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Asked: 2026-06-17T00:13:08+00:00

I’m looking over some exam papers for Operating Systems and I have come across

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I’m looking over some exam papers for Operating Systems and I have come across a question which I simply cannot figure out. The memory management scheme is paging
Here is the question:

An operating system that runs on a CPU with a 16 bit address pointer employs a paging memory management scheme with a page size of 1024 bytes.
a)  At most how many frames can physical memory contain?
b)  Indicate which bits in an address pointer are used for page and offset. 
c)  A process image of size 3.5K resides in memory. You are given the page table of this process in Figure 1 below. What physical address will the hexadecimal logical address 0x0FCE result in? Show your calculations.
d)  How much internal fragmentation does this process produce?

Page    Frame
0             4
1             8
2             9
3             6
Figure 1 Process Page Table

Can anybody help me with this ?

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  1. Editorial Team

    1. A 16bit address bus allows to access 2^16 = 64kB of physical memory. Since on this system you have pages of size 1024B = 2^10, your memory falls into 2^16 / 2^10 = 2^6 physical frames.

    2. Given the previous result, with pages of 1024 = 2^10 bytes, you need 10 bits for accessing any bytes of the page. Thus, the 6 high-order bits ares used for getting the page index from the page table (basically the figure 1 in your homework), and the 10 low-order bits are used for offsetting in that page.

    3. The logical address 0xfce resides in the fourth page because the six high-order bits are 000011b = 3 = 4th page = 0x0c00-0x0fff. Given the page table and assuming the physical memory is sequential, the fourth page maps to the sixth physical frame which start at 1024 * 6 = 0x1800 = 0001100000000000b. The six high-order bits of the page are 000110b, where we add the 10 bits of offset resulting from the previous answer: 000110b << 10 | 0x3ce = 0x1bce.

    4. Since the frame allocation is not sequential (4, 6, 8, 9), the hole between pages 4 and 6 (i.e. 1024B), and the hole between page 6 and 8(i.e. again 1024B), result in physical memory fragmentation.

    Hope this help.

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