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Lecture No. 4 Nominal and Effective Interest Rates 1. General The difference between nominal and effective interest rates is that nominal means once per year and effective means compounding more than once per year. A. Nominal Interest Rate Nominal interest rate, r, is an interest rate that does not include any consideration of compounding. r = interest rate per period x number of periods A nominal rate may be stated for any period: 1 year, 6 months, weekly, daily. r = 1.5% per month x 12 months = 18% Considering 2% per month, all the following are same: 2% per month x 12 months = 24% per year 2% per month x 24 months = 48% per 2 years 2% per month x 6 months = 12% seminanually 2% per month x 3 months = 6% quarterly 2% per month x .231 months = .462 weekly 2% per month x 1/365 months = .005479 daily Note that the nominal rates do NOT make mention of the compounding period, there is no compounding period by definition. Given: 18% per year, compounded monthly Find: Nominal interest rate per a. 2 month b. 6 months c. 2 years a. month i/month = 18/12 = 1.5 r/2months = 1.5 x 2 r/2months = 3% b. 6 months r/6months = 1.5 x 6 r/6months = 9% c. 2 years r/2years = 1.5 x 24 r/years = 36% Lecture No. 4 Nominal and Effective Interest Rates Page No. 2 B. Effective Interest Rate Effective interest rate is the actual rate that applies for a stated period of time. The compounding of interest during the time period of the corresponding nominal rate is accounted for by the effective interest rate. It is commonly expressed on an annual basis as the effective annual rate ia, but any time basis can be used. An effective rate has the compounding frequency attached to the nominal rate statement. If the compounding frequency is NOT stated, it assumed to be the same time period as r meaning that the nominal and effective rates are the same. The following do NOT have the same effective rate over all time periods due to different compounding frequencies. 12% per year, compounded monthly 12% per year, compounded quarterly 3% per quarter, compounded quarterly In the above, the 12% and 3% are nominal interest rates, the effective rate must be calculated by applying the compounding period. The format is r % per time period, compounded m-ly. The m is any time unit. In the last example, compounded quarterly could be omitted because the periods are the same meaning that the nominal rate of 3% per quarter is the same as the effective rate of 3% per quarter compounded quarterly. All calculations must use the effective interest rate. The Annual Percentage Rate, APR, is the same as the nominal interest rate, and the Annual Percentage Yield, APY, is used in lieu of effective interest rate. There are two time units associated with an interest rate statement: Time period – the basic time unit of the interest rate. This is the t in the statement of r % per time period t. The time unit of 1 year is by far the most common and 1 year is assumed unless otherwise stated. Compounding period (CP) – the time unit used to determine the effect of interest. It is defined by the compounding term in the interest rate statement. If not stated, it is assumed to be a year. The compounding frequency is the number m, which is the number of times that compounding occurs within t, the time period. 8% per year compounded monthly has m=12. If 8% is compounded daily, m=365. In the previous chapters t = m = 1 year meaning that the effective and nominal rates were equal. It is common to express the effective rate on the same time basis as the compounding period. Lecture No. 4 Nominal and Effective Interest Rates Page No. 3 r % per time period t r Effective rate per CP (Compounding Period) = m compounding periods per t = m Given: r = 6% per year, compounded monthly Find: CP r 6 CP = = m 12 CP = .50% per month Note that a different time period does NOT alter the compounding period, which is monthly. There are 3 ways to express interest rates p.129, T4-1. 1. 8% per year compounded quarterly, 8% is nominal and the effective must be calculated 2. Effective 8.243% per year compounded quarterly, 8.243% is the effective rate and may be used directly. 