Consider a demand-paging system with a paging disk that has an average access/transfer time of50 ms. Addresses are translated through a page table in main memory, with an access time of 500ns per memory access. Thus, each memory reference through the page table takes two accesses.To improve this time, we have added a TLB that reduces access time to one memory reference ifthe page-table entry is in the TLB. Assume that 80% of the accesses are in the TLB and that, ofthose remaining, 15% (or 3% of the total) cause page faults. We assume that the TLB access timeis 20 ns. What is the effective access time?

In any system that uses demand paging, the operating system copies a disk page into physical memory…

Answered: Consider a demand-paging system with a…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 blocks of cache. If each block contains 8 bytes, determine the size of the offset field, and show your work.
You are given the following data about a virtual memory system:(a)The TLB can hold 1024 entries and can be accessed in 1 clock cycle (1 nsec).(b) A page table entry can be found in 100 clock cycles or 100 nsec.(c) The average page replacement time is 6 msec.If page references are handled by the TLB 99% of the time, and only 0.01% lead to a page fault, what is the effective address-translation time?
Task 6. Consider a paging system with the one-level page table stored in memory. (a) If a memory reference takes 10 ns (nanoseconds), how long does a paged memory reference take? (b) If we add TLBS, and if 90 percent of all page-table references are found in the TLBS, what is the effective memory reference time? (Assume that finding a page-table entry in the TLBS takes 0.5 nanoseconds, if the entry is present.)
In a computer with a 32-bit data-bus, how many 4-bit wide memory components are used? the answer to this part is 32/4 = 8 components (2-bit wide) I need the answer to part two, please If the size of each 4-bit memory component is 4 xn cells where n = 1G (i.e., 4 x n uniquely addressable locations-n: row, 4:2 column/width), what is the total capacity of the memory system? Show your answer in power of 2. (hint: 1000 - 210)
Take into account the following scenario: we have a byte-addressable computer with 2-way set associative mapping, 16-bit main memory addresses, and 32 blocks of cache memory. Since there are 8 bytes in a block, you can use that information to calculate how big the offset field has to be.
please no chatgpt answer . 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 percent of the accesses are in the associative memory and that, of those remaining, 10 percent (or 2 percent of the total) cause page faults. What is the effective memory access time? Consider the following page reference string: 1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6. Assuming demand paging with four frames, Show which pages are resident under the LRU, FIFO, and Optimal replacement algorithms by filling out the following tables. How many page faults would occur…
A paging scheme uses a Translation Look-aside Buffer (TLB). A TLB-access takes 10 ns and a main memory access takes 50 ns. The effective access time(in ns) is_?(if the TLB hit ratio is 90% and there is no page-fault)
Let's pretend for a moment that we have a byte-addressable computer with fully associative mapping, 16-bit main memory addresses, and 32 blocks of cache memory. The following holds true if each block is 16 bits in size: a) Determine how many bytes the offset field is. Measure the tag field's width and height in pixels (b).
Consider a paging system with the page table stored in memory. Memory access takes 200 nanoseconds. If we add associative registers (a TLB), and 75 percent of all page table references are found in associative registers, then the effective memory reference time is__ns Assume that finding a page-table entry in the TLB takes zero time.
Take into account the following scenario: we have a byte-addressable computer with 2-way set associative mapping, 16-bit main memory addresses, and 32 blocks of cache memory. Since there are 8 bytes in a block, you can use that information to calculate how big the offset field has to be.
A computer uses virtual memory, and a new solid-state drive (SSD) as space for paging. Refer to the last ppt file. In the case presented there, the hard disk drive (HDD) required 25 ms to read in a page, and a rate of 1 page fault per 1000 references introduced a 250 slowdown. If the SSD offers a time of only 80 µs, what is the slowdown in performance caused by 1 pf per 1000 references (you are not concerned with dirty vs. clean pages). What is the maximum rate of page faults you can accept if you want no more than a 5% slowdown in execution using virtual memory? Know your metric prefixes and symbols for time: s for seconds, ms for milliseconds, µs for microseconds, ns for nanoseconds.

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