decisions

At the master problem level, given a pattern $p$, the decision will be $\theta_p$ (theta, yes Greek letters are awesome), the number of big rolls which will be used with this pattern. $\theta_p$ is a positive integer.

The decision at the sub-problem level will be to find how many of each cut $j$ to fit onto one big roll, $a_j$.

For a pattern $p$, the number of times a cut $j$ appears is given by $a_{jp}$.

### Constraints

The big roll size constraint is kept in the sub-problem, a pattern built has to respect this constraint: $$\sum_j a_{j} \cdot W_j \leq L$$

The demand $D_j$ is met with all rolls of each pattern so it is kept at the master level. The number of cuts of type $j$ produced is the sum of the number of this cut on each patterns times the number of the pattern in a solution:

$$NumCuts_j = \sum_p a_{jp} \cdot \theta_p \geq D_j$$

### Objective formulation

At the master problem, we minimize the number of rolls, which is simply: $$\sum_{p} \theta_p$$

At the sub-problem, we are trying to maximize the gain associated with the need for the demand + the residual price of the cuts. If we can find a worth using producing compared to its production cost, it is added.

## Implementation

As before, we will formulate the master and sub-problem using Julia with JuMP. Again, we use the Clp and Cbc open-source solvers. We read the problem data (prices, sizes, demand) from a JSON file.

using JuMP
using Cbc: CbcSolver
using Clp: ClpSolver
import JSON

const res = open("data0.json", "r") do f
JSON.Parser.parse(data)
end

const maxwidth = res["maxwidth"]
const cost = res["cost"]
const prices = Float64.(res["prices"])
const widths = Float64.(res["widths"])
const demand = Float64.(res["demand"])
const nwidths = length(prices)


cost is the production cost of a big roll.

### Sub-problem

The subproblem is a function taking reduced costs of each cut and maximizing the utility of the pattern it creates:

"""
subproblem tries to find the best feasible pattern
maximizing reduced cost and respecting max roll width
corresponding to a multiple-item knapsack
"""
function subproblem(reduced_costs, sizes, maxcapacity)
submodel = Model(solver = CbcSolver())
n = length(reduced_costs)
xs = @variable(submodel, xs[1:n] >= 0, Int)
@constraint(submodel, sum(xs. * sizes) <= maxcapacity)
@objective(submodel, Max, sum(xs. * reduced_costs))
solve(submodel)
return round.(Int,getvalue(xs)), round(Int,getobjectivevalue(submodel))
end

### Initial master problem

We saw that the master problem finds a solution and then requires a new pattern from the sub-problem. This is therefore preferable to start from an initial feasible, otherwise we fall into a special case we’re not discussing here. One initial solution would be to build one pattern per cut, with as many cuts as we can, which is $floor(L/w_j)$.

function init_master(maxwidth, widths, rollcost, demand, prices)
n = length(widths)
ncols = length(widths)
patterns = spzeros(UInt16,n,ncols)
for i in 1:n
patterns[i,i] = min(floor(Int,maxwidth/widths[i]),round(Int,demand[i]))
end
m = Model(solver = ClpSolver())
θ = @variable(m, θ[1:ncols] >= 0)
@objective(m, Min,
sum(θ[p] * (rollcost - sum(patterns[j,p] * prices[j] for j=1:n)) for p in 1:ncols)
)
@constraint(m, demand_satisfaction[j=1:n], sum(patterns[j,p] * θ[p] for p in 1:ncols)>=demand[j])
if solve(m) != :Optimal
warn("No optimal")
end
return (m, getvalue(θ), demand_satisfaction, patterns)
end

We can compute the reduced costs from the dual values associated with the demand and the prices of cuts

# getting the model and values
(m, θ, demand_satisfaction, patterns) = init_master(maxwidth, widths, cost, demand, prices);

# compute reduced costs
reduced_costs = getdual(demand_satisfaction)+prices;

# ask sub-problem for new pattern
newcol, newobj = subproblem(reduced_costs, widths, maxwidth)

### Putting it all together

We can now build a column generation function putting all elements together and performing the main iteration:

