Gem Points For IBPS SO (Specialist Officer) – Part -15
Hello Aspirants,as we all know only few days are left for IBPS SO exam.So, we are starting a new series ” Gem points ” in this regard.learn them by heart.Share and discuss in comment section.We will gear up level day by day starting from very basics.So,Put your Seatbelts on and join us in the journey to be a specialist officer.
1) Each request requires that the system consider the :-
resources currently available
resources currently allocated to each process
future requests and releases of each process to decide whether the current request can be satisfied or must wait to avoid a future possible deadlock.
2) Given a priori information about the maximum number of resources of each type that maybe requested for each process, it is possible to construct an algorithm that ensures that the system will never enter a deadlock state.
3) A deadlock avoidance algorithm dynamically examines the resource allocation state, to ensure that a circular wait condition can never exist.
Resource allocation states are used to maintain the availability of the already and current available resources.
4) A state is safe, if :the system can allocate resources to each process in some order and still avoid a deadlock.
5) A system is in a safe state only if there exists a safe sequence.
6) All unsafe states are :not deadlocks.
7) If no cycle exists in the resource allocation graph : then the system will be in a safe state.
8) The resource allocation graph is not applicable to a resource allocation system :with multiple instances of each resource type.
9) The Banker’s algorithm is less efficient than the resource allocation graph algorithm.
10) The data structures available in the Banker’s algorithm are :
11) The content of the matrix Need is :Max – Allocation.
1) The wait-for graph is a deadlock detection algorithm that is applicable when : all resources have a single instance.
2) An edge from process Pi to Pj in a wait for graph indicates that : Pi is waiting for Pj to release a resource that Pi needs.
3) If the wait for graph contains a cycle :then a deadlock exists.
4) If deadlocks occur frequently, the detection algorithm must be invoked considerable overhead in computation time.
5) A deadlock eventually cripples system throughput and will cause the CPU utilization to drop.
6) Every time a request for allocation cannot be granted immediately, the detection algorithm is invoked. This will help identify :
the set of processes that have been deadlocked
the specific process that caused the deadlock
7) A computer system has 6 tape drives, with ‘n’ processes competing for them. Each process may need 3 tape drives. The maximum value of ‘n’ for which the system is guaranteed to be deadlock free is 2.
8) A system has 3 processes sharing 4 resources. If each process needs a maximum of 2 units then, deadlock :can never occur
9) ‘m’ processes share ‘n’ resources of the same type. The maximum need of each process doesn’t exceed ‘n’ and the sum of all their maximum needs is always less than m+n. In this setup, deadlock : can never occur.
1) A deadlock can be broken by :
abort one or more processes to break the circular wait
to preempt some resources from one or more of the deadlocked processes.
2) The two ways of aborting processes and eliminating deadlocks are :
Abort all deadlocked processes
Abort one process at a time until the deadlock cycle is eliminated
3) those processes should be aborted on occurrence of a deadlock, the termination of which incurs minimum cost.
4) The process to be aborted is chosen on the basis of the following factors :
*) priority of the process
*) process is interactive or batch
*) how long the process has computed
*) how much more long before its completion
*) how many more resources the process needs before its completion
*) how many and what type of resources the process has used
5) Cost factors of process termination include :
number of resources the deadlock process is holding
amount of time a deadlocked process has thus far consumed during its execution
6) If we preempt a resource from a process, the process cannot continue with its normal execution and it must be :
7) To roll back the process to a safe state, the system needs to keep more information about the states of processes.
8) If the resources are always pre -empted from the same process, starvation can occur.
9) The solution to starvation is :the number of rollbacks must be included in the cost factor