Consider the virtual memory scheme using paging. The page size is 128 bytes. The entries in the page table for process 3, starting from row 0 of the table onwards, are: (these are shown in binary, spaces are included just for better readability): 000 000 000 100 111 010 011 100 000 000 010 000 100 000 000 101 000 000 Convert the following PHYSICAL ADDRESSES to the corresponding LOGICAL ADDRESSES

Consider the virtual memory scheme using paging. The page size is 128 bytes. The entries in the page table for process 3, starting from row 0 of the table onwards, are: (these are shown in binary, spaces are included just for better readability): 000 000 000 100 111 010 011 100 000 000 010 000 100 000 000 101 000 000 Convert the following PHYSICAL ADDRESSES to the corresponding LOGICAL ADDRESSES

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 a computer using fully associative cache has 224 bytes of byte-addressable main memory and a cache of 128 blocks, where each cache block contains 64 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? c) To which cache block will the memory address 0x01D872 map?
What is the maximum number of 16-bit words that can be stored in a 2^32 byte physical memory?
Problem 0. The following table gives some of the parameters for a number of different hardware caches. Fill in the table with the values of the missing parameters. Recall that m is the number of physical address bits, C is the cache size in bytes, B is the block size in bytes, E is the associativity, i.e., lines per set, S is the number of sets, t is the number of tag bits, s is the number of set index bits, and b is the number of block offset bits. Cache m 48 32030 (b) 48 32 24 24 C B E 16384 32 16 32768 64 65536 128 512 32 1024 8 8 4 14 1 4 S b S t
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the available space list of a computer memory is specified as follows: 9 start address block address in words 100 50 200 150 450 600 1200 400 determine the available space list after allocating the space for the stream of requests consisting of the following block sizes: 25,100,250,200,100,150 use i) first fit ii) best fit and iii) worst fit algorithms
Below shows the following hexidecimal values:Address 1000: 13Address 1001: 03Address 1002: C5Address 1003: 00 Provide the format and assembly language instruction for the following hexadecimal values.
Suppose a byte-addressable computer using set-associative cache has 216 bytes of main memory and a cache of 32 blocks, and each cache block contains 8 bytes.Q.) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache; that is, what are the sizes of the tag, set, and offset fields?
Suppose a computer using direct mapped cache has 232 byte of byte-addressable main memory, and a cache of 1024 blocks, where each cache block contains 32 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag, block, and offset fields? c) To which cache block will the memory address 0x000063FA map?
Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…
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Draw a picture illustrating the contents of memory, given the following data declarations: (You need to draw a diagram showing the memory addresses and their contents.) Assume that your data segment starts at 0x1100 in memory. Your answer should fill the table below and explain why. -- photo attached below Thank you for your help!
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