1 / 54

ENGM 661 Engineering Economics for Managers

ENGM 661 Engineering Economics for Managers. Unit 1 Time Value of Money. Objectives for Today. Given a word problem, be able to set up the corresponding cash flow equivalence diagram. Given a simple cash flow diagram, be able to find:

ivan-dale
Download Presentation

ENGM 661 Engineering Economics for Managers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ENGM 661 Engineering Economics for Managers Unit 1 Time Value of Money

  2. Objectives for Today • Given a word problem, be able to set up the corresponding cash flow equivalence diagram. Given a simple cash flow diagram, be able to find: • Future value given an annuity and/or an initial payment at time 0 • Annuity given a present value • Annuity required to reach some future value • Annuity given beginning of period payments • Present value, future value, or annuity given a gradient series • Present value, future value, or annuity given a geometric growth rate • Explain what Excel functions would be used to find the answer to the above problems.

  3. Time Value-of-Money • Suppose we invest $100. If the bank pays 5% interest, how much will I have after 1 year? F 0 1 t 100

  4. Time Value-of-Money • Suppose we invest $100. If the bank pays 5% interest, how much will I have after 1 year? F1 F1 = 100 + interest = 100 + 100 (.05) = 105 0 1 t 100

  5. Time Value-of-Money • Now suppose I keep that $105 in the bank for another year. Now how much do I have? F2 0 1 2 t F2 = 105 + interest = 105 + 105 (.05) = 110.25 100 105

  6. Time Value-of-Money • In General F1 F1 = P + interest = P + iP = P(1+i) 0 1 2 t P

  7. Time Value-of-Money • In General F2 F1 = P + interest = P + iP = P(1+i) F2 = F1 + interest = F1 + iF1 = F1(1+i) 0 1 2 t P

  8. Time Value-of-Money • In General F2 F1 = P(1+i) F2 = F1 + interest = F1 + iF1 = F1(1+i) = P(1+i)(1+i) = P(1+i)2 0 1 2 t P

  9. Time Value-of-Money • In General Fn Fn = P(1+i)n 0 1 2 3 n P

  10. 20,000 0 1 2 3 n P Example • I would like to have $20,000 for • my son to go to college in 15 years. • How much should I deposit now if • I can earn 10% interest? Fn = P(1+i)n

  11. 20,000 0 1 2 3 n P Example • I would like to have $20,000 for • my son to go to college in 15 years. • How much should I deposit now if • I can earn 10% interest? Recall, Fn = P(1+i)n

  12. 20,000 0 1 2 3 n P Example • I would like to have $20,000 for • my son to go to college in 15 years. • How much should I deposit now if • I can earn 10% interest? Recall, Fn = P(1+i)n Then, P = Fn(1+i)-n

  13. 20,000 0 1 2 3 n P Example I would like to have $20,000 for my son to go to college in 15 years. How much should I deposit now if I can earn 10% interest? Then, P = Fn(1+i)-n = 20,000(1+.1)-15 = 20,000(0.2394) = $2,394

  14. F 1 2 3 4 n 0 . . . . A Future Worth Given Annuity • Suppose I wish to compute annual installments to save for college.

  15. F 1 2 3 4 n 0 . . . . A Annuities Given Future Worth iF = A[(1+i)n - 1] + - n ( 1 i ) 1 F = = F A A ( , i , n ) i A

  16. F i 1 2 3 4 n = A F 0 . . . . + - n ( 1 i ) 1 A Annuities Given Future Worth

  17. F i 1 2 3 4 n = A F 0 . . . . + - n ( 1 i ) 1 A (. 1 ) = A 20 , 000 + - 15 ( 1 . 1 ) 1 = 20 , 000 ( 0 . 0315 ) = $629 . 50 Annuities Given Future Worth

  18. P 1 2 3 4 n 0 . . . . n  k A  P  A ( 1  i ) k k  1 n  k   A ( 1  i ) k  1 Annuities Given Present Worth , for A1 = A2 = A3 = ... = A

  19. n  1 k  F  A ( 1  i ) k k  0 n  1 k   A ( 1  i ) , for A1 = A2 = A3 = . . . = A k  0 Annuities Given Future Worth F 1 2 3 4 n 0 . . . . A1 A2 A3 A4 An

  20. P 1 2 3 4 n 0 . . . . A Annuities Given Present Worth • Suppose we wish to compute the monthly payment of a car if we borrow $15,000 at 1% per month for 36 months.