3. 8% per year, ambiguous because no compounding period is stated. The rate is effective only over the time period of one year; the effective rate for any other time period must be calculated. 2. Effective Annual Interest Rates The most common period is a year by far which is considered in this section. r = nominal interest rate per year m = number of compounding periods per year i = effective interest rate per compounding period CP = r/m ia = effective interest rate per year F = P + Pia = P(1+ia) CP must be compounded through all m periods to obtain the total. F = P(1+i)m Consider of the F value for a present worth P of $1 and equating the two expressions for F and substituting $1 for P: 1+ia = (1+i)m ia = (1+i)m –1 Solving for the effective interest rate: i = (1+ia)1/m –1 Lecture No. 4 Nominal and Effective Interest Rates Page No. 4 If i = r/m r % per year = (i% per CP)(number of CPs per year) = (i)(m) Given: 13% per year compounded monthly Find: effective interest rate per year ia = (1+i)m –1 i=r/m=13/12=1.08333%=.01083333 ia = (1+.0108333)12 –1 ia = 13.80% per year Example p.129, T4.2, T4-3 3. Effective Interest Rates for Any Time Period The payment period, PP, is the frequency of payment or receipts. To evaluate cash flows that occur more frequently than annually, PP<1 year, the effective interest rate over the PP must be used in the engineering economy relations. Substituting r/m for the period interest rate in eq. 4.5 yields Effective i = (1+r/m)m –1 Examples 4.4, 4.5 p.136 Given: Nominal and effective rates of 16 and 16.986 % Find: What is the compounding period Effective i = (1+r/m)m –1 16.986 = (1+.16/m)m – 1 Trial and error, m=4 yields 16.986, therefore, Quarterly Given: An interest rate of 1% is per month Find: What is the equivalent effective rate per 2 months r/2months = 1% x 2months r/2months = 2% Effective i = (1+r/m)m –1 = (1+.02/2)2 –1 Effective i = 2.01% for 2 months Given: An effective rate of 6.707 per semi-annual period, compounded weekly Find: The equivalent weekly interest rate m=26 which is semi-annual, 26 weeks in ½ year. Effective i = (1+r/m)m –1 .06707 = (1+r/26)26 – 1 r/6months = 6.5% i/week = 6.5/26 i/week = .25% per week which is an effective rate Lecture No. 4 Nominal and Effective Interest Rates Page No. 5 4. PP vs CP In a large percentage of equivalency computations, the frequency of cash flow does not equal the frequency of interest compounding. Cash flows often occur monthly whereas compounding may occur annually or quarterly. Consider monthly deposits to a savings account that compounds interest on a quarterly basis. CP is a quarter; PP is a month. To correctly perform any equivalence computation, it is essential that the compounding period and the payment period be placed on the same time basis, and that the interest rate be adjusted accordingly. A. PPCP, Single Amounts Method 1: Determine the effective interest rate over the compounding period CP, and set n equal to the number of compounding periods between P and F. o P = F(P/F, effective i% per CP, total number of periods n) o F = P(F/P, effective i% per CP, total number of periods n) For example assume a credit card advertises a nominal 18% rate compounded monthly. CP is a month. To find P or F over a 3 year period, the effective monthly rate is 18/12=1.5% and the total months is 3x12=36. 1.5% and 36 are used in the P/F and F/P factors. Method 2: Determine the effective interest rate for the time period of t of the nominal rate and set n equal to the total number of periods using the same time period. The same relationships as above, method 1, are used with term effective i% per t substituted for the interest rate. For a credit card rate of 18% per year, compounded monthly, the time period t is 1 year. .18 Effective i% per year = (1 + 12 )12 –1 = 1.196 – 1 = .196 = 19.6%, n = 3 years The P/F factor is the same by both methods: (P/F, 1.5%, 36) = .5851 using T6, p. 707: and (P/F, 19.6%, 3) = .5787 for 20%, T22, p.748. Example 4.6, p.140 Given: A nuclear laser project increased at an average rate of 4% per month over a 5 year