1. Lecture 21 Syed Mansoor Sarwar
Operating Systems
Lecture 21
Syed Mansoor Sarwar

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Agenda for Today
Review of previous lecture
2-Process CS problem solution
N-Process CS problem
The Bakery Algorithm
Hardware solutions
Recap of lecture
8 April 2019
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Review of Lecture 20
2-Process Critical Section Problem Solution
n-Process Critical Section Problem
The Bakery Algorithm
8 April 2019
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Bakery Algorithm
Structure of Pi
do {
choosing[i] = true;
number[i] = max(number[0],
number[1], …,
number [n – 1]) + 1;
choosing[i] = false;
8 April 2019
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Bakery Algorithm
Critical Section
for (j = 0; j < n; j++) {
while (choosing[j]) ;
while ( (number[j] != 0) &&
((number[j], j) < (number[i], i)) ) ; }
8 April 2019
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Bakery Algorithm
remainder section
} while (1);
number[i] = 0;
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Bakery Algorithm
Process
Number P0 3 P1 P2 7 P3 4 P4 8
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Bakery Algorithm P0 P2 P3 P4
(3,0) < (3,0)
(3,0) < (7,2)
(3,0) < (4,3)
(3,0) < (8,4)
Number[1] = 0
(7,2) < (3,0)
(7,2) < (7,2)
(7,2) < (4,3)
(7,2) < (8,4)
(4,3) < (3,0)
(4,3) < (7,2)
(4,3) < (4,3)
(4,3) < (8,4)
(8,4) < (3,0)
(8,4) < (7,2)
(8,4) < (4,3)
(8,4) < (8,4)
8 April 2019
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Bakery Algorithm
P1 not interested to get into its critical section  number[1] is 0
P2, P3, and P4 wait for P0
P0 gets into its CS, get out, and sets its number to 0
P3 get into its CS and P2 and P4 wait for it to get out of its CS
8 April 2019
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Bakery Algorithm
P2 gets into its CS and P4 waits for it to get out
P4 gets into its CS
Sequence of execution of processes:
<P0, P3, P2, P4>
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Bakery Algorithm
Meets all three requirements:
Mutual Exclusion:
(number[j], j) < (number[i], i) cannot be true for both Pi and Pj
Bounded-waiting: At most one entry by each process (n-1 processes) and then a requesting process enters its critical section (First-Come-First-Serve)
8 April 2019
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Bakery Algorithm
Progress:
Decision takes complete execution of the ‘for loop’ by one process
No process in its ‘Remainder Section’ (with its number set to 0) participates in the decision making
8 April 2019
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13. Synchronization Hardware
Normally, access to a memory location excludes other accesses to that same location.
Extension: designers have proposed machine instructions that perform two operations atomically (indivisibly) on the same memory location (e.g., reading and writing).
8 April 2019
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14. Synchronization Hardware
The execution of such an instruction is also mutually exclusive (even on Multiprocessors).
They can be used to provide mutual exclusion but other mechanisms are needed to satisfy the other two requirements of a good solution to the CS problem.
8 April 2019
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15. Test-And-Set (TSL) Instruction
Test and modify a word atomically.
boolean TestAndSet(boolean &target) {
boolean rv = target;
target = true;
return rv; }
8 April 2019
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Solution with TSL
Structure for Pi (‘lock’ is set to false)
do {
while ( TestAndSet(lock) ) ;
Critical Section
lock = false;
Remainder Section
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Solution with TSL
Is the TSL-based solution good? No
Mutual Exclusion: Satisfied
Progress: Satisfied
Bounded Waiting: Not satisfied
8 April 2019
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Swap Instruction
Swaps two variables atomically
void swap (boolean &a, boolean &b) {
boolean temp = a;
a = b;
b = temp; }
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Solution with Swap
Structure for Pi
‘key’ is local and set to false
do {
key = true;
while (key == true) swap(lock,key);
Critical Section
lock = false;
Remainder Section }
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Solution with swap
Is the swap-based solution good? No
Mutual Exclusion: Satisfied
Progress: Satisfied
Bounded Waiting: Not satisfied
8 April 2019
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Recap of Lecture
The Bakery algorithm
Synchronization hardware
The test-and-set and swap instructions
Solutions based on the TSL and swap instructions
8 April 2019
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22. Lecture 21 Syed Mansoor Sarwar
Operating Systems
Lecture 21
Syed Mansoor Sarwar