A firm is choosing between machines that perform the same task in the same time. Assume the minimum attractive return is 8%. Which machine would you choose?
Machine XÂ Â Â Â Â Â Â Â Â Â Machine Y
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First Cost                                                                 $5000               $8000
Esitmated life, in years                                            5                       12
Salvage value                                                              0                  $2000
Annual Maintenance Cost                                        0                       150
Solution
MACHINE X:
total cost of investment= original cost + salvage value= $5000 + 0= $5000
total return= 5000x 8%= 400
average returns= total return/ life of investment= 400/5= $80
average investment= 5000/2= 2500
average rate of returns= (80/2500)x100= 3.2%
MACHINE Y
total cost of investment= original cost+ salvage value= $8000+$2000= $10000
return= 10000x8%= 800
average return= (total returns- annual maintenance cost)/ life of project = (800-150)/12= 54.167
average investment= total investment/ 2= 10000/2= 5000
average rate of return= (54.167/5000)x100= 1.083
the average of return of Machine x is greater than of Machine Y, so Machine X must be purchased
.
A firm is choosing between machines that perform the same task in the.docx
1. A firm is choosing between machines that perform the same task in the same time. Assume the
minimum attractive return is 8%. Which machine would you choose?
Machine XÂ Â Â Â Â Â Â Â Â Â Machine Y
------------------------------------------------------------------------------------------------------
First Cost
                                       Â
                       Â
$5000Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â $8000
Esitmated life, in
years                                     Â
      5                       12
Salvage
value                                     Â
                       Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â $2000
Annual Maintenance Cost
                                      Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 150
Solution
MACHINE X:
total cost of investment= original cost + salvage value= $5000 + 0= $5000
total return= 5000x 8%= 400
average returns= total return/ life of investment= 400/5= $80
average investment= 5000/2= 2500
2. average rate of returns= (80/2500)x100= 3.2%
MACHINE Y
total cost of investment= original cost+ salvage value= $8000+$2000= $10000
return= 10000x8%= 800
average return= (total returns- annual maintenance cost)/ life of project = (800-150)/12= 54.167
average investment= total investment/ 2= 10000/2= 5000
average rate of return= (54.167/5000)x100= 1.083
the average of return of Machine x is greater than of Machine Y, so Machine X must be
purchased
