arch 449 chapter 3

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Financial Mathematics 1

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ARCH 449 Chapter 3. Financial Mathematics. Notation. i = interest rate (per time period) n = # of time periods P = money at present F = money in future After n time periods Equivalent to P now, at interest rate i A = Equal amount at end of each time period on series E.g., annual. - PowerPoint PPT Presentation

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Page 1: ARCH 449 Chapter 3

Financial Mathematics

1

Page 2: ARCH 449 Chapter 3

• i = interest rate (per time period)• n = # of time periods• P = money at present• F = money in future

– After n time periods– Equivalent to P now, at interest rate i

• A = Equal amount at end of each time period on series– E.g., annual

2

Page 3: ARCH 449 Chapter 3

• # on the cash flow means end of the period, and the starting of the next period

0 1 2 3 4 5

500

200

50100

200

500

+

_Time

End of second year

Biggining of third year

Page 4: ARCH 449 Chapter 3

• If P and A are involved the Present (P) of the given annuals is ONE YEAR BEFORE THE FİRST ANNUALS

0 1 2 3 n-1 n

A

P

Page 5: ARCH 449 Chapter 3

• If F and A are involved the Future (F) of the given annuals is AT THE SAME TIME OF THE LAST ANNUAL

• :

0

A

F

0

…………..

n 1 2 3 .. .. n-1

Page 6: ARCH 449 Chapter 3

0

A

F

0

…………..

n 1 2 3 .. .. n-1

P

Page 7: ARCH 449 Chapter 3

7

• Converting from P to F, and from F to P

• Converting from A to P, and from P to A

• Converting from F to A, and from A to F

Page 8: ARCH 449 Chapter 3

Present to Future,

and Future to Present

8

Page 9: ARCH 449 Chapter 3

• To find F given P:

9

P0

Fn

n………….

Fn = P (F/P, i%, n)

Page 10: ARCH 449 Chapter 3

• Invest an amount P at rate i:– Amount at time 1 = P (1+i)– Amount at time 2 = P (1+i)2

– Amount at time n = P (1+i)n

• So we know that F = P(1+i)n – (F/P, i%, n) = (1+i)n – Single payment compound amount factor

Fn = P (1+i)n

Fn = P (F/P, i%, n)

10

Page 11: ARCH 449 Chapter 3

• Invest P=$1,000, n=3, i=10%

• What is the future value, F?

11

0 1 2 3

P = $1,000

F = ??

i = 10%/year

F3 = $1,000 (F/P, 10%, 3) = $1,000 (1.10)3

= $1,000 (1.3310) = $1,331.00

Page 12: ARCH 449 Chapter 3

• To find P given F:– Discount back from the future

12

P

Fn

n………….

(P/F, i%, n) = 1/(1+i)n

Page 13: ARCH 449 Chapter 3

• Amount F at time n:– Amount at time n-1 = F/(1+i)– Amount at time n-2 = F/(1+i)2

– Amount at time 0 = F/(1+i)n

• So we know that P = F/(1+i)n – (P/F, i%, n) = 1/(1+i)n – Single payment present worth factor

13

Page 14: ARCH 449 Chapter 3

• Assume we want F = $100,000 in 9 years. • How much do we need to invest now, if the

interest rate i = 15%?

14

0 1 2 3 8 9…………

F9 = $100,000

P= ??

i = 15%/yr

P = $100,000 (P/F, 15%, 9) = $100,000 [1/(1.15)9]

= $100,000 (0.1111) = $11,110 at time t = 0

Page 15: ARCH 449 Chapter 3

Annual to Present,

and Present to Annual

Page 16: ARCH 449 Chapter 3

• Fixed annuity—constant cash flow

$A per period

P = ??

0

…………..

n 1 2 3 .. .. n-1

Page 17: ARCH 449 Chapter 3

• We want an expression for the present worth P of a stream of equal, end-of-period cash flows A

0 1 2 3 n-1 n

A is given

P = ??

Page 18: ARCH 449 Chapter 3

• Write a present-worth expression for each year individually, and add them

1 2 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

The term inside the brackets is a geometric progression.

This sum has a closed-form expression!

Page 19: ARCH 449 Chapter 3

• Write a present-worth expression for each year individually, and add them

1 2 1

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

(1 ) 1 0

(1 )

n

n

iP A for i

i i

Page 20: ARCH 449 Chapter 3

• This expression will convert an annual cash flow to an equivalent present worth amount:– (One period before the first annual cash flow)

(1 ) 1 0

(1 )

n

n

iP A for i

i i

The term in the brackets is (P/A, i%, n) Uniform series present worth factor

Page 21: ARCH 449 Chapter 3

• Given the P/A relationship:(1 ) 1

0(1 )

n

n

iP A for i

i i

(1 )

(1 ) 1

n

n

i iA P

i

We can just solve for A in terms of P, yielding:

Remember: The present is always one period before the first annual amount!

The term in the brackets is (A/P, i%, n) Capital recovery factor

Page 22: ARCH 449 Chapter 3

Future to Annual,

and Annual to Future

Page 23: ARCH 449 Chapter 3

• Find the annual cash flow that is equivalent to a future amount F

0

$A per period??

$F

The future amount $F is given!

0

…………..

n 1 2 3 .. .. n-1

Page 24: ARCH 449 Chapter 3

• Take advantage of what we know• Recall that:

and

1

(1 )nP F

i

(1 )

(1 ) 1

n

n

i iA P

i

Substitute “P” and simplify!

Page 25: ARCH 449 Chapter 3

• First convert future to present:– Then convert the resulting P to annual

• Simplifying, we get:

1 (1 )

(1 ) (1 ) 1

n

n n

i iA F

i i

(1 ) 1nA

i

iF

The term in the brackets is (A/F, i%, n) Sinking fund factor (from the year 1724!)

Page 26: ARCH 449 Chapter 3

• How much money must you save each year (starting 1 year from now) at 5.5%/year:– In order to have $6000 in 7 years?

Page 27: ARCH 449 Chapter 3

• Solution:– The cash flow diagram fits the A/F factor

(future amount given, annual amount??)

– A= $6000 (A/F, 5.5%, 7) = 6000 (0.12096) = $725.76 per year

– The value 0.12096 can be computed (using the A/F formula), or looked up in a table

Page 28: ARCH 449 Chapter 3

• Given

• Solve for F in terms of A:

(1 ) 1n

iA F

i

)=A

(1 1F

ni

i

The term in the brackets is (F/A, i%, n) Uniform series compound amount factor

Page 29: ARCH 449 Chapter 3

• Given an annual cash flow:

0

$A per period

$F

Find $F, given the $A amounts

0

…………..

n 1 2 3 .. .. n-1

Page 30: ARCH 449 Chapter 3

Single-Payment Compound-Amount Factor

Single-Payment Present-Worth Factor

Equal-Payment-Series Compound-Amount Factor

Equal-Payment-Series Sinking-Fund Factor

Equal-Payment-Series Capital-Recovery Factor

Equal-Payment-Series Present-Worth Factor

, , (1 )nF P i n i

1, ,

(1 )nP F i n

i

(1 ) 1, ,

niF A i n

i

, ,(1 ) 1n

iA F i n

i

(1 ), ,

(1 ) 1

n

n

i iA P i n

i

(1 ) 1, ,

(1 )

n

n

iP A i n

i i

Page 31: ARCH 449 Chapter 3