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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.

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….