Finding Equivalent Values of Cash Flows | California State University, Northridge

University:
California State University, Northridge
Course:
MSE 604 | Engineering Economy and Financial Analysis
Academic year:

2024
Views:

221

Pages:

18
Author:

arterie4b8

Finding Equivalent Values of Cash Flows- Six Scenarios • Given a: • Find its: Present sum of money Equivalent future value Future sum of money Equivalent present value Uniform end-of-period series Equivalent present value Present sum of money Equivalent uniform end-of-period series Uniform end-of-period series Equivalent future value Future sum of money Equivalent uniform end-of-period series 26 Derivation by Recursion: F/P factor • • • • F1 = P(1+i) F2 = F1(1+i)…..but: F2 = P(1+i)(1+i) = P(1+i)2 F3 =F2(1+i) =P(1+i)2 (1+i) = P(1+i)3 In general: F n P 0 FN = P(1+i)n FN = P(F/P,i%,n) … … … …. N Present Worth Factor from F/P • Since FN = P(1+i)n • We solve for P in terms of FN • P = F{1/ (1+i)n} = F(1+i)-n • Thus: P = F(P/F,i%,n) where (P/F,i%,n) = (1+i)-n An Example • How much would you have to deposit now into an account paying 10% interest per year in order to have $1,000,000 in 40 years? • Assumptions: constant interest rate; no additional deposits or withdrawals Solution: P= 1000,000 (P/F, 10%, 40)=... 29 Uniform Series Present Worth and Capital Recovery Factors • Annuity Cash Flow P = ?? 0 1 2 3 ………….. .. $A per period .. n-1 n Uniform Series Present Worth and Capital Recovery Factors • Write a Present worth expression P  1 1 1 1  A + + .. + + 1 2 n −1 n  (1 + i ) (1 + i )   (1 + i ) (1 + i ) [1]  1 P 1 1 1  = A + + .. + + [2] 2 3 n n +1  1+ i (1 + i ) (1 + i )   (1 + i ) (1 + i ) Uniform Series Present Worth and Capital Recovery Factors • Setting up the subtraction  1 P 1 1 1  = A + + .. + + 2 3 n n +1  + + + + 1+ i (1 i ) (1 i ) (1 i ) (1 i )   [2]  1 1 1 1  = - P A  (1 + i)1 + (1 + i)2 + .. + (1 + i)n−1 + (1 + i)n  [1]  1 −i 1  = P A − =  n +1 1+ i (1 + i )   (1 + i ) [3] Uniform Series Present Worth and Capital Recovery Factors • Simplifying Eq. [3] further −i P 1+ i P / A i %, n factor  (1 + i ) − 1   1 1  A − P A for i ≠ 0 =  n +1 n  (1 + i )   (1 + i )  i (1 + i )  n The present worth point of an annuity cash flow is always one period to the left of the first A amount  i (1 + i ) n  A= P  n  (1 + i ) − 1  A/P,i%,n factor Section 3.9 Lotto Example • If you win $5,000,000 in the California lottery, how much will you be paid each year? How much money must the lottery commission have on hand at the time of the award? Assume interest = 3%/year. • Given: Jackpot = $5,000,000, N = 19 years (1st payment immediate), and i = 3% year • Solution: A = $5,000,000/20 payments = $250,000/payment (This is the lottery’s calculation of A P = $250,000 + $250,000(P | A, 3%, 19) P = $250,000 + $3,580,950 = $3,830,950 34 Sinking Fund and Series Compound amount factors (A/F and F/A) • Annuity Cash Flow Find $A given the Future amt. - $F $A per period ………….. 0  1  P=F n  + i (1 )    i (1 + i ) n  A= P  n + i − (1 ) 1   $F   i A= F  n + i − (1 ) 1   N  (1 + i ) n − 1  F=A   i   Example - Uniform Series Capital Recovery Factor • Suppose you finance a $10,000 car over 60 months at an interest rate of 1% per month. How much is your monthly car payment? • Solution: A = $10,000 (A | P, 1%, 60) = $222 per month 36 Example: Uniform Series Compound Amount Factor • Assume you make 10 equal annual deposits of $2,000 into an account paying 5% per year. How much is in the account just after the 10th deposit? 12.5779 • Solution: • F= $2,000 (F|A, 5%, 10) = $25,156 • Again, due to compounding, F>NxA when i>0%. 37 An Example • Recall that you would need to deposit $22,100 today into an account paying 10% per year in order to have $1,000,000 40 years from now. Instead of the single deposit, what uniform annual deposit for 40 years would also make you a millionaire? • Solution: A = $1,000,000 (A | F, 10%, 40) = $ 38 Basic Setup for Interpolation •Work with the following basic relationships Estimating for i = 7.3% • Form the following relationships Interest Rates that vary over time • In practice – interest rates do not stay the same over time unless by contractual obligation. • There can exist “variation” of interest rates over time – quite normal! • If required, how do you handle that situation? 41 Section 3.12 Multiple Interest Factors • Some situations include multiple unrelated sums or series, requiring the problem be broken into components that can be individually solved and then re-integrated. See page 93. • Example: Problem 3-95 • What is the value of the following CFD? 42 Problem 3-95 Solution • F1 = -$1,000(F/P,15%,1) - $1,000 = -$2,150 • F2 = F 1 (F/P,15%,1) + $3,000 = $527.50 • F4 = F 2 (F/P,10%,1)(F/P,6%,1) = $615.07 43