The right way to calculate tax

[vresetnikov]

Hello, I have just bumped into a thing that got me puzzled somewhat. I was trying to get the 22% tax value of 212.41.

The operation: 212.41 / 100 outputs the following:

{
  intValue: 212,
  value: 2.12,
  s: {
    symbol: '',
    separator: '.',
    decimal: '.',
    errorOnInvalid: false,
    precision: 2,
    pattern: '!#',
    negativePattern: '-!#',
    format: [Function: format],
    fromCents: false,
    increment: 0.01,
    groups: /(\d)(?=(\d{3})+\b)/g
  },
  p: 100
}

Now, with the precision set to 2 the last 2 digits 41 are dropped. As a result, after the final multiplication by 22 the result is 46.64 and not 46.73. I somewhat understand the root of the issue (probably the precision set to 2) but I would like to kindly ask if the current calculation and the loss of 9 cents is correct, or am I doing something wrong?

Maybe the following is wrong?

const amount = 212.41;
const result = amount.divide(100).multiply(22)).value; //result has a loss of 9 cents

^ performing the operation above results in a 9 cents wipe, as .41 is dropped during the / 100 operation.

If it is wrong, should I increase the precision to say 4 or is there something else I should do? However, in terms of increasing precision, I really wouldn't want to do that because it messes up the user friendly money representation (2 digits). Thank you!

[scurker]

Are you trying to calculate just the tax or the amount with tax?

You shouldn't need the intermediate divide, you can just multiply by whatever precision you need:

currency(212.41).multiply(0.22)
// {intValue: 4673, value: 46.73, s: {…}, p: 100}

Alternatively the amount with tax:

currency(212.41).multiply(1.22)
// {intValue: 25914, value: 259.14, s: {…}, p: 100}

[vresetnikov]

Hey Jason, thanks for the reply and for your help, much appreciated. Do you think it is a good practice if I increase the precision 5 decimals for math operations and then the last moment, right before storing the value into the database, reinstantiate the value with the precision of 2 decimals? Thank you.

…

On Mon, Aug 1, 2022 at 4:46 PM Jason ***@***.***> wrote: Are you trying to calculate just the tax or the amount with tax? You shouldn't need the intermediate divide, you can just multiply by whatever precision you need: currency(212.41).multiply(0.22) // {intValue: 4673, value: 46.73, s: {…}, p: 100} Alternatively the amount with tax: currency(212.41).multiply(1.22) // {intValue: 25914, value: 259.14, s: {…}, p: 100} — Reply to this email directly, view it on GitHub <#406 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AM6ZMDXGEQRV46GERXAGSW3VW7PKTANCNFSM55D4LUZA> . You are receiving this because you authored the thread.Message ID: ***@***.***>

[scurker]

So currency.multiply is a special case and one of the only methods that does not truncate the passed in value before applying the operation, but the resulting value will always have the correct precision. What you're suggesting would likely work, but you may be doing more work than is actually necessary.

[vresetnikov]

Thank you. Last quick question, and forgive me if I am disturbing you this much, I really appreciate your patience and kindness. As you have concisely explained, there is no reason to do division separately and one should rather do multiplication immediately. Looking to prevent potential future misunderstandings, I would like to kindly ask you why is it that precision suffers when doing it the elaborate way? ( currency(212.41).divide(100). multiply(22) ) Thanks in advance

…

On Mon, Aug 1, 2022 at 7:03 PM Jason ***@***.***> wrote: So currency.multiply is a special case and one of the only methods that does not truncate the passed in value before applying the operation, but the resulting value will always have the correct precision. What you're suggesting would likely work, but you may be doing more work than is actually necessary.

[scurker]

Because currency is immutable, each operation modifies the currency object. Here's what you have broken down:

// This is your starting value
const value1 = currency(212.41)

// The internal value will now be 2.12. Any additional precision beyond 2 is now "lost"
// because the options object has a default precision of 2
const value2 = value1.divide(100)

// The internal value will be 46.64 not 46.73 because it's calculating the value with the
// precision from the previous operation (2.12) 
const value3 = value2.multiply(22)