1) The document provides examples and instructions for modeling and solving linear programming problems in a spreadsheet. It discusses best practices for setting up the spreadsheet model and provides sample problems with solutions.
2) Section 3 provides an example problem involving TV and magazine ads with a maximum profit of $775,000.
3) The document contains 24 sections that provide sample linear programming problems of varying sizes and complexity along with the spreadsheet files containing the solutions.
Spreadsheet Modeling and Decision Analysis A Practical Introduction to Busine...rikujoxizi
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Spreadsheet Modeling and Decision Analysis A Practical Introduction to Business Analytics 8th Edition Ragsdale Test Bank
Spreadsheet Modeling and Decision Analysis A Practical Introduction to Busine...rikujoxizi
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Spreadsheet Modeling and Decision Analysis A Practical Introduction to Business Analytics 8th Edition Ragsdale Test Bank
Investment Strategy Case Analysis (MGT 3050)Afifah Nabilah
A case study of the J.D Williams Investment Strategy Problem. This is an assignment for IIUM students who took Decision Science (MGT 3050) in Semester 2, 2014/2015.
Questions list.
1. Summarize the proposal that Tahki Yazzie submitted to the board.
2. Do a SWOT analysis relating to her proposal.
3. Propose another solution that relies on the reduction of costs and not rebranding the product.
4. Identify the areas that costs can be reduced effectively and calculate the impact it will have on profitability.
Burke: Learning and Growing through Marketing ResearchAsif Mahmood Abbas
Burke is a century-old market research firm that uses a reliable research process and cutting edge technology to help its clients.
Burke works with clients to help them identify what information is needed to make a decision they are facing.
With a series of creative methods, Burke makes sure that what the client thinks the problem is, is really the problem.
1) Use properties of logarithms to expand the following logarit.docxhirstcruz
1) Use properties of logarithms to expand the following logarithmic expression as much as possible.
Log
b
(√xy
3
/ z
3
)
A. 1/2 log
b
x - 6 log
b
y + 3 log
b
z
B. 1/2 log
b
x - 9 log
b
y - 3 log
b
z
C. 1/2 log
b
x + 3 log
b
y + 6 log
b
z
D. 1/2 log
b
x + 3 log
b
y - 3 log
b
z
2) Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, to two decimal places, for the solution.
2 log x = log 25
A. {12}
B. {5}
C. {-3}
D. {25}
3) Write the following equation in its equivalent logarithmic form.
2
-4
= 1/16
A. Log
4
1/16 = 64
B. Log
2
1/24 = -4
C. Log
2
1/16 = -4
D. Log
4
1/16 = 54
4) Use properties of logarithms to condense the following logarithmic expression. Write the expression as a single logarithm whose coefficient is 1.
log
2
96 – log
2
3
A. 5
B. 7
C. 12
D. 4
5) Use the exponential growth model, A = A
0
e
kt
, to show that the time it takes a population to double (to grow from A
0
to 2A
0
) is given by t = ln 2/k.
A. A
0
= A
0
e
kt
; ln = e
kt
; ln 2 = ln e
kt
; ln 2 = kt; ln 2/k = t
B. 2A
0
= A
0
e; 2= e
kt
; ln = ln e
kt
; ln 2 = kt; ln 2/k = t
C. 2A
0
= A
0
e
kt
; 2= e
kt
; ln 2 = ln e
kt
; ln 2 = kt; ln 2/k = t
D. 2A
0
= A
0
e
kt
; 2 = e
kt
; ln 1 = ln e
kt
; ln 2 = kt; ln 2/k = t
oe
6) Find the domain of following logarithmic function.
f(x) = log (2 - x)
A. (∞, 4)
B. (∞, -12)
C. (-∞, 2)
D. (-∞, -3)
7) An artifact originally had 16 grams of carbon-14 present. The decay model A = 16e -0.000121t describes the amount of carbon-14 present after t years. How many grams of carbon-14 will be present in 5715 years?
A. Approximately 7 grams
B. Approximately 8 grams
C. Approximately 23 grams
D. Approximately 4 grams
8) Use properties of logarithms to expand the following logarithmic expression as much as possible.
log
b
(x
2
y) / z
2
A. 2 log
b
x + log
b
y - 2 log
b
z
B. 4 log
b
x - log
b
y - 2 log
b
z
C. 2 log
b
x + 2 log
b
y + 2 log
b
z
D. log
b
x - log
b
y + 2 log
b
z
9) The exponential function f with base b is defined by f(x) = __________, b > 0 and b ≠ 1. Using interval notation, the domain of this function is __________ and the range is __________.
A. bx; (∞, -∞); (1, ∞)
B. bx; (-∞, -∞); (2, ∞)
C. bx; (-∞, ∞); (0, ∞)
D. bx; (-∞, -∞); (-1, ∞)
10) Approximate the following using a calculator; round your answer to three decimal places.
3
√5
A. .765
B. 14297
C. 11.494
D. 11.665
11) Write the following equation in its equivalent exponential form.
4 = log
2
16
A. 2 log
4
= 16
B. 2
2
= 4
C. 4
4
= 256
D. 2
4
= 16
12) Solve the following exponential equation by expressing each side as a power of the same base and then equating exponents.
3
1-x
= 1/27
A. {2}
B. {-7}
C. {4}
D. {3}
13) Use properties of logarithms to expand the followin.
Investment Strategy Case Analysis (MGT 3050)Afifah Nabilah
A case study of the J.D Williams Investment Strategy Problem. This is an assignment for IIUM students who took Decision Science (MGT 3050) in Semester 2, 2014/2015.
Questions list.
1. Summarize the proposal that Tahki Yazzie submitted to the board.
2. Do a SWOT analysis relating to her proposal.
3. Propose another solution that relies on the reduction of costs and not rebranding the product.
4. Identify the areas that costs can be reduced effectively and calculate the impact it will have on profitability.
Burke: Learning and Growing through Marketing ResearchAsif Mahmood Abbas
Burke is a century-old market research firm that uses a reliable research process and cutting edge technology to help its clients.
Burke works with clients to help them identify what information is needed to make a decision they are facing.
With a series of creative methods, Burke makes sure that what the client thinks the problem is, is really the problem.
1) Use properties of logarithms to expand the following logarit.docxhirstcruz
1) Use properties of logarithms to expand the following logarithmic expression as much as possible.
Log
b
(√xy
3
/ z
3
)
A. 1/2 log
b
x - 6 log
b
y + 3 log
b
z
B. 1/2 log
b
x - 9 log
b
y - 3 log
b
z
C. 1/2 log
b
x + 3 log
b
y + 6 log
b
z
D. 1/2 log
b
x + 3 log
b
y - 3 log
b
z
2) Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, to two decimal places, for the solution.
2 log x = log 25
A. {12}
B. {5}
C. {-3}
D. {25}
3) Write the following equation in its equivalent logarithmic form.
