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Operating System Assignment: Memory Management

Added on - 28 May 2020

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Running head: OPERATING SYSTEMOperating SystemName of Student-Name of University-Author’s Note-
OPERATING SYSTEM1BQ1 Memory Management Paginga.The address at the starting of frame 1 is 1025 and the address at the end of frame 1 is 2048.b.1.Yes, page 2 is mapped to frame 22.No, page 3 is nor mapped to frame 3. Frame 3 is kept free.c.Page 3, 6, and 7 are not yet loaded in the memory.d.When Process A needs and address in page 3, then the paging technique finds the mapping ofpage 3 in the frame table. Until the paging process does not respond, the process A has to waitin the waiting queue for its execution [1]. When the frame is found that is mapped to the page,then the data is bought to the CPU for its execution. The CPU generates logical address to searcha particular data. In this example, page 3 is not mapped in the frame table. So, the CPU will notget any data in the main memory.e.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 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
OPERATING SYSTEM2Page 2 is mapped to frame number 2So, the physical address is calculated asFrame number x Page size + offset2 * 1024 + 952 = 3000So, the physical address 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 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 is 0000011110001000BQ2 Fragmentation and Memory MappingRelocation or compaction of memory can be performed before running of any of the programsin the system [2]. The assemblers and the compilers are generated typically which are executed with
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