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Chapter 4 nominal & effective interest rates - students

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Chapter 4 nominal & effective interest rates - students

  1. 1. Chapter 4 Nominal and Effective Interest Rates MS291: Engineering Economy
  2. 2. Content of the Chapter         Interest Rate: important terminologies Nominal and Effective Rate of Interest Effective Annual Interest Rate Converting Nominal rate into Effective Rate Calculating Effective Interest rates Equivalence Relations: PP and CP Continuous Compounding Varying Intrest Rates
  3. 3. Previous Learning  Our learning so for is based “one” interest rate that’s compounded annually  Interest rates on loans, mortgages, bonds & stocks are commonly based upon interest rates compounded more frequently than annually  When amount is compounded more than once annually, distinction need to be made between nominal and effective rate of interests
  4. 4. Interest Rate: important terminologies New time-based definitions to understand and remember Interest period (t) – period of time over which interest is expressed. For example, 1% per month. Compounding period (CP) – The time unit over which interest is charged or earned. For example,10% per year, here CP is a year. Compounding frequency (m) – Number of times compounding occurs within the interest period t. For example, at i = 10% per year, compounded monthly, interest would be compounded 12 times during the one year interest period.
  5. 5. Examples of interest rate Statements  Annual interest rate of 8% compounded monthly …  Here interest period (t) = 1 year  compounding period (CP) = 1 month  compounding frequency (m) = 12  Annual interest rate of 6% compounded weekly …  Here interest period (t) = 1 year  compounding period (CP) = 1 week  compounding frequency (m) = 52
  6. 6. Different Interest Statements Interest rates can be quoted in many ways: • • • • Interest equals “6% per 6-months” Interest is “12%” (annually) Interest is 1% per month “Interest is “12.5% per year, compounded monthly” You must “read” the various ways to state interest and to do calculations.
  7. 7. Nominal Interest Rate • A nominal interest rate is denoted by (r) • It does not include any consideration of the compounding of interest(frequency) • It is given as: r = interest rate per period x number of compounding periods • Nominal rates are all of the form “r % per time period”
  8. 8. Examples: Nominal Interest Rate • 1.5% per month for 24 months – Same as: (1.5%)(24)= 36% per 24 months • 1.5% per month for 12 months – Same as (1.5%)(12 months) = 18% per year • 1.5% per month for 6 months – Same as: (1.5%)(6 months) = 9% per 6 months or semi annual period • 1% per week for 1 year – Same as: (1%)(52 weeks) = 52% per year
  9. 9. Summary: Nominal Interest Rate  A nominal rate do not reference the frequency of compounding  They all have the format “r% per time period”  Nominal rates can be misleading…How? Annual interest of $80 on a $1,000 investment is a nominal rate of 8% whether the interest is paid in $20 quarterly instalments, in $40 semi-annual instalments, or in an $80 annual payment?  We need an alternative way to quote interest rates….  The true Effective Interest Rate is then applied….
  10. 10. Effective Interest Rate (EIR)  Effective interest rates (i) take accounts of the effect of the compounding  EIR are commonly express on an annual basis (however any time maybe used)  EIR rates are mostly of the form: “r % per time period, CP-ly (Compounding Period)” Nominal rates are all of the form “r % per time period”
  11. 11. Examples : Effective Interest Rates Quote: “12 percent compounded monthly” is translated as: 12% is the nominal rate “compounded monthly” conveys the frequency of the compounding throughout the year For this quote there are 12 compounding periods within a year.
  12. 12. Examples: Effective Interest Rate  Some times, Compounding period is not mentioned in Interest statement  For example, an interest rate of “1.5% per month” ……….. It means that interest is compounded each month; i.e., Compounding Period is 1 month.  REMEMBER: If the Compounding Period is not mentioned it is understood to be the same as the time period mentioned with the interest rate.
  13. 13. Example Compounding Period Statement 1. 10% per year 2. 10 % per year compounded monthly 3. 4. 3% per quarter compounded daily 1.5% per month compounded monthly What it is ? 1. CP = not stated but it’s a year 1. Effective rate per year 2. 2. Effective rate per year CP = Stated, CP= month 3. CP= stated, CP= day 3. Effective rate per quarter 4. CP=stated, CP=month 4. Effective rate per month IMPORTANT: Nominal interest rates are essentially simple interest rates. Therefore, they can never be used in interest formulas. Effective rates must always be used hereafter in all interest formulas
  14. 14. Some other names for NIR and EIRs  Interest on Credit Cards, loans and house mortgages sometime use term Annual Percentage Rate(APR) for interest payment….its same as Nominal Interest Rate For example: An APR of 15% is the same as a nominal 15% per year or a nominal 1.25% on a monthly basis.  Returns for investments, certificates of deposits and saving accounts commonly use Annual Percentage Yield (APY) which is same as Effective Interest Rate  Remember: the effective rate is always greater than or equal to the nominal rate, and similarly APY > APR .. Why ?
