1 votes 1 votes Process Pi repeat flag[i] = true; while ( flag[j] ) do no-op; < critical - section > flag[i] = false; < remainder - section > until false; Note : "i" refer0s to current process and "j" refers to another process. How Bounded Wait is satisfied here? Operating System operating-system process-synchronization deadlock-prevention-avoidance-detection semaphore bounded-waiting + – Shubhanshu asked Dec 7, 2017 • edited Dec 7, 2017 by Shubhanshu Shubhanshu 2.0k views answer comment Share Follow See all 16 Comments See all 16 16 Comments reply Show 13 previous comments Shubhanshu commented Dec 8, 2017 reply Follow Share Bounded waiting is : There exists a bound, or limit, on the number of times other processes (P0) are allowed to enter their critical sections after a process(P1) has made request to enter its critical section and before that request is granted. "before that request is granted" But here request never granted to P1. 0 votes 0 votes Kapil commented Dec 8, 2017 reply Follow Share https://gateoverflow.in/74002/doubt-in-some-basic-things-synchronisation 0 votes 0 votes Shubhanshu commented Dec 8, 2017 reply Follow Share If you agree, that the number of times a process P1 may be bypassed is 0 ,i.e, B = 0 then yes, bounded waiting holds . Exactly, if P0 and P1 in busy wait, then no one can bypass other. bounded waiting says => That if a process expresses its desire to enter critical section then there exists a bound on the number of processes that can enter CS, after this request has been made . Thus P0 and P1's desire to enter into Critical section will be postponed indefinitely. Is it will not dissatisfy the BW? 0 votes 0 votes Please log in or register to add a comment.