Engineering Economics Blog

  The  most fundamental factor in engineering economy  is the one that determines the amount of money   F  accumulated after   n  years (or periods) from a   single  present worth P, with interest compounded one time per year (or period). Recall that compound interest refers to interest paid on top of interest. Therefore, if an amount   P  is invested at time t = 0, the amount F1 accumulated 1 year hence at an interest rate...

  The cash fl ow is fundamental to every economic study. Cash fl ows occur in many confi  gurations and amounts - isolated single values, series that are uniform, and series that increase or decrease by constant amounts or constant percentages. This chapter develops derivations for all the commonly used engineering economy factors that take the time value of money into account.    The application of factors is illustrated using their mathematical forms and a standard notation format....

A Japan-based architectural fi rm has asked a United States–based software engineering group to infuse GPS sensing capability via satellite into monitoring software for high-rise structures in order to detect greater than expected horizontal movements. This software could be very benefi  cial as an advance warning of serious tremors in earthquake-prone areas in Japan and the United States. The inclusion of accurate GPS data is estimated to increase annual revenue over that for the current software...

The functions on a computer spreadsheet can greatly reduce the amount of hand work for equivalency computations involving   compound interest  and the terms   P, F, A , i, and n . The use of a calculator to solve most simple problems is preferred by many students and professors as described in Appendix D. However, as cash fl  ow series become more complex, the spreadsheet offers a good alternative.  Microsoft Excel   is used throughout this book because...

  For any investment to be profi  table, the investor (corporate or individual) expects to receive more money than the amount of capital invested. In other words, a fair   rate of return,  or   return on investment,  must be realizable. The defi nition of ROR in Equation [1.4] is used in this discussion, that is, amount earned divided by the principal.    Engineering alternatives are evaluated upon the prognosis that a reasonable ROR can be expected. Therefore,...

Table 1–1 details four different loan repayment plans described below. Each plan repays a $5000 loan in 5 years at 8% per year compound interest.    • Plan 1: Pay all at end.  No interest or principal is paid until the end of year 5. Interest accumulates each year on the total of principal and all accrued interest.    • Plan 2: Pay interest annually, principal repaid at end.  The accrued interest is paid each year, and the entire principal is repaid at the end of year...

Assume an engineering company borrows $100,000 at 10% per year compound interest and will pay the principal and all the interest after 3 years. Compute the annual interest and total amount due after 3 years. Graph the interest and total owed for each year, and compare with the previous example that involved simple interest.   Solution To include compounding of interest, the annual interest and total owed each year are calculated by Equation [1.8].    The repayment plan requires...

GreenTree Financing lent an engineering company $100,000 to retrofi t an environmentally unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money will the fi rm repay at the end of 3 years? Solution The interest for each of the 3 years is Interest  per  year = $100,000(0.10) = $10,000   Total interest for 3 years from Equation [1.7] is Total  interest = $100,000(3)(0.10) = $30,000   The amount due after 3 years...

The  terms    interest,     interest period,  and   interest rate  (introduced in Section 1.4) are useful in calculating equivalent sums of money for one interest period in the past and one period in the future. However, for more than one interest period, the terms   simple interest  and   compound interest   become important.    Simple interest  is calculated using the principal only,...

Howard owns a small electronics repair shop. He wants to borrow $10,000 now and repay it over the next 1 or 2 years. He believes that new diagnostic test equipment will allow him to work on a wider variety of electronic items and increase his annual revenue. Howard received 2-year repayment options from banks A and B. After reviewing these plans, Howard decided that he wants to repay the $10,000 after only 1 year based on the expected increased revenue. During a family conversation, Howard’s...

Manufacturers make backup batteries for computer systems available to Batteries + dealers through privately owned distributorships. In general, batteries are stored throughout the year, and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship owner. Assume you own the City Center Batteries + outlet. Make the calculations necessary to show which of the following statements are true and which are false about battery costs.     (a)   ...

Economic equivalence is a fundamental concept upon which engineering economy computations are based. Before we delve into the economic aspects, think of the many types of equivalency we may utilize daily by transferring from one scale to another. Some example transfers between scales are as follows: Often equivalency involves two or more scales. Consider the equivalency of a   speed  of 110 kilometers per hour (kph) into miles per minute using conversions between distance and...

  A rental company spent $2500 on a new air compressor 7 years ago. The annual rental income from the compressor has been $750. The $100 spent on maintenance the fi rst year has increased each year by $25. The company plans to sell the compressor at the end of next year for $150. Construct the cash fl ow diagram from the company’s perspective and indicate where the present worth now is located. Solution Let now be time t = 0. The incomes and costs for years 7 through 1 (next year) are...

An electrical engineer wants to deposit an amount   P  now such that she can withdraw an equal annual amount of  A1 = $2000 per year for the fi rst 5 years, starting 1 year after the deposit, and a different annual withdrawal of A2 = $3000 per year for the following 3 years. How would the cash fl ow diagram appear if i = 8.5% per year?   Solution The cash fl  ows are shown in Figure 1–8. The negative cash outfl  ow   P  occurs now. The withdrawals...