代写代考 OPIM 5641: Business Decision Modeling – University of Connecticut

Practice Exam

Practice Exam¶
OPIM 5641: Business Decision Modeling – University of Connecticut

Copyright By PowCoder代写 加微信 powcoder

Instructor:

Questions adapted from Chapter 2 of Nagraj et al.

Please add detailed comments to all code so that I can follow your solution. If I cannot understand your code then you will not receive full credit.

Your Name Here
Your NetID Here

%matplotlib inline
from pylab import * # for graphing

import shutil
import sys
import os.path

if not shutil.which(“pyomo”):
!pip install -q pyomo
assert(shutil.which(“pyomo”))

if not (shutil.which(“cbc”) or os.path.isfile(“cbc”)):
if “google.colab” in sys.modules:
!apt-get install -y -qq coinor-cbc
!conda install -c conda-forge coincbc

assert(shutil.which(“cbc”) or os.path.isfile(“cbc”))

from pyomo.environ import *
import numpy as np
import pandas as pd

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Problem 1¶
Solve the following problem using the graphical method.

$Max(Z) = 2X + 4Y$

subject to:

$2X + 4Y \leq 72$
$3X + 6Y \geq 27$
$-3X + 10Y \geq 0$
$X,Y \geq 0$ (nonnegativity)

$-3X + 10Y \geq 0$

$-3X + 10Y = 0$

$X = 40$ –> $-3*40 + 10Y = 0$ –> $-120 + 10Y = 0$ –> $10Y = 120$ –> $Y = 12$

# this is the example from Pyomo cookbook
# pylab makes it easy to make plots

figure(figsize=(6,6))
#subplot(111, aspect=’equal’)
axis([0,40,0,40])
xlabel(‘X’)
ylabel(‘Y’)

# Constraint 1
y = array([18, 0])
x = array([0,36])
plot(x,y,’r’,lw=2)
fill_between([0,36,40], # x area
[18,0,0], # y area
[40,40,40], # upper area
color=’red’, # color
alpha=0.15) # transparency

# Constraint 2
y = array([27/6, 0])
x = array([0,27/3])
plot(x,y,’orange’,lw=2)
fill_between([0,9], # x area
[0,0], # y area
[27/6,0], # upper area
color=’orange’, # color
alpha=0.15) # transparency

# Constraint 3
y = array([0,12])
x = array([0,40])
plot(x,y,’black’,lw=2)
fill_between([0,40], # x area
[0,0], # y area
[0,12], # upper area
color=’black’, # color
alpha=0.15) # transparency

# legend([‘Constraint 1’])

# y axis and orange
A = np.array([[1, 0], [3, 6]])
B = np.array([0, 27])
point_1 = np.linalg.solve(A,B)
print(“Point 1: “, point_1)

# y axis and red
A = np.array([[1, 0], [2, 4]])
B = np.array([0, 72])
point_2 = np.linalg.solve(A,B)
print(“Point 2: “, point_2)

# black and red
A = np.array([[-3, 10], [2, 4]])
B = np.array([0, 72])
point_3 = np.linalg.solve(A,B)
print(“Point 3: “, point_3)

# black and orange
A = np.array([[-3, 10], [3, 6]])
B = np.array([0, 27])
point_4 = np.linalg.solve(A,B)
print(“Point 4: “, point_4)

Point 1: [0. 4.5]
Point 2: [ 0. 18.]
Point 3: [22.5 6.75]
Point 4: [5.625 1.6875]

print(2*point_1[0] + 4*point_1[1])
print(2*point_2[0] + 4*point_2[1])
print(2*point_3[0] + 4*point_3[1])
print(2*point_4[0] + 4*point_4[1])

Point 2: [ 0 , 18]

Is an optimal solution.

Can you prove to me algebraically that this is an optimal solution?

Problem 2¶
Formulate the dual of Problem 1. Can you find a feasible solution that proves that the solution from Problem 1 that you found is optimal?

Solve the following problem using the graphical method.

