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in job stmt i coded time=(2,30) and in step 1 i coded time=(1,30)
and in step i coded time=(1,30), whch one executes first and
what happens if step1 and step2 time executes and wht about the
remaining time if step and step2 executes
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Answer / rakesh jena
As per your question, in this situation, the job executes
step1 first then Step2.So if step1 will take 1 min 30 sec
then remaining 1min will for step2.if step2 will take more
than 1 minute then job will abnormally terminate with S322
abend.most probably it is depending upon the CPU usage time
for the individual step.
| Is This Answer Correct ? | 9 Yes | 1 No |
Answer / kittu
According to your question...
1.If the 2 jobs submitted then the job which is submitted
first excutes first.
2.If the jobs are submitted at the same time then os
decides based on the class parameter.
3.If the class parameter are also same then based on the
priority(PRTY) job will executes.
4.If the priority also same then operator cancells both the
jobs after a wait time.
or
If the region and time specified in both the job & exec
statements then the values specified in job will overwrite
that of Exec statements values.
So job statement coded time will overwrite the step coded
time.
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / srinivas
after step1 executed,for step2 1 minute of job time will be
there,but step2 time is (1,30).In this case,step wont abend
instead, it will override the job time and step2 will aslo
execute successfully..
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / praveen bejjanki
If time parameter is specified in both Job and exec stmt,
then the Time specified for step or the time left out in
the job whichever is small will be the time for that
particular step.
Job = (2,30) means 2mins, 30 seconds.
step1 = (1,30) since step1 time limit is less, step1 will
execute in 1min, 30 seconds.
step2 = (1,30)
Since the time left out in the job is only 1min. so this
1min is assigned to step2
| Is This Answer Correct ? | 0 Yes | 0 No |
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