Equivalence and Compound interest Anastasia Lidya M.

Economic Equivalence What do we mean by “economic equivalence?” Why do we need to establish an economic equivalence? How do we establish an economic equivalence?

Economic Equivalence Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.

Single-Payment Factors (F/P and P/F) Objective: Derive factors to determine the present or future worth of a cash flow Cash Flow Diagram – basic format Fn i% / period 0 1 2 3 n-1 n P0 P0 = Fn1/(1+i)n →(P/F,i%,n) factor: Excel: =PV(i%,n,,F) Fn = P0(1+i)n →(F/P,i%,n) factor: Excel: =FV(i%,n,,P)

Single-Payment Factors (F/P and P/F) If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N. F dollars at the end of period N is equal to a single sum P dollars now, if your earning power is measured in terms of interest rate i.

Equivalence Between Two Cash Flows Step 1: Determine the base period, say, year 5. Step 2: Identify the interest rate to use. Step 3: Calculate equivalence value. $3,000 $2,042 5

Example 2.1: Equivalence Various dollar amounts that will be economically equivalent to $3,000 in 5 years, given an interest rate of 8%

Required: To find P given A Uniform-Series: Present Worth Factor (P/A) and Capital Recovery Factor(A/P) Cash flow profile for P/A factor . . . . 0 1 2 3 n-2 n-1 n $A per interest period i% per interest period Find P Required: To find P given A Cash flows are equal, uninterrupted and flow at the end of each interest period

(P/A) Factor Derivation Setup the following: Multiply by to obtain a second equation… Subtract (1) from (2) to yield… (1) (2) (3)

Example 2.2. Uniform Series: Find A, Given P, i, and N Given: P = $250,000, N = 6 years, and i = 8% per year Find: A Formula to use: Capital Recovery Factor

Example 2.3. – Deferred Loan Repayment Given: P = $250,000, N = 6 years, and i = 8% per year, but the first payment occurs at the end of year 2 Find: A Step 1: Find the equivalent amount of borrowing at the end of year 1: Step 2: Use the capital recovery factor to find the size of annual installment:

Example 2.4. Uniform Series: Find P, Given A, i, and N Present Worth Factor Given: A = $10,576,923, N = 26 years, and i = 5% per year Find: P Formula to use:

Equivalent Future Worth Equal Payment Series F Equivalent Future Worth 1 2 N A A A P N 1 2 N

Equal-Payment Series Compound Amount Factor Formula

An Alternate Way of Calculating the Equivalent Future Worth, F A(1+i)N-2 A A A A(1+i)N-1 N 1 2 1 2 N

Example 2.5. Uniform Series: Find F, Given i, A, and N Given: A = $3,000, N = 10 years, and i = 7% per year Find: F

Example 2.6. Handling Time Shifts: Find F, Given i, A, and N Given: A = $3,000, N = 10 years, and i = 7% per year Find: F Each payment has been shifted to one year earlier, thus each payment would be compounded for one extra year

ANSI Standard Notation for Interest Factors Standard notation has been adopted to represent the various interest factors Consists of two cash flow symbols, the interest rate, and the number of time periods General form: (X/Y,i%,n) X represents what is unknown Y represents what is known i and n represent input parameters; can be known or unknown depending upon the problem

Notation - continued Example: (F/P,6%,20) is read as: To find F, given P when the interest rate is 6% and the number of time periods equals 20. In problem formulation, the standard notation is often used in place of the closed-form equivalent relations (factor) Tables at the back of the text provide tabulations of common values for i% and n

Interpolation in Interest Tables When using tabulated interest tables one might be forced to approximate a factor that is not tabulated

Arithmetic Gradient Factors (P/G) and (A/G) A1+(n-1)G Cash flow profile A1+(n-2)G Find P, given gradient cash flow G A1+2G A1+G Base amount = A1 0 1 2 3 n-1 n CFn = A1 ± (n-1)G

