Financial Mathematics
1
•
•
•
•
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
• # on the cash flow means end of the period,
and the starting of the next period
500
End of second year
+
0
_
500
200
200
1
2
3
4
5
Time
50
100
Biggining of third year
• If P and A are involved the Present (P) of
the given annuals is ONE YEAR BEFORE
THE FİRST ANNUALS
P
0
1
2
3
n-1
A
n
• If F and A are involved the Future (F) of
the given annuals is AT THE SAME
F
TIME OF THE LAST ANNUAL
• :
0
1
0
2
3
..
……
..
……..
A
n-1
n
F
P
0
1
0
2
3
..
……
..
……..
A
n-1
n
• 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
7
Present to Future,
and Future to Present
8
• To find F given P:
Fn
……
…….
P0
n
Fn = P (F/P, i%, n)
9
• 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
• Invest P=$1,000, n=3, i=10%
• What is the future value, F?
F = ??
0
P = $1,000
1
2
3
i = 10%/year
F3 = $1,000 (F/P, 10%, 3) = $1,000 (1.10)3
= $1,000 (1.3310) = $1,331.00
11
• To find P given F:
– Discount back from the future
Fn
……
…….
P
n
(P/F, i%, n) = 1/(1+i)n
12
• 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
• Assume we want F = $100,000 in 9 years.
• How much do we need to invest now, if the
interest rate i = 15%?
F = $100,000
9
i = 15%/yr
0
1
2
3
…………
8
9
P= ??
P = $100,000 (P/F, 15%, 9) = $100,000 [1/(1.15)9]
= $100,000 (0.1111) = $11,110 at time t = 0
14
Annual to Present,
and Present to Annual
• Fixed annuity—constant cash flow
P=
??
1
0
2
3
..
……
..
……..
n-1
$A per period
n
• We want an expression for the present
worth P of a stream of equal, end-ofperiod cash flows A
P=
??
0
1
2
3
n-1
A is given
n
• Write a present-worth expression for each
year individually, and add them
 1
1
1
1 
P  A

 .. 

1
2
n 1
n 
(1

i
)
(1

i
)
(1

i
)
(1

i
)


The term inside the brackets is a geometric progression.
This sum has a closed-form expression!
• Write a present-worth expression for
each year individually, and add them
 1
1
1
1 
P  A

 .. 

1
2
n 1
n 
(1  i)
(1  i) 
 (1  i) (1  i)
 (1  i)n  1
P  A
for i  0
n 
 i(1  i) 
• This expression will convert an annual cash
flow to an equivalent present worth amount:
– (One period before the first annual cash flow)
 (1  i)n  1
P  A
for i  0
n 
 i(1  i) 
 The term in the brackets is (P/A, i%, n)
 Uniform series present worth factor
• Given the P/A relationship:
 (1  i)n  1
P  A
for i  0
n 
 i(1  i) 
We can just solve for A in terms of P, yielding:
 i(1  i)n 
A P

n
(1

i
)

1


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
Future to Annual,
and Annual to Future
$
• Find the annual cash flow that is F
equivalent to a future amount F
1
0
2
3
..
……
..
……..
$A per period??
n-1
n
The future amount $F is
given!
• Take advantage of what we know
• Recall that:
 1 
PF
n 
 (1  i) 
and
 i(1  i)n 
A P

n
 (1  i)  1 
Substitute “P” and
simplify!
• First convert future to present:
– Then convert the resulting P to annual
 1   i(1  i)n 
A F

n 
n
 (1  i)   (1  i)  1
• Simplifying, we get:


i
AF

n
 (1  i )  1 
 The term in the brackets is (A/F, i%, n)
 Sinking fund factor (from the year 1724!)
• 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?
• 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
• Given


i
A F

n
 (1  i )  1 
• Solve for F in terms of A:
 (1  i )  1 
F =A 

i


n
 The term in the brackets is (F/A, i%, n)
 Uniform series compound amount factor
• Given an annual cash flow:
1
0
2
3
..
……
..
……..
$A per period
n-1
$
F
n
Find $F, given the $A
amounts
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
F
P , i, n   (1  i)n
1
 P F , i, n  
(1  i)n
(1  i)n  1
 F A , i, n  
i
i
 A F , i, n  
(1  i)n  1
i(1  i)n
 A P, i, n  
(1  i)n  1
(1  i)n  1
 P A, i, n  
i(1  i)n