# Column generation

This tutorial describes how to implement the Cutting stock problem in JuMP using column generation. It uses the following packages:

```
using JuMP
import GLPK
import SparseArrays
```

## Mathematical formulation

The cutting stock problem is about cutting large rolls of paper into smaller pieces. There is a demand different sizes of pieces to meet, and all large rolls have the same width. The goal is to meet the demand while maximizing the total profit.

We denote the set of possible sized pieces that a roll can be cut into by $i\in 1,\ldots,I$. Each piece $i$ has a width, $w_i$, and a demand, $d_i$. The width of the large roll is $W$.

Here's the data that we are going to use in this tutorial:

```
struct Piece
w::Float64
d::Int
end
struct Data
pieces::Vector{Piece}
W::Float64
end
function Base.show(io::IO, d::Data)
println(io, "Data for the cutting stock problem:")
println(io, " W = $(d.W)")
println(io, "with pieces:")
println(io, " i w_i d_i")
println(io, " ------------")
for (i, p) in enumerate(d.pieces)
println(io, lpad(i, 4), " ", lpad(p.w, 5), " ", lpad(p.d, 3))
end
return
end
function get_data()
data = [
75.0 38
75.0 44
75.0 30
75.0 41
75.0 36
53.8 33
53.0 36
51.0 41
50.2 35
32.2 37
30.8 44
29.8 49
20.1 37
16.2 36
14.5 42
11.0 33
8.6 47
8.2 35
6.6 49
5.1 42
]
return Data([Piece(data[i, 1], data[i, 2]) for i in 1:size(data, 1)], 100.0)
end
data = get_data()
```

```
Data for the cutting stock problem:
W = 100.0
with pieces:
i w_i d_i
------------
1 75.0 38
2 75.0 44
3 75.0 30
4 75.0 41
5 75.0 36
6 53.8 33
7 53.0 36
8 51.0 41
9 50.2 35
10 32.2 37
11 30.8 44
12 29.8 49
13 20.1 37
14 16.2 36
15 14.5 42
16 11.0 33
17 8.6 47
18 8.2 35
19 6.6 49
20 5.1 42
```

To formulate the cutting stock problem as a mixed-integer linear program, we assume that there is a set of large rolls $j=1,\ldots,J$ to use. Then, we introduce two classes of decision variables:

- $x_{ij} \ge 0, \text{integer}, \forall i=1,\ldots,I, j=1,\ldots,J$
- $y_{j} \in \{0, 1\} \forall j=1,\ldots,J.$

$y_j$ is a binary variable that indicates if we use roll $j$, and $x_{ij}$ counts how many pieces of size $i$ that we cut from roll $j$.

Our mixed-integer linear program is therefore:

\[\begin{align} \min & \sum\limits_{j=1}^J y_j \\ \;\;\text{s.t.} & \sum\limits_{i=1}^N w_i x_{ij} \le W y_j & \forall j=1,\ldots,J \\ & \sum\limits_{j=1}^J x_{ij} \ge d_i & \forall i=1,\ldots,I \\ & x_{ij} \ge 0 & \forall i=1,\ldots,N, j=1,\ldots,J \\ & x_{ij} \in \mathbb{Z} & \forall i=1,\ldots,I, j=1,\ldots,J \\ & y_{j} \in \{0, 1\} & \forall j=1,\ldots,J \\ \end{align}\]

The objective is to minimze the number of rolls that we use, and the two constraints ensure that we respect the total width of each large roll and that we satisfy demand exactly.

```
I = length(data.pieces)
J = 1000 # Some large number
model = Model(GLPK.Optimizer)
@variable(model, x[1:I, 1:J] >= 0, Int)
@variable(model, y[1:J], Bin)
@constraint(
model,
[j in 1:J],
sum(data.pieces[i].w * x[i, j] for i in 1:I) <= data.W * y[j],
)
@constraint(model, [i in 1:I], sum(x[i, j] for j in 1:J) >= data.pieces[i].d)
@objective(model, Min, sum(y[j] for j in 1:J))
model
```

```
A JuMP Model
Minimization problem with:
Variables: 21000
Objective function type: AffExpr
`AffExpr`-in-`MathOptInterface.GreaterThan{Float64}`: 20 constraints
`AffExpr`-in-`MathOptInterface.LessThan{Float64}`: 1000 constraints
`VariableRef`-in-`MathOptInterface.GreaterThan{Float64}`: 20000 constraints
`VariableRef`-in-`MathOptInterface.Integer`: 20000 constraints
`VariableRef`-in-`MathOptInterface.ZeroOne`: 1000 constraints
Model mode: AUTOMATIC
CachingOptimizer state: EMPTY_OPTIMIZER
Solver name: GLPK
Names registered in the model: x, y
```

