1. Lecture 28 Syed Mansoor Sarwar
Operating Systems
Lecture 28
Syed Mansoor Sarwar

2. © Copyright Virtual University of Pakistan
Agenda for Today
Review of previous lecture
Deadlock avoidance
Banker’s algorithms
Safety algorithm
Safe Sequence
Recap of lecture
12 January 2019
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Review of Lecture 27
Deadlock handling
Deadlock prevention
Deadlock avoidance
12 January 2019
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RAG Algorithm
Claim edge Pi  Rj indicates that process Pi may request an instance of resource Rj; represented by a dashed line.
Claim edge converts to request edge when a process requests a resource.
12 January 2019
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RAG Algorithm
When a resource is assigned to a process, request edge reconverts to an assignment edge.
When a resource is released by a process, assignment edge reconverts to a claim edge.
Resources must be claimed a priori in the system.
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6. RAG For Deadlock Avoidance
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Unsafe State In RAG
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Banker’s Algorithm
Multiple instances of resources.
Each process must a priori claim maximum use.
When a process requests a resource it may have to wait.
When a process gets all its resources it must return them in a finite amount of time.
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9. Data Structures for the Banker’s Algorithm
n = number of processes
m = number of resource types
Available: Vector of length m. If available[j] = k, there are k instances of resource type Rj available.
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10. Data Structures for the Banker’s Algorithm
Max: n x m matrix. If Max[i,j] = k, then process Pi may request at most k instances of resource type Rj.
Allocation: n x m matrix. If Allocation[i,j] = k then Pi is currently allocated k instances of Rj.
12 January 2019
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11. Data Structures for the Banker’s Algorithm
Need: n x m matrix. If Need[i,j] = k, then Pi may need k more instances of Rj to complete its task.
Need[i,j] =
Max[i,j] – Allocation [i,j]
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Safety Algorithm
Let Work and Finish be vectors of length m and n, respectively. Initialize:
Work = Available
Finish[i] = false; i = 1, 2, …, n.
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Safety Algorithm
Find and i such that both:
(a) Finish[i] = false
(b) Needi  Work
If no such i exists, go to step 4.
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Safety Algorithm
Work = Work + Allocationi Finish[i] = true go to step 2.
If Finish [i] == true for all i, then the system is in a safe state.
12 January 2019
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15. Resource-Request Algorithm for Pi
Requesti = request vector for Pi.
If Requesti [j] = k then process Pi wants k instances of resource type Rj.
If Requesti  Needi go to step 2. Otherwise, report error since process has exceeded its maximum claim.
12 January 2019
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16. Resource-Request Algorithm for Pi
If Requesti  Available, go to step 3. Otherwise Pi must wait, since resources are not available.
Pretend to allocate requested resources to Pi by modifying the state as follows:
12 January 2019
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17. Resource-Request Algorithm for Pi
Available = Available – Requestj;
Allocationi = Allocationi + Requesti;
Needi = Needi – Requesti;
If safe  the resources are allocated to Pi.
If unsafe  Pi must wait, and the old resource-allocation state is restored
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Example System
Five processes: P0 … P4
Three resource types:
A (10 instances)
B (5 instances)
C (7 instances)
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Example System
System state: Is it safe?
Process P0 P1 P2 P3 P4
Max
A B C
7 5 3
3 2 2
9 0 2
2 2 2
4 3 3
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Available A B C 3 2
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Example System
Needi = Maxi – Allocationi
Need matrix
Process A B C P0 7 4 3 P1 1 2 P2 6 P3 P4
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Example System
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 2
Safe Sequence: <>
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Example System
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 2 5
Safe Sequence: < P1>
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Example System
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 2 5 7 4
Safe Sequence: < P1, P3>
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Example System
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 2 5 7 4
Safe Sequence: < P1, P3, P4 >
12 January 2019
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Example System
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 2 5 7 4
Safe Sequence: <P1, P3, P4, P0>
12 January 2019
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Example System
Final safe sequence:
<P1, P3, P4, P0, P2>
Not a unique sequence
Possible safe sequences for the this example:
<P1,P3,P4,P0,P2>, <P1,P3,P4,P2,P0>, <P1,P3,P2,P0,P4>, <P1,P3,P2,P4,P0>, <P1,P3,P0,P2,P4>, <P1,P3,P0,P4,P2>
12 January 2019
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Example
P1 requests (1,0,2). Can this request be granted immediately?
