Given,
Primary Memory Access Time (Tpma) = 82ns
Memory Access Time on Swap (Tsma) = 107ns
Page Fault Rate (Rf) = 0.07.
The Effective Memory Access Time formaulae is given below
Substituting the values,
So the Effective Memory Access Time is 83.75ns.
Question 17 5 pts Assume we have primary memory access time of 82 nanoseconds, and a...
Problem 4 (13 points) Assume we have a demand-paged memory. Assume that the time needed to access the page table is negligible. It requires 8 milliseconds to handle a page fault if an empty page is available or the replaced page is not dirty, and 20 milliseconds if the replaced page is dirty. Memory access time is 100 nanoseconds. Assume that the page to be replaced is dirty 70 percent of the time. What is the maximum acceptable page-fault rate...
8. (5 points) 10.21 Assume we have a demand-paged memory. The page table is held in registers. It takes 8 milliseconds to service a page fault if an empty page is available or the replaced page is not modified, and 20 if the replaced page is modified. Memory access time is 100 nanoseconds. Assume that the page to be replaced is modified 70 percent of the time. What is the maximum acceptable page- fault rate for an effective access time...
Answer the following questions a) If average page-fault service time is 9 milliseconds, a memory access time is 202 nanoseconds, and the page fault rate is 0.2% then what is the effective access time (in nanoseconds) for a demand paging scheme. b) Under what circumstances do page faults occur? Describe the actions taken by the operating system when a page fault occurs.
Regarding demand paging, assume an average page fault service time = 25 ms and a memory access time of 100 ns (ms = milliseconds, 1/1000 sec; ns = nanoseconds, 1/1,000,000,000 sec). Calculate the effective access time in ns. Given a page fault rate of 1/1000, calculate the factor by which the computer would be slowed down over not incurring any page fault.
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Consider a demand-paging system with a paging disk that has an average access and transfer time of 20 milliseconds. Addresses are translated through a page table in main memory, with an access time of 1 microsecond per memory access. Thus, each memory reference through the page table takes two accesses. To improve this time, we have added an associative memory that reduces access time to one memory reference, if the page-table entry is in the associative memory. Assume that 80...
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