Gradients have two components: Gradient Example 0 1 2 3 4 5 6 7 $100 $200 $300 $400 $500 $600 $700 Gradients have two components: 1. The base amount and the gradient 2. The base amount (above) = $100/time period

© 2005 by McGraw-Hill, New York, N.Y All Rights Reserved Gradient Components (n-1)G (n-2)G (n-3)G Find P of gradient series 2G 1G 0G Base amount = A / period …….. 0 1 2 3 n-2 n-1 n Present worth point is 1 period to the left of the 0G cash flow For present worth of the base amount, use the P/A factor (already known) For present worth of the gradient series, use the P/G factor (to be derived) Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved

Example 2.7. – Linear Gradient: Find A, Given A1, G, i, and N Given: A1 = $1,000, G = $300, N = 6 years, and i = 10% per year Find: A

Example 2.8.Declining Linear Gradient Series Given: A1 = $1,200, G = -$200, N = 5 years, and i = 10% per year Find: F Strategy: Since we have no interest formula to compute the future worth of a linear gradient series directly, we first find the equivalent present worth of the gradient series and then convert this P to its equivalent F.

Gradient Decomposition As we know, arithmetic gradients are comprised of two components Gradient component 2. Base amount When working with a cash flow containing a gradient, the (P/G) factor is only for the gradient component Apply the (P/A) factor to work on the base amount component P = PW(gradient) + PW(base amount)

Equivalent A of gradient series Use of the (A/G) Factor A = G(A/G,i,n) (n-1)G (n-2)G Find A, given gradient cash flow G A A A . . . A A 2G Equivalent A of gradient series G 0 1 2 3 n-1 n CFn = (n-1)G

Geometric Gradient Series Factor Cash flow series that starts with a base amount A1 Increases or decreases from period to period by a constant percentage amount This uniform rate of change defines… A GEOMETRIC GRADIENT Notation: g = the constant rate of change, in decimal form, by which future amounts increase or decrease from one time period to the next

Typical Geometric Gradient A1(1+g) A1(1+g)2 . . . . 0 1 2 3 n-2 n-1 n A1(1+g)n-1 Given A1, i%, and g% Required: Find a factor (P/A,g%,i%,n) that will convert future cash flows to a single present worth value at time t = 0

Two Forms to Consider… To use the (P/A,g%,i%,n) factor Case: g = i A1 is the starting cash flow There is NO base amount associated with a geometric gradient The remaining cash flows are generated from the A1 starting value No tables available to tabulate this factor…too many combinations of i% and g% to support tables

Example 2.9. Retirement Plan – Saving $1 Million Given: F = $1,000,000, g = 6%, i = 8%, and N = 20 Find: A1 Solution:

Nominal Versus Effective Interest Rates Nominal Interest Rate: Interest rate quoted based on an annual period Effective Interest Rate: Actual interest earned or paid in a year or some other time period

Nominal Versus Effective Interest Rates Effective interest rate per year, ia, is the annual interest rate taking into account the effect of any compounding during the year. In Example we saw that $100 left in the savings account for one year increased to $105.06, so the interest paid was $5.06. The effective interest rate per year, ia, is $5.06/$100.00 = 0.0506 = 5.06%.

Financial Jargon 18% Compounded Monthly Interest Nominal period interest rate Annual percentage rate (APR) Interest period 18% Compounded Monthly

18% Compounded Monthly What It Really Means? In words, Interest rate per month (i) = 18%/12 = 1.5% Number of interest periods per year (N) = 12 In words, Bank will charge 1.5% interest each month on your unpaid balance, if you borrowed money. You will earn 1.5% interest each month on your remaining balance, if you deposited money. Question: Suppose that you invest $1 for 1 year at 18% compounded monthly. How much interest would you earn?

Effective Annual Interest Rate (Yield) Formula: r = nominal interest rate per year ia = effective annual interest rate M = number of interest periods per year Example: 18% compounded monthly What It really Means 1.5% per month for 12 months or 19.56% compounded once per year

Practice Problem Solution: Suppose your savings account pays 9% interest compounded quarterly. Interest rate per quarter Annual effective interest rate (ia) If you deposit $10,000 for one year, how much would you have?

