Question: Given a system using 8 KB pages, a 48-bit virtual address, and 8 GB of physical memory, determine the amount of space required to store

 Given a system using 8 KB pages, a 48-bit virtual address,

Given a system using 8 KB pages, a 48-bit virtual address, and 8 GB of physical memory, determine the amount of space required to store each of the following types of page tables. Assume each page table entry (PTE) requires 4 bytes unless the problem states otherwise a. A basic, one-level page table that covers the entire virtual address space b. A two-level page table, assuming the currently running process is using the entire first half of 9 its virtual address space. NOTE: Typically, in multi-level paging, the upper address bits are split evenly to index into each level of the page table. For example, given a 32-bit address with a 10 bit page offset, in a two-level page table, the remaining 22 address bits are split so 11 bits are used for the first-level table and 11 bits are used for the second-level tables. However, if the number of remaining bits are odd, assume the extra bit is used to index the first-level table. So, in the example above, if the page offset were 11 bits, not 10, the remaining 21 address bits would be split such that 11 bits were used for the first- level table and 10 bits were used for the second-level tables

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