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ENGM 661 Engineering Economics for Managers

ENGM 661 Engineering Economics for Managers. Multiple/Continuous Compounding. Learning Objectives for tonight:. Understand Effective Interest Rates Figure out how to use Inflation/Deflation in your decisions. Summary of discrete compounding interest factors. Interest Rate Terms….

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ENGM 661 Engineering Economics for Managers

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  1. ENGM 661Engineering Economics for Managers Multiple/Continuous Compounding

  2. Learning Objectives for tonight: • Understand Effective Interest Rates • Figure out how to use Inflation/Deflation in your decisions

  3. Summary of discrete compounding interest factors.

  4. Interest Rate Terms… • Compounding Period(cp) – the time between points when interest is computed and added to the initial amount. • Payment Period(pp) – the shortest time between payments. Interest is earned on payment money once per period (cost of money) • Nominal Rate( r ) – is a simplified expression of the annual cost of money. It means nothing, unless the compounding period is stated along with it. • Annual Percentage Rate(APR) – is the nominal interest rate on a yearly basis (credit cards, bank loans, …). It, too, should have a compounding period stated. • Effective Rate( i ) – is the rate that is used with the table factors or the closed form equations, and it converts the nominal rate taking into account both the compounding period and the payment period so that the blocks match.

  5. Nominal vs. Effective Int. Consider the discrete End-of-Year cash flow tables below: Period Cash Flow Period Cash Flow 0 - $100,000 3 $30,000 1 30,000 4 30,000 2 30,000 5 30,000 Determine the Present Worth equivalent if a. the value of money is 12% compounded annually. b. the value of money is 12% compounded monthly. c. the value of money is 12% compounded continuously.

  6. 30,000 1 2 3 4 5 100,000 Solution; Annual Rate P = -100,000 + 30,000(P/A, ieff, 5) = -100,000 + 30,000(P/A, .12, 5)

  7. 30,000 1 2 3 4 5 100,000 Solution; Nominal/Effective P = -100,000 + 30,000(P/A, ieff, 5) = -100,000 + 30,000(P/A, .12, 5) = -100,000 + 30,000(3.6048) = $8,144

  8. 30,000 1 2 3 4 5 100,000 m r = + - i 1 1 eff m Solution; Compound Monthly

  9. 30,000 1 2 3 4 5 100,000 m r = + - i 1 1 eff m 12 . 12 = + - 1 1 12 = + - 12 ( 1 . 01 ) 1 = = . 1268 12 . 68% Solution; Compound Monthly

  10. 30,000 1 2 3 4 5 100,000 Solution; Compound Monthly P = -100,000 + 30,000(P/A, ieff, 5) = -100,000 + 30,000(P/A, .1268, 5)

  11. 30,000 1 2 3 4 5 100,000 Solution; Compound Monthly P = -100,000 + 30,000(P/A, ieff, 5) = -100,000 + 30,000(P/A, .1268, 5) = -100,000 + 30,000(3.5449) = $6,346

  12. 30,000 1 2 3 4 5 100,000 Solution; Continuous Comp. P = -100,000 + 30,000(P/A, ieff, 5) ieff = ????

  13. Continuous Compounding Now suppose we use an infinite # of compounding periods (continuous). How might we find an answer to our problem of r=12% per year compounded on a continuous basis?

  14. Continuous Compounding Now suppose we use an infinite # of compounding periods (continuous). How might we find an answer to our problem of r=12% per year compounded on a continuous basis F = P(1+.12/9999)9999 (one year period) = P(1.1275) = P(1+.1275)

  15. Continuous Compounding i = e( r )(# of years) – 1 Examples: r = 12% per year compounded continuously ia = e( .12)(1) – 1 = 12.75% What would be an effective six month interest rate for r = 12% per year compounded continuously? i6 month = e( .12)(.5) – 1 = 6.184%

  16. 30,000 1 2 3 4 5 100,000 = - r i e 1 eff = - 1 e 1 = - = 1 . 1275 1 12 . 75% Solution; Continuous Comp.

  17. 30,000 1 2 3 4 5 100,000 Solution; Continuous Comp. P = -100,000 + 30,000(P/A, ieff, 5) = -100,000 + 30,000(P/A, .1275, 5)

  18. 30,000 1 2 3 4 5 100,000 Solution; Continuous Comp. P = -100,000 + 30,000(P/A, ieff, 5) = -100,000 + 30,000(P/A, .1275, 5) = -100,000 + 30,000(3.5388) = $6,164

  19. 30,000 1 2 3 4 5 100,000 Solution; Revisted P = -100,000 + 30,000(P/A, ieff, 5) = -100,000 + 30,000(P/A, .1275, 5) = -100,000 + 30,000(3.5388) = $6,164

  20. Continuous ieff • Example: Suppose a bank pays interest on a CD account at 6% per annum compounded continuously. What is the effective rate?