2
-4
= 1/16
A. Log
4
1/16 = 64
B. Log
2
1/24 = -4
C. Log
2
1/16 = -4
D. Log
4
1/16 = 54
4) Use properties of logarithms to condense the following logarithmic expression. Write the expression as a single logarithm whose coefficient is 1.
log
2
96 – log
2
3
A. 5
B. 7
C. 12
D. 4
5) Use the exponential growth model, A = A
0
e
kt
, to show that the time it takes a population to double (to grow from A
0
to 2A
0
) is given by t = ln 2/k.
A. A
0
= A
0
e
kt
; ln = e
kt
; ln 2 = ln e
kt
; ln 2 = kt; ln 2/k = t
B. 2A
0
= A
0
e; 2= e
kt
; ln = ln e
kt
; ln 2 = kt; ln 2/k = t
C. 2A
0
= A
0
e
kt
; 2= e
kt
; ln 2 = ln e
kt
; ln 2 = kt; ln 2/k = t
D. 2A
0
= A
0
e
kt
; 2 = e
kt
; ln 1 = ln e
kt
; ln 2 = kt; ln 2/k = t
oe
6) Find the domain of following logarithmic function.
f(x) = log (2 - x)
A. (∞, 4)
B. (∞, -12)
C. (-∞, 2)
D. (-∞, -3)
7) An artifact originally had 16 grams of carbon-14 present. The decay model A = 16e -0.000121t describes the amount of carbon-14 present after t years. How many grams of carbon-14 will be present in 5715 years?
A. Approximately 7 grams
B. Approximately 8 grams
C. Approximately 23 grams
D. Approximately 4 grams
8) Use properties of logarithms to expand the following logarithmic expression as much as possible.
log
b
(x
2
y) / z
2
A. 2 log
b
x + log
b
y - 2 log
b
z
B. 4 log
b
x - log
b
y - 2 log
b
z
C. 2 log
b
x + 2 log
b
y + 2 log
b
z
D. log
b
x - log
b
y + 2 log
b
z
9) The exponential function f with base b is defined by f(x) = __________, b > 0 and b ≠ 1. Using interval notation, the domain of this function is __________ and the range is __________.
A. bx; (∞, -∞); (1, ∞)
B. bx; (-∞, -∞); (2, ∞)
C. bx; (-∞, ∞); (0, ∞)
D. bx; (-∞, -∞); (-1, ∞)
10) Approximate the following using a calculator; round your answer to three decimal places.
3
√5
A. .765
B. 14297
C. 11.494
D. 11.665
11) Write the following equation in its equivalent exponential form.
4 = log
2
16
A. 2 log
4
= 16
B. 2
2
= 4
C. 4
4
= 256
D. 2
4
= 16
12) Solve the following exponential equation by expressing each side as a power of the same base and then equating exponents.
3
1-x
= 1/27
A. {2}
B. {-7}
C. {4}
D. {3}
13) Use properties of logarithms to expand the followin.
1) Use properties of logarithms to expand the following logarithm.docxdorishigh
1) Use properties of logarithms to expand the following logarithmic expression as much as possible.
Logb (√xy3 / z3)
A. 1/2 logb x - 6 logb y + 3 logb z
B. 1/2 logb x - 9 logb y - 3 logb z
C. 1/2 logb x + 3 logb y + 6 logb z
D. 1/2 logb x + 3 logb y - 3 logb z
2) Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, to two decimal places, for the solution.
2 log x = log 25
A. {12}
B. {5}
C. {-3}
D. {25}
3) Write the following equation in its equivalent logarithmic form.
2-4 = 1/16
A. Log4 1/16 = 64
B. Log2 1/24 = -4
C. Log2 1/16 = -4
D. Log4 1/16 = 54
4) Use properties of logarithms to condense the following logarithmic expression. Write the expression as a single logarithm whose coefficient is 1.
log2 96 – log2 3
A. 5
B. 7
C. 12
D. 4
5) Use the exponential growth model, A = A0ekt, to show that the time it takes a population to double (to grow from A0 to 2A0 ) is given by t = ln 2/k.
A. A0 = A0ekt; ln = ekt; ln 2 = ln ekt; ln 2 = kt; ln 2/k = t
B. 2A0 = A0e; 2= ekt; ln = ln ekt; ln 2 = kt; ln 2/k = t
C. 2A0 = A0ekt; 2= ekt; ln 2 = ln ekt; ln 2 = kt; ln 2/k = t
D. 2A0 = A0ekt; 2 = ekt; ln 1 = ln ekt; ln 2 = kt; ln 2/k = toe
6) Find the domain of following logarithmic function.
f(x) = log (2 - x)
A. (∞, 4)
B. (∞, -12)
C. (-∞, 2)
D. (-∞, -3)
7) An artifact originally had 16 grams of carbon-14 present. The decay model A = 16e -0.000121t describes the amount of carbon-14 present after t years. How many grams of carbon-14 will be present in 5715 years?
A. Approximately 7 grams
B. Approximately 8 grams
C. Approximately 23 grams
D. Approximately 4 grams
8) Use properties of logarithms to expand the following logarithmic expression as much as possible.
logb (x2 y) / z2
A. 2 logb x + logb y - 2 logb z
B. 4 logb x - logb y - 2 logb z
C. 2 logb x + 2 logb y + 2 logb z
D. logb x - logb y + 2 logb z
9) The exponential function f with base b is defined by f(x) = __________, b > 0 and b ≠ 1. Using interval notation, the domain of this function is __________ and the range is __________.
A. bx; (∞, -∞); (1, ∞)
B. bx; (-∞, -∞); (2, ∞)
C. bx; (-∞, ∞); (0, ∞)
D. bx; (-∞, -∞); (-1, ∞)
10) Approximate the following using a calculator; round your answer to three decimal places.
3√5
A. .765
B. 14297
C. 11.494
D. 11.665
11) Write the following equation in its equivalent exponential form.
4 = log2 16
A. 2 log4 = 16
B. 22 = 4
C. 44 = 256
D. 24 = 16
12) Solve the following exponential equation by expressing each side as a power of the same base and then equating exponents.
31-x = 1/27
A. {2}
B. {-7}
C. {4}
D. {3}
13) Use properties of logarithms to expand the following logarithmic expression as much as possible.
logb (x2y)
A. 2 logy x + logx y
B. 2 logb x + logb y
C. logx - logb y
D. logb x – ...
Please be sure to save at least once every 15 minutes. If you leav.docxrandymartin91030
Please be sure to save at least once every 15 minutes. If you leave this page without saving, or if your session times out, any answers you have not saved will be lost. The Submit for Grading button will become available once you've answered all questions. Exams are not timed; you do not have to finish an exam in one sitting as long as you have saved your answers.
Q1. Find k such that f(x) = x4 + kx3 + 2 has the factor x + 1.
a. -3
b. -2
c. 3
d. 2
Q2. Solve the inequality.
