# Linear Optimization with Python 1

#### Linear Optimization problem

Side Story:  Excel Solver is an excellent tool for Linear Optimization problems, but we need to pay for this platform.
In addition, 5 asked me with an extremely surprised facial expression: “What kind of data scientists would use Excel? “ 😱😱😱

Therefore we will introduce how to use Python to solve a Linear Optimization Problem, so we can all be Real data Scientists.

#### Problem:

Pirelli glass produces two types of fiber optic strands: A-type & H-type (a higher speed variety).

A-type           H-type

Fiber  1                    1

Labor  9 hours        6 hours

Glass  12 lbs            16 lbs

Unit Profit  $350$300

Due to an unusual agreement they made as part of a takeover of a small company next week they are only allowed to produce 200 fibers, there are 1566 hours of labor available, and 2880 lbs of glass available.

#### Question:

We want to determine the best combination of A-type and H-type fibers to produce? pixel2013 / Pixabay

Solution Key:

X1=number of A-type fibers to produce
X2=number of H-type fibers to produce

MAX Revenue : F(x) = 350X1 + 300X2

Constrain:

1X1 + 1X2 <= 200 fibers
9X1 + 6X2 <= 1566 labor
12X1 + 16X2 <= 2880 glass
X1 >= 0
X2 >= 0

Python Code:

from scipy.optimize import linprog
import numpy as np
import matplotlib.pyplot as plt
from numpy.linalg import solve
%matplotlib inline

# Solve
#Note linprog is for min, in order to solve for max,
#we switch the sign
c=[-350, -300]
A=[[1,1],[9,6],[12,16]]
b=[200,1566,2800]
x0_bounds=(0,None)
x_bounds=(0,None)

res = linprog(c, A_ub=A, b_ub=b, bounds=(x0_bounds, x_bounds))

print(res)


Output:

Remember the Trickest point to solve this problem is
Min(f(x))= -(Max(f(x))) ⇔ Max(f(x))= -(Min(f(x)))

Therefore we have our Max Profit = 66100 with A-type fibers=122 and B-type fibers=78

     fun: -66100.0
message: 'Optimization terminated successfully.'
nit: 2
slack: array([  0.,   0.,  88.])
status: 0
success: True
x: array([ 122.,   78.])

Visualization: Python Code:

x= np.linspace(0,200,2000)
y1=200-x
y2=(1566-6*x)/9
y3=(2880-16*x)/12

plt.plot(x, y1, label=r'$y\leq 200-x$')
plt.plot(x, y2, label=r'$9y\leq1566- 6x$')
plt.plot(x, y3, label=r'$12y\leq2880 - 16x$')

plt.xlim((0, 200))
plt.ylim((0, 200))
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')

# Fill feasible region
y5 = np.minimum( y1, y2 )
y6 = np.minimum(y2, y3)
y7= np.minimum(y5,y6)
plt.fill_between(x,y7, color='grey', alpha=0.5)
plt.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.)

plt.show()

Reference:

Bellur Srikar, HU ANLY 705, Linear Optimization Lecture notes
http://www.vision.ime.usp.br/~igor/articles/optimization-linprog.html

http://benalexkeen.com/linear-programming-with-python-and-pulp-part-1/

https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.linprog.html

##### Happy Studying!!! 🤭



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