1) Consider a paging system with the page table stored in memory. a. If a memory
ID: 3702367 • Letter: 1
Question
1) Consider a paging system with the page table stored in memory.
a. If a memory reference takes 200 nanoseconds, how long does a
paged memory reference take?
b. Add TLBs, and assume 75 percent of all page-table references are found in the TLBs, what is the effective memory reference time? (Assume that ?nding a page-table entry in the TLBs takes zero time if the entry is there).
2) Consider the following page reference string: 1, 2, 3, 4, 5, 2, 4, 6, 3, 5, 4, 1, 4, 1, 6, 2, 5, 4, 2, 3. How many page faults would occur if the FIFO replacement algorithm is used? All frames are initially empty.
A) Assume one frame available
B) Assume 2 frames available
C) Assume 4 frames available
D) Assume 8 frames available
3) Consider the following page reference string: 1, 2, 3, 4, 5, 2, 4, 6, 3, 5, 4, 1, 4, 1, 6, 2, 5, 4, 2, 3. How many page faults would occur if the LRU replacement algorithm is used? All frames are initially empty.
A) Assume one frame available
B) Assume 2 frames available
C) Assume 4 frames available
D) Assume 8 frames available
4) Consider the following page reference string: 1, 2, 3, 4, 5, 2, 4, 6, 3, 5, 4, 1, 4, 1, 6, 2, 5, 4, 2, 3. How many page faults would occur if the FIFO replacement algorithm is used? All frames are initially empty.
A) Assume one frame available
B) Assume 2 frames available
C) Assume 4 frames available
D) Assume 8 frames available
5) Consider a system with 64 free frames (1 KB each) and two processes running, one of 15 KB, another of 105 KB. How many more frames will be allocated to the second process if proportional allocation is used instead of equal?
6) Assume that we have a demand-paged memory. The page table is held in registers. It takes 6 ms to handle a page fault if an empty frame is available or if the replaced page is not modified and 15 ms if the replaced page is modified. Memory access time is 80 ns. Assume that 75% of pages are modified. What is the max acceptable page-fault rate for an effective access time to be below 200 ns?
Explanation / Answer
We can answer 4 parts at a time.
I am answering both parts of question 1 and 2 parts of question 2.
1.(a) Effective memory access time = Time to access page table in memory + time to access the page in memory
= 200 ns + 200 ns
= 400 ns
(b) After adding TLB,
Effective memory access= 0.75* TLB access time + 0.25*(memory access time) + memory access time
= 0 + 0.25*200 + 200
= 250 ns
2.
(a) Page faults occur every time whenever there is a change in page.
So no. of page fault = No. of frames - 1
= 20 - 1 = 19
(b) If two frames are available
Frames are filled in the following manner:
Page faults occur every time except first two times.
So no. of faults = 17 - 2 = 15
Post remaining questions separately.
Frames F1 1 1 3 3 5 5 6 6 5 5 1 1 2 2 4 4 3 F2 2 2 4 4 2 2 3 3 4 4 6 6 5 5 2 2Related Questions
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