Given the following:Logical Memory size of 1000Physical Memory size of 2000Page (and frame) size of 100Block A contains data for a programSelect Block A’s size and its starting point in both memories.Then write the page table for Block A based on your selections. See below for the layout of both memories and an example of Block A of size 200. Logical Memory Physical Memorylocation/ page location/frame0 to 99/ 0 0 to 99/ 0100 to 199 /1 Block A100 to 199/ 1200 to 299/ 2 Block A200 to 299/ 2300 to 399/ 3300 to 399/ 3400 to 499/ 4400 to 499/ 4500 to 599/ 5500 to 599/ 5600 to 699/ 6600 to 699/ 6700 to 799/ 7700 to 799/ 7800 to 899/ 8800 to 899/ 8900 to 999/ 9900 to 999/ 9 1000 to 1099/ 101100 to 1199/ 111200 to 1299/ 121300 to 1399/ 13 Block A1400 to 1499/ 14 Block A1500 to 1599/ 151600 to 1699/ 161700 to 1799/ 171800 to 1899/ 181900 to 1999/ 19 PageFrame113214

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e following: Logical Memory size of 1000 Physical Memory size of 2000 Page (and frame) size of 100 Block A contains data for a program Select Block A’s size and its starting point in both memories. Then write the page table for Block A based on your selections. See below for the layout of both memorie

e following: Logical Memory size of 1000 Physical Memory size of 2000 Page (and frame) size of 100 Block A contains data for a program Select Block A’s size and its starting point in both memories. Then write the page table for Block A based on your selections. See below for the layout of both memorie

Microsoft Visual C#

7th Edition

ISBN:9781337102100

Author:Joyce, Farrell.

Publisher:Joyce, Farrell.

Chapter8: Advanced Method Concepts

Section: Chapter Questions

Problem 7RQ

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Please DO NOT respond to this question by copy/pasting the code provided elsewhere on the site, none of those work. Thanks. Virtual Memory Lab This lab project addresses the implementation of page-replacement algorithms in a demand-paging system. Each process in a demand-paging system has a page table that contains a list of entries. For each logical page of the process, there is an entry in the table that indicates if the page is in memory. If the page is in memory, the memory frame number that page is resident in is indicated. Also, for each page, the time at which the page has arrived in memory, the time at which it has been last referenced, and the number of times the page has been referenced since the page arrived in memory are maintained. The page table data structure is a simple array of page-table entries (PTEs). Each PTE contains five fields as defined below: struct PTE { int is_valid; int frame_number; int arrival_timestamp; int…
., which contains temporary data (such as 7. A process generally also includes the process . function parameters, return addresses, and local variables), and a contains global variables. which ..... stack / data section heap / data section stack / code section heap / data section
Student information is being held in a data area, where each student record has the following format: The first nine bytes are the student number, held in ASCII The next byte is the course mark The next word is the section identifier There are well over three hundred such student records that have been loaded sequentially into memory starting at address $10000. The last record loaded is a dummy record with a section identifier of $FFFF, to show the end of the records. If a2 has the address of a student record, what is the location of the next record in the array / data area:
Student information is being held in a data area, where each student record has the following format: The first nine bytes are the student number, held in ASCII The next byte is the course mark The next word is the section identifier There are well over three hundred such student records that have been loaded sequentially into memory starting at address $10000. The last record loaded is a dummy record with a section identifier of $FFFF, to show the end of the records. If a2 has the address of a student record, if using indexed addressing mode, what is the index value X such that X(a2) addresses the course mark?
Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Page Table Physical Memory Physical Address (starting) Oxppppdddd Page Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 PPpp: page number dddd: page offset 1 1 Оxd000 65536 3 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x00011119 What is the physical address for 0x00000001 What is the logical address for Oxd0000001 ? What is the logical address for Oxc0000002 ?
Create a program in C++ which simulates a direct cache. The memory array that contains the data to be cached is byte addressable and can contain 256 single byte entries or lines. The cache has only 8 entries or lines. The Data field in each line of the cache is 8 bits. Since the data stored in each line of the cache is only 8 bits, there is no need for a line field. Only a tag field is needed which is log2(256) = 8 bits. The memory array can be filled with any values of your choice. The program should work by taking user input of a memory address (index). This input represents the memory data that should be cached. Your program will check the cache to see if the item is already cached. If it is not, your program should count a cache miss, and then replace the item currently in the cache with the data from the inputted address. Allow the user to input addresses (in a loop), until they so choose to end the program. The program should output the number of cache misses upon ending.
Create a program in C++ which simulates a direct cache. The memory array that contains the data to becached is byte addressable and can contain 256 single byte entries or lines. The cache has only 8 entriesor lines. The Data field in each line of the cache is 8 bits. Since the data stored in each line of the cache isonly 8 bits, there is no need for a line field. Only a tag field is needed which is log2(256) = 8 bits.The memory array can be filled with any values of your choice. The program should work by taking userinput of a memory address (index). This input represents the memory data that should be cached.Check the cache to see if the item is already cached. If it is not, your program should counta cache miss, and then replace the item currently in the cache with the data from the inputted address.Allow the user to input addresses (in a loop), until they so choose to end the program. The program should output the number of cache misses upon ending.
Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Охрppdddd Page | Frame Frame Size (hex) Size (dec) 2 Охс000 Ox10000 65536 pppp: page number dddd: page offset 1 1 Оxd000 Ox10000 65536 3 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0002ffff What is the physical address for Ox0000abcd ? What is the logical address for Oxf000000f ? What is the logical address for Oxc000bbcc ?
An application loads 100 libraries at start-up. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10 ms. Rotational speed of disk is 6000 rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected.)
Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Oxppppdddd Page | Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 pppp: page number dddd: page offset 1 1 Oxd000 65536 2 2 Oxe000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0000ee00 ? What is the physical address for Ox00020001 ? What is the logical address for Oxe0001234 ? What is the logical address for Oxc0004268 ?
Student information is being held in a data area, where each student record has the following format: The first nine bytes are the student number, held in ASCII The next byte is the course mark The next word is the section identifierThere are well over three hundred such student records that have been loaded sequentially into memory starting at address $10000. The last record loaded is a dummy record with a section identifier of $FFFF, to show the end of the records.Write an assembler subroutine GetMax that will scan the entire list and find the highest mark. If there is a tie, the first student in the list with the highest mark should be found. The subroutine should return the starting address of this record in address register a3.
Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Size (dec) 65536 65536 Frame Size (hex) Oxppppdddd Page Frame 2 Ox10000 Ox10000 Охс000 pppp: page number dddd: page offset 1 1 Oxd000 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0002ffff ? What is the physical address for 0x0000abcd ?

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0 to 99/ 0		0 to 99/ 0
100 to 199 /1 Block A	100 to 199/ 1
200 to 299/ 2 Block A	200 to 299/ 2
300 to 399/ 3	300 to 399/ 3
400 to 499/ 4	400 to 499/ 4
500 to 599/ 5	500 to 599/ 5
600 to 699/ 6	600 to 699/ 6
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800 to 899/ 8	800 to 899/ 8
900 to 999/ 9	900 to 999/ 9
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1300 to 1399/ 13 Block A
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