  21. Continuous ieff Example: Suppose a bank pays interest on a CD account at 6% per annum compounded continuously. What is the effective rate? Soln: ieff = e.06 - 1 = .0618 = 6.18%

  22. Check: Let r=6%, m=999 ieff = ( 1 + r/m)m - 1 = (1+.06/999)999 - 1 = .0618 = 6.18% Continuous ieff Example: Suppose a bank pays interest on a CD account at 6% per annum compounded continuously. What is the effective rate? Soln: ieff = e.06 - 1 = .0618 = 6.18%

  23. Compounding Period is More Frequent than the Payment Perod EFFECTIVE INTEREST RATE ie = effective interest rate per payment period = ( 1 + interest rate per cp)(# of cp per pay period)– 1 = 1 + r me– 1 m Example: r = 12% APR, compounded monthly, payments quarterly imonth = 12% yearly = 1 % compounded monthly 12 months ie = (1 + .01)3 – 1 = .0303 – or – 3.03% per payment

  24. An “APR” or “% per year” statement is a Nominal interest rate – denoted r – unless there is no compounding period stated The EffectiveInterest rate per period is used with tables & formulas Formulas for Effective Interest Rate: If continuous compounding, use y is length of pp, expressed in decimal years If cp < year, and pp = 1 year, use m is # compounding periods per year If cp < year, and pp = cp, use m is # compounding periods per year If cp < year, and pp > cp, use me is # cp per payment period

  25. CRITICAL POINT When using the factors, n and i must always match! Use the effective interest rate formulas to make sure that i matches the period of interest (sum any payments in-between compounding periods so that n matches i before using formulas or tables)

  26. Shows up here on CFD… (End of Period Convention) Returns interest here! Deposit made here … i X Note: Interest doesn’t start accumulating until the money has been invested for the full period! 2 periods 1 0

  27. DIAGRAM: NONE NEEDED! Problem 1 The local bank branch pays interest on savings accounts at the rate of 6% per year, compounded monthly. What is the effective annual rate of interest paid on accounts? GIVEN: r = 6%/yr m = 12mo/yr FIND ia:

  28. DIAGRAM: $2 000 0 5 yrs 1 2 P? Problem 2 What amount must be deposited today in an account paying 6% per year, compounded monthly in order to have $2,000 in the account at the end of 5 years? GIVEN: F5 = $2 000 r = 6%/yr m = 12 mo/yr FIND P:

  29. DIAGRAM: $2 000 0 60mos 1 2 P? Problem 2 – Alternate Soln What amount must be deposited today in an account paying 6% per year, compounded monthly in order to have $2,000 in the account at the end of 5 years? GIVEN: F5 = $2 000 r = 6%/yr m = 12 mo/yr FIND P:

  30. DIAGRAM: $5 000 1 2 yrs 0 A ? Problem 3 A loan of $5,000 is to be repaid in equal monthly payments over the next 2 years. The first payment is to be made 1 month from now. Determine the payment amount if interest is charged at a nominal interest rate of 12% per year, compounded monthly. GIVEN: P = $5 000 r = 12%/yr m = 12 mo/yr FIND A:

  31. DIAGRAM: F? 0 1 2 3 4 yrs $1 000 Problem 4 You have decided to begin a savings plan in order to make a down payment on a new house. You will deposit $1000 every 3 months for 4 years into an account that pays interest at the rate of 8% per year, compounded monthly. The first deposit will be made in 3 months. How much will be in the account in 4 years?

  32. DIAGRAM: F? 1 2 3 0 5 yrs $1 000 Problem 5 Determine the total amount accumulated in an account paying interest at the rate of 10% per year, compounded continuously if deposits of $1,000 are made at the end of each of the next 5 years.

  33. 1 2 3 Problem 6 A firm pays back a $10 000 loan with quarterly payments over the next 5 years. The $10 000 returns 4% APR compounded monthly. What is the quarterly payment amount? DIAGRAM: $10 000 5 yrs = 20 qtrs 0 $A

  34. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1)5 = $1,610.51 But we still only have 1 ton of copper

  35. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1)5 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now

  36. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1)5 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now 10% inflation

  37. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1)5 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now 10% inflation (deflation = neg. inflation)

  38. Combined Interest Rate Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year.

  39. Combined Interest Rate Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year. In today’s dollars $1.00 $1.10 $1.05 $1.10

  40. Combined Interest Rate That is $1.10 = 1.05 (1+d)1 1(1+.10) = 1(1+.05)(1+d)

  41. Combined Interest Rate That is $1.10 = 1.05 (1+d)1 1(1+.10) = 1(1+.05)(1+d) 1+i = (1+j)(1+d) i = d + j + dj

  42. Combined Interest Rate That is $1.10 = 1.05 (1+d)1 1(1+.10) = 1(1+.05)(1+d) 1+i = (1+j)(1+d) i = d + j + dj i = interest rate (combined) j = inflation rate d = real interest rate (after inflation rate)

  43. - i j = d + 1 j Combined Interest Rate Solving for d, the real interest earned after inflation, where i = interest rate (combined) j = inflation rate d = real interest rate (after inflation rate)

  44. Example Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?

  45. Example Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years? Solution: F = 10,000(1+.1)20 = $67,275

  46. Example (cont.) How much is $67,275 20 years from now worth if the inflation rate is 3%?

  47. Example (cont.) How much is $67,275 20 years from now worth if the inflation rate is 3%? Solution: FT = 67,275(P/F,3,20) = 67,275(1.03)-20 = $37,248

  48. i j - = d + 1 j Example (cont.) Alternate: Recall = (.1 - .03)/(1+.03) = .068

  49. - i j = d + 1 j Example (cont.) Alternate: Recall = (.1 - .03)/(1+.03) = .068 FT = 10,000(1+d)20 = 10,000(1.068)20 = $37,248

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