The address of a page table in memory is pointed to by the ____________

A deadlock avoidance algorithm dynamically examines the resource allocation state to ensure that a circular wait condition can never exist.

Which of the following is incorrect?

A process is thrashing if it spends a lot of time paging than executing

Consider a non-negative counting semaphore S. The operation P(S) decrements S, and V(S) increments S. During an execution, 20 P(S) operations and 12 V(S) operations are issued in some order. The largest initial value of S for which at least one P(S) operation will remain blocked is 

The FIFO (First-In, First-Out) algorithm first executes the job that came in first in the queue.

Suppose the functions F and G can be computed in 5 and 3 nanoseconds by functional units U_F and U_G, respectively. Given two instances of UF and two instances of UG, it is required to implement the computation F(G(Xi)) for 1 <= i <= 10. ignoring all other delays, the minimum time required to complete this computation is ________________ nanoseconds 

Consider the following set of processes, assumed to have arrived at time 0. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order P₁, P₂, P₃, P₄.

If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR (round off to 2 decimal places) is

A file system uses an in-memory cache to cache disk blocks. The miss rate of the cache is shown in the figure. The latency to read a block from the cache is 1 ms and to read a block from the disk is 10 ms. Assume that the cost of checking whether a block exists in the cache is negligible. Available cache sizes are in multiples of 10 MB.

The smallest cache size required to ensure an average read latency of less than 6 ms is _______ MB.

Consider the following proposed solution for the critical section problem. There are n processes: P0 ...Pn−1. In the code, function pmax returns an integer not smaller than any of its arguments. For all i, t[i] is initialized to zero.

do { c[i] = 1; t[i] = pmax(t[0],...,t[n-1]) + 1; c[i] = 0; for every j != i in {0, ...., n-1} { while (c[j]); while (t[j] != 0 && t[j] <= t[i]); } Critical Section; t[i] = 0; Remainder Section; } while (true);

Which one of the following is TRUE about the above solution?