$Max(Z) = 2X + 4Y$

subject to:

$2X + 4Y \leq 72$
$3X + 6Y \geq 27$
$-3X + 10Y \geq 0$
$X,Y \geq 0$ (nonnegativity)

$2X + 4Y \leq 72$
$-3X – 6Y \leq -27$
$3X – 10Y \leq 0$
$X,Y \geq 0$ (nonnegativity)

This is the dual. I formulate this by having a new variable for each constraint, and the objective function becomes the right hand side of the equations mutliplied by the variables, and the constraints ensure that I find a combination that is better than the objective function.

$\min 72 z_1 – 27 z_2 + 0 z_3$

Subject to

$2 z_1 – 3 z_2 + 3 z_3 \geq 2$
$4 z_1 – 6 z_2 – 10 z_3 \geq 4$
$z_1, z_2, z_3 \geq 0$

The solution (1,0,0) is feasible. Now let me prove that it is optimal.

First off, what is the objective value in the dual for the solution (1,0,0)?

Plugging in to objective we get:

721−270+0*0 = 72

Therefore, the optimal value can be no more than 72. But how do I know that 72 is the best possible value? Let’s find a combination of the constraints that PROVES it.

($4 z_1 – 6 z_2 – 10 z_3 \geq 4$ ) * 18

$72 z_1 – 108 z_2 – 180 z_3 \geq 72$

It should therefore be clear that any feasible solution to the dual has to satisfy that

$72 z_1 – 108 z_2 – 180 z_3 \geq 72$

If I think about it:

$72 z_1 – 27 z_2 \geq 72 z_1 – 108 z_2 – 180 z_3 \geq 72$

I then know that:

$72 z_1 – 27 z_2 \geq 72$

Claim: z_1 = 1, z_2 = 0, z_3 = 0 is the optimal solution to the dual.

The solution is feasible to the dual, and, the value matches the value of the solution (0,18) to the primal.

A + C VERSUS B + D? NO IDEA

A + C VERSUS B + D? YES!!! greater than or equal

Problem 3¶
A small motor manufacturer makes two types of motor, models A and B. The assembly process for each is similar in that both require a certain amount of
wiring, drilling, and assembly. Each model A takes
3 hours of wiring, 2 hours of drilling, and 1.5 hours
of assembly. Each model B must go through 2 hours of wiring, 1 hour of drilling, and 0.5 hours of assembly. During the next production period, 240 hours of wiring time, 210 hours of drilling time, and 120
hours of assembly time are available. Each model A
sold yields a profit of \$22. Each model B can be sold
for a \$15 profit. Assuming that all motors that are
assembled can be sold, find the best combination of
motors to yield the highest profit.

Formulate the problem as a linear programming model and solve via the graphically method, and also using the brute force method.

$X$: number of type A models
$Y$: number of type B models

$22 \cdot X + 15 \cdot Y$

Constraints

wiring: $3X + 2Y \leq 240$
drilling: $2X + 1Y \leq 210$
assembly: $1.5X + 0.5Y \leq 120$
non-negativity: $X \geq 0, Y \geq 0$

best_profit_so_far = 0
best_x = 0
best_y = 0

for x in range(81):
for y in range(121):
if 3*x + 2*y > 240:
# drilling
if 2*x + 1*y > 210:
# assembly
if 1.5*x + 0.5*y > 120:
candidate_profit = 22*x + 15*y
if candidate_profit > best_profit_so_far:
best_profit_so_far = candidate_profit
best_x = x
best_y = y

print(best_x)
print(best_y)
print(best_profit_so_far)

for y in range(121):

Problem 4¶
The profit per each model is uncertain. Suppose that the profit for model A has a plus/minus 5\% margin, and that the profit for model B has a plus/minus 15\% margin. Produce a histogram that displays the distribution of profits for the best solution found in Problem 3.

generated_profits = []
n_simulations = 1000
for sim_index in range(n_simulations):
profit_from_x = np.random.uniform(1-0.05,1+0.05) * 22 * best_x
profit_from_y = np.random.uniform(1-0.15,1+0.15) * 15 * best_y
total_profit = profit_from_x + profit_from_y
generated_profits.append(total_profit)

plt.hist(generated_profits)

(array([100., 98., 93., 109., 90., 102., 105., 90., 112., 101.]),
array([1530.09218953, 1584.03642399, 1637.98065846, 1691.92489292,
1745.86912738, 1799.81336184, 1853.7575963 , 1907.70183077,
1961.64606523, 2015.59029969, 2069.53453415]),
)

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