Lecture no09 unconventional equivalence calculations

Irregular Payment Series and Unconventional Equivalence Calculations Lecture No Chapter Contemporary Engineering Economics Copyright © 2016 Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 3.23: Uneven Payment Series How much you need to deposit today (P) to withdraw $25,000 at n = 1, $3,000 at n = 2, and $5,000 at n = 4, if your account earns 10% annual interest? $25,000 $3,000 $5,000 P Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Check to see if $28,622 is indeed sufficient Beginning Balance 28,622 6,484.20 4,132.62 4,545.88 Interest Earned (10%) 2,862 648.42 413.26 454.59 Payment +28,622 −25,000 −3,000 −5,000 Ending Balance $28,622 6,484.20 4,132.62 4,545.88 0.47 Rounding error It should be “0.” Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 3.25: Future Value of an Uneven Series with Varying Interest Rates  Given: Deposit series as given over years  Find: Balance at the end of year Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Composite Cash Flows  Situation 1: If you make annual deposits of $100 in your savings account, which earns 10% annual interest, what equal annual amount (A) can be withdrawn over subsequent years?  Situation 2: What value of A would make the two cash flow transactions equivalent if i = 10%? Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Establishing Economic Equivalence Method 1: At n = Contemporary Engineering Economics, th edition Park Method 2: At n = Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 3.26: Cash Flows with Sub-patterns  Given: Two cash flow transactions, and i = 12%  Find: C Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution  Strategy: First select the base period to use in calculating the equivalent value for each cash flow series (say, n = 0) You can choose any period as your base period Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 3.27: Establishing a College Fund  Given: Annual college expenses = $40,000 a year for years, i = 7%, and N = 18 years  Find: Required annual contribution (X) Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution Strategy: It would be computationally efficient if you chose n = 18 (the year she goes to college) as the base period Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Cash Flows with Missing Payments  Given: Cash flow series with a missing payment, i = 10%  Find: P Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution  Strategy: Pretend that we have the 10th missing payment so that we have a standard uniform series This allows us to use (P/A,10%,15) to find P Then, we make an adjustment to this P by subtracting the equivalent amount added in the 10th period Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Example 3.28: Calculating an Unknown Interest Rate  Given: Two payment options o Option 1: Take a lump sum payment in the amount of $192,373,928 o Option 2: Take the 30-installment option ($9,791,667 a year)  Find: i at which the two options are equivalent Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution $192,373,928  $9,791,667(P / A, i ,30) (P / A, i ,30)  22.3965  Excel Solution: Contemporary Engineering Economics, 6th Contemporary Engineering Economics, th edition edition, ©2015 Park Copyright © 2016 by Pearson 15Education, Inc All Rights Reserved Example 3.29: Unconventional Regularity in Cash Flow Pattern  Given: Payment series given, i = 10%, and N = 12 years  Find: P Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Solution  Strategy: Since the cash flows occur every other year, find out the equivalent compound interest rate that covers the two-year period • Equivalence Calculations for a Skipping Cash Flow Pattern Solution Actually, the $10,000 payment occurs every other year for 12 years at 10% We can view this same cash flow series as having a $10,000 payment that occurs every period at an interest rate of 21% over years Contemporary Engineering Economics, th edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved ... other year, find out the equivalent compound interest rate that covers the two-year period • Equivalence Calculations for a Skipping Cash Flow Pattern Solution Actually, the $10,000 payment occurs... edition, ©2015 Park Copyright © 2016 by Pearson 15Education, Inc All Rights Reserved Example 3.29: Unconventional Regularity in Cash Flow Pattern  Given: Payment series given, i = 10%, and N = 12... edition Park Copyright © 2016 by Pearson Education, Inc All Rights Reserved Establishing Economic Equivalence Method 1: At n = Contemporary Engineering Economics, th edition Park Method 2: At n