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
Shwetank Mauria
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 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)]#

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