(x - 5)(x2 + x + 1) > 0
a. (-∞, -1) or (1, ∞)
b. (-1, 1)
c. (-∞, 5)
d. (5, ∞)
Q3. Use the intermediate value theorem to determine whether the polynomial function has a zero in the given interval.
f(x) = 8x3 - 10x2 + 3x + 5; [-1, 0]
a. f(-1) = -16 and f(0) = -5; no
b. f(-1) = -16 and f(0) = 5; yes
c. f(-1) = 16 and f(0) = -5; yes
d. f(-1) = 16 and f(0) = 5; no
Q4. Solve the equation in the real number system.
x4 - 3x3 + 5x2 - x - 10 = 0
a. {-1, -2}
b. {1, 2}
c. {-1, 2}
d. {-2, 1}
Q5. Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value.
f(x) = x2 - 2x - 5
a. maximum; 1
b. minimum; 1
c. maximum; - 6
d. minimum; - 6
Q6. Determine whether the rational function has symmetry with respect to the origin, symmetry with respect to the y-axis, or neither.
f(x) = a. symmetry with respect to the origin
b. symmetry with respect to the y-axis
c. neither
Q7. Find the domain of the rational function.
f(x) = .
a. {x|x ≠ -3, x ≠ 5}
b. {x|x ≠ 3, x ≠ -5}
c. all real numbers
d. {x|x ≠ 3, x ≠ -3, x ≠ -5}
Q8. Find the domain of the rational function.
g(x) = a. all real numbers
b. {x|x ≠ -7, x ≠ 7, x ≠ -5}
c. {x|x ≠ -7, x ≠ 7}
d. {x|x ≠ 0, x ≠ -49}
Q9. Find all zeros of the function and write the polynomial as a product of linear factors.
f(x) = 3x4 + 4x3 + 13x2 + 16x + 4
a. f(x) = (3x - 1)(x - 1)(x + 2)(x - 2)
b. f(x) = (3x + 1)(x + 1)(x + 2i)(x - 2i)
c. f(x) = (3x - 1)(x - 1)(x + 2i)(x - 2i)
d. f(x) = (3x + 1)(x + 1)(x + 2)(x - 2)
Q10. Use the graph to find the vertical asymptotes, if any, of the function.
a. y = 0
b. x = 0, y = 0
c. x = 0
d. none
Q11. Find the power function that the graph of f resembles for large values of |x|.
f(x) = (x + 5)2
a. y = x10
b. y = x25
c. y = x2
d. y = x5
Q12. Determine whether the rational function has symmetry with respect to the origin, symmetry with respect to the y-axis, or neither.
f(x) = a. symmetry with respect to the y-axis
b. symmetry with respect to the origin
c. neither
Q13. Determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find that value.
f(x) = -x2 - 2x + 2
a. minimum; - 1
b. maximum; 3
c. minimum; 3
d. maximum; - 1
Q14. Find the power function that the graph of f resembles for large values of |x|.
f(x) = -x2(x + 4)3(x2 - 1)
a. y = x7.
A possible solution to the struct-hub second design assessment. Inspired by the civic centre building 2018 involving wide slab panels of solid slab construction
Industrial Training at Shahjalal Fertilizer Company Limited (SFCL)MdTanvirMahtab2
This presentation is about the working procedure of Shahjalal Fertilizer Company Limited (SFCL). A Govt. owned Company of Bangladesh Chemical Industries Corporation under Ministry of Industries.
Vaccine management system project report documentation..pdfKamal Acharya
The Division of Vaccine and Immunization is facing increasing difficulty monitoring vaccines and other commodities distribution once they have been distributed from the national stores. With the introduction of new vaccines, more challenges have been anticipated with this additions posing serious threat to the already over strained vaccine supply chain system in Kenya.
Quality defects in TMT Bars, Possible causes and Potential Solutions.PrashantGoswami42
Maintaining high-quality standards in the production of TMT bars is crucial for ensuring structural integrity in construction. Addressing common defects through careful monitoring, standardized processes, and advanced technology can significantly improve the quality of TMT bars. Continuous training and adherence to quality control measures will also play a pivotal role in minimizing these defects.
Student information management system project report ii.pdfKamal Acharya
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
Immunizing Image Classifiers Against Localized Adversary Attacksgerogepatton
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
NO1 Uk best vashikaran specialist in delhi vashikaran baba near me online vas...Amil Baba Dawood bangali
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Overview of the fundamental roles in Hydropower generation and the components involved in wider Electrical Engineering.
This paper presents the design and construction of hydroelectric dams from the hydrologist’s survey of the valley before construction, all aspects and involved disciplines, fluid dynamics, structural engineering, generation and mains frequency regulation to the very transmission of power through the network in the United Kingdom.
Author: Robbie Edward Sayers
Collaborators and co editors: Charlie Sims and Connor Healey.
(C) 2024 Robbie E. Sayers
Courier management system project report.pdfKamal Acharya
It is now-a-days very important for the people to send or receive articles like imported furniture, electronic items, gifts, business goods and the like. People depend vastly on different transport systems which mostly use the manual way of receiving and delivering the articles. There is no way to track the articles till they are received and there is no way to let the customer know what happened in transit, once he booked some articles. In such a situation, we need a system which completely computerizes the cargo activities including time to time tracking of the articles sent. This need is fulfilled by Courier Management System software which is online software for the cargo management people that enables them to receive the goods from a source and send them to a required destination and track their status from time to time.
Automobile Management System Project Report.pdfKamal Acharya
The proposed project is developed to manage the automobile in the automobile dealer company. The main module in this project is login, automobile management, customer management, sales, complaints and reports. The first module is the login. The automobile showroom owner should login to the project for usage. The username and password are verified and if it is correct, next form opens. If the username and password are not correct, it shows the error message.
When a customer search for a automobile, if the automobile is available, they will be taken to a page that shows the details of the automobile including automobile name, automobile ID, quantity, price etc. “Automobile Management System” is useful for maintaining automobiles, customers effectively and hence helps for establishing good relation between customer and automobile organization. It contains various customized modules for effectively maintaining automobiles and stock information accurately and safely.
When the automobile is sold to the customer, stock will be reduced automatically. When a new purchase is made, stock will be increased automatically. While selecting automobiles for sale, the proposed software will automatically check for total number of available stock of that particular item, if the total stock of that particular item is less than 5, software will notify the user to purchase the particular item.
Also when the user tries to sale items which are not in stock, the system will prompt the user that the stock is not enough. Customers of this system can search for a automobile; can purchase a automobile easily by selecting fast. On the other hand the stock of automobiles can be maintained perfectly by the automobile shop manager overcoming the drawbacks of existing system.
Explore the innovative world of trenchless pipe repair with our comprehensive guide, "The Benefits and Techniques of Trenchless Pipe Repair." This document delves into the modern methods of repairing underground pipes without the need for extensive excavation, highlighting the numerous advantages and the latest techniques used in the industry.
