Consider a system with byte-addressable memory, 32-bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is ________.

Given:

• Logical address space (LAS) size = $2^{32}\text{ bytes}$

• Page size = $4\text{ KB} = 4 \times 2^{10}\text{ bytes} = 2^{12}\text{ bytes}$

• Page Table Entry (PTE) size = $4\text{ bytes}$

1. The total number of pages in the logical address space is:

   $$\text{Number of pages} = \frac{\text{Logical Address Space}}{\text{Page Size}} = \frac{2^{32}}{2^{12}} = 2^{20}$$

2. The size of the single-level page table is given by:

   $$\text{Page Table Size} = \text{Number of pages} \times \text{PTE size} = 2^{20} \times 4\text{ bytes} = 4\text{ MB}$$

Therefore, the size of the page table in megabytes is 4.