Question

How to calculate present value?

Let us say that a person gets a job offer with $40,000/yr for the first year and gets a $20,000 raise every year afterward. Assume inflation is a fixed 8% every year.

What is this job offer worth today if the applicant will work for n years?

I can get and expression that includes

sum 1..n of n / 1.08^n but I don't know where to go from there, and i doubt that's even right since I don't think it needs to be that complicated.

Answer 1

Discounted or present value of all his receipts is `20000*1.08/0.08[1/1.08+(1.08^n-1)/(0.08*1.08^n)-(n+1)/1.08^(n+1)]`

Explanation:

The person gets `$40,000` in first year, `60,000` in second year and so on. It is an arithmatic sequence, whose first term is`40000` and common difference is `20000`.

Hence `n^(th)` term is `40000+20000(n-1)=20000n+20000=20000(n+1)`

Now as the person gets this amount after `n` years and rate of inflation is `8%`, discounting `n^(th)` term at `8%` gives us

`20000(n+1)/(1+8/100)^n`

= `20000((n+1))/1.08^n`

and discounted or present value of all his receipts is

`20000*sum_(1->n)((n+1))/1.08^n`

To convert this intoclosed form let us consider

`S=sum_(1->n)((n+1))/1.08^n`

or `S=2/1.08+3/1.08^2+4/1.08^3+.....+(n+1)/1.08^n`

Hence `S/1.08=2/1.08^2+3/1.08^3+4/1.08^4+.....+(n+1)/1.08^(n+1)`

Subtracting latter from former, we get

`S(1-1/1.08)=1/1.08+1/1.08+1/1.08^2+1/1.08^3+.....+1/1.08^n-(n+1)/1.08^(n+1)`

= `1/1.08+[1/1.08+1/1.08^2+1/1.08^3+.....+1/1.08^n]-(n+1)/1.08^(n+1)`

or `S*0.08/1.08=1/1.08+1/1.08((1-1/1.08^n)/(1-1/1.08))-(n+1)/1.08^(n+1)`

or `S=1.08/0.08[1/1.08+(1-1/1.08^n)/0.08-(n+1)/1.08^(n+1)]`

= `1.08/0.08[1/1.08+(1.08^n-1)/(0.08*1.08^n)-(n+1)/1.08^(n+1)]`

and discounted or present value of all his receipts is `20000*1.08/0.08[1/1.08+(1.08^n-1)/(0.08*1.08^n)-(n+1)/1.08^(n+1)]`