Learn about the cost savings, reduced environmental impact, and minimal disruption associated with trenchless technology. Discover detailed explanations of popular techniques such as pipe bursting, cured-in-place pipe (CIPP) lining, and directional drilling. Understand how these methods can be applied to various types of infrastructure, from residential plumbing to large-scale municipal systems.
Ideal for homeowners, contractors, engineers, and anyone interested in modern plumbing solutions, this guide provides valuable insights into why trenchless pipe repair is becoming the preferred choice for pipe rehabilitation. Stay informed about the latest advancements and best practices in the field.
Water scarcity is the lack of fresh water resources to meet the standard water demand. There are two type of water scarcity. One is physical. The other is economic water scarcity.
Hybrid optimization of pumped hydro system and solar- Engr. Abdul-Azeez.pdffxintegritypublishin
Advancements in technology unveil a myriad of electrical and electronic breakthroughs geared towards efficiently harnessing limited resources to meet human energy demands. The optimization of hybrid solar PV panels and pumped hydro energy supply systems plays a pivotal role in utilizing natural resources effectively. This initiative not only benefits humanity but also fosters environmental sustainability. The study investigated the design optimization of these hybrid systems, focusing on understanding solar radiation patterns, identifying geographical influences on solar radiation, formulating a mathematical model for system optimization, and determining the optimal configuration of PV panels and pumped hydro storage. Through a comparative analysis approach and eight weeks of data collection, the study addressed key research questions related to solar radiation patterns and optimal system design. The findings highlighted regions with heightened solar radiation levels, showcasing substantial potential for power generation and emphasizing the system's efficiency. Optimizing system design significantly boosted power generation, promoted renewable energy utilization, and enhanced energy storage capacity. The study underscored the benefits of optimizing hybrid solar PV panels and pumped hydro energy supply systems for sustainable energy usage. Optimizing the design of solar PV panels and pumped hydro energy supply systems as examined across diverse climatic conditions in a developing country, not only enhances power generation but also improves the integration of renewable energy sources and boosts energy storage capacities, particularly beneficial for less economically prosperous regions. Additionally, the study provides valuable insights for advancing energy research in economically viable areas. Recommendations included conducting site-specific assessments, utilizing advanced modeling tools, implementing regular maintenance protocols, and enhancing communication among system components.
Planning Of Procurement o different goods and services
Ch 03
1. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-1
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Chapter 3
Modeling & Solving LP Problems In A Spreadsheet
1. In general, it does not matter what is placed in a variable (changing) cell. Ultimately, Solver will
determine the optimal values for these cells. If the model builder places formulas in changing cells,
Solver will replace the formulas with numeric constants representing the optimal values of the decision
variables. An exception to this general principle is found in Chapter 8 where, when solving nonlinear
programming problems, the values placed in the changing cells represent the initial starting point for the
optimizer.
2. Communication - once the user understands the first formula in a range of copied cells, he or she
should understand all the formulas in the range.
Reliability - assuming the first formula is entered correctly, all the copied formulas should be correct
also.
Auditability - once the user understands the first formula in a range of copied cells, he or she should
understand (and audit) all the formulas in the range
Maintainability - if a change needs to be made, it can be made in one formula and then copied as
necessary.
3. TV ads = 10, Magazine ads =25, Maximum profit = $775,000
See file: Prb3_3.xls
4. Ore 1 = 28, Ore 2 = 8, Minimum cost = $3,480
See file: Prb3_4.xls
5. Beef = 50%, Pork = 50%, Minimum cost per pound = $0.75
See file: Prb3_5.xls
6. Razors = 240, Zoomers = 420, Maximum profit = $33,600
See file: Prb3_6.xls
7. Executive desks = 100, Senator desks = 500, Maximum profit = $59,300
See file: Prb3_7.xls
8. Acres planted in watermelons = 60, Acres planted in cantaloupes = 40, Maximum profit = $26,740
See file: Prb3_8.xls
9. Doors = 20, Windows = 40 , Maximum profit = $26,000
See file: Prb3_9.xls
10. Desktops = 46.15, Laptops = 69.23, Maximum profit = $90,000 (alternate optimal solutions exist)
See file: Prb3_10.xls
11. TV = 20, Managize = 2 , Minimal cost = $3.5 million
See file: Prb3_11.xls
12. a. X1 = Number of country tables to produce
X2 = Number of contemporary tables to produce
MAX 350 X1 + 450 X2
ST 1.5 X1 + 2 X2 ≤ 1,000
3 X1 + 4.5 X2 ≤ 2,000
2.5 X1 + 1.5 X2 ≤ 1,500
2. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-2
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X1/ ( X1 +X2) ≥ 0.20 (implement as X1 ≥ 0.2* ( X1 +X2) )
X2/ ( X1 +X2) ≥ 0.30 (implement as X2 ≥ 0.3* ( X1 +X2) )
Xi ≥ 0
Many students attempt to implement the ratio constraints in their original form; resulting in a
division by zero error at the null solution and a message from Solver that the model is not linear.
The algebraic equivalence of the alternate form of these constraints (given parenthetically above)
should be noted.
b. See file: Prb3_12.xls
c. X1 = 405.80, X2 = 173.91, Maximum revenue = $220,290
13. a. X1,j = Number of dehumidifiers made in Atlanta in month j
X2,j = Number of dehumidifiers made in Phoenix in month j
Bj = Beginning inventory in month j
MIN 400 (X11 + X12 + X13 ) +360 (X21 + X22 + X23 ) + 30 ( B1 + B2 + B3 )
ST B1 + X11 + X21 – 300 ≥ 0
B2 + X12 + X22 – 400 ≥ 0
B3 + X13 + X23 – 500 ≥ 0
Xij ≤ 300
Xij ≥ 0
Where
B1 = 0
B2 = B1 + X11 + X21 – 300
B3 = B2 + X12 + X22 – 400
b. See file: Prb3_13.xls
c. X11 = 0, X12 = 100, X13 = 200, X21 = 300, X22 = 300, X23 = 300, Maximum revenue = $444,000
14. a. X1 = pounds of Whole product to produce
X2 = pounds of Cluster product to produce
X3 = pounds of Crunch product to produce
X4 = pounds of Roasted product to produce
MAX 1.85 X1 + 1.4 X2 + 1.04 X3 + 1.40 X4
ST 1 X1 + 1 X2 + 1 X3 + 1 X4 < 3600
2 X1 + 1.5 X2 + 1 X3 + 1.75 X4 < 3600
1 X1 + 0.7 X2 + 0.2 X3 + 0.00 X4 < 3600
2.5 X1 + 1.6 X2 + 1.25 X3 + 1 X4 < 3600
0.6 X1 + 0.4 X2 + 0.2 X3 + 1 X4 < 1100
0.4 X1 + 0.6 X2 + 0.8 X3 + 0 X4 < 800
1,000 < X1 < 99,999
400 < X2 < 500
0 < X3 < 150
0 < X4 < 200
b. See file: Prb3_14.xls
c. X1= 1000, X2= 500, X3= 80, X4= 200, Maximum profit = $2,913.2
15. a. N1 = number of Newspaper ads to run at $1,000 each
N2 = number of Newspaper ads to run at $900 each
3. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-3
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N3 = number of Newspaper ads to run at $800 each
T1 = number of Television ads to run at $12,000 each
T2 = number of Television ads to run at $10,000 each
T3 = number of Television ads to run at $8,000 each
MAX 900 N1 + 700 N2 + 400 N3 + 10,000 T1 + 7,500 T2 + 5,000 T3
ST 1,000 N1 + 900 N2 + 800 N3 + 12,000 T1 + 10,000 T2 + 8,000 T3 ≤ 145,000
Ni ≤ 10
Ti ≤ 5
Ti, Ni > 0
b. See file: Prb3_12.xls
c. N1 = 10, N2 = 10, N3 = 0, T1 = 5 , T2 = 5 , T3 = 2
New Customers = 113,500
d. N3 is used before N2.
