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I am trying to solve a rate of return question from the book Engineering Economics by R. Paneerselvam. In that particular problem I am given a salvage value along with other factors. Following are the factors:

i) Initial cost

ii) Annual incremental revenue

iii) Life

iv) Life-end Salvage value (Rs.)

Now when salvage value is not given we use the following formula:

$PW_n(i) = -P + A(P/A,i,n)$

Now how will this formula be modified when salvage value is also added? I mean what factor will be introduced alongside the Salvage value?

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    $\begingroup$ I´ve noticed that I didn´t discounted the sum of the annuities. I´ve made an edit of my answer. $\endgroup$ Jul 4 '19 at 15:16
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To obtain the present value of the salvage value ($S$) you discount the value of the salvage $n$ times. Therefore the whole formula is

$$PV_n(i)=-P+A\cdot \frac{(1+i)^n-1}{i\cdot (1+i)^n}+\frac{S}{(1+i)^n}$$

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  • $\begingroup$ won't we use S(F/P) instead of S(P/F)? $\endgroup$ Jul 4 '19 at 22:10
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    $\begingroup$ I don´t know the meaning of S(F/P) or S(P/F). It would be nice if you show what formula it is. Beside this, I´m 100% sure that the Formula I´ve written is right. It is the present value if you have initial cost, n annuities and a salvage. $\endgroup$ Jul 4 '19 at 22:51
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    $\begingroup$ the one you used in the formula is P/F. $S(F/P) = S(1+i)^n$ $\endgroup$ Jul 4 '19 at 22:54
  • $\begingroup$ @AhmadQayyum We get the salvage value (S) after $n$ periods. To obtain the present value we have to discount it $n$ times. $\endgroup$ Jul 4 '19 at 22:59

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