Compounding Semi-annually Compounding Quarterly Nominal and Effective Interest Rates with Different Compounding Periods Effective Rates Nominal Rate Compounding Annually Compounding Semi-annually Compounding Quarterly Compounding Monthly Compounding Daily 4% 4.00% 4.04% 4.06% 4.07% 4.08% 5 5.00 5.06 5.09 5.12 5.13 6 6.00 6.09 6.14 6.17 6.18 7 7.00 7.12 7.19 7.23 7.25 8 8.00 8.16 8.24 8.30 8.33 9 9.00 9.20 9.31 9.38 9.42 10 10.00 10.25 10.38 10.47 10.52 11 11.00 11.30 11.46 11.57 11.62 12 12.00 12.36 12.55 12.68 12.74

Why Do We Need an Effective Interest Rate per Payment Period? Whenever payment and compounding periods differ from each other, one or the other must be transformed so that both conform to the same unit of time. Payment period Interest period Payment period Interest period

Effective Interest Rate per Payment Period (i) Functional Relationships among r, i, and ia, where interest is calculated based on 9% compounded monthly and payments occur quarterly Formula: C = number of interest periods per payment period K = number of payment periods per year CK = total number of interest periods per year, or M r/K = nominal interest rate per payment period

Effective Interest Rate per Payment Period with Continuous Compounding Example: 12% compounded continuously (a) effective interest rate per quarter (b) effective annual interest rate Formula: With continuous compounding

Case 0: 8% compounded quarterly Payment Period = Quarter Interest Period = Quarterly 1st Q 2nd Q 3rd Q 5th Q 1 interest period Given r = 8%, K = 4 payments per year C = 1 interest period per quarter M = 4 interest periods per year

Case 1: 8% compounded monthly Payment Period = Quarter Interest Period = Monthly 1st Q 2nd Q 3rd Q 5th Q 3 interest periods Given r = 8%, K = 4 payments per year C = 3 interest periods per quarter M = 12 interest periods per year

Case 2: 8% compounded weekly Payment Period = Quarter Interest Period = Weekly 1st Q 2nd Q 3rd Q 5th Q 13 interest periods Given r = 8%, K = 4 payments per year C = 13 interest periods per quarter M = 52 interest periods per year

Case 3: 8% compounded continuously Payment Period = Quarter Interest Period = Continuously 1st Q 2nd Q 3rd Q 5th Q  interest periods Given r = 8%, K = 4 payments per year

Summary: Effective Interest Rates per Quarter at Varying Compounding Frequencies Case 0 Case 1 Case 2 Case 3 8% compounded quarterly 8% compounded monthly 8% compounded weekly 8% compounded continuously Payments occur quarterly 2.000% per quarter 2.013% per quarter 2.0186% per quarter 2.0201% per quarter

Exercise A loan shark lends money on the following terms: "If I give you $50 on Monday, you owe me $60 on the following Monday." (a) What nominal interest rate per year (r) is the loan shark charging? (b) What effective interest rate per year (ia)is he charging? (c) If the loan shark started with $50 and was able to keep it, as well as all the money he received, out in loans at all times, how much money would he have at the end of one year?

answer Effective interest rate per year (ia) = F = P(F/P, i, n) Therefore, i= 20% per week Nominal interest rate per year = 52 weeks x 0,20 = 10,40 = 1040% Effective interest rate per year (ia) =      

Exercise On January 1, a woman deposits $5000 in a' credit union that pays 8% nominal annual interest, compounded quarterly. She wishes to withdrawall the money in five equal yearly sums ,beginning December 31 of the first year. How much should she withdraw each year?

Continuous compounding

Continuous compounding A man deposited $500 per year into a credit union that paid 5% interest, compounded annually. At the end of 5 years, he had $2763 in the credit union. How much would he have if the institution paid 5% nominal interest, compounded continuously?