16. a. Xij = Square feet of space rented in month i (i=1, 2, 3, 4, 5) through month j (j=i, i+1, …, 5)
MIN 55X11 + 95X12 + 130X13 + 155X14 + 185X15 + 55X22 + 95X23 + 130X24 + 155X25 + 55X33 +
95X34 + 130X35 + 55X44 + 95X45 + 55X55
ST X11 + X12 + X13 + X14 + X15 > 20000
X12 + X13 + X14 + X15 + X22 + X23 + X24 + X25> 30000
X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 > 40000
X14 + X15 + X24 + X25 + X34 + X35 + X44 + X45 > 35000
X15 + X25 + X35 + X45 + X55 > 50000
Xij > 0
b. See file: Prb3_16.xls
c. X15 = 20000, X25 = 10000, X33 = 5000, X35 = 5000, X55 = 15000
Total leasing cost = $7 million
d. $9.625 million
17. a. X1 = Amount invested in Bonds
X2 = Amount invested in Mortgages
X3 = Amount invested in Car loans
X4 = Amount invested in Personal Loans
MAX 10 X1 + 8.5 X2 + 9.5 X3 + 12.5 X4
ST X1 + X2 + X3 + X4 = 650,000
X4 ≤ .25*(650000)
X4 ≤ X2
X4 ≤ X1
X1 , X2 , X3 , X4 ≥ 0
b. See file: Prb3_17.xls
c. X1 = 325,000, X2 = 162,500, X3 = 0, X4 = 162,500, Maximum return = 10.25%
18. a. X1 = number of HyperLink cards to produce
X2 = number of FastLink cards to produce
X3 = number of SpeedLink cards to produce
X4 = number of MicroLink cards to produce
X5 = number of EtherLink cards to produce
MAX 53 X1 + 48 X2 + 33 X3 + 32 X4 + 38 X5
ST 20 X1 + 15 X2 + 10 X3 + 8 X4 + 5 X5 ≤ 80,000
28 X1 + 24 X2 + 18 X3 + 12 X4 + 16 X5 ≤ 100,000
8 X1 + 8 X2 + 4 X3 + 4 X4 + 6 X5 ≤ 30,000
4. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-4
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0.75 X1 + 0.6 X2 + 0.5 X3 + 0.65 X4 + 1 X5 ≤ 5,000
2 X1 - 1 X2 ≤ 0
Xi ≥ 500
b. See file: Prb3_18.xls
c. X1 = 500, X2 = 1000, X3 = 1500, X4 = 2250, X5 = 500, Total Profit = $215,000
d. No. The assembly constraint is nonbinding.
19. a. A = amount to invest in bond A
B = amount to invest in bond B
C = amount to invest in bond C
D = amount to invest in bond D
E = amount to invest in bond E
MAX 0.095A + 0.08B + 0.09C + 0.09D + 0.09E
ST A + B + C + D + E = 100,000
B + E ≥ 50,000
A + D + E ≤ 50,000
A + B + D ≥ 30,000
0.095A + 0.08B + 0.09D ≥ 0.4* (0.095A + 0.08B + 0.09C + 0.09D + 0.09E)
A, B, C, D, E ≥ 0
b. See file Prb3_19.xls
c. A=20,339, B=20,339, C=29,661, D=0 , E=29,661
Maximum return = $8,898 (or 8.898%)
20. a. M1= number of electric trimmers to make
M2= number of gas trimmers to make
B1= number of electric trimmers to buy
B2= number of gas trimmers to buy
MIN 55M1 + 85 M2 + 67 B1 + 95 B2
ST M1 + B1 = 30,000
M2 + B2 = 15,000
0.20M1 + 0.40M2 ≤ 10,000
0.30M1 + 0.50M2 ≤ 15,000
0.10M1 + 0.10M2 ≤ 5,000
Mi, Bi ≥ 0
b. See file: Prb3_20.xls
c. M1=30,000, M2=10,000, B1=0, B2=5,000
Minimum cost = $2,975,000
21. a. Xij = 1 if component i is assigned to company j; 0, otherwise
MIN 185 X1A +225 X1B +193 X1C +207 X1D
+200 X2A +190 X2B +175 X2C +225 X2D
+330 X3A +320 X3B +315 X3C +300 X3D
+375 X4A +389 X4B +425 X4C +445 X4D
ST X1A + X1B + X1C + X1D = 1
X2A + X2B + X2C + X2D = 1
5. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-5
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X3A + X3B + X3C + X3D = 1
X4A + X4B + X4C + X4D = 1
X1A + X2A + X3A + X4A = 1
X1B + X2B + X3B + X4B = 1
X1C + X2C + X3C + X4C = 1
X1D + X2D + X3D + X4D = 1
b. See file: Prb3_21.xls
c. X1A = X2C = X3D = X4B = 1, Minimum cost = $1,049 (in $1,000s)
22. a. Pi = proportion of compound i to include in the mix
MIN 5.00 P1 + 5.25 P2 + 5.50 P3
ST 0.20 P1 + 0.40 P2 + 0.10 P3 ≥ .20
0.60 P1 + 0.30 P2 + 0.40 P3 ≥ .30
0.20 P1 + 0.30 P2 + 0.50 P3 ≥ .30
0.20 P1 + 0.30 P2 + 0.50 P3 ≤ .45
P1 + P2 + P3 = 1.0
Pi ≥ 0
b. See file: Prb3_22.xls
c. P1=0.5714, P2=0.1429, P3=0.2857
Minimum cost per pound = $5.18
23. a. Bi = pounds of grade i fruit used in baskets
Ji = pounds of grade i fruit used in juice
MAX: $2.50 (B1 + B2 + B3 + B4 + B5 ) + $1.75 (J1 + J2 + J3 + J4 + J5 )
S.T.: B1 + J1 ≤ 90
B2 + J2 ≤ 225
B3 + J3 ≤ 300
B4 + J4 ≤ 100
B5 + J5 ≤ 75
1 B1 + 2 B2 + 3 B3 + 4 B4 + 5 B5 ≥ 3.75 (B1 + B2 + B3 + B4 + B5 )
1 J1 + 2 J2 + 3 J3 + 4 J4 + 5 J5 ≥ 2.50 (J1 + J2 + J3 + J4 + J5 )
Bi , Ji ≥ 0
b. See file: Prb3_23.xls
c. B1 = 0, B2 = 46.67, B3 =0, B4 = 100 B5 = 45.33,
J1 = 90, J2 = 178.33, J3 =300, J4 = 0 J5 = 29.67,
Profit = $1,526,500
24. a. XiR = barrels of input i used to produce regular
XiS = barrels of input i used to produce supreme
MAX: (21-17.25)X1R+(21-15.75)X2R+(21-17.75)X3R+(25-17.25)X1S+(25-15.75)X2S+(25-17.75)X3S
ST: X1R + X1S ≤ 150
