UGC NET CSE | June 2019 | Part 2 | Question: 48

Question

A computer has six tape drives with $n$ processes competing for them. Each process may need two drives. What is the maximum value of $n$ for the system to be deadlock free?

• A. $5$
• B. $4$
• C. $3$
• D. $6$

Answer 1

n processes each requiring 2 tape drives

total no of tape drives = 6 

       Process                          Process1       Process2     Process3     Process4      Process5     Process6

No. of Tape Drives                     1                       1                 1                 1                   1                  1                   

This is a deadlock Situation. So anything less than 6 process will be free from Deadlock because at least 1 tape drive will be in hand which can be used for completion of a  running process.

Here asked for maximum process with deadlock free Hence 5 only.

Option 1) is correct answer.

 

 

Answer 2

We have $6$ drives and we have to find the number of processes.

It is given in question that

Each process may need two drives

So I have $1$ Drive free and alloted all the remaining drives each to one process to maximize number of processes present in the system.

Why keep $1$ Drive free ?

Because in case if a process needs $2$ drives i can allot the drive which is free to that process.

$\therefore$ Option $A.$ is the correct choice.

Answer 3

Option(A): 5

If there will be 6 processes then let each process have 1 drive then this will create a deadlock situation

To avoid this situation,we need a maximum of 5 processes