Transcript Lecture 6: Synchronization

Operating Systems
Review for Chap.6 & 7
Hung Q. Ngo
KyungHee University
Spring 2009
http://uclab.khu.ac.kr/lectures/2009-1-os.html
Exercise 6.1
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Solution to Critical-Section Problem
1.Mutual Exclusion - If process Pi is executing in its
critical section, then no other processes can be
executing in their critical sections
2.Progress - If no process is executing in its critical
section and there exist some processes that wish to
enter their critical section, then the selection of the
processes that will enter the critical section next
cannot be postponed indefinitely
3.Bounded Waiting - A bound must exist on the
number of times that other processes are allowed to
enter their critical sections after a process has made
a request to enter its critical section and before that
request is granted
 Assume that each process executes at a nonzero
speed
 No assumption concerning relative speed of the N
processes
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Exercise 6.4
• Mutual exclusion using Swap( ) with bounded
waiting
• swap (&address, register) { /* x86 */
temp = M[address];
M[address] = register;
register = temp;
}
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Lock solution using Swap
• Shared Boolean variable lock initialized to FALSE;
Each process has a local Boolean variable key.
• Solution:
do {
key = TRUE;
while ( key == TRUE)
Swap (&lock, &key );
//
critical section
lock = FALSE;
//
remainder section
} while ( TRUE);
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Exercise 6.4
• Solution: similar to fig. 6.10 using TestAndSet( )
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Exercise 6.7
• Implement semaphore in multiprocessor
environments using TestAndSet( ) with minimal
busy waiting
• Hint: Similar to “better lock” but here “value” is
more than Free/Busy
• Use a guard shared variable for acquiring critical
section to modify the semaphore value
• Put waiting threads to sleep
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Better Locks using test&set
• Can we build test&set locks without busy-waiting?
– Can’t entirely, but can minimize!
– Idea: only busy-wait to atomically check lock value
int guard = 0;
int value = FREE;
Release() {
Acquire() {
// Short busy-wait time
// Short busy-wait time
while (test&set(guard));
while (test&set(guard));
if anyone on wait queue {
if (value == BUSY) {
take thread off wait queue
put thread on wait queue;
Place on ready queue;
go to sleep() & guard = 0; } else {
} else {
value = FREE;
value = BUSY;
}
guard = 0;
guard = 0;
}
}• Note: sleep has to be sure to reset the guard variable
– Why can’t we do it just before or just after the sleep?
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Exercise 6.8
• The Sleeping-Barber Problem: A barbershop
consists of awaiting room with n chairs and a barber
room with one barber chair. If there are no
customers to be served, the barber goes to sleep.
If a customer enters the barbershop and all chairs
are occupied, then the customer leaves the shop. If
the barber is busy but chairs are available, then the
customer sits in one of the free chairs. If the
barber is asleep, the customer wakes up the barber.
Write a program to coordinate the barber and the
customers.
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Exercise 6.8
Semaphore mutex = 1;
Semaphore customers = 0;
Semaphore haircut = 0;
int waiting = 0
void customer() //khách đến cắt tóc
{
wait( mutex );
if( waiting == N ) //nếu số khách đợi = N thì rời khỏi tiệm
{
signal( mutex ); return ;
}
waiting ++; //tăng số khách đợi
signal( mutex );
signal(customers); //đánh thức barber nếu đang ngủ
wait(haircut); //đang cắt nhưng chưa xong (chờ trên ghế cắt tóc)
}
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Exercise 6.8
Semaphore mutex = 1;
Semaphore customers = 0;
Semaphore haircut = 0;
int waiting = 0
void barber() //thợ cắt tóc
{
while( 1 ) //cắt liên tục, hết khách này đến khách khác
{
wait( customers ); //nếu không có khách, barber sẽ ngủ
wait( mutex );
waiting --; //giảm 1 khách đợi
signal(mutex);
cut_hair();
signal(haircut); //cho khách đã cắt tóc xong rời khỏi tiệm
}
}
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Exercise 6.9 & 6.10
• Write a bounded-buffer monitor in which the
buffers (portions) are embedded within the monitor
itself.
• Problem with previous solution?
• Modify the solution to improve.
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Operating
monitor bounded_buffer {
int items[MAX_ITEMS];
int numItems = 0;
condition full, empty;
void produce(int v) {
while (numItems == MAX_ITEMS)
full.wait();
items[numItems++] = v;
empty.signal();
}
int consume() {
int retVal;
while (numItems == 0)
empty.wait();
retVal = items[--numItems];
full.signal();
return retVal;
}
}Systems
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Producing Water!
MakeH()
{
while (true)
Make-Hydro(); // tạo 1 nguyên tử H
}
MakeO()
{
while (true)
Make-Oxy(); //tạo 1 nguyên tử O
}
/* Tiến trình MakeWater hoạt động đồng hành với các tiến trình MakeH,
MakeO, chờ có đủ 2 H và 1 O để tạo H2O */
MakeWater()
{
while (True)
Make-Water(); //Tạo 1 phân tử H2O
}
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Producing Water!
Semaphore s1=0, s2=0;
MakeH()
{
while (true) {
Make-Hydro(); signal(s1);}
}
MakeO()
{
while (true) {
Make-Oxy(); signal(s2);}
}
/* Tiến trình MakeWater hoạt động đồng hành với các tiến trình MakeH,
MakeO, chờ có đủ 2 H và 1 O để tạo H2O */
MakeWater()
{
while (True) {
wait(s1); wait(s1); wait(s2); Make-Water(); }
Operating
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}
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Exercise 7.10
A single-lane bridge connects the two Vermont villages of
North Tunbridge and South Tunbridge. Farmers in the two
villages use this bridge to deliver their produce to the
neighboring town. The bridge can become deadlocked if both a
northbound and a southbound farmer get on the bridge at the
same time (Vermont farmers are stubborn and are unable to
back up.) Using semaphores, design an algorithm that prevents
deadlock.
Hint: Initially, do not be concerned about starvation (the
situation in which northbound farmers prevent southbound
farmers from using the bridge, or vice versa).
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Simple solution
semaphore ok_to_cross = 1;
void enter bridge() {
ok to cross.wait();
}
void exit bridge() {
ok to cross.signal();
}
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Solution without deadlock & starvation
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Want More? :D
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