The topic is Memory Management in Operating Systems. Please answer but most importantly explain as the concepts are not clear to me

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- If page size =2 bytes - What is the size of the process? Answer: - What is the physical memory size? Answer: - How many bits are required for logical addresses? 4 bytes =22 - How many bits are required for main memory address? 8 bytes =23 Calculating internal fragmentation: - Page size =2,048 bytes - Process size =72,766 bytes - 35 pages (process size / Page size) + 1,086 bytes remaining so we need 36 pages. - Internal fragmentation = page size 2,048 - 1,086 = 962 bytes. - Worst case fragmentation =1 frame is 1 byte (page size +1 byte) extra frame with 2047 bytes free space (internal fragmentation) - On average fragmentation =1/2 frame size (1/21MB or 1/22KB) - So small frame sizes desirable? The problem is the page table will be very big and will use more space in memory. - Each page table entry takes memory to track. - Page sizes growing over time from few MB to several GB) - Solaris supports two-page sizes - 8 KB \&4 MB - Process view \& physical memory now very different. - By implementation process can only access its own memory