rounding to 9's

[jdrott1]

i'm trying to round my currency string to end in 9. it's for a
pricing application.
this is the function i'm using to get the item in currency:
FormatCurrency( BoxCost, , , , TriState.True)

if an item is 41.87 i want the application to bring it up to 41.89

if an item is 41.84 i want the application to round down to 41.79

does anyone know how i could do this?

[rowe_newsgroups]

On Jul 6, 4:06 pm, jdrott1 <jonathandr...@ gmail.comwrote:

i'm trying to round my currency string to end in 9. it's for a
pricing application.
this is the function i'm using to get the item in currency:
FormatCurrency( BoxCost, , , , TriState.True)

if an item is 41.87 i want the application to bring it up to 41.89

if an item is 41.84 i want the application to round down to 41.79

does anyone know how i could do this?

You would have to do it manually - just parse the second digit after
the decimal and adjust the value accordingly.

Thanks,

Seth Rowe

[Mark S. Milley, MCSD (BinarySwitch)]

Hi Jonathan -

Have you tried rounding to the nearest tenth and subtracting one
hundreth?

Dim myVal As Double
myVal = 41.87
myVal = Math.Round(myVa l, 1) - 0.01
Debug.WriteLine (myVal) 'Returns 41.89

myVal = 41.84
myVal = Math.Round(myVa l, 1) - 0.01
Debug.WriteLine (myVal) 'Returns 41.79

....sometimes the simplest approaches work the best.

Happy Coding,

-Mark

[jdrott1]

On Jul 6, 3:50 pm, "Mark S. Milley, MCSD (BinarySwitch)"
<mark.mil...@bi naryswitch.comw rote:

Hi Jonathan -

Have you tried rounding to the nearest tenth and subtracting one
hundreth?

Dim myVal As Double
myVal = 41.87
myVal = Math.Round(myVa l, 1) - 0.01
Debug.WriteLine (myVal) 'Returns 41.89

myVal = 41.84
myVal = Math.Round(myVa l, 1) - 0.01
Debug.WriteLine (myVal) 'Returns 41.79

...sometimes the simplest approaches work the best.

Happy Coding,

-Mark

i'll give it a try. thanks.

[Mark S. Milley, MCSD (BinarySwitch)]

Hi Joregen -

How are you getting 4 down and 5 up? Last time I checked, we were
using base ten.

Numbers ending in 0-4 are rounded down, 5-9 are rounded up; 5 on each
side.

Thanks,

-Mark

[Number 11950 - GPEMC! Replace number with 11950]

"jdrott1" <jo***********@ gmail.comwrote in message
news:11******** ************@m3 6g2000hse.googl egroups.com...

i'm trying to round my currency string to end in 9. it's for a
pricing application.
this is the function i'm using to get the item in currency:
FormatCurrency( BoxCost, , , , TriState.True)

if an item is 41.87 i want the application to bring it up to 41.89

if an item is 41.84 i want the application to round down to 41.79

does anyone know how i could do this?

Set two decimal places correcting the error in the Int function for proper
mathematical rounding we have a very simple function:
X = Int(10*(X+0.005 ))/10

Now set one decimal place mathematical and subtract 1c:
X = Int(10 * (X + 0.06)) / 10 - 0.01
Here, 4 rounds up to 9 as per 5 rounding up to 10
& 3 rounds down to 9 as per 4 rounding down to 0
In the correct mathematical method:
IE +1 inside the brackets compensates -1 outside
just as *10 inside the brackets balances /10 outside.

THUS:

Public Function Nanofy(X as Double) as Double
Nanofy = Int(10 * (X + 0.06)) / 10 - 0.01
End Function

Executing:
Dim x=MsgBox("The Value is: $", CStr(Nanofy(CDb le(InputBox(_
"Enter a Value","Enter a Value")))),MsgB oxStyle.OkOnly, "The answer is")

Nanofy(41.84) Returns 41.89
Nanofy(41.83) Returns 41.79
To bias rounding upwards,
simply increment the inner bracket sum in isolation.

