Consider a 32-bit system with its page size being 16KB. Among the 32 bits that are used for a virtual address (VA), except the bits used for the offsets in a page, all the other bits are used to represent a virtual page number (VPN). Assume that each page directory entry (PDE) or page table entry (PTE) has 4 bytes. (1) (2 pts) How many pages does an address space have? (2) (4 pts) Assume that a linear page table is used for an address space. a) How many bytes are needed to represent the page table? b) If only the lowest-numbered 1000 PTES (i.e., their VPNs are 0, 1, 2, ..., 999) and the highest-numbered 100 PTES are valid, then how many bytes are needed to store this page table? (3) Assume that we use a 2-level page table instead of a linear page table. Assume that we use the top half of the VPN bits (of a virtual address) to represent the number of entries (i.e., PDEs) in the page table directory (PTD), and the other half of the VPN bits to represent the size of a page-table page. a) (2 pts) If we assume that all of the possible PTES are valid, then how many page-table pages are needed, and how many possible PTES are there in each page-table page? b) (4 pts) If we assume that only the 1000 PTES with the lowest VPN numbers (i.e., their VPNs are 0, 1, 2, ..., 999) and the 100 PTES with the highest VPN numbers are valid, then how many bytes are needed to store this 2-level page table? Justify your answer.