COMP 252 Operating Systems

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COMP 252Operating Systems

• Review Question week 9

2
Q. 1

• (a) List and briefly describe the four necessary
  conditions for deadlocks
• Mutual exclusion
• at least one resource must be held in a
  non-sharable mode.
• Hold and wait
• a process must be holding at least one resource
  and waiting to acquire additional resources that
  are currently being held by other processes.

3
Q. 1

• No preemption
• a resource can be released only voluntarily by
  the process holding it.
• Circular wait
• A set P0, P1, , Pn of waiting process must
  exist such that
• P0 is waiting for a resource held by P1,
• P1 is waiting for a resource held by P2,
• Pn-1 is waiting for a resource held by Pn,
• and Pn is waiting for a resource held by P0.

4
Q. 1

• (b) Consider the following snapshot of a system

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Answer the following questions using the bankers
algorithm

• What is the content of the matrix Need denoting
  the number of resources needed by each process?
• Max Allocation Need (maxtrix)

6
Answer the following questions using the bankers
algorithm

• Is the system in a safe state? Why?
• The allocation should be safe right now, with a
  sequence of process execution.
• Yes, with ltP0, P3, P4, P1, P2gt

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Answer the following questions using the bankers
algorithm

• If a request from process P2 arrives for (0, 1,
  2, 0), can the requested be granted immediately?
  Why?
• No, this can not be allocated.
• If this is allocated, the resulting Available()
  is (2, 0, 0, 0), there is no sequence of the
  process execution order that lead to the
  completion of all processes. This is an unsafe
  state.

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Q. 2

• A system is composed of four processes
• P1, P2, P3, P4
• And three types of resources
• R1, R2, R3
• The number of units of the resources are
• C lt3, 2, 2gt

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Q. 2

• System state
• process P1 holds 1 unit of R1 and requests 1 unit
  of R2.
• P2 holds 2 units of R2 and requests 1 unit each
  of R1 and R3.
• P3 holds 1 unit of R1 and requests 1 unit of R2.
• P4 holds 2 units of R3 and requests 1 unit of R1.

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Q. 2

• Show the resource graph to represent the system
  state.
• Consider a sequence of processes executions
  without deadlock.

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• Sequence of processes executions is
• P4 gets the unit of R1 and finishes,
• P2 gets 1 unit of R1 and 1 unit of R3 and
  finishes,
• then P1 and P3 can finish.
• There is no deadlock in this situation.

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Q. 3

• Recall there are four necessary conditions for a
  deadlock to happen, namely
• mutual exclusion
• hold-and-wait
• no preemption
• circular wait
• One way to ensure no deadlock is to break one of
  the conditions.

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Q. 3

• For example, we can design an OS that requires a
  process
• either to acquire all its needed resources before
  execution,
• or wait until all its required resources to
  become available.
• What are the main disadvantages of this approach?

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Q. 3

• The process might have to wait a long time before
  it can acquire all resources.
• The process does not need to use all the
  resources it holds at the same time, thus
  resulting in a waste and other process(es) cannot
  use them.

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Q. 3

• The process might not know prior that all the
  resources it needs before its execution.
• Starvation.

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Q. 4

• Is it possible to have a deadlock involving only
  one single process?
• Explain your answer.

17
Q. 4

• No.
• This follows directly from the hold-and-wait
  condition.

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Q. 5

• Consider a system consisting of four resources of
  the same type that are shared by three processes
• Each process needs at most two resources
• Show that the system is deadlock-free

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Q. 5

• Consider that all three processes simultaneously
  request resources.
• One of them will be able to get 2 resources,
  finish its computation and return the resources
  to the system.
• Then, others can proceed and finish their
  computations also.

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Q. 6

• A system consists of 3 types of resources
• (A, B and C)
• 8 instances of A
• 10 instances of B
• 5 instances of C.
• These resources are used by five processes (P0,
  P1, P2, P3 and P4).

21
Q. 6

• If the current state of the system is given below

22
Q. 6

• Show that the system is in a safe state by
  finding a safe sequence whereby all processes
  would eventually be able to finish (ie. no
  deadlock).
• You should show the allocation strategy and the
  order of process completion.

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Q. 6

• The system is in a safe state
• the safe sequence (P3, P1, P4, P0, P2) or (P3,
  P1, P4, P2, P0) will allow all processes to
  finish execution.
• Initially, available vector
• number of each resource available for
  allocation
• (3, 3, 1)

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Q. 6

• This is greater than the need of P3 so we can
  allocate enough resources to P3 to allow it to
  finish.
• When P3 finishes, it will release all its
  resources. Thus, Available become (5, 3, 1).
• This is greater than the need of P1 which can
  therefore finish and release its resources.

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Q. 6

• Available then becomes (6, 6, 3) which is greater
  then the need of P4.
• P4 can therefore finish and release its
  resources.
• Available then becomes (6, 8, 4) which is greater
  than the need of P0 or P2.
• Either one of them can finish, say P0, making
  Available (8, 8, 5) which will allow P2 to
  finish.

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Q. 7

• Consider the dining-philosophers problem Where
• the chopsticks are placed at the center of the
  table
• and any two of them could be used by a
  philosopher
• Assume that requests for chopsticks are made one
  at a time.

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Q. 7

• Describe a simple rule for determining whether a
  particular request could be satisfied without
  causing deadlock.

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Q. 7

• when a philosopher makes a request for the first
  chopstick, do not satisfy the request only
• if there is no other philosopher with two
  chopsticks and if there is only one chopstick
  remaining.

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Q. 8

• Consider the traffic deadlock

30
Q. 8

• Show that the four necessary conditions for
  deadlock indeed hold in this example.
• State a simple rule for avoiding deadlocks in
  this system.

31
Q. 8

• The mutual exclusion condition holds as only one
  car can occupy a space in the roadway.
• Hold-and-wait occurs where a car holds onto their
  place in the roadway while they wait to advance
  in the roadway.

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Q. 8

• A car cannot be removed (i.e. preempted) from its
  position in the roadway.
• Lastly, there is indeed a circular wait as each
  car is waiting for a subsequent car to advance.

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Q. 8

• A simple rule to avoid this traffic deadlock
• a car may not advance into an intersection if it
  is clear they will not be able to immediately
  clear the intersection.