SO:
Public Function Nanofy(X as Double) as Double
Nanofy = Int(10 * (X + 0.05)) / 10 - 0.01
End Function

Thus executing:
Dim x=MsgBox("The Value is: $", CStr(Nanofy(CDb le(InputBox(_
"Enter a Value","Enter a Value")))),MsgB oxStyle.OkOnly, "The answer is")

Nanofy(41.85) Returns 41.89
Nanofy(41.84) Returns 41.79

Good luck...

--
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Terms & conditions apply. See www.fieldcraft.biz/GPEMC
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& the most advanced speed reading application @ www.fieldcraft.biz

[Number 11950 - GPEMC! Replace number with 11950]

"Number 11950 - GPEMC! Replace number with 11950" <nu****@fieldcr aft.biz>
wrote in message
news:46******** *************** @lon-reader.news.tel stra.net...

ERRATA:

[SNIP]

Set two decimal places correcting the error in the Int function for proper
mathematical rounding we have a very simple function:
X = Int(10*(X+0.005 ))/10

[SNIP]

This should read:

Set ONE decimal place correcting the error in the Int function for proper
mathematical rounding we have a very simple function:
X = Int(10*(X+0.005 ))/10

Cheers

--
Timothy Casey GPEMC! >11950 is the nu****@fieldcra ft.com.au 2email
Terms & conditions apply. See www.fieldcraft.biz/GPEMC
Discover valid interoperable web menus, IE security, TSR Control,
& the most advanced speed reading application @ www.fieldcraft.biz

[Mark S. Milley, MCSD (BinarySwitch)]

Yeah, that was definately much easier than my solution.

So... In what instance would this give a different result than mine?

Apparently Blind,

-Mark

[Joergen Bech]

On Sat, 07 Jul 2007 14:46:36 -0700, "Mark S. Milley, MCSD
(BinarySwitch)" <ma*********@bi naryswitch.comw rote:

>Hi Joregen -

How are you getting 4 down and 5 up? Last time I checked, we were
using base ten.

Numbers ending in 0-4 are rounded down, 5-9 are rounded up; 5 on each
side.

Thanks,

-Mark

I said 5 down, 4 up, and the last could go either way.

Because you are using simple Math.Round, which will round off 5s
to the nearest *even* value, e.g. 2.5 becomes 2, 3.5 becomes 4,
4.5 becomes 4, 5.5 becomes 6, and so on. That is great if that
is what you want, but I presumed it did not apply to the original
poster's needs.

Here is your suggestion wrapped in a test program:
---snip---
Private Sub RoundingTest()
Dim ups As Integer = 0
Dim downs As Integer = 0
Dim totalcount As Integer = 0
For value As Decimal = 1D To 1.99D Step 0.01D
Dim roundedvalue As Decimal = Math.Round(valu e, 1) - 0.01D
Debug.WriteLine (value.ToString ("0.00") & ": " &
roundedvalue.To String("0.00"))
totalcount = totalcount + 1
If roundedvalue < value Then
downs = downs + 1
Else
ups = ups + 1
End If
Next

Debug.WriteLine ("------------------------")
Debug.WriteLine ("Total: " & totalcount.ToSt ring)
Debug.WriteLine ("Ups: " & ups.ToString)
Debug.WriteLine ("Downs: " & downs.ToString)

End Sub
---snip---

The results are below: 55 values rounded down, 45 rounded
up.

You can read about various rounding methods here:
http://en.wikipedia.org/wiki/Rounding

Go watch Superman III for another lesson in rounding.