X2R + X2S ≤ 350
X3R + X3S ≤ 300
X1R + X2R + X3R = 300
X1S + X2S + X3S = 450
(100X1R + 87X2R + 110X3R)/300 ≥ 90
6. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-6
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(100X1S + 87X2S + 110X3S)/450 ≥ 97
Xij ≥ 0
b. See file Prb3_24.xls
c. X1R=0, X2R=260.87, X3R=39.13, X1S=150, X2S=89.13, X3S=210.87 (alternate optimal exist)
Maximum Profit = $5,012.5 (in $1,000s)
25. a. X1 = number of workers starting at 12 am
X2 = number of workers starting at 4 am
X3 = number of workers starting at 8 am
X4 = number of workers starting at 12 pm
X5 = number of workers starting at 4 pm
X6 = number of workers starting at 8 pm
MIN X1+ X2 + X3+ X4 + X5+ X6
ST X6 + X1 ≥ 90
X1 + X2 ≥ 215
X2 + X3 ≥ 250
X3 + X4 ≥ 165
X4 + X5 ≥ 300
X5 + X6 ≥ 125
Xi ≥ 0
b. See file: Prb3_25.xls
c. X1=90, X2 =250, X3=0, X4 =175, X5=125, X6=0 (alternate optimal solutions exist)
Minimum number of employees = 640
Teaching Note: As an interesting extension to this problem, ask students to consider how to minimize
the maximum number of excess employees on any shift while holding the total number of employees
used at its optimal value of 640.
26. a. Xij = number of units of specimen i assigned to machine j
MIN 3 X1A + 4 X2A + 4 X3A + 5 X4A + 3 X5A
+ 5 X1B + 3 X2B + 5 X3B + 4 X4B + 5 X5B
+ 2 X1C + 5 X2C + 3 X3C + 3 X4C + 4 X5C
ST 3 X1A + 4 X2A + 4 X3A + 5 X4A + 3 X5A ≤ 480
5 X1B + 3 X2B + 5 X3B + 4 X4B + 5 X5B ≤ 480
2 X1C + 5 X2C + 3 X3C + 3 X4C + 4 X5C ≤ 480
X1A + X1B + X1C = 80
X2A + X2B + X2C = 75
X3A + X3B + X3C = 80
X4A + X4B + X4C = 120
X5A + X5B + X5C = 60
Xij ≥ 0
b. See file: Prb3_26.xls
c. X1C = 80, X2B = 75, X3A = 75, X3C = 5, X4B = 18.33, X4C = 101.67, X5A = 60
Minimum processing time = 1258.33 minutes. (If an integer solution is needed the LP solution can
be
7. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-7
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rounded to yield the optimal integer solution.)
d. Machine A & C are used all 480 minutes, machine B is used 298.33 minutes
e. A solution exists where all machine are used for an equal amount of time (425.5 minutes each).
This
increases the total time used to 1276.5 minutes.
27. a. Pi = proportion of coal i to include in the mix
MAX 24,000 P1 + 36,000 P2 + 28,000 P3
ST 1,100 P1 + 3,500 P2 + 1,300 P3 ≤ 2,500
1.7 P1 + 3.2 P2 + 2.4 P3 ≤ 2.8
P1 + P2 + P3 = 1.0
Pi ≥ 0
b. See file: Prb3_27.xls
c. P1=0.058, P2=0.5507, P3=0.3913
Maximum steam production = 32,174 pounds per ton
d. 32,174 × 30 = 965,217 pounds of steam
28. a. X1 = number of CD players to produce
X2 = number of tape decks to produce
X3 = number of stereo tuners to produce
MAX 75 X1 + 50 X2 + 40 X3
ST 3 X1 + 2 X2 + 1 X3 ≤ 400,000
50,000 ≤ X1 ≤ 150,000
50,000 ≤ X2 ≤ 100,000
50,000 ≤ X3 ≤ 90,000
b. See file: Prb3_28.xls
c. X1= 70,000, X2 = 50,000, X3 = 90,000
Maximum profit = $11,350,000
29. a. Xij = number of cars shipped from location i to location j
MIN 54 X13 + 17 X14 + 23 X15 + 30 X16 + 24 X23 + 18 X24 + 19 X25 + 31 X26
ST X13 + X14 + X15 + X16 = 16
X23 + X24 + X25 + X26 = 18
5 ≤ X13 + X23 ≤ 10
5 ≤ X14 + X24 ≤ 10
5 ≤ X15 + X25 ≤ 10
5 ≤ X16 + X26 ≤ 10
Xij ≥ 0
b. See file: Prb3_29.xls
c. X23 = 9, X14 = 10, X15 = 1, X25 = 9, X16 = 5
Minimum transportation cost = $730
30. a. See file: Prb3_30.xls
b. Minimum cost = $3,011,360, optimal shipping plan (using all production capacity) is:
8. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-8
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FromTo Tacoma San Diego Dallas Denver St. Louis Tampa Baltimore
Macon 0 0 0 0 0 12000 6000
Louisville 600 0 0 0 14400 0 0
Detroit 400 0 10800 12600 0 0 1200
Phoenix 5800 14200 0 0 0 0 0
31. a. See file: Prb3_31.xls
c. Minimum cost = $44,067.67, recycling plan is:
Newsprint Packaging Print Stock
Newspaper 588.24 11.76 0.00
Mixed Paper 0.00 71.43 428.57
White Office Paper 0.00 300.00 0.00
Cardboard 0.00 397.78 0.00
32. a. Xij = number of bottles produced at vineyard i sold to restaurant j
MAX 39X11 + 36X12 + 34X13 + 34X14 + 32X21 + 36X22 + 37X23 + 34X24
ST X11 + X12 + X13 + X14 = 3,500
X21 + X22 + X23 + X24 = 3,100
X11 + X21 ≤ 1800
X12 + X22 ≤ 2300
X13 + X23 ≤ 1250
X14 + X24 ≤ 1750
Xij ≥ 0
b. See file: Prb3_32.xls
c. X11 = 1,800, X12 = 1,700, X22 = 600, X23 = 1,250, X24 = 1,250, Maximum profit = $241,750
(Alternate optima exist.)
