Exam 3/Final Review

Exam 3

Mean adjusted score: 62% ( C ) Correct multiple choice answers:

1. D2. B3. E4. A5. E6. C7. A8. B or C9. E10. B11. C

Problem 4

Assume a disk has 100 cylinders labeled 0-99, the read-write head is positioned over cylinder 52 heading toward cylinder 0, accessing data requires 1 time unit, and incoming requests arrive as follows:

What is the order of the requests which will be serviced by time 64 (cylinder of request is given) by the LOOK disk scheduling algorithm?

Arrival time

0 10 31 45 51 64

Cylinder 10 26 20 45 2 11

Problem 4

0 10 20 30 40 50 60 70 80 90 99

0

10

20

30

40

50

60

70

(52,0)

(26,27)

(20,34)(10,45)

(29,64)

Time

Cylinder

Problem 5 Assume a disk has 100 cylinders labeled 0-99, the read-

write head is positioned over cylinder 21 heading toward cylinder 0, accessing data requires 1 time unit, and pending requests are as follows: 1, 16, 24, 71, 8, 18, 94, 3 (listed in order of arrival). Which disk scheduling algorithm will finish servicing the requests first? SSTF (break tie by moving toward cylinder 0.) LOOK FCFS C-Scan None of the above: there is a tie.

Problem 5

SSTF  (21 – 18) + 1 Tie between 18 and 24, move to 18+ (18 – 16) + 1+ (16 – 8) + 1 Tie between 8 and 24, move to 8+ (8 – 3) + 1+ (3 – 1) + 1+ (24 – 1) + 1+ (71 – 24) + 1+ (94 – 71) + 1= 121

1 16 24 71 8 18 94 3

Problem 5

Look  (21 – 18) + 1+ (18 – 16) + 1+ (16 – 8) + 1+ (8 – 3) + 1+ (3 – 1) + 1 Do not go to 0, turn around+ (24 – 1) + 1+ (71 – 24) + 1+ (94 – 71) + 1= 121

1 16 24 71 8 18 94 3

Problem 5

FCFS  (21 – 1) + 1+ (16 – 1) + 1+ (24 – 16) + 1+ (71 – 24) + 1+ (71 – 8) + 1 + (18 – 8) + 1+ (94 – 18) + 1+ (94 – 3) + 1= 336

1 16 24 71 8 18 94 3

Problem 5 C-Scan (read/write heading toward zero)

  (21 – 18) + 1+ (18 – 16) + 1+ (16 – 8) + 1+ (8 – 3) + 1+ (3 – 1) + 1+ (1 – 0)+ (99 – 0)+ (99 – 94) + 1+ (94 – 71) + 1+ (71 – 24) + 1+ (24 – 1) + 1= 221

1 16 24 71 8 18 94 3

Final Exam Review

More disk scheduling Exponential moving average for predicting next

burst Banker’s algorithm

SSTF (static) Assume the R/W heads hold

stationary when there is no pending request, and that seek decisions are never reversed on new arrival

What is the total time to service all requests?

(28 – 3) + 1 + (16 – 3) +1 +(58–16) + 1 + (98 – 58) + 1 +(98 – 71) + 1 + (71 – 12) + 1 +(56 – 12) + 1 = 256

Arrival Time Cylinder

0 3

4 16

20 58

51 98

113 71

131 12

185 56

SSTF (static)

3 12 16 28 56 58 71 980102030405060708090100110120130140150160

0

26

83

124

152

Arrival Time Cylinder

0 3

4 16

20 58

51 98

113 71

131 12

185 56

40

SSTF (static)

3 12 16 28 56 58 71 98170180190200210220230240250260270280290300310320330

212

256

Arrival Time Cylinder

0 3

4 16

20 58

51 98

113 71

131 12

185 56

SSTF (dynamic) Assume the R/W heads continue

moving toward cylinder 50 when there is no pending request, and that seek decisions can be changed dynamically

What is the total time to service all requests?

(28 – 3) + 1 + (16 – 3) +1 +

(58–16) + 1 + (98 – 58) + 1 +

(98 – 71) + 1 + (71 – 48) +

(56 – 48 ) + 1 + (56 – 12) + 1 = 229

Arrival Time Cylinder

0 3

4 16

20 58

51 98

113 71

131 12

185 56

SSTF (dynamic)

3 12 16 28 56 58 71 980102030405060708090100110120130140150160

0

26

83

124

152

Arrival Time Cylinder

0 3

4 16

20 58

51 98

113 71

131 12

185 56

40

SSTF (dynamic)

3 12 16 28 56 58 71 98170180190200210220230240250260270280290300310320330

184

229

Arrival Time Cylinder

0 3

4 16

20 58

51 98

113 71

131 12

185 56

Page Replacement

Clock page replacement (2nd chance)

The use bit is shown to the right of the frames

The arrow denotes the position of the clock pointer

0, 1, 3, 5, 2, 1, 4, 5, 0

0

1

3

5

1

1

1

1

Clock page replacement (2nd chance)

The use bit is shown to the right of the frames

The arrow denotes the position of the clock pointer

0, 1, 3, 5, 2, 1, 4, 5, 0

2

1

3

5

1

0

0

0

Clock page replacement (2nd chance)

The use bit is shown to the right of the frames

The arrow denotes the position of the clock pointer

0, 1, 3, 5, 2, 1, 4, 5, 0

2

1

3

5

1

1

0

0

Clock page replacement (2nd chance)

The use bit is shown to the right of the frames

The arrow denotes the position of the clock pointer

0, 1, 3, 5, 2, 1, 4, 5, 0

2

1

4

5

1

0

1

0

Clock page replacement (2nd chance)

