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A system contains three programs and each requires three tape units for its operation. The minimum number of tape units which the system must have such that deadlocks never arise is _________.

  1. $6$
  2. $7$
  3. $8$
  4. $9$
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How A is the ans here ….if for some instance 6 resources are divided equally to every program then deadlock may occur here….so for safe side answere should be 7
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3 Answers

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option B will be correct

three processes require 3 units of tapes for each

Let us assume each process gets 2 units of tapes. Now system will be in deadlock. If one more unit of tape is provided the system will be free from deadlock. Therefore 7 unit tapes are required
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DEADLOCK not to arise,the requirement of taes in this question should be greater than or equal to ( nos of process + peak demand of tapes -1)

here no of process is 3, tape requirement per process is 3 .

so to avoid deadlock the minimum nos of tape should be (3+3-1)=5.

from the given option minimum is found 6 in option (A) which is correct.
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Finding out the minimum no.of tape units…..?

Given programs = 3

Required resources (Tape units) for each process = 3 units

If we give 3 tape units for each process (3 x 3 = 9) ,it is sure that no deadlock will occur...right ! because we satisfied each process’s requirement so that  they can be executed with out any problem.

If we clearly observe the question there he mentioned that the minimum no.of tape units such that no deadlock will occur….

Therefore give resources for each process as many as they want except one ,means no.of resources i give to P1 is (3 – 1) ‘2’….assign for all the processes in the same way...so that we assigned 6 units till now right….still they are in deadlock..

Now the interesting point comes….just give one more tape unit to any one of them (we know max requirement of each process is 3 units) ...so that any one process will use all the 3 resources and release the resources associated to it...now the resources count is increased in such a way it can even fulfill the need of next process...therefore ...in this way all processes will be executed ,finally deadlock will be vanished..

Therefore The the minimum no.of tape units which the system must have such that deadlocks never arise is “7”.(OPTION : B)

Hope You got it….

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