  21. P n ( 1  i )  1 1 2 3 4 n i 0 . . . . A Annuities Given Present Worth Recall: Fn = A and P = Fn(1 + i) -n

  22. P + - n ( 1 i ) 1 = F A 1 2 3 4 n n i 0 . . . . A + - + - n n ( 1 i ) 1 ( 1 i ) 1 - = + = n P A ( 1 i ) A + n i i ( 1 i ) Annuities Given Present Worth • Recall: • and • P = Fn(1 + i) -n

  23. P 1 2 3 4 n 0 . . . . A + n i ( 1 i ) = = A P P ( A / P , i , n ) + - n ( 1 i ) 1 + - n ( 1 i ) 1 = P A + n i ( 1 i ) Annuities Given Present Worth • Inverting and solving for A gives

  24. P 1 2 3 4 36 0 . . . . A Car Example • We have an initial loan of $15,000 at a rate of i=1% per month for n=36 months. The monthly loan payment is then A = 15,000[.01(1.01)36/(1.0136 - 1)] = $498.22

  25. P 1 2 3 4 36 0 . . . . A Car Example; Alternative • We have an initial loan of $15,000 at a rate of i=1% per month for n=36 months. The monthly loan payment is then • A = 15,000(A/P,i,n) • = 15,000(A/P,1,36) • = 15,000(.0032) = $498.21

  26. Car Example; Tables

  27. Example; Home Mortgage • Example: Suppose we borrow $75,000 for house at 9% for 30 years. Find monthly payment. Assume that the monthly interest rate is 9/12 = 3/4%.

  28. 360 . 0075 ( 1  . 0075 ) 360 ( 1  . 0075 )  1 Example; Home Mortgage • Example: Suppose we borrow $75,000 for house at 9% for 30 years. Find monthly payment. Assume that the monthly interest rate is 9/12 = 3/4%. • Using the Formula: • A = $75,000 • = $603.47

  29. Home Mortgage Using the Table: A = P(A/P, i, n) = 75,000(A/P, 9, 30) = 75,000(.0973) = 7,297.5 / year » $608.12 / month

  30. 1600 1200 1100 1000 0 1 2 3 7 P Gradient Series Suppose we have an investment decision which is estimated to return $1,000 in the first year, $1,100 in the second, $1,200 in the third, and so on for the 7 year life of the project.

  31. (n-1)G 2G G A A A A + 0 0 1 2 3 1 2 3 n n P P A G Gradient Equivalent Flows We can replace the cash flow as the equivalent of an annuity and a constant growth

  32. (n-1)G 2G G A A A A + 0 0 1 2 3 1 2 3 n n P P A G Gradient Derivation PW = PA + PG PA = A (P/A, i, n) PG = 0(1+i)-1 + G(1+i)-2 + 2G(1+i)-3 +...

  33. Gradient Derivation • (1+i)PG = G[ (1+i)-1 + 2(1+i)-2 + 3(1+i)-3 +4(1+i)-4 ] • PG = G[ (1+i)-2+ 2(1+i)-3 + 3(1+i)-4 +4(1+i)-5 ]

  34. Gradient Derivation • (1+i)PG = G[ (1+i)-1 + 2(1+i)-2 + 3(1+i)-3 +4(1+i)-4 ] • PG = G[ (1+i)-2+ 2(1+i)-3 + 3(1+i)-4 +4(1+i)-5 ] • iPG = G[ (1+i)-1 + (1+i)-2 + (1+i)-3 +(1+i)-4 - 4(1+i)-5 ]

  35. Gradient Derivation • (1+i)PG = G[ (1+i)-1 + 2(1+i)-2 + 3(1+i)-3 +4(1+i)-4 ] • PG = G[ (1+i)-2+ 2(1+i)-3 + 3(1+i)-4 +4(1+i)-5 ] • iPG = G[ (1+i)-1 + (1+i)-2 + (1+i)-3 +(1+i)-4 - 4(1+i)-5 ] • A Miracle Occurs • PG = G[( 1 - (1+ni)(1+i)-n )/ i2 ]

