IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Problem Set 3
Chapter 5, Problem 9.
Two methods can be used for producing expansion anchors. Method A costs $80,000
initially and will have a $15,000 salvage value after 3 years. The operating cost w
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Problem Set 4
Chapter 5, Problem 30.
Compare the following alternatives on the basis of their capitalized cost at an interest rate
of 10% per year.
PetroleumBased
Feedstock
250,000
130,000
400,000
27
IE 347 FALL 2015 / 2016
OCTOBER 28 2015 - WEEK 4
IE 347 FALL 2015 / 2016
Exam I November 13, Thursday
Exam II December 17, Thursday
17:30
17:30
Quiz I November 4, Wednesday
17:30
Engineering Economy Factors
Single Payment
Compound Factor
(F/P, i, n) = (1+
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Solutions for Recitation 1
Chapter 2, Solution 36.
Convert future to present and then solve for G using P/G factor:
6000(P/F,15%,4) = 2000(P/A,15%,4) G(P/G,15%,4)
6000(0.5718) = 2000(2.8550) G(3.7864
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 1
Chapter 2, Problem 36
For the cash flow shown below, determine the value of G that will make the future worth in
year 4 equal to $6000 at an interest rate of 15% per year.
Year
Cash Flow
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 12
Chapter 16, Problem 19
A warehouse costs $800,000 to construct for Ace Hardware. It has a 15-year life with an
estimated resale value of 80% of the construction cost. However, the build
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 11
Chapter 14, Problem 24
An engineer must recommend one of two machines for integration into an upgraded
manufacturing line. She obtains estimates from two salespeople. Salesman A gives h
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 8
Question 1
Chapter 8, Problem 32
1
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 8
Chapter 8, Problem 33.
The four alternatives described below are being evaluated.
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Solutions for Recitation 5
Part 1
In order to find the maximum amount of A, present worth of the cash flow must be equal to 0.
ia = (1 + 0.1/2)2 1 (effective annual interest rate)
= 10.25%
P = 50,000
Present Worth Analysis
Basics
Economic evaluation of an alternative requires cash ow
estimates over a stated time period and a criterion for selecting
the best alternative
The alternatives are developed from project proposals to
accomplish a stated purpos
Annual Worth Analysis
Basics
So far we learned PW and FW as comparison tools
AW analysis is said to be the best comparison tool.
AW value is the equivalent uniform annual worth of all estimated
receipts and disbursements during the life cycle of the proje
After-Tax Economic Analysis
Basics
Some basic tax terms and relations useful in engineering economy
studies are explained here.
Gross Income, GI: is the total income realized from:
revenue-producing sources, and
sale of assets, royalties, license fees.
In
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Homework 1
Due Date: October 18, 2011
Chapter 2, Problem 5.
Sensotech Inc., a maker of microelectromechanical systems, believes it can reduce
product recalls by 10% if it purchases new software for d
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Solutions for Recitation 12
Chapter 16, Problem 19
B = $800,000; n = 30; S = 0
(a)
Straight line depreciation:
Dt = 800,000 = $26,667
30
(b)
t = 5, 10, 25, and all other years
Double declining balanc
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Solutions for Recitation 11
Chapter 14, Problem 24
Use the inflated rate of return for Salesman A and real rate of return for B
if = 0.15 + 0.05 + (0.15)(0.05) = 20.75%
PWA = -60,000 55,000(P/A,20.75
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Solutions for Recitation 9
Chapter 9, Problem 33
Compare A to DN since it is not necessary to select one of the sites.
A vs DN
AW of Cost = 50(A/P,10%,5) + 3
= 50(0.26380) + 3
= 16.19
AW of Benefits
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Solutions for Recitation 8
Question 1
The solution is available in the Lecture Presentation 8 notes starting from page 3.
Chapter 8, Problem 32
8.32 (a) All machines have ROR > MARR of 12% and all in
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 7
Chapter 7, Problem 32.
A company that makes clutch disks for race cars had the cash flows shown below for one
department. Calculate the composite rate of return for the given cash flows,
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 6
Chapter 5, Problem 15.
Two methods are under consideration for producing the case for a portable hazardous material
photoionization monitor. A plastic case will require an initial invest
IE 347 Engineering Econ. & Cost Analysis-II, Fall 2011, Recitation 5
Part 1
An engineering company has bought a new machine for 5 million dollars now. They estimate an
operating cost of $A each year, starting by the end of the first year. The expected ser
Benet / Cost Analysis
Public Sector Projects
Projects in private sector are owned by corporations, partnerships,
and individuals and used by customers.
Projects in public sector are owned, used and nanced by citizens.
Public sector projects have a primary
Decision Rules
Single Alternative
Based on Sign
Changes of Cash
Flow:
Simple
Investment
Single
Project
i* = IRR
Accept if i* > MARR
start with zero,
one sign change
Non-Simple
Investment
i* ? IRR
Net Investment Test:
Based on Signs of PB
(all need to be n