On a system, a process has 1KB (1024 bytes) of address space, where each byte is
ID: 3709022 • Letter: O
Question
On a system, a process has 1KB (1024 bytes) of address space, where each byte is a separate address. In a pure paging scheme, the process has been allocated 64 pages to hold its address space. The system has 32KB of physical memory, where each byte is a separate address as before, and a frame size is equal to the page size of processes. Coincidentally, the current process’s page table maps every page number i to frame number 3i.
During execution of the process, CPU generates the following logical address: 1 1 0 1 0 1 1 0 0 1
What is the corresponding physical address? (Show how get the answer)
Explanation / Answer
Given address space of the process is = 1 KB (1024 bytes).
Hence total number of bits for the address = log2 1024 = log2 210 = 10 bits.
Number of pages for the process is = 64.
Hence number of bits for the page number is = log2 64 = log2 26 = 6 bits.
Number of bits for offset is = 10 - 6 = 4 bits.
Here in the given logical address 1st 6 bits 1 1 0 1 0 1 is the page number,
which is 53 in decimal.
Next 4 bits 1 0 0 1 is the offset.
Given every page number i maps to frame number 3i.
Hence frame number is = 3 * 53 = 159,
which is 1 0 0 1 1 1 1 1 in binary and offset is = 1 0 0 1.
Here physical memory is = 32 KB = 25 * 210 B = 215 B and
Hence physical memory address size is = 15 bits.
Hence corresponding physical address is = 0 0 1 0 0 1 1 1 1 1 1 0 0 1.