This is a multipart blog article series, and in this series I am going to explain you the concepts of operating system. This article series is divided into multiple modules and this is the forth module which consists of 8 articles.
In this article we will see a question on banker’s algorithm, to get better understanding of the concept.
Question: Check for the deadlock, if deadlock is not present then find the safe sequence.
| Process No. | Allocation of E | Allocation of F | Allocation of G | Max need of E | Max need of F | Max need of G | Current availability of E | Current availability of F | Current availability of G | Remaining need of E | Remaining need of F | Remaining need of G | |
| P1 | 1 | 0 | 1 | 4 | 3 | 1 | 3 | 3 | 0 | 3 | 3 | 0 | |
| P2 | 1 | 1 | 2 | 2 | 1 | 4 | 4 | 3 | 1 | 1 | 0 | 2 | |
| P3 | 1 | 0 | 3 | 1 | 3 | 3 | 5 | 3 | 4 | 0 | 3 | 0 | |
| P4 | 2 | 0 | 0 | 5 | 4 | 1 | 6 | 4 | 6 | 3 | 4 | 1 |
Answer:
- Current availability: (3,3,0)
- With the current availability we can fulfill the request of
P0 - Current availability: (4,3,1)
- With current availability we can fulfill the request of
P2 - Current availability: (5,3,4)
- With current availability we can fulfill the request of
P1 - Current availability: (6,4,6)
- With current availability we can fulfill the request of
P3 Safe sequence = P0 -> P2 -> P1 -> P3- No deadlock will occur. Therefore it is safe.
So this was a question on banker’s algorithm. Hope you liked it and learned something new from it.
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