Download presentation

Presentation is loading. Please wait.

Published byEfrain Littles Modified over 2 years ago

1
CS 603 Process Synchronization: The Colored Ticket Algorithm February 13, 2002

2
Resource Sharing Problem k identical resources, only one process can use a resource at a time Also referred to as critical section –Each process has critical section Exclusion Property: At most k processes can be in their critical section at the same time Impossibility of Deadlock: Processes eventually succeed in entering and leaving critical section Full use: All k can be used simultaneously: –If fewer than k processes in their critical section, another should be able to enter its critical section before others exit theirs

3
Naïve Solutions Global Semaphore –Keeps count –Doesnt prevent starvation –No notion of fairness Reduce to single resource problem –Grocery store solution: choose your line –Doesnt satisfy full use –Failure can cause starvation

4
Fault tolerance Fail-stop model: Process takes no more steps, unannounced –Cant use timeout to tell failed from slow If failed process causes any live process to deadlock/starve/etc., system is deemed to have failed –Failure while in critical section can tie up one resource Must be careful with definition of fair –E.g. Distinction between enter critical section and enabled to enter critical section

5
Colored Ticket Algorithm Assumption: N processors, fixed k resources Meets exclusion, deadlock, starvation, fairness properties –Simulates a queue (FIFO) –Robust: Failure between requesting resource and releasing resource ties up one resource Other failures have no effect Uses O(N 2 ) shared memory –Distributed virtual memory solutions can implement this Optimal in space requirements –Lower bound Ω(N 2 ) for robust algorithm to simulate queue

6
Queue Algorithm Meets all properties –Guarantees exclusion –Fair (FIFO) –No starvation –Robust Problem: Distributed implementation of Queue –True shared memory gives single point of failure –Distributed implementation requires substantial space repeat forever start; ENQUEUE(i) wait until i is in one of the first k positions of QUEUE; finish; { Critical Section } start; REMOVE(i); finish; { Remainder Section } end repeat.

7
Ticket Algorithms Idea: Take ticket when you want a resource –Ticket shows your place in line –If resources available, ticket is valid When done with resource, validate next ticket Numbered ticket: –Two shared variables –TAKE: ISSUE++ –VALIDATE: VALID++ –IS-VALID: VALID ticket Requires infinite space! local variable TICKET; repeat forever start; TICKET := TAKE-NEXT- TICKET; wait until IS-VALID(TICKET); finish; { Critical Section } start; VALIDATE-NEXT- TICKET(TICKET); finish; { Remainder Section }; end repeat.

8
Colored Ticket Key: Not more than N outstanding tickets –Arithmetic mod N gives bounded space –Really only need M 1+max(k, N-k) Separate into all in use and not all in use Can determine if all resources in use with three variables –B = ( VALID/M = ISSUE/M ) –V = VALID mod M –I = ISSUE mod M –If B then unused if VI else unused if V*
{
"@context": "http://schema.org",
"@type": "ImageObject",
"contentUrl": "http://images.slideplayer.com/7/1698622/slides/slide_8.jpg",
"name": "Colored Ticket Key: Not more than N outstanding tickets –Arithmetic mod N gives bounded space –Really only need M 1+max(k, N-k) Separate into all in use and not all in use Can determine if all resources in use with three variables –B = ( VALID/M = ISSUE/M ) –V = VALID mod M –I = ISSUE mod M –If B then unused if VI else unused if V
*

9
Colored Ticket: Simulate Numbered Ticket Divide tickets into blocks –B true iff VALID and ISSUE in same block –V (I) gives position of VALID (ISSUE) in block Replace numbered ticket by colored tickets –T = (t, c); 0 t M-1; c is block –B := VALID.c = ISSUE.c –V := VALID.t –I := ISSUE.t function LEADS(A,B): Boolean { if A.COLOR = B.COLOR return (A.VALUE B.VALUE) else return (A.VALUE < B.VALUE) } function IS-VALID(T): Boolean { if T.COLOR = VALID.COLOR return (T.VALUE VALID.VALUE) else if T.COLOR = ISSUE.COLOR return LEADS(VALID, ISSUE) else return true; }

10
Colored Ticket: Bounding colors Problem: Unbounded number of colors Solution: At most k+1 colors in use –At most k tickets valid –At most (N-k) waiting But these are consecutive! Thus all will fit in one block

11
Colored Ticket Algorithm constant M = 1 + max(k, N - k) global variable ISSUE = (0, 0), VALID = (k, 0); Global variable QUANT[k+1] = 0; function TAKE-NEXT-TICKET: ticket { if (ISSUE.VALUE < M – 1) ISSUE.VALUE++ else { ISSUE.VALUE = 0; if LEADS(ISSUE, VALID) then ISSUE.COLOR = NEW-COLOR Else ISSUE.COLOR = VALID.COLOR, } return ISSUE; } procedure VALIDATE-NEXT-TICKET (T) { if (VALID.VALUE < M – 1) VALID.VALUE++ else { VALID.VALUE = 0; if LEADS(VALID, ISSUE) VALID.COLOR = NEW-COLOR; Else VALID.COLOR = ISSUE.COLOR; } QUANT(VALID.COLOR)++; QUANT(T.COLOR)--; } function NEW-COLOR: integer { local variable C = 0; while QUANT[C] > 0 do C++; return C }

12
Colored Ticket Algorithm: Space Requirements ISSUE, VALID: Range 0..M –M bounded by N QUANT: k variables range 0..M Copies maintained at every site –Guarantees robustness –Various Distributed Shared Memory protocols guarantee consistency Gives total space of O(N 2 ) –Remember that k is constant Paper gives lower bound proof

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google