Unfortunately, we won't attempt to solve this formulation because it takes a very long time to solve. (Try it and see.)

`# optimize!(model)`

However, there is a formulation that solves much faster, and that is to use a column generation scheme.

## Column generation theory

The key insight for column generation is to recognize that the $x$ variables above encode *cutting patterns*. For example, if we look only at the roll $j=1$, then feasible solutions are:

- $x_{1,1} = 1$, $x_{13,1} = 1$ and all the rest $0$, which is 1 roll of piece #1 and 1 roll of piece #13
- $x_{1,20} = 19$ and all the rest $0$, which is 19 rolls of piece #20.

Cutting patterns like $x_{1,1} = 1$ and $x_{2,1} = 1$ are infeasible because the combined length is greater than $W$.

Since there are a finite number of ways that we could cut a roll into a valid cutting pattern, we can create a set of all possible cutting patterns $p = 1,\ldots,P$, with data $a_{i,p}$ indicating how many pieces of size $i$ we cut in pattern $p$. Then, we can formulate our mixed-integer linear program as:

\[\begin{align} \min & \sum\limits_{p=1}^P x_p \\ \;\;\text{s.t.} & \sum\limits_{p=1}^P a_{ip} x_p \ge d_i & \forall i=1,\ldots,I \\ & x_{p} \ge 0 & \forall p=1,\ldots,P \\ & x_{p} \in \mathbb{Z} & \forall p=1,\ldots,P \end{align}\]

Unfortunately, there will be a very large number of these patterns, so it is often intractable to enumerate all columns $p=1,\ldots,P$.

Column generation is an iterative algorithm that starts with a small set of initial patterns, and then cleverly chooses new columns to add to the main MILP so that we find the optimal solution without having to enumerate every column.

## Choosing new columns

For now we assume that we have our mixed-integer linear program with a subset of the columns. If we have all of the columns that appear in an optimal solution then we are done. Otherwise, how do we choose a new column that leads to an improved solution?

Column generation chooses a new column by relaxing the integrality constraint on $x$ and looking at the dual variable $\pi_i$ associated with demand constraint $i$.

Using the economic interpretation of the dual variable, we can say that a one unit increase in demand for piece $i$ will cost an extra $\pi_i$ rolls. Alternatively, we can say that a one unit increase in the left-hand side (for example, due to a new cutting pattern) will *save* us $\pi_i$ rolls. Therefore, we want a new column that maximizes the savings associated with the dual variables, while respecting the total width of the roll:

\[\begin{align} \max & \sum\limits_{i=1}^I \pi_i y_i \\ \;\;\text{s.t.} & \sum\limits_{i=1}^I w_i y_{i} \le W \\ & y_{i} \ge 0 & \forall i=1,\ldots,I \\ & y_{i} \in \mathbb{Z} & \forall i=1,\ldots,I \\ \end{align}\]

If this problem, called the *pricing problem*, has an objective value greater than $1$, then we estimate than adding `y`

as the coefficients of a new column will decrease the objective by more than the cost of an extra roll.

Here is code to solve the pricing problem:

```
function solve_pricing(data::Data, π::Vector{Float64})
I = length(π)
model = Model(GLPK.Optimizer)
set_silent(model)
@variable(model, y[1:I] >= 0, Int)
@constraint(model, sum(data.pieces[i].w * y[i] for i in 1:I) <= data.W)
@objective(model, Max, sum(π[i] * y[i] for i in 1:I))
optimize!(model)
if objective_value(model) > 1
return round.(Int, value.(y))
end
return nothing
end
```

