B1 Introduction
Engineering projects often involve an investment in equipment, buildings, or other assets that are put to productive use. As time passes, these assets lose value, or depreciate. The first part of this chapter is concerned with the concept of depreciation and several methods that are commonly used to model depreciation. Depreciation is taken into account when a firm states the value of its assets in its financial statements, as seen in the second half of this chapter. It also forms an important...
Problems Jgj
For additional practice, please see the problems with selected solutions provided on the Student CD-ROM that accompanies this book. 4.1 IQ Computer assembles Unix workstations at its plant. The current product line is nearing the end of its marketing life, and it is time to start production of one or more new products. The data for several candidates are shown below. The maximum budget for research and development is S300 000. A minimum of S200 000 should be spent on these projects. It is...
Problems Udj
For additional practice, please see the problems with selected solutions provided on the Student CD-ROM that accompanies this book. 6.1 For each of the following, state whether the loss in value is due to use-related physical loss, time-related physical loss, or functional loss a -Albert sold his two-year-old computer for S500, but he paid S4000 for it new. It wasn't fast enough for the new software he wanted. b Beatrice threw out her old tennis shoes because the soles had worn thin. c Claudia...
l vl 1 x J iiiTig
Care must be taken in using the geometric gradient to present worth conversion factor. Four cases may be distinguished 1. i gt g gt 0. Growth is positive, but less than the rate of interest. The growth adjusted interest rate, i , is positive. Tables or functions built into software may be used to find the conversion factor. 2. g gt i gt 0. Growth is positive and greater than the interest rate. The growth adjusted interest rate, i , is negative. It is necessary to compute the conversion factor...
pAgiX V
Capitalized value formula Capital recovery formula Engineering Economics in Action, Part 3B This time it was Naomi who stuck her head in Clem's doorway. Here's the recommendation on the shipping palletizer. Oh, and thanks for the hint on the leasing figures. It cleared up my confusion right away. No problem. What did you figure out Clem had his mentor expression on his face, so Naomi knew he was expecting a clear explanation of the trick used by the leasing company. Well, as you hinted, they...
Compound Interest Factors for Single Disbursements or Receipts
In many situations, a single disbursement or receipt is an appropriate model of cash flows. For example, the salvage value of production equipment with a limited service life will be a single receipt at some future date. An investment today to be redeemed at some future date is another example. Figure 3.1 illustrates the general form of a single disbursement or receipt. Two commonly used factors relate a single cash flow in one period to another single cash flow in a different period. They are...
Conversion Factor for Geometric Gradient Series
A geometric gradient series is a series of cash flows that increase or decrease by a constant percentage each period. The geometric gradient series may be used to model inflation or deflation, productivity improvement or degradation, and growth or shrinkage of market size, as well as manv other phenomena. In a geometric series, the base value of the series is A and the growth rate in the series the rate of increase or decrease is referred to as g. The terms in such a series are given by A, A l...
PAi8 50 00011 000
134 CHAPTER 5 Comparison Methods Part 2 From the interest factor tables, or by trial and error with a spreadsheet, P 4,14 ,8 4.6388 P A,IS ,8 4.4873 By interpolation or further trial and error, i 14.5 The slicer alone is thus economically justified and is better than the do nothing alternative. We now consider the system with the slicer and loader. Its IRR is 12.5 , which may be seen by solving for in P 4,12 ,8 4.9676 P 4,13 ,8 4.7987 12.5 The IRR of the meat slicer and automatic loader is...
Appendix 5A Tests for Multiple IRRs
When the IRR method is used to evaluate projects, we have to test for multiple IRRs. If there are undetected multiple IRRs, an IRR might be calculated that seems correct, but is in error. We consider three tests for multiple IRRs, forming essentially a three-step procedure. In the first test, the signs of the cash flows are examined to see if the project is a simple investment. In the second test, the present worth of the project is plotted against the interest rate to search for interest rates...
Minimum Acceptable Rate of Return MARR
A company evaluating projects will set for itself a lower limit for investment acceptability known as the minimum acceptable rate of return MARR . The MARR is an interest rate that must be earned for any project to be accepted. Projects that earn at least the MARR are desirable, since this means that the money is earning at least as much as can be earned elsewhere. Projects that earn less than the MARR are not desirable, since investing monev in these projects denies the opportunity to use the...
NonStandard Annuities and Gradients
As discussed in Section 3.3, the standard assumption for annuities and gradients is that the payment period and compounding period are the same. If they are not, the formulas given in this chapter cannot be applied directly. There are three methods for dealing with this situation 1. Treat each cash flow in the annuity or gradient individually. This is most useful when the annuity or gradient series is not large. 2. Convert the non-standard annuity or gradient to standard form by changing the...
Pearson
Library and Archives Canada Cataloguing in Publication Global engineering economics financial decision making for engineers Xiall M. Fraser et al. . 4th ed. Previous editions published under title Engineering economics in Canada. 1. Engineering economy. I. Fraser, Xiall M. Xiall Morris , 1952 TA1 77.4.F725 2008 65 8.15 C2008-903776-6 Copvright 2009, 2006, 2000, 1997 Pearson Education Canada, a division of Pearson Canada Inc., Toronto, Ontario. Pearson Prentice Hall. All rights reserved. This...
Present Worth Computations When Ncc
We have until now assumed that the cash flows of a project occur over some fixed, finite number of periods. For long-lived projects, it may be reasonable to model the cash flows as though they continued indefinitely. The present worth of an infinitely long uniform series of cash flows is called the capitalized value of the series. We can get the capitalized value of a series by allowing the number of periods, N, in the series present worth factor to go to infinity The town of South Battleford...
Info Rfb
It will take Clarence four years and four months to pay off the mortgage. He will make 51 full payments of 2000 and will be left with only a fraction of a full payment for his 52nd and last monthly installment. Problem 3.34 asks what his final payment will be. Note also that mortgages can be confusing because of the different terms used. See Close-Up 3.2.B In Example 3.4, it was possible to use the formula for the compound interest factor to solve for the unknown quantity directly. It is not...
P PiF PiX
The second method gives the same result as the first, allowing a small margin for the effects of rounding. It is worth noting here that although it is natural to think about the symbol P as meaning a cash flow at time 0, the present, and F as meaning a cash flow in the future, in fact these symbols can be more general in meaning. As illustrated in the last example, we can consider any point in time to be the present for calculation purposes, and similarly any point in time to be the future,...
What Is Engineering Economics
Just as the role of the engineer in society has changed over the years, so has the nature of engineering economics. Originally, engineering economics was the body of knowledge that allowed the engineer to determine which of several alternatives was economically best the least expensive, or perhaps the most profitable. In order to make this determination properly, the engineer needed to understand the mathematics governing the relationship between time and money Most of this book deals with...
Problems 1
For additional practice, please see the problems with selected solutions provided on the Student CD-ROM that accompanies this book. 2.1 Using 12 simple interest per year, how much interest will be owed on a loan of S500 at the end of two years 2.2 If a sum of 3000 is borrowed for six months at 9 simple interest per year, what is the total amount due principal and interest at the end of six months 2.3 Tiat principal amount will yield S150 in interest at the end of three months when the interest...
Compound Interest Factors for Discrete Compounding
Compound interest factors are formulas that define mathematical equivalence for specific common cash flow patterns. The compound interest factors permit cash flow-analysis to be done more conveniently because tables or spreadsheet functions can be used instead of complicated formulas. This section presents compound interest factors for four discrete cash flow- patterns that are commonly used to model the timing of receipts and disbursements in engineering economic analysis. The four patterns...