Memory Management Assignment

Added on - 28 May 2020

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Running head: MEMORY MANAGEMENTMemory ManagementName of Student-Name of University-Author’s Note-
MEMORY MANAGEMENT1BQ1 Memory Management Paginga.The starting address of frame 1 is 1025 and the end address of frame 1 is 2048.b.1.Yes2.No, because Frame 3 is kept free.c.Page 3, 6, and 7 are not loaded in memory.d.The paging system works by bringing the page from the secondary memory to main memory.CPU generates logical address for the secondary memory. These address are changed to physicaladdress for finding the data in the main memory [4]. When process A reference an address inpage 3, the page is mapped to a frame which has the data that the CPU wants. Then, the data isbought to the CPU for its execution. In this example, as page 3 is not mapped with the frame, sothere cannot be any data in page 3. So, the paging will get a miss after searching the address inpage 3.1.Logical Address- 1023Memory Size of page and frame table- 1024 bytes1024 bytes of logical address with 8 pages. Then, each page is of size 1024/8 = 128 bytesOffset of logical address 1023 % 1024 = 10231023/1024 = 0So, it can be assumed that 1023 is in page 0Page 0 is mapped to frame number 6So, the physical address is calculated asFrame number x Page size + offset6 * 1024 + 1023 = 7167So, the physical address in binary is 00011011111111112.Logical Address = 3000Memory Size of page and frame table= 1024 bytes1024 bytes of logical address with 8 pages. Then, each page is of size 1024/8 = 128 bytesOffset = 3000 % 1024 = 9523000/1024 = 2.929So, it can be assumed that 1023 is in page 2
MEMORY MANAGEMENT2Page 2 is mapped to frame number 2So, the physical address is calculated asFrame number x Page size + offset2 * 1024 + 952 = 3000So, the physical address in binary is 00001011101110003.Logical Address- 4120Memory Size of page and frame table- 1024 bytes1024 bytes of logical address with 8 pages. Then, each page is of size 1024/8 = 128 bytesOffset of logical address 4120 % 1024 = 241023/1024 = 4.023So, it can be assumed that 1023 is in page 4Page 4 is mapped to frame number 1So, the physical address is calculated asFrame number x Page size + offset1 * 1024 + 24 = 1048So, the physical address in binary is 00000100000110004.Logical Address- 5000Memory Size of page and frame table- 1024 bytes1024 bytes of logical address with 8 pages. Then, each page is of size 1024/8 = 128 bytesOffset of logical address = 5000 % 1024 = 9045000/1024 = 4.882So, it can be assumed that 1023 is in page 4Page 4 is mapped to frame number 1So, the physical address is calculated asFrame number x Page size + offset1 * 1024 + 904 = 1028So, the physical address in binary is 0000011110001000BQ2 Fragmentation and Memory MappingCompaction or relocation can be performed when there is internal fragmentation in the system[3]. Compaction is done to rearrange all the data together taking all the empty space together. This
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