ACCRINTM

ACCRINTM(issue, settlement, rate [,par] [,basis])

Returns the accrued interest for a security that pays interest at maturity.

issue	The date the security is issued.
settlement	The settlement date (or maturity date) of the security.
rate	The annual coupon rate of the security.
par	(Optional) The par value of the security (1000).
basis	(Optional) The type of day counting to use. 0 = 30/360 (default, US convention) 1 = Actual/Actual 2 = Actual/360 3 = Actual/365 4 = 30/360 (European convention)

REMARKS

* Dates must be entered as text strings within quotation marks or as serial numbers.
* The maturity date is the date that the security expires.
* If "issue" is not an integer, it is truncated.
* If "issue" is not a valid date, then #VALUE! is returned.
* If "issue" >= "settlement", then #NUM! is returned.
* The "settlement" date is when the security is traded to the buyer.
* If "settlement" < "issue", then
* If "settlement" is not an integer, it is truncated.
* If "settlement" is not a valid date, then #VALUE! is returned.
* If "rate" <= 0, then #NUM! is returned.
* If "par" <= 0, then #NUM! is returned.
* The "par" is not optional, but can be left blank which is very confusing. Example 9.
* If "par" is left blank, then 1000 is used.
* If "basis" is left blank, then 0 is used.
* If "basis" is not an integer, it is truncated.
* If "basis" < 0, then #NUM! is returned.
* If "basis" > 4, then #NUM! is returned.
* The difference between the US convention and the European convention is how they handle the last day of the month. more
* You can use the ACCRINT function to return the accrued interest for a security that pays interest periodically.
* You can use the DATE function to return the date as a date serial number given a year, month, day.
* For the Microsoft documentation refer to support.microsoft.com

	A	B
1	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, 1000, 0) = 80	01-Jan-2022
2	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, , 0) = 80	31-Dec-2023
3	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, 1000, ) = 80
4	=ACCRINTM("1 Jan 2022", "1 Jul 2023", 0.04, 1000, 0) = 60
5	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.08, 1000, 0) = 160
6	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, 2000, 0) = 160
7	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, , 1) = 80
8	=ACCRINTM("1/1/2022", "31 Dec 2023", 0.04, 1000, 0) = 80
9	=ACCRINTM(DATE(2022, 1, 1), "31 Dec 2023", 0.04, ) = 80
10	=ACCRINTM(B1, B2, 0.04, 1000, 0) = 80
11	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, -1000, 0) = #NUM!
12	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, , -1) = #NUM!
13	=ACCRINTM("1 Jan 2022", "31 Dec 2023", 0.04, , 5) = #NUM!

1 - How much interest has accrued at maturity on a US treasury bond with a par value of $1,000 issued on 1 Jan 2022 with a settlement date of 31 Dec 2023 with an annual coupon of 4% with a day count of 30/360.
2 - The same as Example 1 except the par value has been left blank.
3 - The same as Example 1 except the basis has been left blank.
4 - The same as Example 1 except the settlement date is 1 Jul 2023.
5 - The same as Example 1 except the annual coupon is 8%.
6 - The same as Example 1 except the par value is 2000.
7 - The same as Example 1 except the basis is Actual/Actual instead of 30/360.
8 - The same as Example 1 except the issue date has been provided in the "dd/mm/yyyy" format.
9 - The same as Example 1 except the issue date has been provided using the DATE function.
10 - The same as Example 1 except the issue date and settlement dates are coming from cells.
11- The same as Example 1 except the par value is negative.
12 - The same as Example 1 except the basis is -1 which is less than 0.
13 - The same as Example 1 except the basis is 5 which is greater than 4.