Suppose that the logical address space of a process consist of32 pages and each
ID: 3612233 • Letter: S
Question
Suppose that the logical address space of a process consist of32 pages and each page consist of 2 bytes. The main memory consistsof 32 frames and the size of each frame is also 2 bytes. Calculatethe following from the given information:
a) Calculate the size of logicaladdress space of the process?
b) Calculate the physical addressspace of the main memory?
c) Calculate the no. of bitsrequired for the Page Number ‘p’?
d) Calculate the no. of bitsrequired for offset ‘d’?
e) Find the number of entries inthe page table?
Explanation / Answer
Dear User, a) The size of thelogical address space of the process = 32 x 2 bytes = 64 bytes =26 x 23 =29 m = 9 bits m = 9 bits b) The size of thephysical address space of the main memory = 32 x 2 bytes = 64 bytes c) The number of bits requiredfor the page number P = m- n bits = 9 - 4 =5 bits pagesize = 2 bytes = 21 x 23 = 24 So, n=4 bits d) The number ofbits required for offset (d) = n=4 bits e) The number ofentries in the page table = number of pages = 32 I hope this will helps toyou So, n=4 bits d) The number ofbits required for offset (d) = n=4 bits e) The number ofentries in the page table = number of pages = 32 I hope this will helps toyouRelated Questions
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