MOT SO BAI TAP CAN CHU Y !!!

Round-Robin, SJFS, B nh o, B ng FAT, Thu t gi i Nh bng, S n xu t-Tiu th (semFull-semEmpty), Dining-Philosopers (deadlock, khng deadlock).

Thu t gi i Nh bng ........................................................................................................... 1 Round-Robin, SJFS, ............................................................................................................ 3 bi t p phn o n, tnh a ch v t l cho a ch logic ...................................................... 7

bi t p phn o n, tnh a ch v t l cho a ch logic c tr ng h p khng h p l . . 9 B ng FAT, ......................................................................................................................... 11 RAG ................................................................................................................................... 13 B nh o ......................................................................................................................... 14 S n xu t-Tiu th (semFull-semEmpty), ......................................................................... 18 Dining-Philosopers (deadlock, khng deadlock). .............................................................. 21

BI T P H I U HNH

Thu t gi i Nh bng Cu 1:M t h th ng c 3 bng t v 3 ti n trnh P1, P2, P3 v i tr ng thi c p pht ti nguyn th i i m Ti th hi n b ng vc-t Allocation = (1, 0, 1) v Max = (1, 2, 2): Dng thu t gi i nh bng : a. Ch ng minh tr ng thi ny an ton. (1 i m) b. Xc nh c nn p ng hay khng yu c u xin thm 1 n a c a c a P3 ? (1 i m)Gi i:a. Xt t i th i i m Ti m 3 ti n trnh c c p pht nh bi ta c:

V i: Need[i] = Max[i] Allocation[i] v Available = 3 (1 + 0 + 1) = 1Tm chu i an ton:

V y t i th i i m T0 t n t i chu i an ton {P1, P2, P3}. Suy ra, h th ng t i th i i m Ti tr ng thi an ton.

b. Ta th y, yu c u thm 1 n a c a P3 tho cc i u ki n: o Request3

o H n n a vi c c p pht thm 1 n a cho P3 th h th ng v n tr ng thi an ton v t n t i chu i an ton {P1, P3, P2} trong khi ti nguyn trong h th ng khng cn n a. Th t v y:

Do v y ta c th c p thm cho yu c u xin thm 1 c a P3 t i th i i m ny.

Cu 2M t h th ng c 3 bng t v 3 ti n trnh P1, P2, P3 v i tr ng thi c p pht ti nguyn t i th i i m T i th hi n b ng cc vc-t Allocation=(0, 2, 1) v Max=(2, 2, 2). Dng thu t gi i Nh bng : a. Ch ng minh tr ng thi ny an ton (1,0 i m) b. Xc nh c p ng c hay khng yu c u xin thm 1 n a c a P2 (1,0 i m)

Tr l i: a. Ch ng minh tr ng thi t i th i i m T i an ton:- Tnh Need = Max Allocation = (2, 0, 1)- Tnh Available=3-(0+2+1)=0- Theo thu t gi i Nh bng, tm c 2 chu i an ton l:

Do t n t i t nh t 1 chu i an ton (chu i no cng c), tr ng thi h th ng t i th i i m T i l an ton.b. Xc nh c p ng c hay khng yu c u xin thm 1 n a c a P2 :Khng c v:- Need2=(2-2)=0, ngha l h t h n m c n nh cho P2. - M t khc, Available=0, ngha l h khng cn bng no.

Cu 3.M t h th ng c 5 ti n trnh v i tnh tr ng ti nguyn nh sau:

Process Allocation Max AvailableA B C D A B C D A B C DP0 0 0 1 2 0 0 1 2 1 5 2 0P1 1 0 0 0 1 7 5 0

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P2 1 3 5 4 2 3 5 6P3 0 6 3 2 0 6 5 2P4 0 0 1 4 0 6 5 6

Dung thuat giai Nha bang e:a. Ch ng minh tr ng thi ny an ton. (1

i m)b. Xc nh c nn p ng yu c u (0, 4, 3, 0) c a P1 ? (1

i m)Gi i:a. Xt t i th i i m T0 m 5 ti n trnh c c p pht nh bi ta c:

Need[i] = Max[i] Allocation[i]

Process NeedA B C DP0 0 0 0 0P1 0 7 5 0P2 1 0 0 2P3 0 0 2 0P4 0 4 4 2

Tm chu i an ton:Work >= Need[i] P[i] Allocation[i]

A B C D A B C D A B C D1 5 2 0 0 0 0 0 P0 0 0 1 2 1 5 3 2 1 0 0 2 P2 1 3 5 42 8 8 6 0 0 2 0 P3 0 6 3 22 14 11 8 0 4 4 2 P4 0 1 1 42 15 12 12 0 7 5 0 P1 1 0 0 0V y t i th i i m T 0 t n t i chu i an ton {P 0, P2, P3, P4, P1}. Suy ra, h

th ng t i th i i m T 0 tr ng thi an ton.

b. Ta th y, yu c u thm (0, 4, 3, 0) c a P 1 tho i u ki n Request 1 Need1, nh ng khng tho i u ki n: Request 1 Available v ti nguyn C trong h th ng ch cn 2 m yu c u 3. Do v y, khng th c p pht thm (0, 4, 3, 0) cho P1 c.

