Class notes ENGINEERING ECONOMY Engineering Economy

, chapter 2



factor of &F.
single payment only consists
calculate it tables
To calculate it
manually
: To
using
>
-
F
=
p(l + i)m F P/FIB i-, n) ,




>
-
P =
E(1)( + i(h) P = F(pIf if in)
,
.




Example :




P = 5000 i =
8 % F =? n = 10
years
F 5000 (1 + 0 00)10 10799 is C
manually : =
.
= answer


table : F =
5000 (2 1589)
.
=
10794S



Example :




F-50000 n = 9 i = 10 % P =
?

50000(1/(1 + 0 15) 31046 1 is
Manually 4 =
: = . .
answer


p :
50000 (0 .




6209) = 31045

,uniform consists of PR A
series
only
calculate it To calculate it tables
To
using
:

manually
:




P
=
A .
(( + is" -
1/ istim] P =
A(p(A , i) - (n)

A =
p .

[iltil/ (Iti)" -
1] A =
P(AIP , 1 % n)

that present worth located
period first
Note
always
is one prior to A
p .




Example :




A = 5000 1 = 10 % P =? R = S

5000 [(8 + 0 113-1101(10 15) 18933 9 so answer is D
manually
: -
.
= .




table : 5000(3 7908) .



=
18994

, of
onlyConsisting
uniform series FRA
To calculate it calculate it tables
manually
: To
using
:




f= A .


[(Hi)- 1/i] = A(FIA , i % in)

A = F .

[i)(1 i) 17 + -

A = F (Alf ,
i %, n)
that there's last period f
Note
always an A with




# =
10000 i =
8 % n =
7 F = ?

F 100005 00 -1/0 08]
Manually (1 + 0 89228 answer is
= =
: .
.




table : F = 10000 (8 9228).

=
89228



factor values for untabulated i

> use manual formulas
interpolate doesn't exact
>
-
>
-



give answer




usingFormula
>
-
: (1 + 0 .
08331 = 2 2197
.




>
interpolating : z 2 1989
.

-




83 -
X



9 2 .


3674

3674- 2 1989 approximation .
&
2 .
.
=
2 . 2215 - >


2 3674 .
-
X