As described in 5.7, virtual memory uses a page table to track the mapping of virtual...
The following data constitutes a stream of virtual addresses as seen on a system. Assume 4 KiB pages, a 4-entry fully associative TLB, true LRU replacement, and that the TLB is initially empty. Please show the contents of the TLB and the age of the contents (LRU) during the process.
Consider the following system: Byte addressable 16-bit addresses 256B pages Single level Page Table System The system utilizes a fully-associative TLB with 4 entries and an LRU replacement policy. Given the access pattern and timings below, complete the following table. Assume the TLB begins initially empty and the cache is physically addressed NOTE: Access times are not inclusive and components are accessed sequentially. Time to update the TLB is negligible, everything else that needs to be updated will require a...
4. Assume it take 50 nanoseconds to resolve a memory reference when accessing the physical memory address directly. a) We designed a system using virtual addresses with page tables without a TLB. In other words, when fetching data from memory, the page table is accessed to get the PTE for translating an address, a translation is completed, and finally, a memory reference to the desired data is resolved. In this system, what is the effective memory reference time. Assume the...
Problem 6 (13 points) The page table below is for a system with 16-bit virtual as well as physical addresses and with 4,096-byte pages. The reference bit is set to 1 when the page has been referenced. Periodically, a thread zeroes out all values of the reference bit. A dash for a page frame indicates the page is not in memory. The LRll pagg-replacement algorithm is used. The numbers are given in decimal Page Frame eferepceit 14 10 13 15...
please answer $5 UXIF map in the computer uses direct mapping Question 18 5 pts Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 blocks of cache. Suppose also that each block contains 8 bytes. The size of the block offset field is bits, the bits. size of the set field is bits, and the size of the tag field is 5 pts Question 19 Suppose we have a byte-addressable computer...
Problem 6 (13 points) The page table below is for a system with 16-bit virtual as well as physical addresses and with 4,096-byte pages. The reference bit is set to 1 when the page has been referenced. Periodically, a thread zeroes out all values of the reference bit. A dash for a page frame indicates the page is not in memory. The LRll pagg-replacement algorithm is used. The numbers are given in decimal Page Frame eferepceit 14 10 13 15...
A computer uses a byte-addressable virtual memory system with a four-entry TLB and a page table for a process P. Pages are 16 bytes in size. Main memory contains 8 frames and the page table contains 16 entries. a. How many bits are required for a virtual address? b. How many bits are required for a physical address?
3. Virtual Memory (20 points) An ISA supports an 8 bit, byte-addressable virtual address space. The corresponding physical memory has only 256 bytes. Each page contains 32 bytes. A simple, one-level translation scheme is used and the page table resides in physical memory. The initial contents of the frames of physical memory are shown below. VALUE address size 8 bit byte addressable each byte of addressing type memory has its own address 32 B page size physical memory size 256...
1) Given a virtual memory system with: virtual address 36 bits physical address 32 bits 32KB pages (15 bit page offset) Each page table entry has bits for valid, execute, read and dirty (4 bits total) and bits for a physical page number. a) How many bits in the page table? (do not answer in bytes!) Three digit accuracy is good enough. The exponent may be either a power of 2 or a power of 10. b) The virtual address...
Assume you have: 32-bit addresses, 4KB Page size, 4MB Physical Memory Space, 4KB Cache with 4-way set associative and LRU replacement, 32 Byte Cache block size, 4-entry fully associative TLB. A program to be run on this machine begins as follows: double A[1024]; int i, j; double sum = 0; for( i = 0; i < 1024; i++ ) // first loop A[i] = i; for( j = 0; j < 1024; j += 16 ) // second loop ...