Engineering Economy. Time Value of Money-3



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Engineering Economy Time Value of Money-3 Prof. Kwang-Kyu Seo 1

Chapter 2 Time Value of Money Interest: The Cost of Money Economic Equivalence Interest Formulas Single Cash Flows Equal-Payment Series Dealing with Gradient Series Composite Cash Flows. 2009. Fall 2

Capital Recovery Factor Find A, given P, I, N Commonly used in consumer loans. You borrow an amount P, to be paid back in equal installments over N periods, at an interest rate of i per period. What would be your installment payments per period? 3

Practice Problem 2 Car Loan Sticker Price = $20,000 Down payment = 10% or $2000 Rest financed over 5 years (60 months) at 6% APR (0.5% per month) compounded monthly. What is the monthly payment? What is the loan balance after one year? 4

Decision Dilemma Take a Lump Sum or Annual Installments A suburban Chicago couple won the Powerball. They had to choose between a single lump sum $104 million, or $198 million paid out over 25 years (or $7.92 million per year). The winning couple opted for the lump sum. Did they make the right choice? 5

Annuity Value of Lump Sum P = $104M 0 1 2 3 N A? Example 2.14:Powerball Lottery Given: P= $104M, N = 25 years, and i = 8% Find: A A = $104M(A/P,8%,25)=104(.0937)=$9.74M 6

Excel Solution Given: P = $104M i = 8% N = 25 Find: A = PMT(8%,25,-104) = $9.74M 0 1 2 P = 104 A = $9.74 million i = 8% 25 7

Powerball Lottery The worth of the annuity series for a lump sum of $104M is $9.74M per year for 25 years. If the couple can get 8% return on their investment, then opting for the lump sum of $104M now is the right decision. Even at 6% return, the annual payments will be $8.14M, making the lump sum option attractive. 8

Power of Compounding Example 4 Consider two investment options at age 21: 1. Save $2000 a year for 10 years, make no further investments, but let the funds accumulate till age 65. 2. No savings for first 10 years (spend and enjoy!). Then start saving $2000 year for 34 years, until 65. Given 8% return on savings, which option would have more money at age 65? 9

Example 4: Early Savings Plan 8% interest? Option 1: Early Savings Plan 0 1 2 3 4 5 6 7 8 9 10 44 $2,000? Option 2: Deferred Savings Plan 0 1 2 3 4 5 6 7 8 9 10 11 12 44 $2,000 10

Example 4 Comparison of Options $396,644 Option 1: Early Savings Plan 0 1 2 3 4 5 6 7 8 9 10 44 $2,000 $317,253 Option 2: Deferred Savings Plan 0 1 2 3 4 5 6 7 8 9 10 11 12 44 $2,000 11

Free Investment Advice! Start saving early to maximize the power of compounding! 12

Practice Problem 3 Finding a Loan s Interest Rate If $10,000 is borrowed and payment of $2000 are made each year for 9 years to pay off the loan, what is the interest rate? 13

Solution Using Tables To solve the Equation: 2000 = 10000(A/P, i, 9) for i 0.2 = (A/P,i,9) From Tables, the interest rate is between 13% (.1949) and 14% (.2022) By Extrapolation: i=0.13 +(0.2 -.1949)(.14 -.13)/(.2022 -.1949) =.13699 or ~13.7%. 14

Summary Annuities 15

Future Value of an Annuity (F / A, i, N ) = (1 + i) i N 1 Uniform Series Compound Factor The future value F of an annuity in which $A is invested (or received) at the end of each of the next N periods: 16

Other Annuity Factors Sinking Fund Factor i ( A / F,i,N ) = N (1 + i) 1 Uniform Series Present Worth Factor (P / A,i,N ) (1 + i) i(1 + i) N = N 1 Capital Recovery Factor N i(1 + i) ( A / P,i,N ) = N (1 + i) 1 17

Present Value of a Perpetuity A perpetuity is an annuity that continues forever. lim N (P / A,i, N ) = lim N (1 + i) i(1 + N i) N 1 = 1 i Capitalization Factor Example: A university wants to establish a scholarship fund that will give 100 students a $10,000 stipend each year. How much do they need today to establish the fund if interest can be earned at 10% per year? 18

Interest Formulas (Gradient Series) 19

Linear Gradient Series A linear gradient series is a series of cash flows spaced equally in time that increase (or decrease) by a constant amount. 0 G 2G...... 0 1 2 3 N-2 N-1 N N = Number of payments G = Amount of increase (N-3)G (N-2)G (N-1)G First payment is made at the END of the SECOND period 20

Linear Gradient Series N P G i ( 1+ i ) = in 1 i 2 ( 1+ i) N = GP ( / GiN,, ) P 21

$1,000 $1,250 $1,500 $1,750 $2,000 0 P =? 1 2 3 4 5 How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure? 22

Method 1: $1,000 $1,250 $1,500 $1,750 $2,000 0 P =? 1 2 3 4 5 $1,000(P/F, 12%, 1) = $892.86 $1,250(P/F, 12%, 2) = $996.49 $1,500(P/F, 12%, 3) = $1,067.67 $1,750(P/F, 12%, 4) = $1,112.16 $2,000(P/F, 12%, 5) = $1,134.85 $5,204.03 23

Method 2: P1 = $1, 000( P/ A, 12%,5 ) = $3, 604. 80 P2 = $250( P/ G, 12%,5 ) = $1, 599. 20 P= $3, 604. 08+ $1, 59920. = $5,204 24

Linear Gradient Series Future Value Note: P = G (P /G,i,N) Since F = P (1+i)^N, F = G [ P /G,i,N) (1+i)^N]. Hence, (F/G,i,N) = (P/G,i,N) (1+i)^N 25

Geometric Gradient Series A geometric gradient series is a series of cash flows spaced equally in time that increase (or decrease) by a constant percentage. 0 A 1 A 1 (1+g)...... 0 1 2 N-2 N-1 N N = Number of payments g = Rate of increase A 1 (1+g) N-1 A 1 (1+g) A N-3 1 (1+g) N-2 First payment A 1 is made at the END of the FIRST period 26

Geometric Gradient Series N N P A 1 ( 1+ g) ( 1+ i) 1, if i g = i g NA /( 1+ i), if i = g 1 27

Retirement Savings Practice Problem 4 You would like to have a retirement income of $50,000 a year for 25 years. What should be your total savings at the time of retirement, if you can get 7% interest on your savings during retirement? 28

Retirement Savings Practice Problem 5 Suppose the cost of living is expected to increase 5% per year during retirement. How much more in savings should you have to cover the cost of living increase? 29

Given: g = 5% i = 7% N = 25 years A 1 = $50,000 Find: P Practice Problem 5: Find P, Given A 1,g,i,N 30

Retirement Savings Practice Problem 6 Given that you ll work for 30 years before retirement, what annual contribution you should make in a retirement account to generate enough savings to cover your retirement income, under the following options: 1. Without COLA 2. With COLA Assume zero savings now and investment earnings of 10% a year on contributions until retirement. 31

If you make 4 annual deposits of $100 in your savings account which earns a 10% annual interest, what equal annual amount can be withdrawn over 4 subsequent years? Practice Problem 7 32

33

Key Ideas to Remember Series of cash flows can be compared by evaluating them at the same point in time. Cash flows are equivalent (with respect to a particular interest rate) if at any point in time they have the same economic value. For any single amount, annuities, gradient series, and combinations thereof, we can calculate: Present Value Future Value Equivalent Annuity See Table 2.10 (Park s Text) for Summary. 34