Regards,

Joergen Bech

---snip---
1,00: 0,99
1,01: 0,99
1,02: 0,99
1,03: 0,99
1,04: 0,99
1,05: 0,99
1,06: 1,09
1,07: 1,09
1,08: 1,09
1,09: 1,09
1,10: 1,09
1,11: 1,09
1,12: 1,09
1,13: 1,09
1,14: 1,09
1,15: 1,19
1,16: 1,19
1,17: 1,19
1,18: 1,19
1,19: 1,19
1,20: 1,19
1,21: 1,19
1,22: 1,19
1,23: 1,19
1,24: 1,19
1,25: 1,19
1,26: 1,29
1,27: 1,29
1,28: 1,29
1,29: 1,29
1,30: 1,29
1,31: 1,29
1,32: 1,29
1,33: 1,29
1,34: 1,29
1,35: 1,39
1,36: 1,39
1,37: 1,39
1,38: 1,39
1,39: 1,39
1,40: 1,39
1,41: 1,39
1,42: 1,39
1,43: 1,39
1,44: 1,39
1,45: 1,39
1,46: 1,49
1,47: 1,49
1,48: 1,49
1,49: 1,49
1,50: 1,49
1,51: 1,49
1,52: 1,49
1,53: 1,49
1,54: 1,49
1,55: 1,59
1,56: 1,59
1,57: 1,59
1,58: 1,59
1,59: 1,59
1,60: 1,59
1,61: 1,59
1,62: 1,59
1,63: 1,59
1,64: 1,59
1,65: 1,59
1,66: 1,69
1,67: 1,69
1,68: 1,69
1,69: 1,69
1,70: 1,69
1,71: 1,69
1,72: 1,69
1,73: 1,69
1,74: 1,69
1,75: 1,79
1,76: 1,79
1,77: 1,79
1,78: 1,79
1,79: 1,79
1,80: 1,79
1,81: 1,79
1,82: 1,79
1,83: 1,79
1,84: 1,79
1,85: 1,79
1,86: 1,89
1,87: 1,89
1,88: 1,89
1,89: 1,89
1,90: 1,89
1,91: 1,89
1,92: 1,89
1,93: 1,89
1,94: 1,89
1,95: 1,99
1,96: 1,99
1,97: 1,99
1,98: 1,99
1,99: 1,99
------------------------
Total: 100
Ups: 45
Downs: 55

[Number 11950 - GPEMC! Replace number with 11950]

"Mark S. Milley, MCSD (BinarySwitch)" <ma*********@bi naryswitch.comw rote
in message news:11******** *************@k 79g2000hse.goog legroups.com...

Yeah, that was definately much easier than my solution.

So... In what instance would this give a different result than mine?

Apparently Blind,

-Mark

I don't know about easier. Just a different way of getting to the same
place.
Also, sometimes doing the math helps centre one's perspective. I call it
'simple' because it's nothing like an algorithm for nailing the trig
function bugs in a 3D polar rotation. Normally a fairly strightforward piece
of math that on paper produces a single formula; The coding requires four
separate formulas because the canned trig functions cannot even correctly
sign the appropriate quadrant. As you can imagine, I'm a little cautious of
invoking canned subroutines because:

1. I don't know what's in them,
2. Sometimes they are slower than the derived subroutine or function,
3. Occasionally they are just plain old dead wrong.

I have to admit, I'd be disappointed if there was any difference in output,
or if your example was slower than mine. Your example did one thing for me,
and that was to emphasise the fact that there are many more improved
functions than I realised. However, reworking canned functions is good
practice for when there is no canned function or method to call on...

By the way, the debug.writeline (myfunction(myv ariable)) doesn't seem to work
the way it used to as debug.print(myf unction(myvaria ble)) did in VB6. Am I
missing a setting or reference?

--
Timothy Casey GPEMC! >11950 is the nu****@fieldcra ft.com.au 2email
Terms & conditions apply. See www.fieldcraft.biz/GPEMC
Discover valid interoperable web menus, IE security, TSR Control,
& the most advanced speed reading application @ www.fieldcraft.biz