33. a. X1 = Cases of Extra Hot sauce to produce
X2 = Cases of Hot sauce to produce
X3 = Cases of Mild sauce to produce
A1 = Advertising dollars spent promoting Extra Hot sauce
A2 = Advertising dollars spent promoting Hot sauce
A3 = Advertising dollars spent promoting Mild sauce
MAX 4 X1 + 4.5 X2 + 4.75 X3 - A1 - A2 - A3
ST X1 = 8,000 + 10 A1
X2 = 10,000 + 8 A2
X3 = 12,000 + 5 A3
A1 + A2 + A3 ≤ 25,000
Ai ≥ 5,000
Note that the Xi can be computed directly from the Ai. Therefore, the Ai are the only decision
variables (changing cells) in the model. The Xi can be computed in the spreadsheet using the
conditions imposed by the first three constraints. Therefore, it is not necessary to indicate these as
constraints cells for Solver.
9. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-9
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b. See file: Prb3_33.xls
c. X1 = 158,000, X2 = 50,000, X3 = 37,000, A1 = 15,000, A2 = A3 = 5000
Maximum profit = $1,007,750
34. a. Pi = Number of units to produce in month i
Ii = Inventory held at the end of month i
MIN 49 X1 + 45 X2 + 46 X3 + 47 X4 - 1.5 (120 + 2I1 + 2I2 + 2I3 + I4 )/2
ST I1 =120 + P1 - 420
I2 = I1+ P2 - 580
I3 = I2+ P3 - 310
I4 = I3+ P4 - 540
400 ≤ P1 ≤ 500
400 ≤ P2 ≤ 520
400 ≤ P3 ≤ 450
400 ≤ P4 ≤ 550
Ii ≥ 50
Note that the Ii can be computed directly from the Pi. Therefore, the Pi are the only decision
variables (changing cells) in the model. The Ii can be computed in the spreadsheet using the
conditions imposed by the first four constraints. Therefore, it is not necessary to indicate these as
constraints cells for Solver. However, note that lower bounds of 50 must be indicated for these
cells.
b. See file: Prb3_34.xls
c. X1 = 410, X2 = 520, X3 = 400, X4 = 450, I1 = 110, I2 = 50, I3 = 140, I4 = 50
Minimum cost = $83,617
35. a. Xij = tons of commodity i stored in hold j
MAX 70(X11+X12+X13) + 50(X21+X22+X23) + 60(X31+X32+X33) + 80(X41+X42+X43)
ST X11 + X12 + X13 < 4800
X21 + X22 + X23 < 2500
X31 + X32 + X33 < 1200
X41 + X42 + X43 < 1700
X11 + X21 + X31 + X41 < 3000
X12 + X22 + X32 + X42 < 6000
X13 + X23 + X33 + X43 < 4000
40X11 + 25X21 + 60X31 + 55X41 < 145000
40X12 + 25X22 + 60X32 + 55X42 < 180000
40X13 + 25X23 + 60X33 + 55X43 < 155000
0.9(X13 + X23 + X33 + X43 ) < X11 + X21 + X31 + X41 < 1.1(X13 + X23 + X33 + X43)
0.4 × Total < X12 + X22 + X32 + X42 < 0.6 × Total
Xij > 0
b. See file: Prb3_35.xls
c. Profit = $669,000
1 2 3 4
Forward X11=1287.5 X21=0 X31=0 X41=1700
10. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-10
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Center X12=1580 X22=2500 X32=0 X42=0
Rear X13=1932.5 X23=0 X33=1200 X43=0
36. a. Xij = Square feet (in 1000s) leased at beginning of month i for j months
MIN 300 (X11 + X21 + X31 + X41 + X51) + 525 (X12 + X22 + X32 + X42 ) + 775 (X13 + X23 + X33 )
+ 850 (X14 + X24) + 975 X15
ST X11 + X12 + X13 + X14 + X15 > 25
X21 + X22 + X23 + X24 + X12 + X13 + X14 + X15 > 10
X31 + X32 + X33 + X22 + X23 + X24 + X13 + X14 + X15 > 20
X41 + X42 + X32 + X33 + X23 + X24 + X14 + X15 > 10
X51 + X42 + X33 + X24 + X15 > 5
Xij > 0
b. See file: Prb3_36.xls
c. X11 = 15, X14 = 5, X15 = 5, X31 = 10
Total cost = $16,625
37. a. Pi = Number of tons purchased in month i
Si = Number of tons sold in month i
Ii = Inventory held at the end of month i
MAX 135 S1 + 110 S2 + 150 S3 + 175 S4 + 130 S5 + 145 S6
-135 P1 - 110 P2 - 150 P3 - 175 P4 - 130 P5 - 145 P6
- 10 ( 70 + 2I1 + 2I2 + 2I3 + 2I4 + 2I5 + 2I6)/2
ST I1 = 70 + P1 - S1
I2 = I1 + P2 - S2
I3 = I2 + P3 - S3
I4 = I3 + P4 - S4
I5 = I4 + P5 - S5
I6 = I5 + P6 - S6
I6 = 0
0 ≤ Ij ≤ 400
Pj ≥ 0
Sj ≥ 0
Note that the Ij can be computed directly from the Pj and Sj . Therefore, the Pj and Sj are the only
decision variables (changing cells) in the model. The Ij can be computed in the spreadsheet using
the conditions imposed by the first six constraints. Therefore, it is not necessary to indicate these
as constraints cells for Solver. However, note that lower bounds of 0 and upper bounds of 400
apply to these cells. Some students will want to know how much Earl paid for the 70 tons of
soybeans in the beginning inventory. This represents a sunk cost that is irrelevant for the problem
at hand.
