Algorithms For Dummies

  Employing Linear Programming 

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a = np.linspace(0, 30, 30)

c1 = ((res['res_1']['t'] * res['res_1']['n'])-

      res['res_1']['A']*a) / res['res_1']['B']

c2 = ((res['res_2']['t'] * res['res_2']['n'])-

      res['res_2']['A']*a) / res['res_2']['B']

c3 = ((res['res_3']['t'] * res['res_3']['n'])-

      res['res_3']['A']*a) / res['res_3']['B']

plt.plot(a, c1, label='constrain #1')

plt.plot(a, c2, label='constrain #2')

plt.plot(a, c3, label='constrain #3')

axes = plt.gca()

axes.set_xlim([0,30])

axes.set_ylim([0,30])

plt.xlabel('qty model A')

plt.ylabel('qty model B')

border = np.array((c1,c2,c3)).min(axis=0)

plt.fill_between(a, border, color='yellow', alpha=0.5)

plt.scatter(*zip(*[(0,0), (20,0),

                   (0,20), (16,6), (6,16)]))

plt.legend()

plt.show()

The constraints turn into three lines on a chart, as shown in Figure 19-2. The lines 

intersect among themselves, showing the feasible area. This is the area delimited 

by the three lines whose A and B values are always inferior or equal compared to 

the values on any of the constraint lines. (The constraints represent a frontier; you 

can’t have A or B values beyond them.)

According to the simplex method, the optimal solution is one of the five vertexes 

of the polygon (which are (0,0), (20,0), (0,20), (16,6), and (6,16)). You can dis-

cover which one is the solution by setting the necessary functions from the PuLP 

package. First, define the problem and call it model. By doing so, you determine 

that it’s a maximization problem and that both A and B should be positive.

model = pulp.LpProblem("Max profit", pulp.LpMaximize)

A = pulp.LpVariable('A', lowBound=0)

B = pulp.LpVariable('B', lowBound=0)

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