Suppose we have two Jobs A and B. A and B carry data of size 1 Kilo Byte each. Data carried by Jobs A and B has to be processed.
Let us further suppose that there are two slots ( Slot I and Slot II ) available. The important condition is that processing of Jobs A and B can occur, only if they appear in Slot I. When these jobs appear in Slot II, they are not processed.
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Case I:
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The permutations of Jobs A and B are as:
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Slot I Slot II
Perm I A B
Perm II B A
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//Permutation- A Great Fun//
//Titbits Under the Sun//
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Clearly job A would be in Slot I for once. Job B too would be in Slot I for once.
Let us suppose the data processing speed of our imaginary Processor is 1 Kilo Byte ( 1 KB ) per second ( 1 KB/s ). Job A whose size is 1 KB would be completely processed in 1 second. Equally sized Job B ( of 1 KB ) would also be finished in 1 second. Slot I is occupied once each by Job A and Job B in the twin possible permutations of A and B. Total time dedicated to Slot I by the Processor is 2 seconds.
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Case II:
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Now, Jobs A, B, and C are with us. Job C is new. So, the three Jobs produce permutations. There is a restriction. Now, we would consider only two jobs at a time.The following permutations are possible:
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Slot I Slot II
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Job Job
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A B
A C
B A
B C
C A :Ignore
C B :Ignore
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//Permutations Tutored to Permit//
//The Crispy Cakewalk- Boys Admit//
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The permutations show that Slot I is occupied by Job C in the last two rows. Job C is not under consideration. So, the last two rows are ignored.
In the new permutations, we see that Job A appears twice in Slot I. Job B too appears twice in Slot I. As mentioned before, we are interested in Job A and Job B. From the start, we have considered only these two ( A and B ).
The same imaginary Processor having a data processing speed of 1 Kilo Byte per second is used. The same number of Bytes- 1 Kilo Byte each- is associated with Job A and Job B.
Number appearances of Job A and Job B, in Slot I, is two for both ( Look at the given Permutations of Job A, Job B, and Job C listed above ). So, the size of data brought by Job A in Slot I is ( 1 KB + 1 KB = ) 2 KB. Similarly, the size of data that is handed over to the Processor by Job B is ( 1 KB + 1 KB = ) 2 KB. Total data to be processed is:
Job A ------> 1 KB + 1 KB = 2 KB
Job B -------> 1 KB + 1 KB = 2 KB
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Total Size of Data = 4 KB
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The data processing speed of the Processor is 1 KB/second . Requirement of time is 4 seconds to process the data carried by Job A and Job B in Slot I.
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Observation:
Permutations of two unique objects ( Job A and Job B ) puts each of them in the high priority slot ( Slot I ) only once.
So, Requirement of Processor Time = 2 Seconds
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Permutations of three unique objects ( Job A, Job B, Job C ) puts Job A and Job B twice in the high priority slot ( Slot I ).
So, Requirement of Processor Time = 4 Seconds
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Conclusion: Requirement of Processor's time may depend on permutations of tasks.
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