Making Weapons

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Problem Definition

Din Djarin is a retired bounty hunter that now runs a small factory specialised in making classic weapons in a colony recently re-established in Mandalore. The factory manufactures two different types of weapons, swords and spears. Din uses special alloys, mixtures of metals and rare materials, which have made a name for the endurance of his weapons across the colony. The table below shows the weight in pounds of each material needed to make each type of weapon:

Metal	spear	sword
Beskar	1	0.5
Tungsten	1	1
Titanium	3	1

And the table below shows the total amount of each metal available, also in pounds:

Metal	Amount
Beskar	125
Tungsten	225
Titanium	300

The price of a spade is 40 credits and the price of a sword is 5 credits. Din has already sold 100 swords, so he needs to manufacture at least that many.

Find the optimal number of spades and swords that maximise revenues

Model

The objective is to maximise the revenues obtained with the sales of both types of weapons. The objective function can be modeled as:

\(\max z = 40x_{1} + 5x_{2}\)

where z represents the objective variable (profits) and the decision variables are:

• \(x_{1}\): units of spears

• \(x_{2}\): units of swords

Let us for now consider that they are real and non-negative.

The objective function is subject to the following constraints:

Metal availability constraints:

\(x_{1} + 0.5x_{2} \leq 125\)

\(x_{1} + x_{2} \leq 225\)

\(3x_{1} + x_{2} \leq 300\)

And the demand on swords:

\(x_{2} \geq 100\)

# Import the numpy and pyplot libraries, we set the aliases np and plt so that it is easier to use
import numpy as np
import matplotlib.pyplot as plt

# We set the mode inline of matplotlib to get the result at the output of the cell code
%matplotlib inline
# Construct lines, in our coordinate system x represents our decision variable x1 and y represents our decision variable x2

x=np.linspace(0,300,2000) #2000 numbers from 0 to 300

y1=(125-x)/0.5 # Beskar constraint
y2=x*0         # non negativity constraint
y3=225-x       # Tungsten constraint
y4=100+x*0     # Demand constraint
y5=300-3*x     # Titanium constraint

# Solution with objective value
y8=(3166.67-40*x)/5 # we clear the x2 with the value of z

#1. Plot the lines
plt.plot(x,y1,label=r'$x_{1}+0.5x_{2}\leq125$')
plt.plot(x,y2,label=r'$x_{2}\greater0$')
plt.plot(x,y3,label=r'$x_{1}+x_{2}\leq225$')
plt.plot(x,y4,label=r'$x_{2}\geq100$')
plt.plot(x,y5,label=r'$3x_{1}+x_{2}\leq300$')
plt.plot(x,y8,label=r'$z=3166.67$')

#2. Adjust axis
plt.xlim((0,150))
plt.ylim((0,250))
plt.xlabel(r'$x_{1}$')
plt.ylabel(r'$x_{2}$')

#3. Fill feasible region
y6=np.minimum(y1,y3) #Line representing the minimum between y6 and y2
y7=np.minimum(y6,y5) #Line representing the minimum between y6 and y5
plt.fill_between(x,y7,y4,where=y7>y4,color='grey',alpha=0.5)

#4. Plot legend
plt.legend(bbox_to_anchor=(1.05,1),loc=2,borderaxespad=0.)

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