With paging there is no ________ fragmentation.
Answer: B
Correct answer: external
Paging assigns any free frame to any page, so free memory never needs to be contiguous and external fragmentation cannot arise. Internal fragmentation remains, in the last page of each process.
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The operating system maintains a ______ table that keeps track of how many frames have been allocated, how many are there, and how many are available.
Answer: C
Correct answer: frame
The frame table records the state of every physical frame - whether it is free or allocated and, if allocated, to which process and page.
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Paging increases the ______ time.
Answer: C
Correct answer: context – switch
Each process has its own page table, so switching processes means reloading the page table registers or the base register, and flushing the TLB. That work is added to every context switch.
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Smaller page tables are implemented as a set of _______
Answer: D
Correct answer: registers
A small page table can be held entirely in a set of dedicated registers, which makes translation as fast as possible. Only when the table grows too large for that must it be moved into main memory.
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The page table registers should be built with _______
Answer: B
Correct answer: very high speed logic
The page table is consulted on every memory reference, so any delay in it is added to every instruction. That is why those registers are built with the fastest logic available.
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For larger page tables, they are kept in main memory and a __________ points to the page table.
Answer: A
Correct answer: page table base register
When the table is too large for registers it is kept in memory and the page table base register holds its address, so a context switch need only reload that one register.
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For every process there is a __________
Answer: A
Correct answer: page table
Each process has its own page table, which is what gives it a private address space: the same page number in two processes maps to different frames.
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Time taken in memory access through PTBR is :
Answer: D
Correct answer: slowed by a factor of 2
With the table in memory, each reference needs two accesses: one to read the page table entry and one to reach the datum. The effective access time is therefore doubled, which is precisely why the TLB exists.
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Each entry in a Translation look-aside buffer (TLB) consists of :
Answer: A
Correct answer: key
A TLB is associative memory: each entry is searched by its key, the page number, and returns the value stored with it, the frame number. The search is on the key, which is what the question is asking about.
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If a page number is not found in the TLB, then it is known as a :
Answer: A
Correct answer: TLB miss
A page number not found in the TLB is a TLB miss, and the page table in memory must then be consulted; the entry is usually added to the TLB afterwards.
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