操作系统课程2.4_3_死锁的处理策略—避免死锁

Requesti[j]≤Available[j] (0≤j≤m)
”①
Pi
Available = Available - Requesti; Allocation[i, j] = Allocation[i, j] + Requesti[j]; Need[i, j] = Need[i, j] – Requesti[j]
④
①
④
①
m n*m Max
Available ①
n*m Allocation
Max – Allocation = Need
m
Request
③ ①
”
① ①
Allocation
(57, 34, 23) (1, 2, 2) (6, 0, 0) (0, 1, 1) (4, 3, 1)
Need
①
③
④ ① Pi “
①
m n*m Max
Available ①
n*m Allocation
Max – Allocation = Need
m
Request
③ ①
”
① ①
① ①
(0, 1, 0) (2, 0, 0) (3, 0, 2) (2, 1, 1) (0, 0, 2)
(7, 4, 3) (1, 2, 2) (6, 0, 0) (0, 1, 1) (4, 3, 1)
3, 3, 2
④ P1 P3
④
P1 P3
(2, 0, 0) + (2, 1, 1) + (3, 3, 2) = (7, 4, 3)
P0 P2 P4
④
5
① ①
(10, 5, 7)
① P1 P3
① (37, 34, 23)
①
P0 (8, 5, 3) P1 (3, 2, 2) P2 (9, 5 ,2) P3 (2, 2, 2) P4 (4, 3, 6)
(0, 1, 0) (2, 0, 0) (3, 0, 2) (2, 1, 1) (0, 0, 2)
(7, 4, 3) (1, 2, 2) (6, 0, 0) (0, 1, 1) (4, 3, 1)
① > (53, 3, 2) < (3, 3, 2) > (5, 3, 2) < (5, 3, 2)
④P1 ④P3 ……
… {P1} P3} P0 P2 P4}
① (3, 3, 2)
④
P1
①
① (5, 3, 2)
(10, 5, 7)
① P1 P3
① (37, 34, 23)
n
Max Rj
Need
n*m ①
Max[i, j]=K
①
Pi
①
Requesti
m① ①
Pi n*m
K① Allocation
Max – Allocation =
①
m
Available
①
m ①
Requesti[j]≤Need[i, j] (0≤j≤m)
④
P3
①
① (7, 4, 3)
④
5
(10, 5, 7)
① (357, 34, 23)
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①
P1
P1
P1 ①
①
①
P3
(2,P03, 0) + (3, 3, 2) = (5, 3, 2)
P3
①
①
(2, 1, 1) + (5, 3, 2) = (7, 4, 3)
(5, 3, 2)
(7, 4, 3) ……
P0 (7, 5, 3) P1 (3, 2, 2) P2 (9, 0 ,2) P3 (2, 2, 2) P4 (4, 3, 3)
B
70
A
40
T
50
②
20
50
10
30
30
20
(7, 2, 5)
(3, 3, 2)
①
① —— ①
5 (10, 5, 7)
P0~P4 3
P0 (7, 5, 3) P1 (3, 2, 2) P2 (9, 0 ,2) P3 (2, 2, 2) P4 (4, 3, 3)
(0, 1, 0) (2, 0, 0) (3, 0, 2) (2, 1, 1) (0, 0, 2)
Available = (13, 23, 12)
Request0 = (2, 1, 1)
P0 (7, 5, 3) P1 (3, 2, 2) P2 (9, 0 ,2) P3 (2, 2, 2) P4 (4, 3, 3)
Max
(02, 12, 01) (2, 0, 0) (3, 0, 2) (2, 1, 1) (0, 0, 2)
(7, 4, 3) (1, 2, 2) (6, 0, 0) (0, 1, 1) (4, 3, 1)
P0 (7, 5, 3) P1 (3, 2, 2) P2 (9, 0 ,2) P3 (2, 2, 2) P4 (4, 3, 3)
(0, 1, 0) (2, 0, 0) (3, 0, 2) (2, 1, 1) (0, 0, 2)
A
40
T
50
B
70
A
40
T
50
”100 B
70 … 40 … 50 …
①…
A
T
20 10 30 … ②
20
50
10
30
30
20
②
20 10+20=30
30
50 30-20=10
20
BAT
…
40
…A
20
A
……
TàBàA
AàTàB
OK OK
20
A 20 A 10 … A 20+30 = 50TàBàA ATà20TàB T
(8, 4, 3) (1, 2, 2) (6, 5, 0) (0, 1, 1) (4, 3, 4)
3, 3, 2
④ P1 P3
④
P1 P3
(2, 0, 0) + (2, 1, 1) + (3, 3, 2) = (7, 4, 3)
P0 (8, 4, 3) P2 (6, 5, 0) P4 ④
① (4, 3, 4)
A
40
T
50
B
70
A
40
T
50
”100 B
70 … 40 … 50 …
①…
A
T
20 10 30 … ②
20
50
10
30
30
20
②
20 10+20=30
30
50 30-20=10
20
BAT
…
40
…A
20
A
……
TàBàA
OK
20
20 50+20=70
TT 20+30=50 BB
A
B A T
…
BAT
B
70
50+30 = 80
… B…
②
B
70
20+30=60 50-30=20
A
40
10
30
T
50
30
20
②
B
70
20
50
A
40
10+20=30 30-20=10
T
50
30
20
B 30 BAT
…
20 ④…
10
A 20
①
TàBàA
① ①
” ①
A
10
TàB àA
①
①
Dijkstra ④
①
BAT ①
① R0~R2
——
B A T
…
BAT
B
70
A
40
T
50
B
70
A
40
T
50
”100 B
70 … 40 … 50 …
①…
A
T
20 10 30 … ②
20
50
10
30
30
20
②
20+30=50 10 30
50-30=20 30 20
BAT
…
40
…B
30
B
……
10 20
B 30
10 BAT ④……
B A T
…
BAT
B
70