Question: Consider the following snapshot of a system
0

enter image description here

i. What is the content of the matrix need?

ii. Is the system in safe state?

iii. If the request for P1 arrives for (1,0,2) can the request be granted immediately?

Marks: 10 M

Year: May 2015

ADD COMMENTlink
modified 3.4 years ago by gravatar for Sagar Narkar Sagar Narkar1.6k written 3.5 years ago by gravatar for Sayali Bagwe Sayali Bagwe2.5k
2

enter image description here

enter image description here enter image description here

ADD COMMENTlink
written 3.4 years ago by gravatar for Sagar Narkar Sagar Narkar1.6k
0

Need matrix

It is calculated by subtracting Allocation Matrix from the Max matrix

enter image description here

To check if system is in a safe state

  • The Available matrix is [3 3 2].
  • A process after it has finished execution is supposed to free up all the resources it hold.
  • We need to find a safety sequence such that it satisfies the criteria need Need ≤ Available.
  • Since Need (P1) ≤ Available , we select P1.[Available]=[Available]+[Allocation(P1)]

Available= [3 3 2] + [2 0 0]=[5 3 2

  • Need(P3) ≤ Available→Available=[ 5 3 2 ]+[ 2 1 1 ]=[7 4 3 ]
  • Need(P4) ≤ Available→Available=[ 7 4 3 ]+[ 0 0 2]=[ 7 4 5]
  • Need(P0) ≤ Available→Available=[ 7 4 5 ]+[0 1 0 ]=[ 7 5 5]
  • Need(P2) ≤ Available→Available=[ 7 5 5 ]+[ 3 0 2 ]=[10 5 7]
  • So it is in safe state and the safe sequence is <p1,p3,p4,p0,p2>

A request from process P1 arrives for (1,0,2)

  • System receives a request for P1 for Req(P1)[ 1 0 2 ]
  • First we check if Req(P1) is less than Need(P1) → [ 1 0 2 ]< [ 1 2 2] is true
  • Now we check if Req(P1) is less than Available →[1 0 2]<[ 3 3 2 ] is true.
  • So we update the values as:
    • Available=Available-Request=[ 3 3 2 ]- [ 1 0 2 ]=[ 2 3 0 ]
    • Allocation=allocation(P1)+Request= [ 2 0 0 ]+ [ 1 0 2 ]=[ 3 0 2 ]
    • Need= Need(P1)-Request=[ 1 2 2 ]- [ 1 0 2 ]=[ 0 2 0 ]

enter image description here

  • This is the modified table

  • So it is still in safe state and the safe sequence is < P1,P3,P4,P0,P2 >

ADD COMMENTlink
modified 3.4 years ago  • written 3.5 years ago by gravatar for Sayali Bagwe Sayali Bagwe2.5k
Please log in to add an answer.