  15. 15. Converting Nominal rate into Effective Rate per CP • So for, we always used t and CP values of 1 year so compounding frequencies was always m=1, which make them all effective rate of interest (For EIR , Interest period and Compounding period should be same) • But that’s not always the case, we may have situation in which t has different value than CP in that case we need to find effective rate of interest per compounding period • Effective rate per CP can be determined from nominal rate by using following relation Effective rate per CP = r % per time period t = r m compounding periods per t m Where: CP is compounding period, t is the basic time unit of the interest, m is the frequency of compounding and r is nominal interest rate
  16. 16. Example: Calculating Effective Interest rates  Three different bank loan rates for electric generation equipment are listed below. Determine the effective rate on the basis of the compounding period for each rate (a) 9% per year, compounded quarterly (b) 9% per year, compounded monthly (c) 4.5% per 6 months, compounded weekly
  17. 17. Example: Calculating Effective Interest rates per CP a. 9% per year, compounded quarterly. b. 9% per year, compounded monthly. c. 4.5% per 6 months, compounded weekly. The principle amount change in each period for EIR here, …in case of nominal it will remain same in each case
  18. 18. More About Interest Rate Terminology  Sometimes it is not obvious whether a stated rate is a nominal or an effective rate.  Basically there are three ways to express interest rates
  19. 19. Effective Annual Interest Rates When we talk about “Annual” we consider year as the interest period t , and the compounding period CP can be any time unit less than 1 year Nominal rates are converted into Effective Annual Interest Rates (EAIR) via the equation: ia = (1 + i)m – 1 where ia = effective annual interest rate i = effective rate for one compounding period (r/m) m = number times interest is compounded per year
  20. 20. Example For a nominal interest rate of 12% per year, determine the nominal and effective rates per year for (a) quarterly, and ia = (1 + i)m – 1 (b) monthly compounding Solution: where ia = effective annual interest rate i = effective rate for one compounding period (r/m) m = number times interest is compounded per year (a) Nominal r per year = 12% per year Nominal r per quarter = 12/4 = 3.0% per quarter Effective i per year = (1 + 0.03)4 – 1 = 12.55% per year (b) Nominal r per month = 12/12 = 1.0% per month Effective I per year = (1 + 0.01)12 – 1 = 12.68% per year
  21. 21. r = 18% per year, compounded CP-ly
  22. 22. Effective Interest Rates for any Time Period  The following relation of Effective Annual Interest Rates ia = (1+i)m – 1 can be generalize for determining the effective interest rate for any time period (shorter or longer than 1 year). i = (1 + r / m)m – 1 where i = effective interest rate for any time period r = nominal rate for same time period as i m = no. times interest is comp’d in period specified for i
  23. 23. Example: Effective Interest Rates For an interest rate of 1.2% per month, determine the nominal and effective rates (a) per quarter, and i = (1 + r / m)m – 1 (b) per year Solution: (a) Nominal rate (r) per quarter = (1.2)(3) = 3.6% per quarter Effective rate (i) per quarter = (1 + 0.036/3)3 – 1 = 3.64% per quarter (b) Nominal rate (r) per year = (1.2)(12) = 14.4% per year Effective rate (i) per year = (1 + 0.144 / 12)12 – 1 = 15.39% per year
  24. 24. Economic Equivalence: From Chapter 1  Different sums of money at different times may be equal in economic value at a given rate Rate of return = 10% per year $110 Year 0 1 $100 now 1  $100 now is economically equivalent to $110 one year from now, if the $100 is invested at a rate of 10% per year  Economic Equivalence: Combination of interest rate (rate of return) and time value of money to determine different amounts of money at different points in time that are economically equivalent ….. Compounding/Discounting (F/P, P/F, F/A, P/G etc.)