b. See file: Prb3_37.xls
c. P1 = 0, P2 = 400, P3 = 0, P4 = 0, P5 = 400, P6 = 0
S1 = 70, S2 = 0, S3 = 0, S4 = 400, S5 = 0, S6 = 400
Maximum profit = $29,100
38. a. A = Amount to invest in investment A
11. Chapter 3 - Modeling & Solving LP Problems In A Spreadsheet : S-11
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B = Amount to invest in investment B
C = Amount to invest in investment C
D = Amount to invest in investment D
E = Amount to invest in investment E
S98 = Amount to invest in savings in 1998
S99 = Amount to invest in savings in 1999
S00 = Amount to invest in savings in 2000
MAX 1.25 B + 1.35 C + 1.13 D + 1.08 S00
ST A + C + E + S98 = 1,000,000
0.5 A + 1.08 S98 - B - S99 = 0
0.8 A + 1.27 E + 1.08 S99 - D - S00 = 0
0 ≤ A ≤ 500,000
0 ≤ B ≤ 500,000
0 ≤ C ≤ 500,000
0 ≤ D ≤ 500,000
0 ≤ E ≤ 500,000
50,000 ≤ S98 ≤ 500,000
50,000 ≤ S99 ≤ 500,000
50,000 ≤ S00 ≤ 500,000
b. See file: Prb3_38.xls
c. A=500,000, B= 275,685, C=0, D=500,000, E=429,921, S98=70,079, S99=50,000, S00=500,000
Maximum amount of money at the beginning of 2001 = $1,449,606
39. a. MIN A12 + B13 + C14 + D18
ST 1.06A12 - A23 = 0
1.06A23 + 1.14B13 - A34 - B35 = 0
1.06A34 + 1.18C14 - A45 - C47 = 0
1.06A45 + 1.14B35 - A56 - B57 = 0
1.06A56 - A67 = 12
1.06A67 + 1.14B57 + 1.18C47 - A78 - B79 = 14
1.06A78 + 1.65D18 - A89 = 16
1.06A89 + 1.14B79 = 18
Aij, Bij, Cij, Dij ≥ 0
b. See file Prb3_39.xls
c. A56 = $11,321, A89 = $16,981, B13 = $18,161, B35 = $20,703, B57 = $12,281, D18 = $19,989
Minimum investment = $38,149
40. a. Same as in problem 38 above with the following additional constraints:
-3A12 -1B13 + 2C14 + 4D18 ≤ 0
-3A23 -1B13 + 2C14 + 4D18 ≤ 0
-3A34 -1B35 + 2C14 + 4D18 ≤ 0
-3A45 -1B35 + 2C47 + 4D18 ≤ 0
-3A56 -1B57 + 2C47 + 4D18 ≤ 0
-3A67 -1B57 + 2C47 + 4D18 ≤ 0
-3A78 -1B79 + 4D18 ≤ 0
-3A89 -1B79 ≤ 0
b. See file Prb3_40.xls
c. A56 = $11,321, A78 = $4,945, A89 = $1,345, B13 = $31,306, B35 = $35,688, B57 = $29,364,
B79 = $14,531, D18 = $7,341
Minimum investment = $38,647
41. a. See file Prb3_41.xls
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b. Borrow $220,000 for 5 months in month 1.
Borrow $275,000 for 4 months in month 2.
Borrow $49,505 for 1 months in month 3.
Borrow $270,270 for 2 months in month 3.
Borrow $5,225 for 3 months in month 3.
Total Finance Charge = $22,878.
c. There is no feasible solution if the company is restricted to borrowing no more than $100,000 at
each level in the term / rate structure. Ask students to determine what the borrowing limit would
need to be increase to in order to obtain a feasible solution. The answer (of $101,257) can be
obtained by: 1) making the borrowing limit a changing cell, and 2) also making it the set cell and
minimizing its value.
42. a. Ti = Number of people in group i surveyed by telephone
Wi = Number of people in group i surveyed in “person” vai web-cam
MIN: 18 T1 + 14T2 + 25T3 + 20T4 + 40W1 + 35W2 + 60W3 + 45W4
ST
2000 < T1 + T2 + W1 + W2 < 4000
1000 < W1 + W2 + W3 + W4 < 4000
1000 < W1 + W2 + W3 + W4 < 4000
-4000 < T1 – W1 < 0
0 < W2 + W4 + T2 + T4 < 1600
0 < 0.25W2 + 0.25W4 – 0.75T2 – 0.75T4 < 4000
400 < Ti , Wi < 2000
b. See file: Prb4_42.xls
c. T1 = 1000, T2 = 1200, T3 = 400, T4 = 400, W1 = 1000, W2 =W3=W4= 0 ;
Minimum cost = $92,800
43. See file: Prb3_43.xls
Take a six month loan for $48,000.
Borrow $27,200 against receivables in February and $105,000 in March.
Defer $3,000 in payments in March.
44. a. See file: Prb3_44.xls
b. Purchase 79.5337 units on bond 1, 82.8987 units of bond 2, and 35.023 units of bond 3 and invest
$52,482 in the savings account. Total investment = $246,769.
45. a. See file: Prb3_45.xls
b. Total Profit = $1,309,900
Thousand cubic feet
Day Bought Sold
1 200.00 0.00
2 0.00 170.00
3 0.00 180.00
4 160.00 0.00
5 200.00 0.00
6 0.00 180.00
7 0.00 180.00
8 0.00 0.00
9 180.00 0.00
10 0.00 180.00
c. The prices and inventory level should be updated and re-solved everyday with the decisions for
"Day 1" implemented each day.
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Teaching note: Some students may want to know how much money the company paid for the
150,000 cf of gas in storage. Because this represents a sunk cost, it is irrelevant for decision
making purposes.
46. a. See file: Prb3_44.xls
b. Sheritown Inn, Merrylot, FairPrice Inn, and Western Hotels are efficient.
c. Target values:
Satisfact. Value Price Conv. Comfort Climate Service Food
88 86.1474 64.8421 1.38947 1.01895 0.18526 0.46316 0.46316
47. a. See file: Prb3_45.xls
b. Branches 1, 2, 6 & 8 are efficient.
c. Target values:
ROA New Loans Satisfaction Labor Hrs Op. Costs
6.4000 928.9375 98.0000 5.5692 6.8260
Case 3-1: Putting the Link in the Supply Chain
a. See file: Case3_1.xls; Maximum profit = $1,304,544
b. The solution uses all the aluminum in Daytona and Memphis and all the wood in Tempe.
c. Maximum profit = $1,349,439
d. If 80% of the demand must be met the company could earn $1,329,986. This is $25,422 more than
under the original 90% scenario. Thus, after paying the $10,000 penalty the company would still be
$15,422 ahead under the 80% scenario. However, the increased profit may or may not offset the
potential good will that may be lost if the customers are not happy having less of their desired order
quantities met.
Case 3-2: Baldwin Enterprises
a. See file: Case3_2.xls
b. In millions:
Sell Buy USD EUR GBP HKD JPY
USD 0.000 2.945 0.000 0.000 0.000
EUR 0.000 0.000 0.000 0.000 0.000
GBP 0.000 0.000 0.000 0.000 0.840
HKD 9.328 0.000 0.000 0.000 0.000
JPY 0.000 0.000 0.000 131.314 0.000
c. Transaction cost: $27,867
d. The transaction cost barely changes to $27,860.
e. This creates an unbounded solution – or an arbitrage opportunity.
Case 3-3: The Wolverine Retirement Fund
See file: Case3_3.xls
a. Buy 411 share of AC&C, 169 share of MicroHard (fractional solution rounded up), Total Cost =
$495,892
b. Cost of stipulation (1): $506,590, Cost of stipulaiton (2): $500,736
Case 3-4: Saving the Manatees
a. See file: Case3_4.xls. Maximum impact rating = 23,523
b. Minimum Full-Page ads in the daily papers, minimum full-page Sunday ads, minimum evening TV
spots, maximum highway billboards, maximum 15 & 30 second radio ads, minimum full-page magazine
ads, maximum number of total daily paper ads, maximum Sunday paper ads, maximum magazine ads.
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c. Not at all. These constraints are not binding.
d. Maximum impact rating = 29,289