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Look up one number based on mulitple criteria!!!!



 
 
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Old August 5th, 2008, 05:22 PM posted to microsoft.public.excel.misc
Bob I
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Posts: 10,698
Default Look up one number based on mulitple criteria!!!!

If you chart/graph the table, the resultant lines are not smooth but
jerk around as opposed to the nice continuous line a mathematical
expression will create.

scottgorilla wrote:

Bob,

You lost me with the reply what do you mean by????
"If you make a chart of the two lines you will have a feel for why the table
and formula are
dissimilar."

Thanks,


Scottgorilla.
"Bob I" wrote:


Glad you have the solution that works for you. If you make a chart of
the two lines you will have a feel for why the table and formula are
dissimilar.

scottgorilla wrote:


In case anyone was interested I found the solution it looks like this

=INDEX($AP$3:$AP$36,MATCH(1,(AL3=FORMULAS!$AQ$ 3:$AQ$36)*(FORMULAS!AG3=$AR$3:$AR$36),0)

followed by CONTROL+SHIFT+ENTER

Well thanks all for the help


Scottgorilla
"scottgorilla" wrote:



Bob,

You deserve a medal for even messing with this.

Ok so I tried your formula based on the sizes actually kicked out by my
program and it looked like this

IF(((AL6^3)*AG6*0.0000007)1000,100,70))

AL6=127.50 which is actually determined by width
AG6=308.464 which is determined by sq. ft. times product weight

mathematically your formula returns 447.54 which is less than 1000 so it did
return 70 but since the 1000 number is a constant and my numbers must
fluctuate because of the span how can I get that to happen???

Thank You again


"Bob I" wrote:



This doesn't check for maximums, only whether 70 can be used.

Span is A1
Weight is B1

=IF(((A1^3)*B1*0.0000007)1000,100,70)

See if it does what you intended.

scottgorilla wrote:



Bob,

Thank you again for your patience and assistance. The weight and span vary
constantly. We are building a custom product that constantly changes based
size of the application. So the only constant is the table because these are
maximums.


Scottgorilla




"Bob I" wrote:




Ok, then I understand the end result correctly, the issue is that the
table is actually discrete points on the line created by that formula.
That's the only other way to get there other than listing and comparing
values. Are you looking up an arbitrary weight and then a span on the
list? Or are the weight and the span both allowed to be any number at all?


scottgorilla wrote:




Sorry Bob,

The list was given to me in PDF format and was done by an engineer that I
will not be able to get the calculations from. What I am trying to accomplish
is not try and find the fomula for the table but based on results that are
determined in another part of my program. I'm am trying to get Excel to pick
the appropriate size 70 or 100. If the results are as follows say up to 71.5"
and up to 1397.00lbs then pick 70 and again if the results are up to 71.5"
and greater than 1397.001 then pick 100 and so on....

It must be possible but I just can't seem to find a way unless I write every
possioble scenario like this.


IF(and(d1=a,e1=b),70,if(and(d1=a,e1b), 100 and so on......


Thank You again,


Scott

"Bob I" wrote:





I'm Bob not Mike,

I am just looking at what you say you are trying to accomplish. IF the
weight is X and the span is less than Y you may use 70 otherwise you
must use 100. Correct? So, this requirement comes from the strength
required and the span times the weight sets the minimum tube size to do
that. Who made the table up? Where did it come from, there is a
mathematical formula that it was created from. You are attempting to
figure that formula out by trial and error. Can you get that formula

from the creator of the table? Otherwise you will need to play with the



numbers to find the line that creates the breakpoint between needing 100
instead of 70.



scottgorilla wrote:





Mike,


If you call it reverse engineering I'll except that. You say determine that
number (what number????) and use it as the choice between 100 v 70. I don't
think it is weight times span .... so much as span can't exceed weight and
then choose between the two.


Thank You for your time and response.

Scottgorilla

"Bob I" wrote:






It looks to me like you are attempting to "reverse engineer" the
requirements. The 100 vs.70 is going to be required due to the "weight
times the span". Determine that number and use it as the choice between
70 or 100.

scottgorilla wrote:






Dave,

Unfortunately it would be nice to say ok use the 100 on anything wider than
72" but it is not cost prohibitive.

Thus the reason for trying to create the formula so anyone whom enters the
data in the program would not have to have all the knowledge or have to
continually look the data up based on size and weight. Have excel read the
data and give the solution. I have tried with smaller formulas but
unfortunately I'm still getting the same result. It will not pick the correct
answer and I can't look over all the employees to make sure they didn't make
a mistake because mistakes cost money

"Dave Peterson" wrote:







It sure looks like you should be changing the table.

Maybe...

70 24 100000.000
70 36 3628.000
70 48 2712.950
70 60 2162.080
100 72 10133.09
100 84 6366.37
100 96 4249.74
100 108 2969.17
100 120 2148.71
100 132 1598.32
100 144 1214.89
100 156 939.16
100 168 735.42
100 180 581.28
100 192 462.22
100 204 368.51
100 216 293.52

So that any weight greater than or equal to 72 has to return 100.

But I'm not sure what happens at 71.99999

Did you try creating the smaller formula to see how that evaluated.

scottgorilla wrote:






I'm using Excel 2007 and those are named cell references so no need for the
double quotations and secondly AQ and AR are on the same sheet.

What I am trying to do based on criteria that I have set like as follows

if length is for argument sake is 72 and weight is less than 1397.550 the
result would be 70

if length is 72 and weight greater than 1397.550 the result would be 100

70 and 100 are diameter of a tube I must use to build my product based on
weights and span

if length is 133 and weight is less than 122.940 then I can use 70

if the length is 133 and weight is greater than 122.940 then it must be 100.

I truly appreciate your assistance and apologize that I seem not to be doing
a great job explaining what I'm trying to accomplish.

"Dave Peterson" wrote:







First, the strings in your formula have to be surrounded by double quotes:

=IF(OR(Master!C18="ALUM.050DADE",Ma ster!C18="ALUM.050PALM",

Second, it still looks like AQ3:AQ36 and AR3:AR36 are on different sheets.

Third, if all the data and formulas are on the same sheet, then the formula
doesn't need the sheet references.

Fourth, try this simplified formula:
=MATCH(1,(AL5=$AQ$3:$AQ$36)*(AG5= $AR$3:$AR$36),0)
(still an array formula)

This returns 11. That's 11 rows into the table--or this line of data.
70 144 122.94

The first time that 139.5 is less than or equal to the values in AQ3:AQ36 and at
the same time 336.72 is greater than or equal to the values in AR3:AR36.

So that 70 is the correct result for your formula.

=====
But I think you've got other problems.

In your table, you have:






70 24 100000.000

....






100 24 100000.000

And there's never going to be a formula that uses =index(match()) that will
return 100. The first row found that matches those identical conditions will
always be the 70.

I'm not sure what you're doing, but maybe you could use two tables (one for 70
and one for 100) or add an extra indicator to know what "group" you want to
inspect.

But it sure looks like you're deciding to bring back 70 or 100. I don't
understand what you really want.


scottgorilla wrote:






 




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