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bundyloco
 
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Default Dividing by unknown to get a number


I am trying to determine the amount of people it will take to complete a
task. The problem is, I have calculated how big the task will be, but I
would like a function to determine how many people are needed.
For example:

I know 1 person can move 700 items without risk of injury (so I would
like to stay around this number...it's arbitrary). So I would like a
function to calculate how many people it will take to move say 5000
items?
ie 5000/x=700 where x is the number of workers. Does anyone have an
idea of an if statement that could do this. Obvisously, if the number
is less than 700, I would have a default of 1.

Thanks in advance.


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Paul B
 
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bundyloco, with 5000 in A1, =IF(A1<701,1,A1/700) Am I missing something??

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"bundyloco" wrote
in message ...

I am trying to determine the amount of people it will take to complete a
task. The problem is, I have calculated how big the task will be, but I
would like a function to determine how many people are needed.
For example:

I know 1 person can move 700 items without risk of injury (so I would
like to stay around this number...it's arbitrary). So I would like a
function to calculate how many people it will take to move say 5000
items?
ie 5000/x=700 where x is the number of workers. Does anyone have an
idea of an if statement that could do this. Obvisously, if the number
is less than 700, I would have a default of 1.

Thanks in advance.


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BenjieLop
 
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bundyloco Wrote:
I am trying to determine the amount of people it will take to complete a
task. The problem is, I have calculated how big the task will be, but I
would like a function to determine how many people are needed.
For example:

I know 1 person can move 700 items without risk of injury (so I would
like to stay around this number...it's arbitrary). So I would like a
function to calculate how many people it will take to move say 5000
items?
ie 5000/x=700 where x is the number of workers. Does anyone have an
idea of an if statement that could do this. Obvisously, if the number
is less than 700, I would have a default of 1.

Thanks in advance.


If you want, you can put "700" (without the quotes or any other number
since you said this is arbitrary) in Cell A1 and the total number of
items that you would like to move in Cell B1 (in your example, this is
5000).

In Cell C1, enter this formula : =if((B1/A1)<1,1,B1/A1)

Another formula would be : =if(B1<A1,1,B1/A1)

Regards.


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R.VENKATARAMAN
 
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intrigued by the question
i thought if
5000/x=700-by cross multiplication
then x=5000/700
it may give fraction 7.142
appoint 8 people


bundyloco wrote in
message ...

I am trying to determine the amount of people it will take to complete a
task. The problem is, I have calculated how big the task will be, but I
would like a function to determine how many people are needed.
For example:

I know 1 person can move 700 items without risk of injury (so I would
like to stay around this number...it's arbitrary). So I would like a
function to calculate how many people it will take to move say 5000
items?
ie 5000/x=700 where x is the number of workers. Does anyone have an
idea of an if statement that could do this. Obvisously, if the number
is less than 700, I would have a default of 1.

Thanks in advance.


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Ken Wright
 
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=ROUNDUP(A1/700,0)

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"bundyloco" wrote
in message ...

I am trying to determine the amount of people it will take to complete a
task. The problem is, I have calculated how big the task will be, but I
would like a function to determine how many people are needed.
For example:

I know 1 person can move 700 items without risk of injury (so I would
like to stay around this number...it's arbitrary). So I would like a
function to calculate how many people it will take to move say 5000
items?
ie 5000/x=700 where x is the number of workers. Does anyone have an
idea of an if statement that could do this. Obvisously, if the number
is less than 700, I would have a default of 1.

Thanks in advance.


--
bundyloco
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bundyloco's Profile:
http://www.excelforum.com/member.php...o&userid=23386
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bundyloco
 
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Do'h...thanks everyone, I knew I was making this way too difficult!


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