4330 THIRD ASSIGNMENT

1. 4330 THIRD ASSIGNMENT Summer 2014

2. The problem • The owners of a club want to avoid an excessive number of male patrons • They instruct the bouncer to refuse admission to male patrons whenever there would otherwise be more than three male patrons foe each female patron inside the club

3. A very simplified view I can let three male patrons in for each female patron who shows up

4. Our approach • A main program will fork processes representing either male or female patrons • No bouncer process • Its role will be simulated by the patron processes themselves

5. A naïve solution • Have one semaphore admit controlling the admission of male patrons • Initial value is zero • When a male patron arrives, he does P(&admit) • When a female patron arrives, she does three V(&admit) • When a male patron leaves, he does V(&admit)

6. Underlying idea (I) • The current value of the semaphore admit represents the current number of available admission tokens Admit one male patron

7. Underlying idea (II) • Male patron must have a token to enter the club • Handed out FCFS by the bouncer • Each time a female patron arrives, the bouncer gets three new tokens • Each time a male patron leaves, he gives back his token to the bouncer

8. The problem • It does not work because female patrons can leave at any time: • Alice arrives • Bob, Charles and Dean get in • Alice leaves • Emily arrives • Cannot admit Fred, George and Henry

9. A better solution (I) • Two constraints: • Cannot prevent female patrons from leaving • Cannot quick out male patrons

10. A better solution • Will restrict • the creation of new tokens when a female patron arrives • the recycling of tokens when a male patron leaves • Will sometimes delete available tokens when a female patron leaves.

11. Female patron arrival • Updae the number of female patrons • Check how many male patrons can enter the club without having more than three male patrons for each female patrons • Do as many do V() operations on the admit semaphore • Zero, one, two, or three • We limit the creation of new tokens

12. Pseudocode • nfemales +=1 # inside a critical sectionncanadmit = 3*nfemales – nmalesfor i in range(0, ncanadmit) : V(&admit)

13. Female patron departure • Update number of female patrons • Check the value of the admit semaphore • Represents the number of pending tokens • If and only if admitting these male patrons would cause result in the admission of too many male patrons, do the required number of P() operations on the admit semaphore. • Will delete the required number of tokens

14. Pseudocode • nfemales –=1 # inside a critical sectionntoomany = nmales – 3*nfemalesfor i in range(0, ntoomany): if value(&admit) > 0 : # decrement value(&admit) P(&admit)

15. Male patron arrival • Do P(&admit) • Increment number of male patrons

16. Pseudocode • P(&admit)nmales +=1

17. Male patron departures • Update the number of male patrons • Do a V() operation on the admit semaphore if and only it there are enough female patrons in the club • Limit token recycling

18. Pseudocode • nmales –= 1 # inside a critical sectionncanadmit = 3*nfemales – nmales if ncanadmit > 0 : V(&admit)

19. Important • Do not forget the –lrt option when you compile your code • Use unique names for your semaphores and shared memory segments • Delete them after each run of your program • Or expect unexpected errors • Terminate promptly your processes using _exit( )

20. Compiling your program • If the compiler returns error messages such as undefined reference to `sem_open' • You forgot the –lrt flag • Your Linux system does not support Posix semaphores

21. Debugging your program (I) • If your program crashes after informing you that it cannot create a semaphore or a shared memory segment • Check that nobody else uses the same semaphore names and shared memory key

22. Debugging your program (II) • If your program does not work anymore as it did before your last change • Delete your semaphores and shared memory segments then retry