 # Tackling the cutting stock problem: part 1, problem exploration Integer optimization often feels weird (at least to me). Simple reformulations of a (mixed) integer optimization problem (MIP) can make it way easier to solve. We’re going to explore one well-known example of such integer problem in two blog posts. This first part introduces the problem and develops a naive solution. We’re going to see why it’s complex to solve and why this formulation does not scale.

In a second post, we will see a reformulation of the problem which makes it easier to solve and scales to bigger instances.

## Integer optimization reminder

An optimization problem takes three components: decisions variables $x$, a set of constraints telling you if a decision is feasible or not and a cost function $c(x)$ giving a total cost of a decision. Optimization is a domain of applied mathematics consisting in finding the best feasible decision for a problem. Lots of decision problems come with integrality constraints: if $x$ is the decision, then it can only take integer values 0,1,2… or even only binary values ${0,1}$. Think of problems involving number of units produced for a good, yes/no decisions, etc… If a problem has lots of variables, naive enumerations of feasible solutions becomes impossible: even problems with 50 variables can make your average laptop crash.

## The cutting stock problem

The problem is not new and has been given quite some thoughts because of its different industrial applications, it has been one of the first applications of the column generation method we are going to use. The key elements of the problems are: given some large rolls (metal, paper or other), we need to cut smaller portions of given lengths to satisfy a demand for the different small lengths. Find more details here. A small instance might be: given rolls of size $100cm$, we want to cut at least 7 rolls of size $12cm$ and 9 rolls of size $29cm$. The objective is to minimize the number of big rolls to satisfy this demand.

How do we formulate this mathematically?

### Decisions

$Y_i$ is a binary decision indicating if we use the big roll number $i$. $X_{ij}$ is an integer giving the number of times we cut a small roll $j$ in the big roll $i$.

### Constraints

$Y$ are binary variables, $X$ are integer. Now the less trivial constraints:

• Demand satisfaction constraint: the sum over all $i$ big rolls of the cut $j$ has to satisfy the demand for that cut: $$\sum_{i} X_{ij} \geq D_j$$

For the two-cut example with the demand of $7 \times 12cm$ and $9 \times 29cm$, let’s suppose we have 10 big rolls $i \in {1…10}$, the demand for the first 12cm cut is 7 cuts, the number of cuts of this size produced is: $$\sum_i X_{i1} = X_{1,1} + X_{2,1} + … + X_{10,1}$$

This total must at least match the demand, so: $$X_{1,1} + X_{2,1} + … + X_{10,1} \geq 7$$

• Roll size constraint: if a roll $i$ is used, we cannot fit more width onto it than its total width: $$\sum_{j} X_{ij} \cdot W_j \leq L \cdot Y_i$$

For the two-cut example with the demand of $7 \times 12cm$ and $9 \times 29cm$, let’s suppose we have one roll $i$:

• If $Y_i = 0$, the roll size constraint becomes:

$$\sum_{j} X_{ij} \cdot W_j = 12 \cdot X_{i1} + 29 \cdot X_{i2} \leq 0$$

The only feasible solution for this roll $i$ is ($X_{i1} = 0,X_{i2} = 0$).

• If $Y_i = 1$, the roll size constraint becomes: $$12 \cdot X_{i1} + 29 \cdot X_{i2} \leq 100$$

Which means we can fit as many cuts as the roll size allows for.

## A first naive implementation

Let’s first import the necessary packages: we’re using JuMP as a modeling tool, which is an optimization-specific language embedded in Julia (compare it to AMPL, GAMS, Pyomo, PuLP). As I consider it an embedded language, I’ll do a full import into my namespace with using (unlike what I usually do with packages). We also use Cbc, an open-source solver for integer problems from the Coin-OR suite.

using JuMP
using Cbc: CbcSolver

We can define our optimization problem within a function taking the parameters of the cutting stock problem, namely a maxwidth of the big rolls, scalar assuming all of them have the same width, a widths vector, one element for each cut size $j$ and a demand vector, again, one for each cut size.

