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3. Pertimbangkan jika skim pembahagian ingatan secara dinamik digunakan, dan berikut adalah gambar konfigurasi memori semasa: Consider a dynamic partitioning scheme is being used, and

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3. Pertimbangkan jika skim pembahagian ingatan secara dinamik digunakan, dan berikut adalah gambar konfigurasi memori semasa: Consider a dynamic partitioning scheme is being used, and the following is snapshot of the memory configuration: 20 10 20 40M 40M 10 10 20M 60M 40M 20M 30M 40M 40M M M MM M a) Tunjukkan permintaan penempatan ingatan untuk 20M, 10M dan 40M dengan menggunakan dua(2) algoritma yang sedia ada. Nyatakan nama algoritma-algoritma tersebut. Show the memory placement request of 20M, 40M and 10M using the two(2) available algorithms. State the name of the algorithms. (3 M) b) Andaikan saiz satu ingatan adalah 256 MB. Dengan menggunakan skema perwakilan Buddy Tree System. Gambarkan kemasukan ruangan ingatan bagi permintaan ruangan ingatan seperti berikut: Assume the size of one memory is 256 MB. Using the Buddy Tree System representation scheme. Describe the memory space entry for the memory space request as follows: A-40MB, B=50MB, C=60MB, D=30MB. (4M) c) Pertimbangkan sistem penghalaman mudah yang menggunakan skema pengalamatan 16 bit. Saiz halaman ialah 1K setiap satu. Diberi jadual halaman proses seperti berikut: Consider a simple paging system which uses 16 bits addressing scheme. The page size is IK each. Given a process page table as follows: 0 000101 1 000111 2 001001 3 001011 4 001100 Tentukan alamat fizikal untuk alamat logik berikut 0001000001111110. Determine the physical address of the following logical address 0001000001111110. (1.5M) d) Pertimbangkan sistem segmen mudah yang menggunakan skema pengalamatan 16 bit. Saiz segmen ialah 1K setiap satu. Diberi jadual segmen proses seperti berikut: Consider a simple segment system which uses 16 bits addressing scheme. The segment size is IK each. Given a process segment table as follows: Segment Length Base 0 11110 11111 0000000010100000 1 11100 11111 0000000001111000 2 11111 000000000000010011010 3 10000 000000000000010010000 4 11111 11111 0000000010101000 3 Tentukan alamat fizikal untuk alamat logik berikut, 0001001110001111. Determine the physical address of the following logical address, 0001001110001111. (1.5M)

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