TM 661 Engineering Economics for Managers Unit 1 Time Value of Money

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

Time Value-of-Money Suppose we invest $100. If the bank pays 5% interest, how much will I have after 1 year? 100 F1F1 0 1 t F 1 = interest = (.05) = 105

Time Value-of-Money Now suppose I keep that $105 in the bank for another year. Now how much do I have? t F 2 = interest = (.05) = F2F2

Time Value-of-Money In General P t F 1 = P + interest = P + iP = P(1+i) F1F1

Time Value-of-Money In General P t F 1 = P + interest = P + iP = P(1+i) F 2 = F 1 + interest = F 1 + iF 1 = F 1 (1+i) F2F2

Time Value-of-Money In General P t F 1 = P(1+i) F 2 = F 1 + interest = F 1 + iF 1 = F 1 (1+i) = P(1+i)(1+i) = P(1+i) 2 F2F2

Time Value-of-Money In General P n F n = P(1+i) n FnFn

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? F n = P(1+i) n P n 20,000

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, F n = P(1+i) n P n 20,000

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, F n = P(1+i) n Then, P= F n (1+i) -n P n 20,000

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= F n (1+i) -n = 20,000(1+.1) -15 = 20,000(0.2394) = $2,394 P n 20,000

Future Worth Given Annuity Suppose I wish to compute annual installments to save for college n A F

Annuities Given FutureWorth F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n A(1+i) + A n A F

Annuities Given Future Worth (1+i)F = A(1+i) n + A(1+i) n-1 + A(1+i) n A(1+i) F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n A(1+i) + A n A F

Annuities Given Future Worth iF = A(1+i) n A (1+i)F = A(1+i) n + A(1+i) n-1 + A(1+i) n A(1+i) F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n A(1+i) + A n A F

Annuities Given Future Worth iF = A(1+i) n A = A[(1+i) n - 1] (1+i)F = A(1+i) n + A(1+i) n-1 + A(1+i) n A(1+i) F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n A(1+i) + A n A F

Annuities Given Future Worth iF = A[(1+i) n - 1] n A F FA i i A F A in n    ) (,,) 11(

Annuities Given Future Worth n A F AF i i n   ()11

Annuities Given Future Worth n A F AF i i n   ()11 A     , (.) (.),(.) $629.

Annuities Given Present Worth PAi k k n k     1 1()     Ai k k n ()1 1, for A 1 = A 2 = A 3 =... = A n A P

Annuities Given Future Worth n A1A1 F A2A2 A3A3 AnAn A4A4 FAi k k n k     0 1 1()     Ai k k n ()1 0 1, for A 1 = A 2 = A 3 =... = 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 n A P

Recall: F n = A and P = F n (1 + i) -n Annuities Given Present Worth n A P ()11  i i n

Recall: and P = F n (1 + i) -n Annuities Given Present Worth n A P FA i i n n   ()11 PA i i iA i ii n n n n       () () () ()

Annuities Given Present Worth Inverting and solving for A gives n A P P A i ii n n    () () 11 1 AP ii i PAPin n n     () () (/,,) 1 11

A= 15,000[.01(1.01) 36 /( )] = $ 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 P

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) = $ A P

Car Example; Tables

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%.

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 = $ Example; Home Mortgage.(.) (.)  

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

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. P

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

P W = P A + P G P A = A (P/A, i, n) P G = 0(1+i) -1 + G(1+i) G(1+i) Gradient Derivation P A n AAAA + P G n (n-1)G G 2G

Gradient Derivation (1+i)P G = G[ (1+i) (1+i) (1+i) -3 +4(1+i) -4 ] P G = G[ (1+i) (1+i) (1+i) -4 +4(1+i) -5 ]

Gradient Derivation (1+i)P G = G[ (1+i) (1+i) (1+i) -3 +4(1+i) -4 ] P G = G[ (1+i) (1+i) (1+i) -4 +4(1+i) -5 ] iP G = G[ (1+i) -1 + (1+i) -2 + (1+i) -3 +(1+i) (1+i) -5 ]

Gradient Derivation (1+i)P G = G[ (1+i) (1+i) (1+i) -3 +4(1+i) -4 ] P G = G[ (1+i) (1+i) (1+i) -4 +4(1+i) -5 ] iP G = G[ (1+i) -1 + (1+i) -2 + (1+i) -3 +(1+i) (1+i) -5 ] A Miracle Occurs P G = G[( 1 - (1+ni)(1+i) -n )/ i 2 ]

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. P G = A(P/A,10,7) +G(P/G,10,7) = 1000(4.8684) +100( ) = $6, P

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% Depreciation Example PWPW

Example; (cont.) PW= 1000(P/A, 10, 10) - 100(P/G, 10, 10) = 1000(6.1446) - 100( ) = $3, PAPA PGPG

Gradient Alternative A = G(A/G, 10, 10) = 100(3.7255) = $ P W = ( )(P/A, 10, 10) = (6.1446) = $3, PAPA PGPG

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

Geometric Series P= A 1 (1 + i) -1 + A 2 (1 + i) A n (1 + i) -n = A(1 + i) -1 + A(1 + j)(1 + i) -2 + A(1 + j) 2 (1 + i) A(1 + j) n - 1 (1 + i) -n = A ()()     ji tt t n P n A A(1+j) A(1+j) 2 A(1+j) n-1

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

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

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) ()()     ji tt t n Miracle 2 Occurs  PA ji ij nn     111()()

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 ()()     ji tt t n Miracle 2 Occurs  PA ji ij nn     111()()       A j j i t n t (1) 1 1 1

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%,

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,

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?

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

Summary F n = P(1 + i) n = P(F/P,i,n) P = F n (1 + i) -n = F(P/F,i,n) F n 12340n 100..

Summary = A(F/A,i,n) = F n (A/F,i,n) n A F n FA i i n n   ()11 AF i i n n   ()11

Summary = A(P/A,i,n) = P(A/P,i,n) n A P PA i ii n n    () () 11 1 AP ii i n n    () () 1 11

Summary P G = G = G(P/G,i,n) A= G = G (A/G, i, n) n (n-1)G G 2G 1 i 11 n ii n   [()   ()()nii i n

Summary Case i j P = A = A(P/A,i,j,n) F = A = A(F/A,i,j,n) P n A A(1+j) A(1+j) 2 A(1+j) n-1 ()()11   ij ij nn 111    ()()ji ij nn 

Break Time

Geometric Derivation n A F n     Ai k k n ()1 0 1 F

Geometric Derivation n A F n     Ai k k n ()1 0 1 F Recall the geometric series x x x k k n n       Letting x = (1+i) k

Geometric Derivation n A F n     Ai k k n ()1 0 1 F Recall the geometric series x x x k k n n       Letting x = (1+i) k  Fx k  k  0 n  1

Geometric Derivation n A F n    1 1 x x n  Fx k  k  0 n  1

Geometric Derivation n A F n    1 1 x x n  Fx k  k  0 n  1     () () i i n   1()i i n A

1  Geometric Derivation n A F n    1 1 x x n  Fx k  k  0 n  1     () () i i n   1()i i n A FA i i A F A in n   () (,,) 1