Margaret has a project with a $30,000 first cost that returns $5,600 per year over its 10-year life. It has salvage value of $3,800 at the end of 10 years. If the MARR is 15 %.
(Use 5 significant figures for your calculations, and round your answers to the nearest dollar. Indicate losses as a negative value.)
(a) What is the present worth of this project?
(b) What is the annual worth of this project?
(c) What is the future worth of this project after 10 years?
a ) Ans: the present worth of this project is -$954
Explanation:
P = $30,000
Annual return (A) = $5,600
Rate of interest ( i ) = 15%
Number of interest period ( n ) = 10
Salvage value ( F ) = $3,800
PW = - 30,000 + 5600 ( P / A , i ,n ) + 3800( P / F , i ,n )
= -30,000 + 5600 ( P / A , 15% , 10 ) + 3800( P / F , 15% ,10 )
= -30,000 + 5600 ( 5.019 ) +3800( 0.2472 )
= -30,000 + 28106.4 + 939.36
= - 30,000 + 29045.76
= - $954.24
b ) Ans: the annual worth of this project is -$192
Explanation:
P = $30,000
Annual return (A) = $5,600
Rate of interest ( i ) = 15%
Number of interest period ( n ) = 10
Salvage value ( F ) = $3,800
AW = -30,000 ( A / P , i ,n ) + 5600 + 3800 ( A / F , i ,n )
= -30,000 ( A / P , 15% , 10 ) + 5600 + 3800 ( A / F , 15% ,10 )
= -30,000 ( 0.1993 ) + 5600 + 3800 ( 0.0493)
= - 5979 + 5600 +187.34
= - 5979 + 5787.34
= - $191.66
c ) Ans: the future worth of this project is -$3878
Explanation:
P = $30,000
Annual return (A) = $5,600
Rate of interest ( i ) = 15%
Number of interest period ( n ) = 10
Salvage value ( F ) = $3,800
FW = -30,000 ( F / P , i ,n ) + 5600 ( F / A , i ,n ) + 3800
= -30,000 ( F / P , 15% ,10 ) + 5600 ( F / A , 15% ,10 ) + 3800
= -30,000 ( 4.046 ) + 5600 ( 20.304 ) + 3800
= -121380 + 113702.4 + 3800
= -121380 + 117502.4
= - $3877.6
Margaret has a project with a $30,000 first cost that returns $5,600 per year over its...
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per year. Now that the investment has been in operation for 1 year,
revised figures indicate that it actually cost $228,000, will have
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completed), and...