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rasprasads · Posted: Thu Sep 11, 2003 4:26 am Post subject: Accuracy in Computing

I have defined 3 variables :

somuk · Beginner Joined: 04 Feb 2003 Posts: 113 Topics: 37

Hi Rasprasad ,
Let us take an equation F=(A/B)*C.

Compute performs the following operations in sequence.
1.Value A is divided by B and is stored in a temporary register, say R1.
2.Value C is multiplied with the register value R1.
Now the problem is with the picture clause of the (space allocation) register that stores the intermediate result.

Try the following method (I got this information from a friend).
Declare another variable

kolusu · Posted: Thu Sep 11, 2003 8:59 am Post subject:

Rasprasad,

As you already guessed the culprit is the INTERMEDIATE result of (A/B).check this link for Intermediate results and arithmetic precision.

http://publibz.boulder.ibm.com/cgi-bin/bookmgr_OS390/BOOKS/igy3pg10/APPENDIX1.1?DT=20020923143836#HDRWQ955

Now with rounding,after decimal point alignment, the number of places in the fraction of the result of an arithmetic operation is compared with the number of places provided for the fraction of the resultant identifier.

When the size of the fractional result exceeds the number of places provided for its storage, truncation occurs unless ROUNDED is specified. When ROUNDED is specified, the least significant digit of the resultant identifier is increased by 1 whenever the most significant digit of the excess is greater than or equal to 5.

When the resultant identifier is described by a PICTURE clause containing rightmost Ps, and when the number of places in the calculated result exceeds the number of integer positions specified, rounding or truncation occurs, relative to the rightmost integer position for which storage is allocated.

In a floating-point arithmetic operation, the ROUNDED phrase has no effect; the result of a floating-point operation is always rounded

When the ARITH(EXTEND) compiler option is in effect, the ROUNDED phrase is not supported for arithmetic receivers with 31 digit positions to the right of the decimal point.

Solutions:

Let us take your example.

COMPUTE WS-C = ( WS-A / WS-B ) * 100.

Define the decimal part of anyone of those variables WS-A or WS-C (not the divider or exponentiation) in such a way that it is greater than (No. of decimals defined for A + No. of integers defined for B) . This works with any pic clause and the % error is very less.

Since you did not want to change the pic cluse of the variables you can try as suggested by somu.Since you are adding a zero it is NOT going to change the actual value , but nullifies the Truncation errors.

Hope this helps...

cheers

kolusu

rasprasads · Posted: Fri Sep 12, 2003 1:27 am Post subject:

I am not still clear about the rounded option. In CASE - 2 the result is 49.90.

Here 900.03/1800.33 would give 0.499925. This would be stored as 0.49 if rounded option is not sepcified. But as Rounded is specified, the no. of decimal place will be increased by 1 (to 3). So value should have been (0.500 ---> as 0.4999 would have been rounded so) and finally should be 50.00. But this is not the case and the result is 49.90. Can you explain why this is happening ?
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Rasprasad S

Dibakar · Posted: Fri Sep 12, 2003 5:51 am Post subject:

Try "COMPUTE WS-C = ( WS-A * 100)/ WS-B )". You might get beter result.

rasprasads · Posted: Fri Sep 12, 2003 6:12 am Post subject:

I am ok with the solution(s). But want to know the reason behind the value generated when using ROUNDED option. Please help...
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Rasprasad S

rasprasads · Posted: Fri Sep 12, 2003 8:57 am Post subject:

How will this solve the issue :

slade · Posted: Sat Sep 13, 2003 12:13 am Post subject:

Hi Rasp,

WS-C..-F.. forces the compiler to create an intermediate result field with more than enough decimal places to accommodate the result and, therefore, no erroneous truncation results.

The attraction of it is that it's a "one size fits all" solution, unless you're using the ARITH=EXTEND option.

Regards, Jack.

neilxt · Beginner Joined: 01 Mar 2004 Posts: 23 Topics: 1

js01 · Posted: Thu Mar 26, 2009 5:10 pm Post subject: Rounded in cobol

Hello friends,

could you please help me in below question

declared

05 VAR1 S9(5)V99
05 VAR2 S9(5)V99 value 2070

Compute VAR1 ROUNDED = VAR2 * 1.4

what is the value of VAR1

thank you

Terry_Heinze · Posted: Thu Mar 26, 2009 10:19 pm Post subject:

As suggested, please refrain from asking the same question in 2 different threads.
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....Terry

Nic Clouston · Posted: Fri Mar 27, 2009 11:37 am Post subject:

simple - write a little program to test it
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