A computer has two processors, $M1$ and $M2$. Four processes $P1, P2, P3, P4$ with CPU bursts of $20, 16, 25,$ and $10$ milliseconds, respectively, arrive at the same time and these are the only processes in the system. The scheduler uses non-preemptive priority scheduling, with priorities decided as follows: - $M1$ uses priority of execution for the processes as, $P1 > P3 > P2 > P4$, i.e., $P1$ and $P4$ have highest and lowest priorities, respectively. - $M2$ uses priority of execution for the processes as, $P2 > P3 > P4 > P1$, i.e., $P2$ and $P1$ have highest and lowest priorities, respectively. A process $Pi$ is scheduled to a processor $Mk$, if the processor is free and no other process $Pj$ is waiting with higher priority. At any given point of time, a process can be allocated to any one of the free processors without violating the execution priority rules. Ignore the context switch time. What will be the average waiting time of the processes in milliseconds?

Consider a demand paging memory management system with $32$-bit logical address, $20$-bit physical address, and page size of $2048$ bytes. Assuming that the memory is byte addressable, what is the maximum number of entries in the page table?

A disk of size $512$M bytes is divided into blocks of $64$K bytes. A file is stored in the disk using linked allocation. In linked allocation, each data block reserves $4$ bytes to store the pointer to the next data block. The link part of the last data block contains a $NULL$ pointer (also of $4$ bytes). Suppose a file of $1$M bytes needs to be stored in the disk. Assume, $1\text{K} = 2^{10}$ and $1\text{M} = 2^{20}$. The amount of space in bytes that will be wasted due to internal fragmentation is . (Answer in integer)

A computer has two processors, $M1$ and $M2$. Four processes $P1, P2, P3, P4$ with CPU bursts of $20, 16, 25,$ and $10$ milliseconds, respectively, arrive at the same time and these are the only processes in the system. The scheduler uses non-preemptive priority scheduling, with priorities decided as follows: - $M1$ uses priority of execution for the processes as, $P1 > P3 > P2 > P4$. i.e., $P1$ and $P4$ have highest and lowest priorities, respectively. - $M2$ uses priority of execution for the processes as, $P2 > P3 > P4 > P1$. i.e., $P2$ and $P1$ have highest and lowest priorities, respectively. A process $Pi$ is scheduled to a processor $Mk$, if the processor is free and no other process $Pj$ is waiting with higher priority. At any given point of time, a process can be allocated to any one of the free processors without violating the execution priority rules. Ignore the context switch time. What will be the average waiting time of the processes in milliseconds?

Suppose in a multiprogramming environment, the following C program segment is executed. A process goes into I/O queue whenever an I/O related operation is performed. Assume that there will always be a context switch whenever a process requests for an I/O, and also whenever the process returns from an I/O. The number of times the process will enter the ready queue during its lifetime (not counting the time the process enters the ready queue when it is run initially) is . (Answer in integer) !image(https://uadmin.udsfgecw.tech/file/download/public/public:76f5a863-90b1-4238-8100-4663a549ae70.png)

Consider the following two threads T1 and T2 that update two shared variables a and b. Assume that initially a = b = 1. Though context switching between threads can happen at any time, each statement of T1 or T2 is executed atomically without interruption. $$\begin{array}{cc} \textbf{T1} & \textbf{T2} \\ \text{a} = \text{a} + 1; & \text{b} = 2 \text{b}; \\ \text{b} = \text{b} + 1; & \text{a} = 2 \text{a}; \end{array}$$ Which one of the following options lists all the possible combinations of values of a and b after both T1 and T2 finish execution?

Consider a memory management system that uses a page size of $2\text{ KB}$. Assume that both the physical and virtual addresses start from $0$. Assume that the pages $0, 1, 2,$ and $3$ are stored in the page frames $1, 3, 2,$ and $0$, respectively. The physical address (in decimal format) corresponding to the virtual address $2500$ (in decimal format) is