A demand paging system takes $100$ time units to service a page fault and $300$ time units to replace a dirty page. Memory access time is $1$ time unit. The probability of a page fault is $p$. In case of a page fault, the probability of page being dirty is also $p$. It is observed that the average access time is $3$ time units. Then the value of $p$ is $0.194$ $0.233$ $0.514$ $0.981$
Consider a system with a two-level paging scheme in which a regular memory access takes $150$ $nanoseconds$, and servicing a page fault takes $8$ $milliseconds$. An average instruction takes $100$ nanoseconds of CPU time, and two memory accesses. The TLB ... average instruction execution time? $\text{645 nanoseconds}$ $\text{1050 nanoseconds}$ $\text{1215 nanoseconds}$ $\text{1230 nanoseconds}$