This question was previously asked in

GATE CS 2013 Official Paper

Option 2 : X: P(b)P(a)P(c) Y: P(b)P(c)P(d) Z: P(a)P(c)P(d)

IBPS SO IT Officer Mains: Full Mock Test

5486

60 Questions
60 Marks
45 Mins

** Answer**: Option 2

** Explanation**:

Given

Process X will perform down operation on a, b, c

Process Y will perform down operation on b, c, d

Process Z will perform down operation on c, d, a

all the binary semaphores initialized to 1.

**Option 1**: X: P(a)P(b)P(c) Y: P(b)P(c)P(d) Z: P(c)P(d)P(a)

Consider the following sequence

X: P(a) // a = 0

Y: P(b) // b = 0

Z: P(c) // c = 0

X: P(b) // unsuccessful down operation hence X will be blocked.

Y: P(c) // again unsuccessful down operation hence Y will be blocked.

Z: P(d) // d=0

Z: P(a) // unsuccessful down operation hence Z will be blocked. Hence all 3 processes are blocked. so this option does not represent the deadlock-free order.

**Option 2**: X: P(b)P(a)P(c) Y: P(b)P(c)P(d) Z: P(a)P(c)P(d)

In this Order, you execute in any sequence deadlock not possible. Hence this is the **correct **Option**.**

**Option 3: **X: P(b)P(a)P(c) Y: P(c)P(b)P(d) Z: P(a)P(c)P(d)

X: P(b) // b = 0

Y: P(c) // c = 0

Z: P(a) // a = 0

X: P(a) // unsuccessful down operation hence X will be blocked.

Y: P(b) // again unsuccessful down operation hence Y will be blocked.

Z: P(c) // unsuccessful down operation hence Z will be blocked. Hence all 3 processes are blocked. so this option does not represent the deadlock-free order.

**Option 4**: X: P(a)P(b)P(c) Y: P(c)P(b)P(d) Z: P(c)P(d)P(a)

X: P(a) // a = 0

Y: P(c) // c = 0

Z: P(c) // unsuccessful down operation hence Z will be blocked.

X: P(b) // b = 0

Y: P(b) // unsuccessful down operation hence Y will be blocked.

X: P(c) // unsuccessful down operation hence X will be blocked. Hence all 3 processes are blocked. so this option does not represent the deadlock-free order.