Round-Robin, SJFS,Cu 2M t h th ng c 3 ti n trnh v i th i i m n v th i gian s d ng CPU nh sau:

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Ti n trnh Th i i m n (ms) CPU-Burst (ms)

P1 3 37

P2 10 20

P3 24 14

Dng thu t gi i Round-Robin v i th i l ng 10 ms i u ph i CPU: a. Th hi n b ng bi u Gantt (1,0 i m) b. Tnh th i gian ch trung bnh c a cc ti n trnh (1,0 i m)

Tr l i: a. Th hi n b ng bi u Gantt :

b. Th i gian ch trung bnh c a cc ti n trnh :(34+13+29)/3 = 76/3 = 25,3 msCu 4

M t h th ng c 3 ti n trnh v i th i i m n v th i gian s d ng CPU nh sau:

Ti n trnh Th i i m n (ms) CPU-Burst (ms)

P1 5 47

P2 23 15

P3 45 28

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Dng thu t gi i Round-Robin v i th i l ng b ng 20 ms i u ph i CPU:

a. Th hi n b ng bi u Gantt (0,5 i m)

b. Tnh th i gian ch trung bnh c a cc ti n trnh (0,5 i m)

Tr l i:

a. Th hi n b ng bi u Gantt:

b. Tnh th i gian ch trung bnh c a cc ti n trnh:

- Th i gian ch c a cc ti n trnh:

P1 = 35 ms

P2 = 2 ms

P3 = 22 ms

- Th i gian ch trung bnh = ( 35 + 2 + 22 ) / 3 = 59 / 3 = 19,66 ms

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bi t p phn o n, tnh a ch v t l cho a ch logic

sau khi tm hi u v bi t p ny, mnh post ln cho m i ng i cng trao i

GI I1. V vng b nh V t l d ng cc o n segment T b ng d li u bi

Ta v c vng b nh v t l nh sau:

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Cc b n nhn vo hnh mnh h ng d n cch tnh v cch v , cc b n ch ph n mu ch mnh s d ng nh n ra d h n. V i segment 0: ta c + a ch v t l c s (basic) l 300+ Limit l 700==> a ch v t l c a segment 0 l t 300 -> 1000 V i segment 1:+ a ch v t l c s (basic) l 1200, nn ta s v b t u t 1200, nh v y t 1000- >1200 l tr ng, khng c segment no+ Limit l 500==> a ch v t l c a segment 1l t 1200 -> 1700

v segment 2 cc b n tnh t ng t

2. Cch tnh a ch logic

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Tnh a ch v t l + V i d li u bi cho l (1,200), ta xc nh: tnh a ch v t l c a segment 1, a ch logic l 200 (l u : gi tr X tnh c ny n m trong segment 1 hay ko (1200 khng h p l v 800 > 700 (limit c a segment 0) + (2,650) --> khng h p l v 650 > 600 (limit c a segment 2) + (1,501) --> khng h p l v 501 > 500 (limit c a segment 1) Tm l i n u nhn th y a ch logic m bi cho > limit c a segment th k t lu n ngay l khng h p l

bi t p phn o n, tnh a ch v t l cho a ch logic c tr ng h p khng h p l

Gi s c B ng o n sau:

Hy tnh a ch v t l cho m i a ch l-gic sau: a. 0430

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b. 1010c. 2500d. 3400e. 4112gi i

Tnh a ch v t l + V i d li u bi cho l (0,430) = 219 +430 = 649 ( h p l ) V n n m trong o n Segment 0.+ (1,010) = 2300+ 10 = 2310 (h p l ) + (2,500) = 90 + 500 = 1400 (khng h p l ) + (3,400) = 1327+ 400 = 1727 (h p l ) + (4,112) = 1952 + 112 = 2064 (khng h p l )

=>V i cc a ch logic (0,430); (1,010); (1,500); (3,400); (4,112) ta c cc a ch v t l t ng ng l 649; 2310; khng h p l ;1727; khng h p l .

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