Question

A machine is purchased for $100,000 and is depreciated under straight-line for 6 years, with a depreciation salvage value of $20,000. It generated net income per year of $45,000 for six years. After the sixth year it is disposed of for $25,000. Suppose that the firm assembles the $100,000 capital required by borrowing $35,000 and putting the balance from its own funds. Furthermore, assume that money is borrowed at 9% with repayment of principal and interest in 3 equal annual amounts. Assuming a 40-percent income tax-rate, what are the after-tax flows? What is the annual worth of these flows if the after-tax MARR is 10% percent?

Please answer without using excel or a graphic calculator. Thanks!

Answer 1

Solution : Computation of cash flow after tax ( amount in $)

Year. EBDIT. (D) (I)    EBT. EBT(1-t) Cash flow. after tax

1.    45,000 12,500 3150. 29,350 17,610. 30,110

2 . 45,000. 12,500 . 2189. 30,311. 18,187 . 30,687

3 . 45,000. 12,500. 11,41. 31,359 18,815. 31,315

4. 45,000. 12,500. ------ 32,500 19,500. 32,000

5. 45,000. 12,500 . ------- 32,500. 19,500. 32,000

6. 45,000. 12,500. ------- 32,500. 19,500. 32,000

Present value of after tax cash flow : ( Annual worth)

Year. Cash flow after tax. P. V. Factor . P. V. of cash flow  

1. 30,110. .909. 27,370

2. 30,687.    .826.    25,347

3. 31,315. .751. 23,527

4. 32,000 .    .683. 21.856

5. 32,000. .621. 19,872

6. 32,000.    .565 . 18,080

Working :

Depreciation on SLM :- $ 1,00,000 - $25,000/ 6 = $ 12,500

D = Depreciation

I = Interest

EBDIT = Earning before depreciation, interest and taxes

Calculation of equal installment :

E = P X. I X [( 1+i)^n / (1+i)^n -1 ]

E = Equal installment

P = Loan amount

I = interest rate

n = number of years of the loan

E = 35,000 X. 09 X (1.09)^3 / (1.09)^3 -1

= $ 13,828

Instalment. Principal. Interest. Balance

13,828. 10,678. 3,150 . 24,322

13,828 . 11,639. 2,189. 12,683

13=828. 12,687 . 1,141. ------------