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Operating Systems, 371-1-1631 Winter Semester 2011

Operating Systems, 371-1-1631 Winter Semester 2011. Practical Session 5, Synchronization. 1. Motivation. Multiprocessing needs some tools for managing shared resources Printers Files Data Bases. 2. Conditions for a good Solution.

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Operating Systems, 371-1-1631 Winter Semester 2011

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  1. Operating Systems, 371-1-1631Winter Semester 2011 Practical Session 5, Synchronization 1

  2. Motivation • Multiprocessing needs some tools for managing shared resources • Printers • Files • Data Bases 2

  3. Conditions for a good Solution • Mutual Exclusion – No two processes are in the critical section(CS) at the same time • Deadlock Freedom – If two or more processes are trying to enter a CS onewill eventually enter it. • Starvation Freedom – A process trying to enter a CS will eventually manage to enter it. 3

  4. Solution Archetypes • Busy wait • Wastes CPU • Priority inversion:Task L (low priority) runs and gains exclusive use of resource R. H (high priority) task is introduced and attempts to acquire R – blocked. M (medium priority) becomes runnable before L releases R. Result: L can’t run and release R, H is waiting for R (and L) as long as M keeps running. • Sleep & Wake up 4

  5. Peterson’s Solution N=2; interested[0]=interested[1]=FALSE; int turn; /* Should be named NOT_MY_TURN */ int interested[N]; /* all initially 0 (FALSE) */ voidenter_region(int process){ /* who is entering 0 or 1 ? */int other = 1- process; /* opposite of process */interested[process] = TRUE; /* signal that you're interested */turn = process; /* set flag – NOT my turn */while (turn == process && interested[other] == TRUE); /* null statement */ } voidleave_region(int process){ /* who is leaving 0 or 1 ? */interested[process] = FALSE; /* departure from critical region */ } 5

  6. Peterson’s Solution Process = 0 Process = 1 interested[0]=interested[1]=FALSE int turn; int interested[2]; voidenter_region(int process){ int other = 1; interested[0] = TRUE; turn = 0; while (turn == 0 && interested[1] == TRUE); } voidleave_region(int process){ interested[0] = FALSE;} interested[0]=interested[1]=FALSE int turn; int interested[2]; voidenter_region(int process){ int other = 1- process; interested[1] = TRUE; turn = 1; while (turn == 1 && interested[0] == TRUE); } voidleave_region(int process){ interested[1] = FALSE;} 6

  7. Question 1 N=1; interested[0]=interested[1]=FALSE; int turn; /* Should be named NOT_MY_TURN */ int interested[N]; /* all initially 0 (FALSE) */ void enter_region(int process){ /* who is entering 0 or 1 ? */int other = 1- process; /* opposite of process */ turn = process; /* set flag – NOT my turn */interested[process] = TRUE; /* signal that you're interested */while (turn == process && interested[other] == TRUE); /* null statement */ } void leave_region(int process){ /* who is leaving 0 or 1 ? */interested[process] = FALSE; /* departure from critical region */ } What happens when the two red lines are in this order? 7

  8. Peterson’s Solution Process = 0 Process = 1 voidenter_region(int process){ int other = 1; turn = 0;interested[0] = TRUE; while (turn == 0 && interested[1] == TRUE); } voidleave_region(int process){ interested[0] = FALSE;} interested[0]=interested[1]=FALSE int turn; int interested[2]; voidenter_region(int process){ int other = 1- process; turn = 1; interested[1] = TRUE;while (turn == 1 && interested[0] == TRUE); } voidleave_region(int process){ interested[1] = FALSE;} 8

  9. Answer We will get a mutual exclusion violation: P1 does turn:=1; P0 does turn:=0; P0 does interested[0]:=true; P0 does while(turn == 0 && interested [1]); // (#t and #f)  #f P0 enters the CS. P1 does interested [1]:=true; P1 does while(turn == 1 && interested [0]); // (#f and #t)  #f P1 enters the CS. now both processes are in the critical section. 9

  10. Question 2 Assume the while statement in Peterson's solution is changed to:while (turn != process && interested[other] == TRUE) Describe a scheduling scenario in which we will receive a Mutual Exclusion and one in which we will NOT receive a Mutual Exclusion violation. 10

  11. Question 2 N=2; interested[0]=interested[1]=FALSE; int turn; /* Should be named NOT_MY_TURN */ int interested[N]; /* all initially 0 (FALSE) */ voidenter_region(int process){ /* who is entering 0 or 1 ? */int other = 1- process; /* opposite of process */interested[process] = TRUE; /* signal that you're interested */turn = process; /* set flag – NOT my turn */ while (turn != process && interested[other] == TRUE) /* null statement */ } voidleave_region(int process){ /* who is leaving 0 or 1 ? */interested[process] = FALSE; /* departure from critical region */ } 11

  12. Peterson’s Solution Process = 0 Process = 1 voidenter_region(int process){ int other = 1; interested[0] = TRUE; turn = 0; while (turn != 0 && interested[1] == TRUE); } voidleave_region(int process){ interested[0] = FALSE;} interested[0]=interested[1]=FALSE int turn; int interested[2]; voidenter_region(int process){ int other = 1- process; interested[1] = TRUE; turn = 1; while (turn != 1 && interested[0] == TRUE); } voidleave_region(int process){ interested[1] = FALSE;} 12

  13. Answer for Q.2 No Mutex violation: • One enters and exits and only afterwards the second receives a time quanta. • P0 – interested[0]=true, P0 – turn = 0, P1- interested[1]=true, P1 – turn =1, P1 passes while loop and enters CS, P0 – stuck in while loop. 13

  14. Answer for Q.2 Mutex violation: Process A executes: `interested [process] = TRUE', `turn = process‘ and falls through the while loop shown above. While in the critical section, the timer fires and process B executes: `interested [process] = TRUE', `turn = process' and falls through the while loop shown above. Both processes are in the critical section. 14

  15. Bakery Algorithm int value[N]={0, 0,…,0}; /* processes get “waiting numbers” */ int busy[N]={FALSE,…, FALSE}; /* just a flag... */ voidenter_region(inti ) /* process i entering.. */ { busy[i] = TRUE; /* guard the value selection */ value[i] = max(value[0], value[1], …, value[N-1]) + 1; /* LAST in line … */ busy[i] = FALSE; for(k=0; k < N; k ++){ while (busy[k] == TRUE) ; /* wait before checking */ while((value[k] != 0) && ((value[k], k) < (value[i], i))); /* wait */ } } void leave_region(inti ) { value[i ] = 0;} 15

  16. Semaphores Two atomic operations: up & down. down(S) if S≤0 block process. S-- up(S) S++ if there are blocked processes wake one up. NOTE: Semaphore’s interface doesn’t enforce the implementation of starvation freedom. 16

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