Engineering Fundamentals: An Introduction to Engineering, 4th ed.c

procedure is demonstrated, step-by-step, next. The relation between a present value and a future value is given by Equation (20. ...

666 Chapter 20 Engineering Economics year. For this situation, Equation (20.9) is modified to incorporate the frequency of com- ...

Example 20.8 Determine the monthly payments for a five-year, $10,000 loan at an interest rate of 8% com- pounding monthly. To ca ...

668 Chapter 20 Engineering Economics TABLE 20.7 A Summary of Formulas for Situations wheniCompounds Annually and the Uniform Ser ...

For example, when evaluating the series payment equivalence of a present principal, instead of writing, we writeAP(A/P,i,n), wh ...

670 Chapter 20 Engineering Economics TABLE 20.9 The Interest–Time Factors fori8% n (F/P,i,n)(P/F,i,n)(P/A,i,n)(A/P,i,n)(F/A,i,n ...

Additional values of interest – time factors for other interest rates can be created using Excel. Keep in mind that you can use ...

672 Chapter 20 Engineering Economics We can use Table 20.8 to look up the interest – time factor values, which leads to Therefor ...

cost or choose the alternative with the highest present worth or profit. To employ this method, you begin by calculating the equ ...

674 Chapter 20 Engineering Economics Note that we have determined the equivalent present worth of all future cash flow, includin ...

Alternative A: or Alternative B: or 20.10 Excel Financial Functions You also can use Excel Financial Functions to solve engineer ...

676 Formula F  P(1  i) n F  P(1  i/m) nm F Future value(Future worth) i Interest per year (%) m Interest compoundingfrequenc ...

Formula A Annuity, or Payment fora loan based on uniformpayments and a con-stant interest rate i Interest per year (%) n Period ...

678 Chapter 20 Engineering Economics introduce you to engineering economics, but keep in mind that we have just scratched the su ...

Problems 679 Problems 20.1. Compute the future value of the following deposits made today: a. $10,000 at 6.75% compounding annua ...

680 Chapter 20 Engineering Economics 10 $1000 $2000 $2000 021 3456789 $3000 Problem 20.15 Problem 20.16^10 $1000 $2000 021 34567 ...

Problems 681 20.22.You need to borrow $12,000 to buy a car, so you visit two banks and are given two alternatives. The first ban ...

682 Chapter 20 Engineering Economics 20.25.Your future company has purchased a machine and has entered into a contract that requ ...

Problems 683 20.34.Solve Problem 20.3 using Excel. 20.35.Solve Problem 20.4 using Excel. 20.36.Solve Problem 20.5 using Excel. 2 ...

APPENDIX Traffic flow: the relationship between linear and angular speed: mass flow rate mass time specific weight weight volu ...