Official PYQ · 2007Q57DeadlockMCQ1 mark

GATE CS 2007 Q57 — Deadlock

A single processor system has three resource types $X, Y$ and $Z$ , which are shared by three processes. There are $5$ units of each resource type. Consider the following scenario, where the column alloc denotes the number of units of each resource type allocated to each process, and the column request denotes the number of units of each resource type requested by a process in order to complete execution. Which of these processes will finish LAST?

$\begin{array}{|l|lll|lll|}\hline &\text{} \rlap{\textbf{alloc}} &&&\rlap{ \textbf{request}} \\ \hline &\text{X} & \text{Y} & \text{Z} &\text{X} & \text{Y} & \text{Z} \\ \hline \textbf{P0} & \text{1} & \text{2} & \text{1}& \text{1} & \text{0} & \text{3} \\ \hline \textbf{P1} & \text{2} & \text{0} & \text{1} & \text{0} & \text{1} & \text{2} \\ \hline \textbf{P2} & \text{2} & \text{2} & \text{1} & \text{1} & \text{2} & \text{0} \\ \hline \end{array}$

View worked solution

Answer: C

To find the order in which the processes finish, we first calculate the total allocated resources and the initially available resources.

1. Calculate Available Resources:

Total resources: $X = 5, Y = 5, Z = 5$
Total allocated resources:
- $X = 1 + 2 + 2 = 5$
- $Y = 2 + 0 + 2 = 4$
- $Z = 1 + 1 + 1 = 3$
$\text{Available} = \text{Total} - \text{Allocated} = [5-5, 5-4, 5-3] = [0, 1, 2]$

2. Step-by-Step Execution:

Compare requests with the current available resources $[0, 1, 2]$ $[0, 1, 2]$ :
- $P0$ request $[1, 0, 3]$ cannot be satisfied.
- $P1$ request $[0, 1, 2]$ can be satisfied.
- $P2$ request $[1, 2, 0]$ cannot be satisfied.
Thus, $P1$ $P 1$ runs first. After completion, it releases its allocated resources $[2, 0, 1]$ $[2, 0, 1]$ .
$\text{New Available} = [0, 1, 2] + [2, 0, 1] = [2, 1, 3]$ $New Available = [0, 1, 2] + [2, 0, 1] = [2, 1, 3]$ .
Compare remaining requests with available resources $[2, 1, 3]$ $[2, 1, 3]$ :
- $P0$ request $[1, 0, 3]$ can be satisfied.
- $P2$ request $[1, 2, 0]$ cannot be satisfied ( $2 > 1$ for resource $Y$ ).
Thus, $P0$ $P 0$ runs next. After completion, it releases its allocated resources $[1, 2, 1]$ $[1, 2, 1]$ .
$\text{New Available} = [2, 1, 3] + [1, 2, 1] = [3, 3, 4]$ $New Available = [2, 1, 3] + [1, 2, 1] = [3, 3, 4]$ .
Finally, $P2$ request $[1, 2, 0]$ can be satisfied with $[3, 3, 4]$ . $P2$ runs and finishes.

The safe sequence of execution is $P1 \to P0 \to P2$ . Therefore, $P2$ finishes last.

Free practice · Your answers on this page are not saved.

Published 10 Oct 2026

Report a correction

More Deadlock questions

GATE CS 2007 Q57 — Deadlock

Related questions