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shalem
New User
Joined: 11 Apr 2005 Posts: 8
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Hi, can you please help me ROUNDED issue in COMPUTE statement
AVERAGE-RATE PIC S9(3)V999 COMP-3.
COMPUTE AVERAGE-RATE (S1, S2, S3) ROUNDED =
(((TOP-VALUE (S1, S2, S3) /
BOTTOM-VALUE (S1, S2, S3)) - 1.0) * 100).
TOP VALUE/BOTTOM VALUE = 12024546.51/11674330.16= 1.029998
1.029998-1.0=0.029998
0.029998*100= 2.9998
When I gave rounded option why I am getting 2.99 instead of 3.00 |
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Terry Heinze
JCL Moderator
Joined: 14 Jul 2008 Posts: 1249 Location: Richfield, MN, USA
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COBOL does not carry the precision of intermediate results that you might expect. Study "Appendix A. Intermediate results and arithmetic precision" in the Programming Guide. Also, how are S1, S2, and S3 defined? |
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Terry Heinze
JCL Moderator
Joined: 14 Jul 2008 Posts: 1249 Location: Richfield, MN, USA
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Also, what is TOP VALUE and BOTTOM VALUE? Do you mean the Intrinsic functions MAX and MIN? |
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Robert Sample
Global Moderator
Joined: 06 Jun 2008 Posts: 8696 Location: Dubuque, Iowa, USA
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First of all, your post is inconsistent -- if AVERAGE-RATE is defined PIC S9(3)V999 then there is NO WAY it will ever have a value of 2.99 or 3.00; 2.999 or 2.990 or 3.000 but not with just 2 decimal digits.
Second, since you haven't posted accurate information (if TOP-VALUE is a function you wrote, why not post it -- and if it is NOT a function, why did you not post the EXACT code you are using) this is no way we can tell you why the rounding is not correct. |
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shalem
New User
Joined: 11 Apr 2005 Posts: 8
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Thanks for quick reply. Both Top and Bottom values are defined IN ARRAY
TOP-VALUE PIC S9(14)V99 COMP-3.
BOTTOM-VALUE PIC S9(14)V99 COMP-3.
If COBOL does not carry the precision of intermediate results then what is alternate for me. |
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Robert Sample
Global Moderator
Joined: 06 Jun 2008 Posts: 8696 Location: Dubuque, Iowa, USA
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Have you considered making TOP-VALUE and BOTTOM-VALUE V999 instead of V99 and see if that makes a difference? As Terry indicated, the COBOL rules for intermediate results are NOT easy to understand without a LOT of experience. If some of your intermediate results only have 2 decimal places, you're going to get 2 decimal places in the result in most cases.
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If COBOL does not carry the precision of intermediate results then what is alternate for me. |
Possible solution 1: carry more decimal digits after the decimal point so the intermediate results have more digits.
Possible solution 2: break up the calculation into pieces. |
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Bill Woodger
Moderator Emeritus
Joined: 09 Mar 2011 Posts: 7309 Location: Inside the Matrix
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You must multiply first, divide last.
Your final action is to multiply by 100. What do you think there could possibly be to "round" after that? |
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Terry Heinze
JCL Moderator
Joined: 14 Jul 2008 Posts: 1249 Location: Richfield, MN, USA
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My mistake. I didn't realize you had a 3-dimensional table. Have you read thoroughly the Appendix I mentioned? I'd break the COMPUTE statement up into individual DIVIDEs and MULTIPLYs paying close attention to the precision of the intermediate results. |
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shalem
New User
Joined: 11 Apr 2005 Posts: 8
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Thanks Terry , I will try with Break COMPUTE into individual DIVIDEs and MULTIPLYs. |
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Terry Heinze
JCL Moderator
Joined: 14 Jul 2008 Posts: 1249 Location: Richfield, MN, USA
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Hint:
When I calculate a percentage, I use the following:
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05 ws-size pic s999 comp-3.
05 ws-max pic s999 comp-3.
05 ws-part pic s9v99 comp-3.
05 ws-pct pic zz9.
divide ws-size by ws-max giving ws-part rounded
multiply 100 by ws-part giving ws-pct |
Hope this helps. |
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Bill Woodger
Moderator Emeritus
Joined: 09 Mar 2011 Posts: 7309 Location: Inside the Matrix
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Terry, you have two surplus decimal places (surplus to your final result). shalem could achieve the same in the COMPUTE by adding two redundant decimal-places to his fields :-)
shalem,
Simply rearrange your COMPUTE to do the multiplication first, and the division last.
Work your example through, existing COMPUTE and rearranged COMPUTE, with the Appendix Terry suggested earlier, and you'll see the difference in action.
If you want to split it up, that is fine, but you have to look after sizes yourself (because what is otherwise an intermediate value, you have to save for the next stage).,
You either have to make you intermediate 100 times larger than the source field (MULTIPLY first), or have the two excess decimal places (DIVIDE first).
Using COMPUTE you need neither excess size nor excess decimal places. You just have to do things in the correct order. In COPUTE, multiply, then divide, will get your result.
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COMPUTE C = ( 100 * A ) / B |
is what you want.
Algebraically that is the same as 100 * ( A / B ). It just isn't the same, because the intermediates are not like a calculator.
A = 33
B = 70
A / B = 0.47142857142857142857142857142857 with a calculator. In the intermediates (with ROUNDED you get an extra decimal place, which is how the compiler will know to round or not) it is 0.471.
When you multiply that by 100, you get 47.100, and the ROUNDED operates on the low-order digit, which is 100% guaranteed to always be zero (because of the multiply).
Multiply first, divide last, and the precision is preserved for the correct rounding. |
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shalem
New User
Joined: 11 Apr 2005 Posts: 8
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Thanks guys for your timely help. I got right value 3.0 by following your procedure. Appreciate your help. |
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