  25. 25. Equivalence Relations: Payment Period(PP) & Compounding Period(CP)  The payment period (PP) is the length of time between cash flows (inflows or outflows) r = nominal 8% per year, compounded semi-annually CP 6 months 0 1 │PP │ 1 month 2 3 CP 6 months 4 5 6 7 8 9 10 11 12 Months
  26. 26. Equivalence Relations: Payment Period(PP) and Compounding Period • It is common that the lengths of the payment period and the compounding period (CP) do not coincide • To do correct calculation …Interest rate must coincide with compounding period • It is important to determine if PP = CP, PP >CP, or PP<CP Length of Time Involves Single Amount (P and F Only) Involves Gradient Series (A, G, or g) PP = CP P/F , F/P P/A, P/G, P/g F/A etc. P/F, F/P P/A, P/G, F/A etc. PP > CP PP < CP
  27. 27. Case I: When PP>CP for Single Amount for P/F or F/P  Step 1: Identify the number of compounding periods (M) per year  Step 2: Compute the effective interest rate per payment period (i)  i = r/M  Step 3: Determine the total number of payment periods (n)  Step 4: Use the SPPWF or SPCAF with i and N above
  28. 28. Example Determine the future value of $100 after 2 years at credit card stated interest rate of 15% per year, compounded monthly. Solution: P = $100, r = 15%, m = 12 EIR /month = 15/12 = 1.25% n = 2 years or 24 months Alternative Method i = (1 + r/m)m – 1 = (1+0.15/12)12 – 1 = 16.076 F = P(F/P, i, n) F = P(F/P, i, n) F = P(F/P, 0.0125, 24) F = P(F/P, 0.16076, 2) F = 100(F/P, 0.0125, 24) F = 100(1.3474) F = 100(1.3474) F = $134.74 F = 100(1.3456) F = $134.56 The results are slightly different because of the rounding off 16.076% to 16.0%
  29. 29. Factor Values for Untabulated i or n There are 3 ways to find factor values for untabulated i or n values 1. Use formula 2. Use spreadsheet function 3. Linearly interpolate in interest tables  Formula or spreadsheet function is fast & accurate  Interpolation is only approximate
  30. 30. Factor Values for Untabulated i or n  Factor value axis f2 f Linear assumption unknown f1 X1 Required X X2 i or n axis Absolute Error = 2.2215 – 2.2197 = 0.0018
  31. 31. Case II: When PP >CP for Series for P/A or F/A  For series cash flows, first step is to determine relationship between PP and CP  Determine if PP ≥ CP, or if PP < CP  When PP ≥ CP, the only procedure (2 steps) that can be used is as follows: First, find effective i per PP Example: if PP is in quarters, must find effective i/quarter Second, determine n, the number of A values involved Example: quarterly payments for 6 years yields n = 4×6 = 24  You can use than the standard P = A(P/A, i , n) or F = A(F/A, i, n) etc.
  32. 32. Example: PP >CP for Series for P/A or F/A For the past 7 years, Excelon Energy has paid $500 every 6 months for a software maintenance contract. What is the equivalent total amount after the last payment, if these funds are taken from a pool that has been returning 8% per year, compounded quarterly? Solution: Compounding Period (CP) = Quarter & PP = 6 months r = 8 % per year or 4% per 6 months & m=2/ quarter PP > CP Effective rate (i) per 6 months = (1+r/m)m -1 i= (1+0.04/2)2 – 1 => 4.04% Since, total time is 7 years and PP is 6 months we have total 7x2=14 payments F = A(F/A, i, n) F = 500(F/A, 0.0404, 14) F = 500(18.3422) => $9171.09
  33. 33. Case III: Economic Equivalence when PP< CP  If a person deposits money each month into a savings account where interest is compounded quarterly, do all the monthly deposits earn interest before the next quarterly compounding time?  If a person's credit card payment is due with interest on the 15th of the month, and if the full payment is made on the 1st, does the financial institution reduce the interest owed, based on early payment? Anyone ? CP: 3 months = 1 quarter 0 1 │PP │ 1 month 2 3 4 5 6 7 8 9 10 11 12 Months
  34. 34. Case III: Economic Equivalence when PP< CP Two policies: 1. Inter-period cash flows earn no interest (most common) 2. inter-period cash flows earn compound interest  positive cash flows are moved to beginning of the interest period (no interest earned) in which they occur and negative cash flows are moved to the end of the interest period (no interest paid) cash flows are not moved and equivalent P, F, and A values are determined using the effective interest rate per payment period
  35. 35. Example 4.11: Example: Clean Air Now (CAN) Company Last year AllStar Venture Capital agreed to invest funds in Clean Air Now (CAN), a start-up company in Las Vegas that is an outgrowth of research conducted in mechanical engineering at the University of Nevada–Las Vegas. The product is a new filtration system used in the process of carbon capture and sequestration (CCS) for coal-fired power plants. The venture fund manager generated the cash flow diagram in Figure in $1000 units from AllStar’s perspective. Included are payments (outflows) to CAN made over the first year and receipts (inflows) from CAN to AllStar. The receipts were unexpected this first year; however, the product has great promise, and advance orders have come from eastern U.S. plants anxious to become zero-emission coal-fueled plants. The interest rate is 12% per year, compounded quarterly, and AllStar uses the no-inter period-interest policy. How much is AllStar in the “red” at the end of the year?