  36. 1600 1200 1100 1000 0 1 2 3 7 P Example • Estimated return of $1,000 in the first year, $1,100 in the second, $1,200 in the third, and so on for the 7 year life of the project. • PG = A(P/A,10,7) +G(P/G,10,7) • = 1000(4.8684) +100(12.7631) • = $6,144.71

  37. 1000 900 800 100 0 1 2 3 10 PW Depreciation Example • Determine PW of a depreciation scheme which saves $1000 in the first year and declines by $100 per year for the next 10 years if i = 10%

  38. PG 1000 1000 1000 1000 1 2 3 10 + 0 1 2 3 0 10 100 200 PA 900 Example; (cont.) • PW = 1000(P/A, 10, 10) - 100(P/G, 10, 10) • = 1000(6.1446) - 100(22.8913) • = $3,855.47

  39. PG 1000 1000 1000 1000 1 2 3 10 + 0 1 2 3 0 10 100 200 PA 900 Gradient Alternative • A = G(A/G, 10, 10) • = 100(3.7255) • = $372.55 • PW = (1000 - 372.55)(P/A, 10, 10) • = 627.45(6.1446) • = $3,855.43

  40. 1331 1210 1100 1000 0 1 2 3 4 Geometric Series Suppose we start a new computer consultant business. We assume that we will start our business with a modest income but that the business will grow at a rate of 10% per year. If we assume an initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows.

  41. A(1+j)n-1 A(1+j)2 A(1+j) A 0 1 2 3 n P n  t  1  t ( 1  j ) ( 1  i ) t  1 Geometric Series P = A1(1 + i)-1 + A2(1 + i)-2 +......+An(1 + i)-n = A(1 + i)-1 + A(1 + j)(1 + i)-2 + A(1 + j)2(1 + i)-3 +..... + A(1 + j)n - 1(1 + i)-n = A

  42. n  t  1  t ( 1  i ) ( 1  i ) t  1 n   1 ( 1  i ) t  1 Geometric Series Special Case: For the special case where i = j, we have P = A = A

  43. n  t  1  t ( 1  i ) ( 1  i ) t  1 n   1 ( 1  i ) t  1 nA ( 1  i ) Geometric Series Special Case: For the special case where i = j, we have P = A = A P =

  44. n  t  1  t ( 1  j ) ( 1  i ) t  1 - - + + n n 1 ( 1 j ) ( 1 i ) = P A - i j Geometric Series Case i j: For the case when the interest rate i and the growth rate j are not equal, we have P = A = A(P/A,i,j,n)  Miracle 2 Occurs

  45. t n A 1  j n  t  1  t ( 1  j ) ( 1  i )   (1  j ) 1  i t  1 t  1 - - + + n n 1 ( 1 j ) ( 1 i ) = P A - i j Geometric Series Case i j: For the case when the interest rate i and the growth rate j are not equal, we have P = A  Miracle 2 Occurs

  46. 1331 1210 1100 1000 0 1 2 3 4 Example; Geometric • Computer business has initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows. If i = 10%,

  47. 1331 1210 1100 1000 0 1 2 3 4 Example; Geometric Computer business has initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows. If i = 10%, P = 1000(P/A,i,j,n) = 1000(P/A,10,10,4) = 1000(3.6363) = $3,636

  48. Example Example: Individual deposits of $1000 in end of year 1 and increases by 10% each year for 30 years. If the account earns 10% per year, how much will he have at end of 30 years?

  49. Example (Cont). Solution: F = A(F/A, i, j, n) = 1000((F/A, 10, 10, 30) = 1000(475.8934) = $475,893 Alternative Solution: F = 1000(30)(1 + .10)29 = $475,892

  50. F n Summary • Fn = P(1 + i)n = P(F/P,i,n) • P = Fn(1 + i)-n = F(P/F,i,n) 0 1 2 3 4 . . . . n 100

More Related