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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:

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ENGM 661 Engineering Economics for Managers

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  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

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