The following table shows the core map of a virtual memory system at time t, which has a page size of 1000 bytes. In this table, column "Counter" indicates the number of references on the corresponding page till time t [small value indicates the least recently accessed page]. Process ID Page # Frame # Counter 1 1 lo 1 1 2 4 2 3 3 1 3 2 3 4 lo 2 6 6 3 7 17 Let's assume the given system has memory of 8000 bytes. Right now (@time t), process 1 issues a reference to its logical address 4500, there will be a page fault. If the system deploys LRU algorithm for page replacement with local replacement policy, what is the corresponding physical address being referenced? O a. 5500 O b. 500 O c. 2500 O d. 1500 O e. 4500
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- Oeew s Sat DO D 4 Hene et e H P E Dei vi At D Metig eta O t TE Fie • REC In Net ASSIGNMENT 1. The table below presents a list of devices that are to be addressed in a certain memory space. They have been ordered in the manner in which S S they are to be addressed with the first component being placed on the upper end of memory, starting at address $000000. By considering the size each component, provide the start and end address using the appropriate hexadecimal value. l Device Description Device Name Amount of memory to address ROM Chip ROM 1 RAM 1 4KB RAM Chip 8KB ROM Chip ROM 2 4KB Peripheral PER 1 4 bytes Peripheral PER 2 2 bytes 2. Assume a very simple microprocessor with 12 address lines Let's assume we wish to implement all its memory space and we use 517x8 memory chigs. a. What is the size of the largest addressable memory? 1aa H Q O B CEOEMemory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [r0, #-4] is executed as a result of r1 = ?(data overlay big endian)? Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = 0X7F O B. R1 = Oxffffffd4 O C. R1 = Oxffffff7F O D. R1=0XD4000000 O E. R1 = 0XD4Student 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.
- Implementation of a Disk-based Buffer Pool A buffer pool is an area of main memory that has been allocated for the purpose of caching datablocks as they are read from disk. The main purpose of a buffer pool is to minimize disk I/O.Tasks: Implement a disk-based buffer pool application based on the below three buffer poolreplacement schemes: FIFO (first in, first out), LRU (least recently used), and LFU (Leastfrequently used) buffer pool replacement strategies.FIFO (First In, First Out): the oldest data block in the buffer pool is replaced.LRU (Least Recently Used): the least recently accessed data block in the buffer pool is replaced.LFU (Least Frequently Used): the least frequently accessed data block in the buffer pool isreplaced.Initially, the buffer pool is free. Disk blocks are numbered consecutively from the beginning ofthe file with the first block numbered as 0. When I execute your program,First, it asks the user to input the following parameters:o ?: number of blocks in the…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 ?Student information is stored in a data area, and each student entry is formatted as follows:The first nine bytes are the student ID in ASCII format.The following byte is the course score.The following phrase is the section identifier. There are over 300 such student records in memory, which have been loaded consecutively beginning at address $10000. To indicate the conclusion of the records, the last record loaded is a dummy record with a section identifier of $FFFF. Create an assembler subroutine GetMax that searches the complete list for the highest mark. If there is a tie, the student with the highest score on the list should be found. The beginning address of the subroutine should be returned.
- Assignment/Lab 9 In this lab, your task is to develop a simple address book tool. The tool maintains contact information of people. For each person, the tool maintains name and date of birth. Contact information consists of physical address and phone number. For physical addresses, the system must be able to maintain current and previous addresses. Current/active address information should have starting date and no end date, previous addresses must have a starting date and an ending date. A person cannot have multiple active physical address. For phone number, the system must maintain active phone numbers only, no previous phone numbers. For implementation, use MySQL database system. MySQL is open source and is available for Linux as well as Windows. For software application, use Python. The tool/application must support the following functionality: 1- Search current contact information by last name, the user enters last name, the system must locate active physical address and phone…Given the 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 memories and an example of Block A of size 200. Logical Memory Physical Memory location/ page location/frame 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/ 6 700 to 799/ 7 700 to 799/ 7 800 to 899/ 8 800 to 899/ 8 900 to 999/ 9 900 to 999/ 9 1000 to 1099/ 10 1100 to 1199/ 11 1200 to 1299/ 12 1300 to 1399/ 13 Block A 1400 to 1499/ 14 Block…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?Program binaries in many systems are typically structured as follows. Code is stored starting with a small fixed virtual address such as 0. The code segment is followed by the data segment that is used for storing the program variables. When the program starts executing, the stack is allocated at the other end of the virtual address space and is allowed to grow towards lower virtual addresses. What is the significance of the above structure on the following schemes: a. contiguous-memory allocation b. pure segmentation c. pure pagingCreate 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.