Final answer:
The number of physical address bits needed can be calculated using the formula: Number of physical address bits = log2(Number of page frames). In this case, the computer has 4 M page frames, so there must be 22 bits in the physical memory address bus.
Step-by-step explanation:
The number of physical address bits needed in the physical memory address bus can be calculated using the formula:
Number of physical address bits = log2(Number of page frames)
In this case, there are 4 M (i.e., 4 million) page frames in physical memory, which can be represented as 2^22. So, the number of physical address bits required would be:
Number of physical address bits = log2(2^22) = 22 bits
Therefore, there must be 22 bits in the physical memory address bus for this computer.
The computer system requires 34 bits for the physical memory address bus, calculated by accounting for the 4 Mbyte page frames and the 4 Kbyte page size, and including 22 bits to address the 4 million frames and an additional 12 bits to address each byte within the pages.
To determine the number of physical address bits for the physical memory address bus in a computer system, we must consider the size of physical memory in terms of page frames and the page size. The computer has 4 Mbyte page frames, which means there are 4 million frames for pages to be stored. Since the page size is 4 Kbytes, we can calculate the number of pages that can be stored in one frame.
Firstly, we should understand the units of measurement for computer memory. A kilobyte (KB) is 1024 bytes, and a megabyte (MB) is 1024 KB. Thus, a 4 Kbyte page is 4 * 1024 bytes, and 4 Mbyte of page frames is 4 * 1024 * 1024 pages.
To find out the number of bits needed to address each page frame, we need to convert 4 million frames into binary. This will give us the number of bits required to differentiate each frame. As 222 is 4194304, which is just over 4 million, 22 bits are required to address each of the 4 million frames. However, as the system is byte-addressed and each page has 4096 (4 Kbytes) addressable units, we need to add 12 bits to address each byte within a page (since 212 = 4096).
Therefore, we need a total of 22 (to address the frame) + 12 (to address the byte within each page) = 34 bits for the physical memory address bus in this computer system.