Complete GATE CSE 2019 Paper 1 overview, original pages and practice · Timed previous-paper practice

Questions 49–51 of 65. Imported answers require review unless explicitly marked reviewed.

Paper practice 65 questionsPage 17 of 22

Question 39

Open question
CSMCQ2 marksNeeds review
Consider the following snapshot of a system running

concurrent processes. Process

is holding

instances of a resource R,

. Assume that all instances of R are currently in use. Further, for all

, process

can place a request for at most

additional instances of R while holding the

instances it already has. Of the

processes, there are exactly two processes

and

such that

. Which one of the following conditions guarantees that no other process apart from

and

can complete execution?
Your answerChoose one option.

Log in to check the answer

You can attempt this question now. Sign in to view the answer key, explanations and discussion.Log in to check answer

Question 40

Open question
CSMCQ2 marksNeeds review
Consider the following statements: I. The smallest element in a max-heap is always at a leaf node II. The second largest element in a max-heap is always a child of the root node III. A max-heap can be constructed from a binary search tree in

time IV. A binary search tree can be constructed from a max-heap in

time Which of the above statements are TRUE?
Your answerChoose one option.

Log in to check the answer

You can attempt this question now. Sign in to view the answer key, explanations and discussion.Log in to check answer

Question 41

Open question
CSNAT2 marksNeeds review
Consider the following for four processes with arrival times (in milliseconds) and their length of CPU bursts (in milliseconds) as shown below:
Process
P1
P2
P3
P4
Arrival time
0
1
3
4
CPU burst time
3
1
3
Z
These processes are run on a single processor using preemptive Shortest Remaining Time First scheduling algorithm. If the average waiting time of the processes is 1 millisecond, then the value of Z is ________.

Log in to check the answer

You can attempt this question now. Sign in to view the answer key, explanations and discussion.Log in to check answer