  36. 36. Example 4.11: Example: Clean Air Now (CAN) Company The venture fund manager generated the cash flow diagram in $1000 units from AllStar’s perspective as given below. Included are payments (outflows) to CAN made over the first year and receipts (inflows) from CAN to AllStar. The receipts were unexpected this first year; however, the product has great promise, and advance orders have come from eastern U.S. plants anxious to become zero-emission coal-fueled plants. The interest rate is 12% per year, compounded quarterly, and AllStar uses the no-inter period-interest policy. How much is AllStar in the “red” at the end of the year?
  37. 37. Example: Clean Air Now (CAN) Company Given cash flows Positive Cash flows (inflows) at the start of CP period Negative Cash flows (outflows) at the end of CP period
  38. 38. Example: Clean Air Now (CAN) Company Solution: Effective rate per quarter = 12/4 = 3% Now F = 1000[-150(F/P, 3%, 4)-200(F/P, 3%, 3) +(180-175 )(F/P, 3%, 2)+ 165(F/P, 3%, 1)-50] F = $ (-262111) Investment after one year
  39. 39. Continuous Compounding  If compounding is allowed to occur more and more frequently, the compounding period becomes shorter and shorter, and m (the number of compounding periods per payments) increases  Continuous compounding is present when the duration of CP, the compounding period, becomes infinitely small and m , the number of times interest is compounded per period, becomes infinite.  Businesses with large numbers of cash flows each day consider the interest to be continuously compounded for all transactions.
  40. 40. Continuous Compounding  We have effective interest rate generalize formula as follows: i = (1 + r/m)m – 1  Taking “m” limits tends to infinity… and simplifying the equation we get the following expression for continuous effective interest rate i = er – 1
  41. 41. Example: Continuous Compounding Example: If a person deposits $500 into an account every 3 months at an interest rate of 6% per year, compounded continuously, how much will be in the account at the end of 5 years? Solution: Payment Period: PP = 3 months Nominal rate per three months: r = 6% /4 = 1.50% Continuous Effective rate per 3 months: i = e0.015 – 1 = 1.51% F = 500(F/A,1.51%,20) = $11,573 Practice: Example 4.12 & 4.13
  42. 42. Varying Interest Rates  Interest rate does not remain constant full life time of a project  In order to do incorporate varying interest rates in our calculations, normally, engineering studies do consider average values that do care of these variations.  But sometimes variation can be large and having significant effects on Present or future values calculated via using average values  Mathematically, varying interest rates can be accommodated in engineering studies
  43. 43. Varying Interest Rates When interest rates vary over time, use the interest rates associated with their respective time periods to find P  The general formula for varying interest rate is given as: P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + ….. + Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1)  For single F or P only the last term of the equation can be used.  For uniform series replace “F” with “A”
  44. 44. Example: Varying Interest Rates Given below the cash flow calculate the Present value. $70,000 i=7% $35,000 i=7% i=9% 0 P=? 1 2 $25,000 i=10% 3 4 Year P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + ….. + Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1) P = [ 70,000(P/A, 7%, 2) + 35,000 (P/F, 9%, 1) (P/F, 7%, 2) + 25000(P/F, 10%, 1) (P/F, 7%, 2) (P/F, 9%, 1) = 70,000 (1.8080) + 35,000 (0.9174)(0.8734) + 25,000 (0.9091)(0.8734)(0.0.9174) = $172,816
  45. 45. Problem 4.57: Varying Interest Rates Calculate (a) the Present value (b) the uniform Annual worth A of the following Cash flow series P=? i=10% 1 2 3 4 5 6 7 i=14% 8 Year 0 $100 $100 $100 $100 $100 $160 $160 $160 P = F1(P/F, i1, 1) + F2(P/F, i1, 1)(P/F, i2, 1) + ….. + Fn (P/F, i1, 1)(P/F, i2, 1) …(P/F, in, 1) P = 100(P/A, 10%, 5) + 160 (P/A, 14%, 3) (P/F, 10%, 5) = 100(3.7908) + 160(2.3216)(0.6209) = $609.72
  46. 46. Problem 4.57: Varying Interest Rates (b) the uniform Annual worth A of the following Cash flow series P = 609.72 i=10% 1 2 3 4 5 6 7 i=14% 8 Year 0 A=? P = 100(P/A, 10%, 5) + 150 (P/A, 14%, 3) (P/F, 10%, 5) = 100(3.7908) + 160(2.3216)(0.6209) = $609.72 609.72 = A(3.7908) + A(2.3216)(0.6209) A = 609.72 / 5.2323 A = $ 116.53 per year
  47. 47. THANK YOU

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