We are now going to demonstrate the use of PuLP with a slightly larger problem. For this notebook, we will be considering the explicit transportation model formulation that we discussed in Part 2 of our session.

Since we `pip`

-installed `pulp`

while executing our previous notebook, there is no reason to repeat the process again.

`LpProblem()`

function to initialise our model.

In [1]:

```
from pulp import *
prob = LpProblem('problem',LpMinimize)
```

`lowBound`

parameter when initialising each variable.

In [2]:

```
x_s1d1 = LpVariable("x_s1d1", lowBound = 0)
x_s1d2 = LpVariable("x_s1d2", lowBound = 0)
x_s1d3 = LpVariable("x_s1d3", lowBound = 0)
x_s2d1 = LpVariable("x_s2d1", lowBound = 0)
x_s2d2 = LpVariable("x_s2d2", lowBound = 0)
x_s2d3 = LpVariable("x_s2d3", lowBound = 0)
x_s3d1 = LpVariable("x_s3d1", lowBound = 0)
x_s3d2 = LpVariable("x_s3d2", lowBound = 0)
x_s3d3 = LpVariable("x_s3d3", lowBound = 0)
```

We now begin populating our optimization problem.

Our objective function is:

$6x_{S1D1}+3x_{S1D2}+8x_{S1D3}+7x_{S2D1}+5x_{S2D2}+9x_{S2D3}+4x_{S3D1}+5x_{S3D2}+5x_{S3D3}$

In [3]:

```
prob += 6 * x_s1d1 + 3 * x_s1d2 + 8 * x_s1d3 + 7 * x_s2d1 + 5 * x_s2d2 + 9 * x_s2d3 + 4 * x_s3d1 + 5 * x_s3d2 + 3 * x_s3d3
```

We can now add the constraints.

In [4]:

```
prob += x_s1d1 + x_s1d2 + x_s1d3 == 300
prob += x_s2d1 + x_s2d2 + x_s2d3 == 300
prob += x_s3d1 + x_s3d2 + x_s3d3 == 100
prob += x_s1d1 + x_s2d1 + x_s3d1 == 200
prob += x_s1d2 + x_s2d2 + x_s3d2 == 200
prob += x_s1d3 + x_s2d3 + x_s3d3 == 300
```

You might not be able to tell by simply reading this notebook, but typing all these equations is incredibly tedious and repetitive.

It is also a very error-prone process - it took me 3 attempts to type the model correctly, as I had made several mistakes in quite a few variable names.

We can now print the model formulation to check that it corresponds to the one that we were trying to implement.

In [5]:

```
print(prob)
```

It all appears to be correct, and we can now solve oue problem instace using the `solve()`

command.

In [6]:

```
status = prob.solve()
print(f'STATUS\n{LpStatus[status]}\n')
```

STATUS Optimal

As before, we are using the following block of code to print the result of our model.

In [7]:

```
print('SOLUTION:')
for v in prob.variables():
print(f'\t\t{v.name} = {v.varValue}')
print('\n') # Prints a blank line
print(f'OBJECTIVE VALUE: {prob.objective.value()}')
```