Update loan balance every month

malaki1974

New Member
Joined
Feb 14, 2006
Messages
46
Hello,
Assume a loan start amount, current loan amount, and a start date and todays date. I’d like to see in cell b2 (Current loan amount) an updated remaining amount every month until 0.
The monthly loan payment amount can be in the formula or in another column. Assume 1000 dollars monthly payment for the below to be showing correctly with this new formula someone can help me write I hope!

abcd
100005000Jan 1 2022May 1 2022

thank you,

me
 

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Not sure if this is what you're looking for...

but if it isn't please explain in detail what you expect from the solution

Table1 is the little table to the right - this will serve as the master table where you will be able to input information about the loans handed out.

Table 2 will automatically calculate each loan's payment plan

P.S - Please ignore/delete column F, unless you require it


Cell Formulas
RangeFormula
A2:A31A2=IF(B1="","",ROW()-1)
B2:B31B2=IFERROR(IF(VLOOKUP(A2,Table1[[Helper]:[Applicant]],2)=0,"",VLOOKUP(A2,Table1[[Helper]:[Applicant]],2)),"")
C2:C31C2=IFERROR(INDEX(Table1[Total Loaned],MATCH(B2,Table1[Applicant],0)),"")
D2:D31D2=IFERROR(MIN(INDEX(Table1[Monthly Payments],MATCH(B2,Table1[Applicant],0))+IFERROR(LOOKUP(2,1/($B$1:B1=B2),$D$1:D1),0),C2),"")
E2:E31E2=IFERROR(D2-LOOKUP(2,1/($B$1:B1=B2),$D$1:D1),D2)
I3:I6I3=I2+N2
N2:N6N2=IF([@Applicant]="","",ROUNDUP([@[Total Loaned]]/[@[Monthly Payments]],0))
G2:G31G2=IF([@[Loan Applicant ]]="","",DATE(YEAR(INDEX(Table1[Date],MATCH(B2,Table1[Applicant],0))),MONTH(IFERROR(LOOKUP(2,1/($B$1:B1=B2),$G$1:G1),INDEX(Table1[Date],MATCH(B2,Table1[Applicant],0))))+1,DAY(1)))
 
Last edited:
Upvote 0
Thank you for the direction help. In looking at yours it made me think something like this: 100000−(1000×(DATEDIF(B3,TODAY(),"M"))) so that every month the number next to my mortgage would be reduced by $1000. b3 represents the date the mortgage began. So Jan 1, 2022 would be the date, $100,000 is loan amount, and $1000 is mortgage payment (fake data). In my liability column it would show $95,000 as soon as we hit June. This way I have a semi-accurate representation (ignoring interest etc) of the current liability for a bunch of loans I have without having to manually update whenever I open the sheet to check.
 
Upvote 0
Thank you for the direction help. In looking at yours it made me think something like this: 100000−(1000×(DATEDIF(B3,TODAY(),"M"))) so that every month the number next to my mortgage would be reduced by $1000. b3 represents the date the mortgage began. So Jan 1, 2022 would be the date, $100,000 is loan amount, and $1000 is mortgage payment (fake data). In my liability column it would show $95,000 as soon as we hit June. This way I have a semi-accurate representation (ignoring interest etc) of the current liability for a bunch of loans I have without having to manually update whenever I open the sheet to check.
Ah yeah! That is a clever way to handle it too.. good on you

Be sure to add the MIN function in it to prevent it from subtracting more than the total amount and going in the negative.

hope that helped!
 
Upvote 0
Good idea. I might do it but I was thinking if it goes negative the loan is paid off and then I delete the row! But seriously, not sure how to add the MIN so I might just try it.
 
Upvote 0

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