Lecture 2

ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Types of Cash Flows (a) Single cash flow (b) Equal (uniform)

payment series (c) Linear gradient series (d) Geometric gradient series (e) Irregular payment series

Future Given Present † P is the present value at Time 0 † F is the future value at Time N „

(N periods in the future)

† i is the effective interest rate F? 0

1

2

3

N

P

F = P(F/P,i,N)

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Single Cash Flow Formula † Single payment

F = P(1 + i) N F = P( F / P, i, N)

compound amount factor (growth factor) † Given: i = 10% N = 8 years P = $ 2 ,0 0 0 † Find: F

F

0 N

F = $2,000(1 + 010 . )8 = $2,000( F / P,10%,8 ) = $4,28718 .

P 2.1436

Example 1 If $1,600 is deposited (today) into an account earning 12% compounded annually, what amount of money will be in the account at the end of 17 GIVEN: years? P = $1,600 F17?

DIAGRAM: 0

1

2

3 N=17

$1,600

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i = 12% N = 17 FIND F17: F17 = P(F|P,i,N) = 1,600(F|P,12%,17) = 1,600(6.8660) = $10,986

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 1 - Concept Assuming a 12% per year return on my investment, I am indifferent between having $1,600 today and $10,986 in 17 years. DIAGRAM: 0

1

$10,986

2

3 N=17

$1,600

Present Given Future † P is the present value at Time 0 † F is the future value at Time N „

(N periods in the future)

† i is the effective interest rate for each period F 0

1

2

3

N

P?

P = F(P/F,i,N)

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Single Cash Flow Formula † Single payment present

worth factor (discount factor) † Given: i = 1 2 % N = 5 y e a rs F = $ 1,0 0 0 † Find: P

P = F(1 + i)− N P = F(P / F, i, N)

F

0 N P

P = $1, 000 (1 + 0 .12 ) − 5 = $1, 000 ( P / F ,12% ,5 )

0.5674

= $567.40

Example 2 What is the present value of having $6,200 fifty-three years from now at 12% compounded annually? GIVEN: DIAGRAM: 0

1

2

FIND P:

3 N=53

P?

F53 = $6,200 i = 12% N = 53

$6,200

P = F53(P|F,i,N) = F53(1+ i)–N = 6,200(1+ .12)–53 = 6,200(0.00246) = $15.27

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 2 - Concept Having $15.27 today is equivalent to having $6,200 in 53 years assuming that I will invest that $15.27 and earn 12% per year on my money. P50 = F53(P|F,i,N) ALTERNATIVE: 0

1

2

$6,200

F50 = P50

3

51

52

1

2

N=53

= 6,200(P|F,12%,3) = 6,200(0.7118) = F50 P = F50(P|F,i,N)

3

= P50(P|F,12%,50)

P50

P?

= 6,200(0.7118)(P|F,12%,50) = 6,200(0.7118)(0.0035) = $15.45

Equal Payment Series A

0

1

2

3

4

5

N-1

N

F

P

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Future Given Annual † A is the equal annual value over the time period „

(time period: Time 0 to Time N, 1st flow at Time 1)

† F is the future value at Time N „

(N periods in the future)

† i is the effective interest rate for each period F? 0

1

2

3

N A

F = A(F/A,i,N) †Note:

cash flow A does not have to be annual, just periodic

Equal Payment Series Compound Amount Factor F

0

1

2

3 N

A

(1 + i ) N − 1 F=A i = A( F / A, i , N )

Example 4.13: † Given: A = $3,000, N = 10 years, and i = 7% † Find: F † Solution: F = $3,000(F/A,7%,10) = $41,449.20

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 3 If $1,400 is deposited at the end of each year (every year) for 10 years, what is the accumulated value at the end of 10 years at 18% compounded GIVEN: annually? A = $1,400 F10 ?

DIAGRAM: 0

1

2

3

9 N=10 $1,400

i = 18% N = 10 FIND F10: F10 = A(F|A,i,N) = 1,400(F|A,18%,10) = 1,400(23.5213) = $32,930

Example 3 - Concept Assuming I can earn 18% on my funds, I am indifferent between 10 yearly payments of $1,400 and having $32,930 at the end of 10 years.

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Annual Given Future † A is the equal annual value over the time period „

(time period: Time 0 to Time N, 1st flow at Time 1)

† F is the future value at Time N „

(N periods in the future)

† i is the effective interest rate for each period F 0

1

2

3

N A?

