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Stels [109]
2 years ago
7

You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydr

aulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 15% per year? Why is yours the correct choice?
Alternative X Y
First Cost $-45,000 $-58,000
Maintenance cost, per Year $-8000 $-4000
Salvage Value $2,000 $12,000
Life 5 years 5 years

The present worth of alternative X is( ______$ )and that of alternative Y is(_______ $ ).
Business
1 answer:
seraphim [82]2 years ago
7 0

Answer:

Present value of Project X is $41990.6

Present value of Project Y is $34605.2

Explanation:

                           <u>PROJECT X   </u>                                  

PARTICULARS    YEAR      Cost/Value   Present Value factor 15% Present Value

Initial Cost                0              45000         1                  45000

Maintenance           1- 5             8000         3.352           26816

cost

Annual Depreciation 1-5           (8600)       3.352           (28827.2)

Salvage Value              5           (2000)       0.4971         <u> (994.2)    </u>

                      Present value of cash outflows             41,990.6

                             <u>PROJECT Y</u>

<u>PARTICULARS </u>   YEAR   COST/VALUE Present value factor 15% PRESENT VALUE

Initial Cost                 0           58000             1                    58000

Maintenance           1- 5           4000            3.352               13409  

cost

Annual Depreciation  1-5        (9200)           3.352             (30838.4)  

Salvage Value              5         12000            0.4971          <u>  (5965.2) </u>    

Present Value of cash outflow                                          34605.2  

Note: Figures in parenthesis denote cash inflow

Working Notes

Depreciation for project X  = \frac{45000\ -\ 2000}{5}  = $8600 p.a

Depreciation for project Y = \frac{58000\ -\ 12000}{5}  = $9200 p.a

Decision: Since present value of cash outflows is lesser for Project Y, it should be taken up.

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Answer:

Side opposite the river = 120 ft

Other sides = 240 ft

Explanation:

Let 'R' denote the length of fence opposite to the river and 'L' denote the length of the other two sides.

The cost as a function of R is:

L*R = 28,800\\L=\frac{28,800}{R}\\ C = 40R+10*2*\frac{28,800}{R} \\C(R) = 40R+576,000R^{-1}

The value of R for which the derivate of the cost function is zero is the length that minimizes cost:

C'(R) =0= 40 -576,000R^{-2}\\R=\sqrt{\frac{576,000}{40}}\\R=120\ ft\\

If R is 120 ft, then the value of L is:

L = \frac{28,800}{120}\\L=240\ ft

The dimensions that will minimize costs are:

Side opposite the river = 120 ft

Other sides = 240 ft

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