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■ Q: Reader and writers don't deadlock or livelock (safety)? ■ Q: Every reader or writers will eventually succeed (liveness)? ■ Q: Fairness: • Multiple readers can eventually read? • One writer can eventually write? ■ Yes, all readers can eventually read. ◉ No, if there are always at least one readers, the writer cannot write forever. ■ How to solve it? • Propose your solution in Assignment 2

■ Q: Reader and writers don't deadlock or livelock (safety)? ■ Q: Every reader or writers will eventually succeed (liveness)? ■ Q: Fairness: • Multiple readers can eventually read? • One writer can eventually write? ■ Yes, all readers can eventually read. ◉ No, if there are always at least one readers, the writer cannot write forever. ■ How to solve it? • Propose your solution in Assignment 2

Operations Research : Applications and Algorithms
Operations Research : Applications and Algorithms
4th Edition
ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Wayne L. Winston
Chapter17: Markov Chains
Section: Chapter Questions
Problem 12RP
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Question

Given a shared data set, we allow multiple readers to read at the same time, and only one single writer can access the shared data at the same time. In the lecture slides, a solution is given. However, the problem is that the write cannot write forever, if there are always at least one reader.

How to ensure that the writer can eventually write? Propose your solution by using semaphores and implemented in Python

from threading import Thread, Semaphore
from time import sleep
from sys import stdout

class Reader(Thread):
def__init__(self, name):
self.n=name; Thread.__init__(self)
defrun(self):
globalnr, nw, dr, dw
whileTrue:
# ⟨await nw == 0 then nr += 1⟩
e.acquire()
ifnw>0:
#if nw > 0 or dw > 0 :
dr+=1; e.release(); r.acquire()
nr+=1
ifdr>0: dr-=1; r.release()
else: e.release()
# read data
stdout.write(self.n+' reading\n')
sleep(1)
# ⟨nr -= 1⟩
e.acquire()
nr-=1
ifnr==0anddw>0:
dw-=1 ; w.release()
else: e.release()

class Writer(Thread):
def__init__(self, name):
self.n=name; Thread.__init__(self)
defrun(self):
globalnr, nw, dr, dw
whileTrue:
# ⟨await nr == 0 and nw = 0 then nw += 1⟩
e.acquire()
ifnr>0ornw>0:
dw+=1; e.release(); w.acquire()
nw+=1
e.release()
# write data
stdout.write(self.n+' writing\n')
sleep(2)
# ⟨nw -= 1⟩
e.acquire()
nw-=1
ifdr>0: dr-=1; r.release()
elifdw>0: dw-=1; w.release()
else: e.release()
#if dw > 0: dw -= 1; w.release()
#elif dr > 0: dr -= 1; r.release()
#else: e.release()

e = Semaphore(1)
r, w = Semaphore(0), Semaphore(0)
nr, nw = 0, 0
dr, dw = 0, 0

r1 = Reader('R1'); r2 = Reader('R2')
w1 = Writer('W1'); w2 = Writer('W2')
r1.start(); r2.start(); w1.start(); w2.start() 

 

■ Q: Reader and writers don't deadlock or livelock (safety)? ■ Q: Every reader or writers will eventually succeed (liveness)? ■ Q: Fairness: • Multiple readers can eventually read? • One writer can eventually write? ■ Yes, all readers can eventually read. ◉ No, if there are always at least one readers, the writer cannot write forever. ■ How to solve it? • Propose your solution in Assignment 2
Transcribed Image Text:■ Q: Reader and writers don't deadlock or livelock (safety)? ■ Q: Every reader or writers will eventually succeed (liveness)? ■ Q: Fairness: • Multiple readers can eventually read? • One writer can eventually write? ■ Yes, all readers can eventually read. ◉ No, if there are always at least one readers, the writer cannot write forever. ■ How to solve it? • Propose your solution in Assignment 2
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