period and the cost now, after 5 years, is $5B. Find: Initial cost that is 5 years ago. Months = 5yrs x 12 = 60 months P = 5(P/F, 4%,60) = 5(.0951) P = .4755B = $475.5million Given: The Patriot missile program cost 6.9B over 10 years. The original estimate was 3.9B. Find: The monthly interest rate n = 10 years x 12 = 120 months 6.9=3.9(F/P, i, 120) (F/P, i, 120) = 1.7692 =(1+i)120 i = .477% per month Lecture No. 4 Nominal and Effective Interest Rates Page No. 6 B. PPCP, Series Amounts Basically the same as Method 2 above except that PP is now defined by the frequency of the cash flows. When cash flows involve a series (A, G, g) and the payment period equals or exceeds the compounding period in length. Find the effective i per payment period. Determine n as the total number of payment periods. Examples 4.7, 4.8, 4.9 p.142 Given: An odor company made deposits of $10,000 at the end of year 2, $25,000 at the end of year 3 and $30,000 at the end of year 5. The interest rate is 16% per year, compounded semi-annually. Find: The future worth at year 5 i/6 months =16/2 = 8% F = 10,000(F/P, 8%, 6) + 25,000(F/P, 8%,4) + 30,000 F = 10,000(1.5869) + 25,000((1.3605) + 30,000 F = $79,881 Or Effective i = (1+.16/2)2 –1 Effective i = 16.64% F = 10,000(F/P, 16.64%, 3) + 25,000(F/P, 16.64%,2) + 30,000 F = $79,881 Given: A water utility buys water at $100,000 per month Feb thru Sept. The utility desires to make a single payment at the end of Nov. The interest rate is .5% per month. Find: What is the subsidy ie the difference between the single and monthly payments. The subsidy is equal to the foregone interest F = $100,000(F/A, .5%,8)(F/P,.5%,2) = 100,000(8.1414)(1.0100) F = $822,281 Actual cost = $100,000 x 8 months Actual cost = $800,000 Subsidy = $822,281 - $800,000 Subsidy = $22,281 Given: A manufacturer borrowed 2M and repaid it over 5 years in equal quarterly payments of principal and interest at an interest rate of 1% per month. Find: What was the size of each quarterly payment? PP>CP; find effective i per PP = quarter i/quarter = ( 1+.03/3)3 –1 i/quarter = 3.03% A=2(A/P, 3.03%,20) = 2(.0674) A = .134805M = $134,805/quarter C. PP<CP, Series Amounts There are two policies: interperiod cash flows earn no interest or they earn compound interest. For no interest, deposits are regarded as deposited at the end of the compounding period and withdrawals are regarded as withdrawn at the beginning. This means if you make a payment on the 15th that is due on the 30th, you do not get interest credit. It also means that if you withdraw money from a quarterly compounded savings account 1 day before the end Lecture No. 4 Nominal and Effective Interest Rates Page No. 7 of the quarter, you get no interest for that quarter. These policies obviously favor the financial institution. If money is earned, m<1 because there is only a fractional part of the CP within one PP. Given: At t=0, an engineer deposits $10,000 which pays 8% per year, compounded quarterly. She withdrew $1000 in months 2, 11 and 23. Find: What was the total value of the account at the end of 2 years assuming no interperiod compounding. Move $1000 withdrawal in month 2 to time 0. Move $1000 withdrawal in month 11 to month 9. Move $1000 withdrawal in month 23 to month 21. F = (10,000-1000)(F/P,2%,8) – 1000(F/P,2%,5) – 1000(F/P,2%,1) F = 9000(1.1717) – 1000(1.1041) – 1000((1.0200) F = $8421 5.Effective Interest Rate for Continuous Compounding Continuous compounding realistically occurs in businesses that have that have a very large number of cash flows every day. 1 h lim (1 + h )h = e = 2.71828… r m lim i =(1 + m )m -1 1/h = r/m or m=hr 1 1 h lim (1 + h )hr -1 = h lim [(1 + h )h ] r -1 i = er – 1 Given: The nominal interest rate is 18% Find: The effective continuous rate per year i = er – 1 = e.18 –1 = 1.1972 – 1 i = .1972 = 19.72% Given: $20,000 deposited for 10 years compounded continuously at 10% Find: What is the effective rate and how much money will accumulate i/year = er – 1 = e.1– 1 i/year = 10.517% F = 20,000(F/P, 10.517%,10)= 20,000(2.7183) F = $54,365 Examples 4.11, 4.12, p.149

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