April 7th, 2015, 12:44 PM  #11 
Newbie Joined: Apr 2015 From: miami Posts: 18 Thanks: 0 
I was trying to use your formula in a ms access table calculated field but found out that these fields can not accept brackets [] or braces {} only parentheses() so I tried to substitute () for the brackets and braces but the formula didn't work correctly. Any help is appreciated. [1  x^(1/p)]/(x1) * [ (A+g*ng) * [y^n  1]/(y1) ]  [ g/(y1) * { [y^n  1]/(y1)  n } ] Last edited by skipjack; April 9th, 2015 at 02:59 PM. 
April 7th, 2015, 04:35 PM  #12 
Newbie Joined: Apr 2015 From: miami Posts: 18 Thanks: 0 
Here is what I ended up with: ((1x^26)/(x1)*((A+g*ng)*(y^n1)/(y1))(g/(y1)*((y^n1)/(y1)n))) Not sure why this is not working MS Access is similar to Excel re order of operations: o Brackets calculated first o Indices (^) calculated next o Multiplication and division calculated next o Addition and subtraction calculated last Last edited by cb123; April 7th, 2015 at 04:41 PM. 
April 7th, 2015, 09:08 PM  #13 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5  Code: PV = (1  x^(1/p))/(x1) * ((A+g*ng)*(y^n  1)/(y1)  g/(y1)*((y^n  1)/(y1)  n)) The above formula only works for whole number of years such as 3 years 
April 7th, 2015, 11:12 PM  #14 
Newbie Joined: Apr 2015 From: miami Posts: 18 Thanks: 0 
Yes thank you, that would be the formula I need. (Will it also work when the periods are whole years?)
Last edited by skipjack; April 9th, 2015 at 03:02 PM. 
April 7th, 2015, 11:45 PM  #15 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5  Yes  give me 5 minutes, please.
Last edited by skipjack; April 9th, 2015 at 03:03 PM. 
April 8th, 2015, 12:22 AM  #16  
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5  Quote:
When n is complete year such as 3, then n_2 will be 0, thus the sum to the right of the formula will be 0, giving you present value for complete years. When n has a fractional part such as 3.5, then n_2 will be 0.5, thus the sum to the right of the formula will add to the present value up till complete years of 3, giving you the total present value for 3.5 years. Please note this is a generic formula, therefore you may use any sort of payment period. In your example the payments were biweekly, yet you can define payments such as quarterly by using p=1/4. Any type of interest compounding is also allowed. Let me know if it worked out in Access code. Code: A=100 g=15 i=13% annual cpd biweekly n=3.5 years n_1=FLOOR(n)=3 n_2=nFLOOR(n)= 3.53 = 0.5 c=1/26 biweekly interest compounding p=1/26 period is a biweek Find PV PV = (1  x^(1/p))/(x1) * ((A+g*ng)*(y^n  1)/(y1)  g/(y1)*((y^n  1)/(y1)  n)) + (A+g*n_1) * (y^(n_1)) * (1x^((n_2/p)))/(x1) aey(i,c) = (1 + i*c)^(1/c)  1 i=0.13 c=1/26 aey(13%,1/26) = (1 + 0.13/26)^(26)  1 aey(13%,1/26) = 0.138459553 aey(13%,1/26) = 13.85% x=[1 + aey(i,c)]^p p=1/26 x=[1 + 13.85%]^(1/26) x=[1.1385]^(0.038462) x=1.005 y=x^(1/p) y=(1.005)^(26) y=0.87837991 PV = (1  x^(1/p))/(x1) * ((A+g*n_1g)*(y^n_1  1)/(y1)  g/(y1)*((y^n_1  1)/(y1)  n_1)) + (A+g*n_1) * (y^(n_1)) * (1x^((n_2/p)))/(x1) PV = (1  1.005^(26))/(1.0051) * ((100+15*315)*(0.87837991^3  1)/(0.878379911)  15/(0.878379911)*((0.87837991^3  1)/(0.878379911)  3)) + (100+15*3) * (0.87837991^3) * (11.005^((0.5*26)))/(1.0051) PV = (1  1.005^(26))/(0.005) * (30*(0.87837991^3  1)/(0.12162009)  15/(0.878379911)*((0.87837991^3  1)/(0.12162009)  3)) + 145 * (0.87837991^3) * (11.005^(13))/(0.005) PV = $7,329.20 + $1,233.88 PV = $8,563.08 Last edited by skipjack; April 9th, 2015 at 03:17 PM. Reason: added explanation  
April 8th, 2015, 05:51 AM  #17 
Newbie Joined: Apr 2015 From: miami Posts: 18 Thanks: 0 
I will try this out tonight after I get off work and post back. Again, can't thank you enough for all your help and effort.

April 8th, 2015, 05:53 PM  #18 
Newbie Joined: Apr 2015 From: miami Posts: 18 Thanks: 0 
WORKS PERFECTLY!!!! You the man Abraham 
April 8th, 2015, 06:11 PM  #19 
Member Joined: May 2014 From: Rawalpindi, Punjab Posts: 69 Thanks: 5  You're welcome, If you ever need other financial formulas, post here. I have plenty of these formulas in my head, the current count is around 9,000+. Last edited by skipjack; April 9th, 2015 at 03:18 PM. 
April 8th, 2015, 06:48 PM  #20 
Newbie Joined: Apr 2015 From: miami Posts: 18 Thanks: 0 
9000+ formulas in your head. Must be hard to sleep at night. Thanks for sharing your knowhow.
Last edited by skipjack; April 9th, 2015 at 03:18 PM. 

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