0 votes 0 votes A system has $3$ processes sharing $4$ resources. If each process needs a maximum of $2$ units then, deadlock Can never occur Has to occur May occur None of the options Operating System nielit2017dec-scientistb operating-system resource-allocation + – admin asked Mar 30, 2020 retagged Oct 27, 2020 by Krithiga2101 admin 1.1k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Anu007 commented Dec 18, 2017 reply Follow Share lets take worst case . each need 2 resource = then assign 1,1,1 then 1 is left give it to any one , so atleast one process is always run. hence no deadlock. A 1 votes 1 votes Avdhesh Singh Rana commented Dec 18, 2017 reply Follow Share https://gateoverflow.in/55698/isro2016-47 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes (A) Can never occur Dileep kumar M 6 answered Dec 18, 2017 Dileep kumar M 6 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Min requirement of all processes is 3x1 + 1=4 which is not exceeding the available number of resources, which is 4 in the system, and therefore satisfies the necessary & condition for deadlock-free execution, hence option A is right s_dr_13 answered Mar 30, 2022 s_dr_13 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done. Hence,Option(A)deadlock can never occur. Shefalee Chaudhary answered Oct 7, 2023 Shefalee Chaudhary comment Share Follow See all 0 reply Please log in or register to add a comment.