Formula needed

[Terry Olsen]

I'm trying to calculate insurance fees for declared value rates. The
definition in the rate guide states this:

Declared Value Insurance Fee
<= $100.00 $0.00

$100 $0.60 for each $100 of the total value declared, with

a minimum charge of $1.80.

So here's the first few ranges:

0-100:0
100.01-300:1.80
300.01-400:2.40
400.01-500:3.00

I'm at a loss on creating a formula to figure this. Can anyone help?

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[=?Utf-8?B?U3VydHVyWg==?=]

RE: Formula needed

This isn't elegant, but it should work:

Function InsuranceFee(By Val declaredValue As Double) As Double
If declaredValue <= 100 Then Return 0
Dim dblTemp As Double = 0
dblTemp = 0.6 * declaredValue / 100
If dblTemp < 1.8 Then dblTemp = 1.8
dblTemp = Math.Round(dblT emp, 2) 'round to nearest cent
Return dblTemp
End Function

You might need to change the variable type from Double to Decimal or
whatever you are using.
--
David Streeter
Synchrotech Software
Sydney Australia


"Terry Olsen" wrote:

I'm trying to calculate insurance fees for declared value rates. The
definition in the rate guide states this:
>
Declared Value Insurance Fee
<= $100.00 $0.00

$100 $0.60 for each $100 of the total value declared, with

a minimum charge of $1.80.
>
So here's the first few ranges:
>
0-100:0
100.01-300:1.80
300.01-400:2.40
400.01-500:3.00
>
I'm at a loss on creating a formula to figure this. Can anyone help?
>
*** Sent via Developersdex http://www.developersdex.com ***
>

[=?Utf-8?B?U3VydHVyWg==?=]

RE: Formula needed

Sorry, if you only add 60c each time, use this:

Function InsuranceFee(By Val declaredValue As Double) As Double
If declaredValue <= 100 Then Return 0
Dim dblTemp As Double = 0
dblTemp = 0.6 * CDbl(CLng(decla redValue) \ 100)
If declaredValue Mod 100 0 Then
dblTemp += 0.6
End If
If dblTemp < 1.8 Then dblTemp = 1.8
dblTemp = Math.Round(dblT emp, 2) 'round to nearest cent
Return dblTemp
End Function

[Brian Gideon]

Re: Formula needed

On Jul 7, 12:15 pm, Terry Olsen <tolse...@hotma il.comwrote:

I'm trying to calculate insurance fees for declared value rates. The
definition in the rate guide states this:
>
Declared Value     Insurance Fee
<= $100.00             $0.00 $100            $0.60 for each $100 of the total value declared, with
>
a minimum charge of $1.80.
>
So here's the first few ranges:
>
0-100:0
100.01-300:1.80
300.01-400:2.40
400.01-500:3.00
>
I'm at a loss on creating a formula to figure this. Can anyone help?
>
*** Sent via Developersdexht tp://www.developersd ex.com***

Public Function CalculateRate(B yVal value as Double) As Double
If value 100 Then
Return Math.Max(((valu e \ 100) + 1) * 0.60, 1.80)
End If
Return 0
End Function

[Steve Gerrard]

Re: Formula needed

Terry Olsen wrote:

I'm trying to calculate insurance fees for declared value rates. The
definition in the rate guide states this:
>
Declared Value Insurance Fee
<= $100.00 $0.00

> $100 $0.60 for each $100 of the total value declared,
>with a minimum charge of $1.80.

>
So here's the first few ranges:
>
0-100:0
100.01-300:1.80
300.01-400:2.40
400.01-500:3.00
>
I'm at a loss on creating a formula to figure this. Can anyone help?
>

One more. This one works for 599.99 and 600.00. :)

Public Function GetFee(ByVal Value As Double) As Double
Dim n As Integer
If Value <= 100 Then
n = 0
ElseIf Value <= 300 Then
n = 3
Else
n = CInt(Fix(Value / 100))
If Value Mod 100 0 Then
n += 1
End If
End If
Return Math.Round(n * 0.6, 2)
End Function

[cfps.Christian]

Re: Formula needed

In my tests I found that the rounding is going to be the most complex.

Dim DeclaredValue as Double = [...]
Dim InsuranceFee as Double = 0

If DeclaredValue 100 AND DeclaredValue < 300.01 Then
InsuranceFee = 1.80
Else If DeclaredValue 300
Dim Multiplier as Integer = Convert.ToInt32 (DeclaredValue / 100)
If DeclaredValue Multiplier * 100 Then
Multiplier += 1
End If
InsuranceFee = .6 * Multiplier
End If

[Terry Olsen]

Re: Formula needed

Thanks everybody. I have some things to try now.

"cfps.Christian " <ge0193387@otc. eduwrote in message
news:ca840ca6-6e48-45fc-a131-63d11e6fdb68@k1 3g2000hse.googl egroups.com...

In my tests I found that the rounding is going to be the most complex.
>
Dim DeclaredValue as Double = [...]
Dim InsuranceFee as Double = 0
>
If DeclaredValue 100 AND DeclaredValue < 300.01 Then
InsuranceFee = 1.80
Else If DeclaredValue 300
Dim Multiplier as Integer = Convert.ToInt32 (DeclaredValue / 100)
If DeclaredValue Multiplier * 100 Then
Multiplier += 1
End If
InsuranceFee = .6 * Multiplier
End If
>