Discussion
Consider a system having m resources of the same type. These resources are shared by 3 processes A, B and C which have peak demands of 3, 4 and 6 respectively. For what value of m deadlock will not occur ?
Answer: D
Each process can be made to hold one resource fewer than its peak demand, and if the system still has one more resource, some process can complete. The processes A, B and C have peak demands of 3, 4 and 6, so the worst case holds 2 + 3 + 5 = 10 resources with every process waiting for one more. A deadlock cannot happen if m is at least
\[ (3 - 1) + (4 - 1) + (6 - 1) + 1 = 11 \]Hence m = 7, 9 and 10 can lead to a deadlock, and m = 13 is safe. The keyed answer is 13.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development