A = F(A/F,i,N) †

Note: cash flow A does not have to be annual, just periodic

Sinking Fund Factor F

i (1 + i) N − 1 = F ( A / F , i, N )

A= F 0

1

2

3 N

A

Example 4.15: † Given: F = $5,000, N = 5 years, and i = 7% † Find: A † Solution: A = $5,000(A/F,7%,5) = $869.50

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 4 What series of equal yearly payments must be made into an account to accumulate $90,000 in 72 years at 6.3% compounded annually? GIVEN:

F72 = $90,000 i = 6.3% N = 72 $90 000

DIAGRAM: 0

1

2

FIND A: A = F72(A|F,i,N) = F72(A|F,6.3%,72)

3 N=72 A?

⎡ ⎤ i = F72 ⎢ ⎥ N ⎣ (1 + i) − 1⎦ ⎡ ⎤ 0.063 = 90000 ⎢ ⎥ = $70.56 72 + − ( 1 0 . 063 ) 1 ⎦ ⎣

Example 4 - Concept If $70.56 was deposited in an account earning 6.3% yearly, $90,000 would be in the account at the end of 72 years.

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Present Given Annual † A is an equal annual flow over the time period „

(time period: Time 0 to Time N, 1st flow at Time 1)

† P is the present value at Time 0 „

(N periods in the past)

† i is the effective interest rate for each period P? 0

1

2

3

N A

P = A(P/A,i,N) †

Note: cash flow A does not have to be annual, just periodic

Equal Payment Series Present Worth Factor P 1

2

3 N

0

A

(1 + i ) N − 1 P= A i (1 + i ) N = A( P / A, i , N )

Example 4.18: † Given: A = $32,639, N = 9 years, and i = 8% † Find: P † Solution: P = $32,639(P/A,8%,9) = $203,893

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 5 What is the present worth of deposits of $1,000 at the end of each of the next 9 years at 8% compounded annually? GIVEN:

A = $1,000 i = 8% N=9

DIAGRAM:

P? 1 0

FIND P:

2

3

N=9

P = A(P|A,i,N) = 1,000(P|A,8%,9)

$1,000

= 1,000(6.2469) = $6,247

Example 5 - Concept Assuming an 8% annual return on investment, nine yearly payments of $1,000 are equivalent to $6,247 today.

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Annual Given Present † A is the equivalent annual flow over the time period „

(time period: Time 0 to Time N, 1st flow at Time 1)

† P is the present value at Time 0 „

(N periods in the past)

† i is the effective interest rate for each period P 0

1

2

3

N A?

A = P(A/P,i,N) †

Note: cash flow A does not have to be annual, just periodic

Capital Recovery Factor P 1

2

3 N

0

A

i(1 + i) N A= P (1 + i) N − 1 = P( A / P, i, N )

Example 4.16: † Given: P = $250,000, N = 6 years, and i = 8% † Find: A † Solution: A = $250,000(A/P,8%,6) = $54,075

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 6 What series of equal, annual payments is necessary to repay $50,000 in 10 years at 8.5% compounded annually? GIVEN:

P = $50,000 i = 8.5% N = 10

DIAGRAM: FIND A:

$50,000

A = P(A|P,i,N) 1 0

2

3

N=10 A?

= 50,000(A|P,8.5%,10) ⎡ i(1 + i)N ⎤ = P⎢ ⎥ N ⎣ (1 + i) − 1⎦ ⎡ 0.085(1 + 0.085 )10 ⎤ = 50000 ⎢ ⎥ = $7620 10 ⎣ (1 + 0.085 ) − 1 ⎦

Example 6 - Concept I am indifferent between having $50,000 today and 10 annual payments of $7,620, given a return rate of 8.5% per year.

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Present Given Gradient (Linear) † G is the linear gradient over the time period „

(time period: Time 0 to Time N, 1st flow at Time 2)

† P is the present value of the flow at Time 0 „

(N periods in the past)

† i is the effective interest rate for each period P? 0

1

2

3

N

G

P = G(P/G,i,N) †

Note: cash flow is periodic, no flow at Time 1, flow of G at Time 2

Linear Gradient Series

F

i(1+i) N −iN −1 P=G 2 i (1+i) N = G( P / G,i, N) F = P(1+i)N

P

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= G(F|G,i,N)

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Gradient Series as a Composite Series

Example 7 What is the present worth of a series of 30 end of the year payments that begin at $250 and increase at the rate of $50 a year if the interest rate is 9% compounded annually? DIAGRAM:

P? 1

GIVEN:

2

3

Payments i = 9% N = 30

N=30

0 $50

$250

FIND P:

$100 $1,450

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 7 This can be broken into an Annual flow and a Linear Gradient flow DIAGRAM:

