w=3 x=8 y=8 z=6 Determine the I- Future worth in actual and constant dollar 2- A

Question

w=3
x=8
y=8
z=6

Determine the I- Future worth in actual and constant dollar 2- Annual equivalent in constant dollar for the following projects. The nominal annual market interest rate is (w+x+y+2 % (compounded annually). The project life is 10 years and the base year is at year zero. Value Inflation rate wxyz*5000 | (w+x+2+y)/4% wxyz0 |(w+X+2+y)/100 Initial Investment every year 70% Monthly maintenance cost (e every month | ( w +x+2+y)/2% % | (vw+x+2+y)/3% % Annual material and labor cost (actual dollar)wxyz* 100 every year Annual Revenue (constant dollar) z*500 Future salvage value (actual dollar) wxyz" 1000 |(w+X+2+y)/5% % every year

Explanation / Answer

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Under the Discount Adjusted method:

Pn = An/(1+i)^n

i = i dash + f bar + i dash * f bar

Cite: Courtesy Professor C S Park of the department of Economics Lecture Notes PDF Document released in the year 2005

Steps to calculate Equivalence:

1. Interest Rate types:

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Initial capital outlayed or invetsed wxyz*5000 5760000 (w+x+z+y)/4% =(3+8+8+6)/0.04 625 =(3+8+8+6)/0.04 625 Cost per month to maintain wxyz*10 11520 (w+x+z+y)/10 % % =(3+8+8+6)/0.0001 250000 =(3+8+8+6)/0.01 2500 Cost per annum for labor wages and Raw materials wxyz*100 115200 (w+x+z+y)/2% % =(3+8+8+6)/0.0002 125000 =(3+8+8+6)/0.02 1250 Cash inflow or Revenue per annum wxyz*500 576000 (w+x+z+y)/3% % =(3+8+8+6)/0.0003 83333.33333 =(3+8+8+6)/0.03 833.333333 Scrap value or Salvage value after the life of the asset wxyz*1000 1152000 (w+x+z+y)/5% % =(3+8+8+6)/0.0005 50000 =(3+8+8+6)/0.05 500

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