`solve_pricing (generic function with 1 method)`

## Choosing the initial set of patterns

For the initial set of patterns, we create a trivial cutting pattern which cuts as many pieces of size $i$ as will fit, or the amount demanded, whichever is smaller.

```
patterns = Vector{Int}[]
for i in 1:I
pattern = zeros(Int, I)
pattern[i] = floor(Int, min(data.W / data.pieces[i].w, data.pieces[i].d))
push!(patterns, pattern)
end
P = length(patterns)
```

`20`

We can visualize the patterns by looking at the sparse matrix of the non-zeros:

`SparseArrays.sparse(hcat(patterns...))`

```
20×20 SparseArrays.SparseMatrixCSC{Int64, Int64} with 20 stored entries:
⠑⢄⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠑⢄⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠑⢄⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠑⢄⠀⠀
⠀⠀⠀⠀⠀⠀⠀⠀⠑⢄
```

## Solving the problem

First, we create our initial linear program:

```
model = Model(GLPK.Optimizer)
set_silent(model)
@variable(model, x[1:P] >= 0)
@objective(model, Min, sum(x))
@constraint(model, demand[i = 1:I], patterns[i]' * x == data.pieces[i].d)
model
```

```
A JuMP Model
Minimization problem with:
Variables: 20
Objective function type: AffExpr
`AffExpr`-in-`MathOptInterface.EqualTo{Float64}`: 20 constraints
`VariableRef`-in-`MathOptInterface.GreaterThan{Float64}`: 20 constraints
Model mode: AUTOMATIC
CachingOptimizer state: EMPTY_OPTIMIZER
Solver name: GLPK
Names registered in the model: demand, x
```

Then, we run the iterative column generation scheme:

```
while true
# Solve the linear relaxation
optimize!(model)
# Obtain a new dual vector
π = dual.(demand)
# Solve the pricing problem
new_pattern = solve_pricing(data, π)
# Stop iterating if there is no new pattern
if new_pattern === nothing
break
end
push!(patterns, new_pattern)
# Create a new column
push!(x, @variable(model, lower_bound = 0))
# Update the objective coefficients
set_objective_coefficient(model, x[end], 1.0)
# Update the non-zeros in the coefficient matrix
for i in 1:I
if new_pattern[i] > 0
set_normalized_coefficient(demand[i], x[end], new_pattern[i])
end
end
end
```

Let's have a look at the patterns now:

`SparseArrays.sparse(hcat(patterns...))`

```
20×41 SparseArrays.SparseMatrixCSC{Int64, Int64} with 88 stored entries:
⠑⢄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠈⠀⠀
⠀⠀⠑⢄⠀⠀⠀⠀⠀⠀⡀⣀⡀⠠⠒⡀⠐⠤⠁⠊⠁
⠀⠀⠀⠀⠑⢄⠀⠀⠀⠀⢈⢀⡨⠡⠀⠚⠣⢀⠀⠂⠀
⠀⠀⠀⠀⠀⠀⠑⢄⠀⠀⠡⠄⡀⠐⢌⠂⡀⠀⠠⡠⠁
⠀⠀⠀⠀⠀⠀⠀⠀⠑⢄⢁⣈⢊⠅⠱⠐⠤⠌⣄⠈⠀
```

Nice! We found over 20 new patterns.

Here's pattern 21:

```
for i in 1:I
if patterns[21][i] > 0
println(patterns[21][i], " unit(s) of piece $i")
end
end
```

```
1 unit(s) of piece 8
2 unit(s) of piece 13
1 unit(s) of piece 17
```

## Looking at the solution

Since we only solved a linear relaxation, some of our columns have fractional solutions. We can create a integer feasible solution by rounding up the orders:

```
for p in 1:length(x)
v = ceil(Int, value(x[p]))
if v > 0
println(lpad(v, 2), " roll(s) of pattern $p")
end
end
```

```
31 roll(s) of pattern 1
44 roll(s) of pattern 2
30 roll(s) of pattern 3
41 roll(s) of pattern 4
15 roll(s) of pattern 5
15 roll(s) of pattern 21
23 roll(s) of pattern 22
16 roll(s) of pattern 25
1 roll(s) of pattern 26
26 roll(s) of pattern 28
11 roll(s) of pattern 31
10 roll(s) of pattern 32
3 roll(s) of pattern 33
19 roll(s) of pattern 34
11 roll(s) of pattern 36
4 roll(s) of pattern 37
8 roll(s) of pattern 38
15 roll(s) of pattern 39
12 roll(s) of pattern 40
8 roll(s) of pattern 41
```