function cutting_stock_model(maxwidth, widths, demand, N = sum(demand))
# Define the JuMP model
m = Model(solver = CbcSolver())
# define the two groups of variables over their respective indices
Y = @variable(m, Y[1:N],Bin)
X = @variable(m, X[i=1:N,j=1:length(widths)],Int)
# define both constraints and objective
demand_satisfac = @constraint(m, [j=1:length(widths)],
sum(X[i,j] for i in 1:N) >= demand[j]
)
roll_size_const = @constraint(m, [i=1:N],
sum(X[i,j] * widths[j] for j in 1:length(widths)) <= Y[i] * maxwidth
)
@objective(m, Min, sum(Y[i] for i in 1:N))
# return the model formulation to solve later
return m
end

Here $N$ has to be an upper bound on the number of big rolls to use, otherwise the problem will be infeasible (not enough big rolls to find a solution satisfying the demand). An initial naive value for this could be the total demand, after all one small cut per roll can be considered a worst-case solution.

Note that we don’t call solve on the model yet, the function simply builds the model, this will help us see how it evolves with various entry parameters. In Julia REPL, or by using the @show macro, we can have more details on the model. Using println(m) instead of @show will build a mathematical formulation of the model in a LateX-like style, which can be valuable to ensure your implementation matches the initial formulation.

julia> println(cutting_stock_model(100, [12,10], [3,4]))
Min Y + Y + Y + Y + Y + Y + Y
Subject to
X[1,1] + X[2,1] + X[3,1] + X[4,1] + X[5,1] + X[6,1] + X[7,1] ≥ 3
X[1,2] + X[2,2] + X[3,2] + X[4,2] + X[5,2] + X[6,2] + X[7,2] ≥ 4
12 X[1,1] + 10 X[1,2] - 100 Y ≤ 0
12 X[2,1] + 10 X[2,2] - 100 Y ≤ 0
12 X[3,1] + 10 X[3,2] - 100 Y ≤ 0
12 X[4,1] + 10 X[4,2] - 100 Y ≤ 0
12 X[5,1] + 10 X[5,2] - 100 Y ≤ 0
12 X[6,1] + 10 X[6,2] - 100 Y ≤ 0
12 X[7,1] + 10 X[7,2] - 100 Y ≤ 0
Y[i] ∈ {0,1} ∀ i ∈ {1,2,…,6,7}
X[i,j], integer, ∀ i ∈ {1,2,…,6,7}, j ∈ {1,2}

Let’s see what the model looks like for different instances:

julia> cutting_stock_model(100, [12,10], [85,97], 200)
(Minimization problem with:
* 602 linear constraints
* 600 variables: 200 binary, 400 integer
Solver is CbcMathProg,
X[i,j], integer, ∀ i ∈ {1,2,…,199,200}, j ∈ {1,2},
Y[i] ∈ {0,1} ∀ i ∈ {1,2,…,199,200})

julia> cutting_stock_model(100, [12,10,25], [85,97,52], 300)
(Minimization problem with:
* 1203 linear constraints
* 1200 variables: 300 binary, 900 integer
Solver is CbcMathProg,
X[i,j], integer,∀ i ∈ {1,2,…,299,300}, j ∈ {1,2,3},
Y[i] ∈ {0,1} ∀ i ∈ {1,2,…,299,300})

julia> cutting_stock_model(100, [12,10,25,40,30,41], [85,97,52,63,77,31], 500)
(Minimization problem with:
* 3506 linear constraints
* 3500 variables: 500 binary, 3000 integer
Solver is CbcMathProg,
X[i,j], integer, ∀ i ∈ {1,2,…,499,500}, j ∈ {1,2,…,5,6},
Y[i] ∈ {0,1} ∀ i ∈ {1,2,…,499,500})

We see the number of variables and constraints explode as we add more possible cut sizes. More precisely:

• Number of variables: $size(X) + size(Y) = Nrolls \cdot Ncuts + Nrolls$
• Number of constraints: $size(DemandConstr) + size(WidthConstr) = Ncuts + Nrolls$

Without going into details on the solving process, two things make the problem difficult to solve:

1. Symmetry: if we place cuts on a roll $Y_1$ and leave another $Y_2$ unused, the resulting solution is concretely the same as using $Y_2$ and leaving $Y_1$ unused.
2. Bad relaxation: integer solvers mostly work by solving a “relaxed” version of the problem without the integrality constraint, and then iteratively restricting the problem to find the best integer solution. If the relaxed version of the problem yields solutions far away from an integer one, the solver will have more work to get there.