P? 1

2

PA ? 1 3

N=30

2

3

N=30

0 $250

0 $50

$250 PG ? 1

$100

2

3

$1,450 0

$50

N=30

$100 $1,450

Example 7 - Concept Annual payments that begin at $250 and increase with a linear gradient of $50 each year for 30 years are equivalent to $7,020 today, assuming 9% return on investment. DIAGRAM:

P? 1

2

3

N=30 P = PA + PG = A(P|A,i,N) + G(P|G,i,N)

0 $50

$250 $100

= 250(10.2737) + 50(89.0280) $1,450

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= 250(P|A,9%,30) + 50(P|G,9%,30)

= 2,568.43 + 4,451.40 = $7,020

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Future Given Gradient (Linear) † G is the linear gradient over the time period „

(time period: Time 0 to Time N, 1st flow at Time 2)

† F is the future value of the flow at Time N „

(N periods in the future)

† i is the effective interest rate for each period F? 0

1

2

3

N

G

F = G(F/G,i,N) †

Note: cash flow is periodic, no flow at Time 1, flow of G at Time 2

Example 8 What is the future worth of a series of 42 deposits that begin at $10,000 and decrease at the rate of $100 a year with 8% interest compounded annually? GIVEN:

Payments i = 8% N = 42 FIND F42 :

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 8 What is the future worth of a series of 42 deposits that begin at $10,000 and decrease at the rate of $100 a year with 8% interest compounded annually? FA ? 1

2

FG ? 1

3

2

3 N=42

N=42 0

0 $10 000

$100

$200

F42 = FA – FG = A(F|A,i,N) – G(F|G,i,N)

$4 100

= 10 000(F|A,8%,42) – 100(F|G,8%,42)

Example 8 - Concept $2,714,631 would be accumulated if yearly payments are made that begin at $10,000 and decrease yearly with a linear gradient of $100, given an interest rate of 8%. ⎛ 1 ⎡ (1 + i)N − 1 ⎤ ⎞ ⎡ (1 + i)N − 1⎤ − N⎥ ⎟ F42 = A(F|A,i,N) – G(F|G,i,N) = A ⎢ ⎥ − G⎜⎜ ⎢ ⎟ i i ⎣ ⎦ ⎦⎠ ⎝i⎣ ⎛ 1 ⎡ (1 + 0.08)42 − 1 ⎡ (1 + 0.08)42 − 1⎤ ⎤⎞ = 10000 ⎢ − 42⎥ ⎟ ⎥ − 100⎜⎜ ⎢ ⎟ 0.08 0.08 ⎣ ⎦ ⎦⎠ ⎝ 0.08 ⎣ = 10000(304.24352 ) − 100[12.5(304.24352 − 42)] = 3042435 − 327804 = $2714631

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Reviewing … You may add or subtract cash flows or equivalents only at the same point in time! G = Linear Gradient g = Geometric Gradient The geometric gradient will help us deal with inflation.

Geometric Gradient Series

1− (1+ g) N (1+ i)− N , if i ≠ g A P= 1 i−g if i = g NA1 / (1+ i),

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example: Geometric Gradient: Find P, Given A1,g,i,N † Given:

P

g = 7% i = 12% N = 5 years A1 = $54,440 † Find: P

. )5 (1+ 012 . )−5 1− (1+ 007 . − 007 . 012 = $151,109

P = $54,440

Example 9 Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses?

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Example 9 † Given:

g = 8% i = 5% N = 4 years A1 = $10,000 † Find: P DIAGRAM: $10,000

0 P?

1

2

3

N=4

P = A1(P|A1,g,i,N)

1 − [(1 + g ) /(1 + i )]N P = A1{ } (i – g) 1 − [(1 + 0.08) /(1 + 0.05)]4 P = 10,000{ } (0.05 – 0.08)

P = 10,000{3.9759} = $39,759

Example 9 - Concept If your rich Aunt Edna wanted to put a sum of money in the bank today to pay for your next for years of tuition, that sum would be $39,759 assuming 5% return on investment and tuition that begins at $10,000 increasing by 8% per year. This problem assumes tuition is due at the end of the year.

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ENGR 390 Winter 2006 Section 1 Lecture 2 Time is Money: Problems

Standard Factors Used to Solve ECON Problems ( F | P, i, N) Î ( P | F, i, N) Î ( F | A, i, N) Î ( A | F, i, N) Î ( P | A, i, N) Î ( A | P, i, N) Î ( P | G, i, N) Î ( A | G, i, N) Î ( F | G, i, N) Î

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Find F Given P Find P Given F Find F Given A Find A Given F Find P Given A Find A Given P Find P Given G Find A Given G Find F Given G

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