Material Thickness Range or Lookup?

[materialperson]

I am calculating a minimum thickness to use for a steel part and need to have the output in standard thicknesses.

Minimum Thickness (In.) Calculated	Actual Thickness of part to be used
0.3794	0.5
1.6784	1.75
3.2493	3.375
4.7854	4.875

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The standard sizes of material used are 0.25, 0.375, 0.5, 0.625, 0.75, 0.875, 1.00, 1.25, 1.375, ....6 inches.

What would be the best formula to use? IF/AND, VLOOKUP, RANGE_LOOKUP, or is there a better one to use?
The calculated thickness cells contain Roundup and IF functions, the length of the numbers is dictated by ASME code/ specific figures involved in the calculation.
Is there a way to have <= and >= in a lookup function?

Thanks

 

[SpillerBD]

VLOOKUP.
You may need to control the fractional part with a MROUND.

 

[thisoldman]

In the third line, why does 3.2493 require 3.375? Is 3.25 too close or not available?

This rounds up to the next one-eighth, if 3.2493 can take 3.25 as the solution:
=(QUOTIENT(A2,0.125)+1)*0.125

This works for the numbers as given:
=(QUOTIENT(ROUNDUP(A2,1),0.125)+1)*0.125

 

[Taul]

Hi,
Am I correct in thinking there is a missing value in the material available sequence, between 1 and 1.25 (1.125 missing) based on the difference of 0.125 interval used elsewhere.
If so, you could try

=CEILING(A2,0.125)

Paul.

 

[materialperson]

Hi,
Am I correct in thinking there is a missing value in the material available sequence, between 1 and 1.25 (1.125 missing) based on the difference of 0.125 interval used elsewhere.
If so, you could try

=CEILING(A2,0.125)

Paul.

Yes, there is 1.125 material, however, it is more cost efficient to buy the 1.25. 1 1/8" plate is typically a custom order from the mill, whereas 1 1/4 is stock.

 

[materialperson]

You are correct, typo on my part. Should read 3.3493.

 

[Taul]

Hi,
In that case, post #3 from thisoldman should give you what you need.
Paul