Difficulty (1) is pretty intuitive, but we could get some insight on (2). Let’s define our relaxed problem. We’re going to use the Clp solver, which will solve the same problem, but without the Int restriction for $X$ nor the Bin restriction for $Y$:

function relaxed_cutting_stock(maxwidth, widths, demand, N = sum(demand))
m = Model(solver = ClpSolver())
Y = @variable(m, 0 <= Y[1:N] <= 1)
X = @variable(m, X[1:N,1:length(widths)] >= 0)
demand_satisfac = @constraint(m, [j=1:length(widths)], sum(X[i,j] for i in 1:N) >= demand[j])
roll_size_const = @constraint(m, [i=1:N], sum(X[i,j] * widths[j] for j in 1:length(widths)) <= Y[i] * maxwidth)
@objective(m, Min, sum(Y[i] for i in 1:N))
return (m,Y,X)
end

Let’s see the results:

julia> res = [(i,getvalue(Y[i])) for i in 1:N if getvalue(Y[i]) ≉ 0]
33-element Array{Tuple{Int64,Float64},1}:
(1, 1.0)
(2, 1.0)
(3, 1.0)
(4, 1.0)
(5, 1.0)
(6, 1.0)
(7, 1.0)
(8, 1.0)
(9, 1.0)
(10, 1.0)
(11, 1.0)
(12, 1.0)
(13, 1.0)
(14, 1.0)
(15, 1.0)
(16, 1.0)
(17, 1.0)
(18, 1.0)
(19, 1.0)
(20, 1.0)
(21, 1.0)
(22, 1.0)
(23, 1.0)
(24, 1.0)
(25, 1.0)
(26, 1.0)
(27, 1.0)
(28, 1.0)
(29, 1.0)
(30, 1.0)
(31, 1.0)
(32, 0.9)
(84, 1.0)

idxs = [i for (i,_ ) in res]
julia> [getvalue(X)[i,:] for i in idxs]
33-element Array{Array{Float64,1},1}:
[0.0, 7.0, 1.2]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 10.0, 0.0]
[0.0, 10.0, 0.0]
[0.0, 0.0, 4.0]
[0.0, 10.0, 0.0]
[0.0, 10.0, 0.0]
[0.0, 10.0, 0.0]
[0.0, 10.0, 0.0]
[0.0, 10.0, 0.0]
[0.0, 10.0, 0.0]
[0.0, 10.0, 0.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[0.0, 0.0, 4.0]
[8.0, 0.0, 0.16]
[8.0, 0.0, 0.16]
[8.0, 0.0, 0.16]
[8.0, 0.0, 0.16]
[8.0, 0.0, 0.16]
[8.0, 0.0, 0.16]
[8.0, 0.0, 0.16]
[8.0, 0.0, 0.16]
[5.8, 0.0, 1.216]
[7.2, 0.0, 0.144]
[8.0, 0.0, 0.16]

We notice the $Y$ variables are overall pretty saturated and almost integer, but the $X$ variables are highly fractional: the linear cuts are divided such that they fit perfectly the big rolls. This will make the variable hard to get to an integer solution.

## Conclusion

This was a quick intro to the cutting stock problem to get a grasp of its structure and difficulty, the goal was not to get too technical and keep a broad target audience.

Hope you enjoyed it, if that’s the case, I’ll see you on the next article, we’ll implement a column generation algorithm from scratch to solve it. If you have any question/remarks, feel free to get in touch.

## 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.

## Thanks

Special thanks to Soham and Aristide for their feedback, these helped me a great deal simplify the structure and add details and explanations where needed. ##### Mathieu Besançon
###### Postdoctoral researcher, mathematical optimization

Mathematical optimization, scientific programming and related.