Question

Consider a logical address space of 4 pages of 1024 words each, mapped onto a physical memory of 16 frames. a. How many bits are there in the logical address? b. How many bits are there in the physical address? te: Show the calculation.)

Operating systems

Answer 1

To represent 4 page we need 2 bits (4=2^2) and each page contains 1024 words therefore offset d can be represented by 10 bit (1024=2^10) .

so total number of bits in logical address= 10+2=12 bits.

similarly number of frames=16

so number of bits required =4 (2^4=16)

and number of offset bits remain same=10.

So number of bits in physical address=10+4= 14 bits.

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