Is Request1  Need1?
(1,0,2)  (1,2,2)  true
Is Request1  Available?
(1,0,2)  (3,3,2)  true
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Available A B C 3 2
12 January 2019
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
0 2 0
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
3 0 2
2 1 1
0 0 2
Work A B C 2 3
Safe Sequence: <>
12 January 2019
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30. © Copyright Virtual University of Pakistan
Example
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
0 2 0
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
3 0 2
2 1 1
0 0 2
Work A B C 2 3 5
Safe Sequence: < P1 >
12 January 2019
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
0 2 0
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
3 0 2
2 1 1
0 0 2
Work A B C 2 3 5 7 4
Safe Sequence: < P1, P3 >
12 January 2019
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
0 2 0
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
3 0 2
2 1 1
0 0 2
Work A B C 2 3 5 7 4
Safe Sequence: < P1, P3, P4 >
12 January 2019
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
0 2 0
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
3 0 2
2 1 1
0 0 2
Work A B C 2 3 5 7 4
Safe Sequence: <P1, P3, P4, P0>
12 January 2019
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Example
Executing safety algorithm shows that sequence <P1,P3,P4,P0,P2> satisfies safety requirement.
Yes, P1’s request may be granted immediately.
12 January 2019
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More Examples
Can request for (0,2,0) by P0 be granted?
Is Request0  Need0?
(0,2,0)  (7,4,3)  true
Is Request1  Available?
(0,2,0)  (3,3,2)  true
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 4 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 1 0
2 0 0
3 0 2
2 1 1
0 0 2
Available A B C 3 2
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 2 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 3 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 1 2
Safe Sequence: <>
12 January 2019
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Example
Process P0 P1 P2 P3 P4
Need
A B C
7 2 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 3 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 1 2 5
Safe Sequence: < P3 >
12 January 2019
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39. © Copyright Virtual University of Pakistan
Example
Process P0 P1 P2 P3 P4
Need
A B C
7 2 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 3 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 1 2 5 7
Safe Sequence: < P3,P1 >
12 January 2019
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40. © Copyright Virtual University of Pakistan
Example
Process P0 P1 P2 P3 P4
Need
A B C
7 2 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 3 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 1 2 5 7 10
Safe Sequence: < P3,P1,P2 >
12 January 2019
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41. © Copyright Virtual University of Pakistan
Example
Process P0 P1 P2 P3 P4
Need
A B C
7 2 3
1 2 2
6 0 0
0 1 1
4 3 1
Allocation
A B C
0 3 0
2 0 0
3 0 2
2 1 1
0 0 2
Work A B C 3 1 2 5 7 10
Safe Sequence: <P3,P1,P2,P0,P4>
12 January 2019
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Example
Executing safety algorithm shows that sequence <P3,P1,P2,P0,P4> satisfies safety requirement.
Yes, P0’s request may be granted immediately.
12 January 2019
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More Examples
Can request for (3,3,0) by P4 be granted?
Do it yourself!
12 January 2019
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44. © Copyright Virtual University of Pakistan
Recap of Lecture
Deadlock avoidance
Banker’s algorithm
Safety algorithm
Safe state
Safe sequence
12 January 2019
© Copyright Virtual University of Pakistan

45. Lecture 28 Syed Mansoor Sarwar
Operating Systems
Lecture 28
Syed Mansoor Sarwar