This requires 343 rolls:

`sum(ceil.(Int, value.(x)))`

`343`

Alternatively, we can re-introduce the integrality constraints and resolve the problem:

```
set_integer.(x)
optimize!(model)
for p in 1:length(x)
v = round(Int, value(x[p]))
if v > 0
println(lpad(v, 2), " roll(s) of pattern $p, each roll of which makes:")
for i in 1:I
if patterns[p][i] > 0
println(" ", patterns[p][i], " unit(s) of piece $i")
end
end
end
end
```

```
33 roll(s) of pattern 1, each roll of which makes:
1 unit(s) of piece 1
44 roll(s) of pattern 2, each roll of which makes:
1 unit(s) of piece 2
30 roll(s) of pattern 3, each roll of which makes:
1 unit(s) of piece 3
41 roll(s) of pattern 4, each roll of which makes:
1 unit(s) of piece 4
2 roll(s) of pattern 5, each roll of which makes:
1 unit(s) of piece 5
3 roll(s) of pattern 7, each roll of which makes:
1 unit(s) of piece 7
1 roll(s) of pattern 9, each roll of which makes:
1 unit(s) of piece 9
11 roll(s) of pattern 21, each roll of which makes:
1 unit(s) of piece 8
2 unit(s) of piece 13
1 unit(s) of piece 17
22 roll(s) of pattern 22, each roll of which makes:
1 unit(s) of piece 9
1 unit(s) of piece 12
1 unit(s) of piece 15
1 unit(s) of piece 20
1 roll(s) of pattern 24, each roll of which makes:
1 unit(s) of piece 8
1 unit(s) of piece 12
1 unit(s) of piece 17
2 unit(s) of piece 20
17 roll(s) of pattern 25, each roll of which makes:
1 unit(s) of piece 8
1 unit(s) of piece 12
1 unit(s) of piece 16
1 unit(s) of piece 18
2 roll(s) of pattern 26, each roll of which makes:
1 unit(s) of piece 9
1 unit(s) of piece 11
1 unit(s) of piece 17
2 unit(s) of piece 20
24 roll(s) of pattern 28, each roll of which makes:
1 unit(s) of piece 7
1 unit(s) of piece 11
1 unit(s) of piece 14
12 roll(s) of pattern 31, each roll of which makes:
1 unit(s) of piece 8
1 unit(s) of piece 10
1 unit(s) of piece 14
9 roll(s) of pattern 32, each roll of which makes:
1 unit(s) of piece 9
1 unit(s) of piece 10
2 unit(s) of piece 18
1 roll(s) of pattern 33, each roll of which makes:
1 unit(s) of piece 9
1 unit(s) of piece 10
1 unit(s) of piece 16
1 unit(s) of piece 19
18 roll(s) of pattern 34, each roll of which makes:
1 unit(s) of piece 6
1 unit(s) of piece 11
2 unit(s) of piece 19
9 roll(s) of pattern 36, each roll of which makes:
1 unit(s) of piece 7
1 unit(s) of piece 12
2 unit(s) of piece 17
4 roll(s) of pattern 37, each roll of which makes:
1 unit(s) of piece 5
3 unit(s) of piece 19
1 unit(s) of piece 20
5 roll(s) of pattern 38, each roll of which makes:
1 unit(s) of piece 1
1 unit(s) of piece 15
2 unit(s) of piece 20
15 roll(s) of pattern 39, each roll of which makes:
1 unit(s) of piece 6
1 unit(s) of piece 10
1 unit(s) of piece 16
15 roll(s) of pattern 40, each roll of which makes:
1 unit(s) of piece 5
1 unit(s) of piece 15
1 unit(s) of piece 17
15 roll(s) of pattern 41, each roll of which makes:
1 unit(s) of piece 5
1 unit(s) of piece 13
```

This now requires 334 rolls:

`total_rolls = sum(ceil.(Int, value.(x)))`

`334`

This tutorial was generated using Literate.jl. View the source `.jl`

file on GitHub.