The use bit is shown to the right of the frames

The arrow denotes the position of the clock pointer

0, 1, 3, 5, 2, 1, 4, 5, 0

2

1

4

5

1

0

1

1

Clock page replacement (2nd chance)

The use bit is shown to the right of the frames

The arrow denotes the position of the clock pointer

0, 1, 3, 5, 2, 1, 4, 5, 0

2

0

4

5

0

1

1

0

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

0

1

3

5

100

100

100

100

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

0

1

3

5

010

010

110

010

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

2

1

3

5

100

010

110

010

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

2

1

3

5

010

001

011

001

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

2

1

3

5

110

101

011

001

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

2

1

3

4

110

101

011

100

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

2

1

3

4

011

010

001

010

LRU approximation (shift bits)

Each ‘c’ denotes a clock shift event The reference bit and clock bits are

shown to the right of the frame

0, 1, 3, 5, c, 3, 2, c, 1, 2, 4, c, 5

2

1

5

4

011

010

100

010

Banker’s Algorithm

Example of Banker’s Algorithm 5 processes P0 through P4; 3 resource types A (4 instances),

B (5 instances), and C (6 instances). Snapshot at time T0:

Allocation Max AvailableA B C A B C A B C

P0 0 1 0 3 5 3 1 3 2

P1 2 0 0 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

Example of Banker’s Algorithm Is the system currently safe? Suppose a request for <1, 1, 1> came from P1.

Allocation Max AvailableA B C A B C A B C

P0 0 1 0 3 5 3 1 3 2

P1 2 0 0 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 1 2 2 2 0 1 2 1 13 3 1

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Is the request legal? Are the resources available?

Allocation Max AvailableA B C A B C A B C

P0 0 1 0 3 5 3 1 3 2

P1 2 0 0 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 1 2 2 2 0 1 2 1 13 3 1

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Make a hypothetical grant of the resources Test to see if the state is safe

Allocation Max AvailableA B C A B C A B C

P0 0 1 0 3 5 3 0 2 1

P1 3 1 1 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 0 1 1 2 0 1 2 1 13 3 1

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Make a hypothetical grant of the resources Test to see if the state is safe

Allocation Max WorkA B C A B C A B C

P0 0 1 0 3 5 3 0 2 1

P1 3 1 1 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 0 1 1 2 0 1 2 1 13 3 1

P1

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Make a hypothetical grant of the resources Test to see if the state is safe

Allocation Max WorkA B C A B C A B C

P0 0 1 0 3 5 3 3 3 2

P1 3 1 1 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 0 1 1 2 0 1 2 1 13 3 1

P1

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Make a hypothetical grant of the resources Test to see if the state is safe

Allocation Max WorkA B C A B C A B C

P0 0 1 0 3 5 3 3 3 4

P1 3 1 1 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 0 1 1 2 0 1 2 1 13 3 1

P1

P4

P3

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Make a hypothetical grant of the resources Test to see if the state is safe

Allocation Max WorkA B C A B C A B C

P0 0 1 0 3 5 3 3 4 5

P1 3 1 1 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 0 1 1 2 0 1 2 1 13 3 1

P1

P4

P3

P0

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Make a hypothetical grant of the resources Test to see if the state is safe

Allocation Max WorkA B C A B C A B C

P0 0 1 0 3 5 3 3 5 5

P1 3 1 1 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 0 1 1 2 0 1 2 1 13 3 1

P1

P4

P3

P0

P2

Example of Banker’s Algorithm Suppose a request for <1, 1, 1> came from P1

Make a hypothetical grant of the resources Test to see if the state is safe

Allocation Max WorkA B C A B C A B C

P0 0 1 0 3 5 3 4 5 6

P1 3 1 1 3 2 2

P2 1 0 1 3 0 2

P3 0 1 1 2 2 2

P4 0 0 2 3 3 3

NeedA B C3 4 3 0 1 1 2 0 1 2 1 13 3 1

P1

P4

P3

P0

P2

Detection Algorithm

1.Let Work and Finish be vectors of length m and n, respectively Initialize:(a) Work :- Available

(b) For i = 1,2, …, n, if Allocationi 0, then Finish[i] := false;otherwise, Finish[i] := true.

2.Find an index i such that both:(a) Finish[i] = false

(b) Requesti Work

If no such i exists, go to step 4.

Detection Algorithm

3. Work := Work + Allocationi

Finish[i] := truego to step 2.

4. If Finish[i] = false, for some i, 1 i n, then the system is in deadlock state. Moreover, if Finish[i] = false, then Pi is deadlocked.

Algorithm requires an order of m x n2 operations to detect whether the system is in deadlocked state.

Example of Detection Algorithm Five processes P0 through P4; three resource types

A (7 instances), B (2 instances), and C (6 instances). Snapshot at time T0:

Allocation Request AvailableA B C A B C A B C

P0 0 1 0 0 0 0 0 0 0

P1 2 0 0 2 0 2

P2 3 0 3 0 0 0

P3 2 1 1 1 0 0

P4 0 0 2 0 0 2 Sequence <P0, P2, P3, P1, P4> will result in Finish[i] = true for

all i.

Example (Cont.) P2 requests an additional instance of type C.

RequestA B C

P0 0 0 0

P1 2 0 1

P2 0 0 1

P3 1 0 0

P4 0 0 2 State of system?

Can reclaim resources held by process P0, but insufficient resources to fulfill other processes; requests.

Deadlock exists, consisting of processes P1, P2, P3, and P4.