function column_generation(maxwidth, widths, rollcost, demand, prices; maxcols = 5000)
(m, θ, demand_satisfaction, patterns) = init_master(maxwidth, widths, rollcost, demand, prices)
ncols = nwidths
while ncols <= maxcols
reduced_costs = getdual(demand_satisfaction) + prices
newcol, newobj = subproblem(reduced_costs, widths, maxwidth)
netcost = cost - sum(newcol[j] * (getdual(demand_satisfaction)[j]+prices[j]) for j in 1:nwidths)
println("New reduced cost: $netcost") if netcost >= 0 return (:Optimal, patterns, getvalue(θ)) end patterns = hcat(patterns, newcol) ncols += 1 m = Model(solver = ClpSolver()) θ = @variable(m, θ[1:ncols] >= 0) @objective(m, Min, sum(θ[p] * (rollcost - sum(patterns[j,p] * prices[j] for j=1:nwidths)) for p in 1:ncols) ) @constraint(m, demand_satisfaction[j=1:nwidths], sum(patterns[j,p] * θ[p] for p in 1:ncols)>=demand[j]) if solve(m) != :Optimal warn("No optimal") return (status(m), patterns, getvalue(θ)) end end return (:NotFound, patterns, :NoVariable) end We’ve printed information along the computation to see what’s going on more clearly, now launching it: status, patterns, θ = column_generation(maxwidth, widths, cost, demand, prices, maxcols = 500); New reduced cost: -443.18181818181824 New reduced cost: -375.0 New reduced cost: -264.0 New reduced cost: -250.0 New reduced cost: -187.5 New reduced cost: -150.0 New reduced cost: -150.0 New reduced cost: -107.14285714285711 New reduced cost: -97.5 New reduced cost: -107.14285714285734 New reduced cost: -72.0 New reduced cost: -53.571428571428555 New reduced cost: -53.125 New reduced cost: -50.0 New reduced cost: -43.40625 New reduced cost: -36.0 New reduced cost: -34.625 New reduced cost: -41.5 New reduced cost: -21.8515625 New reduced cost: -22.159090909090878 New reduced cost: -20.625 New reduced cost: -16.304347826086314 New reduced cost: -16.304347826086996 New reduced cost: -20.310344827586277 New reduced cost: -18.0 New reduced cost: -8.837209302325732 New reduced cost: -6.060606060606119 New reduced cost: 0.0 While the cost of a new pattern is negative, we can add it to the master and keep running. This seems to make sense. Now, one thing to note, we have not yet specified the integrality constraints, meaning that we don’t have integer number of patterns. We can see that on the$\theta\$ variable:

println(θ)
[0.0, 0.0, 0.0, ... 70.0, 0.0, 0.0, 0.0, 12.56, 46.86, 0.0, 0.0, 0.0, 0.0,
3.98, 0.0, 0.0, 21.5, 5.0, 31.12, 61.12, 33.58, 0.0, 0.0, 32.2, 44.0,
46.88, 19.0, 1.88, 16.42]
println(sum(θ))
446.1000000000001

We saw in the last post that the problem without integrality constraints is a relaxation and therefore, can only yield a better result. This means that we cannot have an integer solution using 446 big rolls or less, the minimum will be 447 rolls. Let’s solve the problem with the same patterns, but adding the integrality:

# compute initial integer solution:
# take worse case from linear solution, round up
intial_integer = ceil.(Int,θ);

"""
From patterns built in the column generation phase, find an integer solution
"""function branched_model(patterns, demand, rollcost, prices; npatts = size(patterns)[2], initial_point = zeros(Int,npatts))
npatts = size(patterns)[2]
m = Model(solver = CbcSolver())
θ = @variable(m, θ[p = 1:npatts] >= 0, Int, start = initial_point[p])
@objective(m, Min,
sum(θ[p] * (rollcost - sum(patterns[j,p] * prices[j] for j=1:nwidths)) for p in 1:npatts)
)
@constraint(m, demand_satisfaction[j=1:nwidths], sum(θ[p] * patterns[j,p] for p in 1:npatts) >= demand[j])
status = solve(m)
return (status, round.(Int,(getvalue(θ))))
end

Let’s see what the results look like:

status, θ_final = branched_model(patterns, demand, cost, prices; initial_point = intial_integer)
println(status)
:Optimal
println(sum(θ_final))
447

Given that we cannot do better than 447, we know we have the optimal number of rolls.

## Conclusion

After seeing what a mess integer problems can be in the first part, we used a powerful technique called Dantzig-Wolfe decomposition, splitting the problem into master and sub-problem, each handling a subset of the constraints.

Column generation is a technique making this decomposition usable in practice, by adding only one or few columns (patterns) at each iteration, we avoid an exponentially growing number of variables. The fact that JuMP is built as an embedded Domain Specific Language in Julia makes it a lot easier to specify problems and play around them. Most optimization specific modeling languages are built around declarative features and get messy very quickly when introducing some logic (like column generation iterations). Developers could relate this technique to lazy value computation: we know all values are there, but we just compute them whenever needed.

Hope you enjoyed reading this second post on the cutting stock problem. A Jupyter notebook summing up all code snippets can be found at this repository, feel free to ping me for feedback.

## Code and citation

Found this post useful for your work? The corresponding repository is available on GitHub, consider citing it using the following DOI 10.5281/zenodo.3329389, the BibTeX entry is available on Zenodo.

### Note on performance

The column generation approach we just saw scales well to huge problems, but this particular implementation can feel a bit slow at first. One recommended thing is to do in such case is “warm-starting” the solver: give it a good initial solution to start from. Since we built both the master and subproblem as stateless functions, the model is being re-built from scratch each time. The advantage is that any solver can be used, since some of them don’t support warm starts.

Thanks to Aristide for his very sharp ideas and views on this article which contributed to its improvement!

##### Mathieu Besançon
###### Researcher in mathematical optimization

Mathematical optimization, scientific programming and related.