EXAMPLE 1.9 - Cash Flows: Estimation and Diagramming

Each year Exxon-Mobil expends large amounts of funds for mechanical safety features throughout its worldwide operations. Carla Ramos, a lead engineer for Mexico and Central American operations, plans expenditures of $1 million now and each of the next 4 years just for the improvement of fi eld-based pressure-release valves. Construct the cash fl ow diagram to fi  nd the equivalent value of these expenditures at the end of year 4, using a cost of capital estimate for safety-related funds of...

Cash Flows: Estimation and Diagramming

As mentioned in earlier sections, cash fl ows are the amounts of money estimated for future projects or observed for project events that have taken place. All cash fl ows occur during specifi c time periods, such as 1 month, every 6 months, or 1 year. Annual is the most common time period. For example, a payment of $10,000 once every year in December for 5 years is a series of 5 outgoing cash flows. And an estimated receipt of $500 every month for 2 years is a series of 24 incoming cash flows. Engineering...

EXAMPLE 1.8 - Terminology and Symbols

Last year Jane’s grandmother offered to put enough money into a savings account to generate $5000 in interest this year to help pay Jane’s expenses at college. (  a ) Identify the symbols, and (  b ) calculate the amount that had to be deposited exactly 1 year ago to earn $5000 in interest now, if the rate of return is 6% per year. Solution  (a) Symbols    P  (last year is 1) and   F  (this year) are needed. (b) Let F = total amount now and...

EXAMPLE 1.7 - Terminology and Symbols

You plan to make a lump-sum deposit of $5000 now into an investment account that pays 6% per year, and you plan to withdraw an equal end-of-year amount of $1000 for 5 years, starting next year. At the end of the sixth year, you plan to close your account by withdrawing the remaining money. Defi ne the engineering economy symbols involved. Solution All  fi  ve symbols are present, but the future value in year 6 is the unknown.     P = $5000 A = $1000 per year for 5...

EXAMPLE 1.6 - Terminology and Symbols

Today, Julie borrowed $5000 to purchase furniture for her new house. She can repay the loan in either of the two ways described below. Determine the engineering economy symbols and their value for each option.  (a)    Five equal annual installments with interest based on 5% per year.   (b)    One payment 3 years from now with interest based on 7% per year.     Solution   (a)    The repayment schedule requires...

Terminology and Symbols

The equations and procedures of engineering economy utilize the following terms and symbols. Sample units are indicated.      P = value or amount of money at a time designated as the present or time 0. Also   P   is referred to as present worth (PW), present value (PV), net present value (NPV), discounted cash fl  ow (DCF), and capitalized cost (CC); monetary units, such as dollars      F = value or amount of money at some future...

EXAMPLE 1.5 Interest Rate and Rate of Return

   (a)    Calculate the amount deposited 1 year ago to have $1000 now at an interest rate of 5% per year.     (b)    Calculate the amount of interest earned during this time period.    Solution   (a)    The total amount accrued ($1000) is the sum of the original deposit and the earned interest. If   X  is the original deposit,    Total  accrued = deposit + deposit(interest...

EXAMPLE 1.4 Rate of Return and Rate of Return

Stereophonics, Inc., plans to borrow $20,000 from a bank for 1 year at 9% interest for new recording equipment. (  a ) Compute the interest and the total amount due after 1 year. (  b )  Construct a column graph that shows the original loan amount and total amount due after 1 year used to compute the loan interest rate of 9% per year. Solution   (a)    Compute the total interest accrued by solving Equation [1.2] for interest accrued.   Interest...

EXAMPLE 1.3 Interest Rate and Rate of Return

An employee at LaserKinetics.com borrows $10,000 on May 1 and must repay a total of  $10,700 exactly 1 year later. Determine the interest amount and the interest rate paid.  Solution  The perspective here is that of the borrower since $10,700 repays a loan. Apply Equation [1.1] to determine the interest paid.  Interest  paid = $10,700 - 10,000 = $700  Equation [1.2] determines the interest rate paid for 1 year. ...

Interest Rate and Rate of Return

Interest  is the manifestation of the time value of money. Computationally, interest is the difference between an ending amount of money and the beginning amount. If the difference is zero or negative, there is no interest. There are always two perspectives to an amount of interest—interest paid and interest earned. These are illustrated in  Figure 1–2 . Interest is   paid  when a person or organization borrowed money (obtained a loan) and repays a larger amount over time....

EXAMPLE 1.2 - Economic Decisions

Jamie is an engineer employed by Burris, a United States–based company that develops sub- way and surface transportation systems for medium-sized municipalities in the United States and Canada. He has been a registered professional engineer (PE) for the last 15 years. Last year, Carol, an engineer friend from university days who works as an individual consultant, asked Jamie to help her with some cost estimates on a metro train job. Carol offered to pay for his time and talent, but Jamie saw no...

Professional Ethics and Economic Decisions

Many of the fundamentals of engineering ethics are intertwined with the roles of money and economics-based decisions in the making of professionally ethical judgments. Some of these integral connections are discussed here, plus sections in later chapters discuss additional aspects  of ethics and economics. For example, Benefit/Cost Analysis and Public Sector Eco-nomics, includes material on the ethics of public project contracts and public policy. Although it  is very limited in scope and...

Steps in an Engineering Economy Study - Problem Description and Objective Statement, Alternatives, Cash Flows, Engineering Economy Analysis, Selection of the Best Alternative.

Problem Description and Objective Statement      A succinct statement of the problem and  primary objective(s) is very important to the formation of an alternative solution. As an illustration, assume the problem is that a coal-fueled power plant must be shut down by 2015 due to the  production of excessive sulfur dioxide. The objectives may be to generate the forecasted  electricity needed for 2015 and beyond, plus to not exceed all the projected emission...

Performing an Engineering Economy Study

An engineering economy study involves many elements: problem identification, definition of the objective, cash flow estimation, financial analysis, and decision making. Implementing a structured procedure is the best approach to select the best solution to the problem. The steps in an engineering economy study are as follows:     1.    Identify and understand the problem; identify the objective of the project.      2.    Collect...

Example 1.1 Illustrates the Fundamental Nature of Variation in Estimates

An engineer is performing an analysis of warranty costs for drive train repairs within the fi  rst  year of ownership of luxury cars purchased in the United States. He found the average cost (to  the nearest dollar) to be $570 per repair from data taken over a 5-year period. What range of repair costs should the engineer use to ensure that the analysis is sensitive to  changing warranty costs? Solution At  first glance the range should be approximately –25% to 15%...

Engineering Economics: Description and Role in Decision Making

Decisions are made routinely to choose one alternative over another by individuals in everyday life; by engineers on the job; by managers who supervise the activities of others; by corporate presidents who operate a business; and by government officials who work for the public good. Most decisions involve money, called   capital  or   capital funds , which is usually limited in  amount. The decision of where and how to invest this limited capital is motivated by a...