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.

1 Answer
Feb 9, 2018

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 is40000 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)]