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C 3CF
F
L
E E
7
L B
A 1
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L :
. :
1. D , F, A
2. C , F, A .
3.
M .
4. D M
.
5. D
.
2
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C
I C 2, , , I ,
. F ,
1, /A ,
0, . F ,
. , , F .
, F, A .
3
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LEA NING C ME
4
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A
2, 1
5
Remember: / / ,
/ / ,
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E /A F :
6
The present worth of the cash flow shown below at i = 10% is:
(a) $25,304 (b) $29,562 (c) $34,462 (d) $37,908
P0 = ?
A = $10,000
0 1 2 3 4 5 6
i = 10%
Solution: (1) Use P/A factor with n = 5 (for 5 arrows) to get P1 in year 1(2) Use P/F factor with n = 1 to move P1 back for P0 in year 0
0 1 2 3 4 5
P1 = ?
Actual year
Series year
0 1( / ,10%,1) ( / ,10%,5)( / ,10%,1) 10,000(3.7 08) (0. 0 1) $34,462
( )
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E F/A F :
How much money would be available in year 10 if $8000 is depositedeach year in years 3 through 10 at an interest rate of 10% per year?
7
0 1 2 3 4 5 6 7 8 9 10
FA = ?
A = $8000
i = 10%
: Re-number diagram to determine n = 8 (number of arrows)
0 1 2 3 4 5 6 7 8
Cash flow diagram is:
FA= 8000(F/A,10%,8)= 8000(11.4359)= $91,487
Actual year Series year
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A
F :
8
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E : A
F 0 10% .
9
0 1 2 3 4 5 6 7 8 9 10
PT = ?
A = $5000
i = 10%
First, re-number cash flow diagram to get n for uniform series:n = 8
$2000
0 1 2 3 4 5 6 7 8 9 10
PT = ?
A = $5000
i = 10%
$2000
0 1 2 3 4 5 6 7 8
Solution:
Actual year Series year
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E : A
/A A 2: A = 5000( /A,10%,8) = 5000(5.3349) = $26,675M A 0 /F: 0 = 26,675( /F,10%,2) = 26,675(0.8264) = $22,044
M $2000 0: 2000 = 2000( /F,10%,8) = 2000(0.4665)
= $933
N , 0 2000 : = 22,044 + 933 = $22,977
10
A = $5000
i = 10%
0 1 2 3 4 5 6 7 8
Actual year 0 1 2 3 4 5 6 7 8 9 10
$2000
PT = ?
PA
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E D
( F/A /A )
:
F/A FA 10: FA = 5000(F/A,10%,8) = 5000(11.4359) = $57,180
M FA 0 /F: 0 = 57,180( /F,10%,10) = 57,180(0.3855) = $22,043
M $2000 0: 2000 = 2000( /F,10%,8) = 2000(0.4665) = $933
N , : = 22,043 + 933 = $22,976 .. A ,
11
0 1 2 3 4 5 6 7 8 9 10
PT = ?
A = $5000
i = 10%
0 1 2 3 4 5 6 7 8
FA = ?
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A G
1 21
G 2 2
M 03
A , 0 (A/ , , )4
13
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E : A G J D $5
4 . I 1 3 $60, 10 12% .
2 $5 ($60 ) 2 2 = 60( /A,12%,8) + 5( /G,12%,8) = $370.41
N , 2 0 0 = 2( /F,12%,2) = $295.29
N , A $60 1 2 A = 60( /A,12%,2) = $101.41
F , 0 A 0 = 0 + A = $396.70
14
0 1 2 3 104 5
60 60 6065
70
95
PT = ?i = 12%
G = 5
0 1 2 3 8 Gradient years
Actual years
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G G
15
1 2
( 1
( ) 1 1 (1+ )/(1+ ) /( )
( ) 1 1 (1+ )/(1+ ) /( )
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E : G G 5
$7000 . , 12%
8 . I $3 0
= 15% .
16
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E : G G
G 5 6; 1 2.
= 7000 1 (1+0.12)/(1+0.15) 9/(0.15 0.12) = $49,401
M 0
= 35,000 + 7000( /A,15%,4) + 49,401( /F,15%,4) = $83,232 17
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N G F , G +
G :
F ,
1 1 (1 )/(1+ ) /( + )
A
18
Changed from + to -
Changed from + to-
Changed from - to +
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E : N A G F ,
= 10%
: G G 2
1 2
G 0 ( 1);
$700 1 6 G = 700( /A,10%,6) 50( /G,10%,6) = 700(4.3553) 50(9.6842) = $2565
F = G(F/ ,10%,6) = 2565(1.7716) = $4544 19
F = ?
G = $-50
0 1 2 3 4 5 6 7
700 650
500 450550600
0 1 2 3 4 5 6
Actual years
Gradient years
i = 10%PG = ?
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A
21
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A
22