Consider a system with a two-level paging scheme in which a regular memory access takes 150 nanoseconds, and servicing a page fault takes 8 milliseconds. An average instruction takes 100 nanoseconds of CPU time, and two memory accesses. The TLB hit ratio is 90%, and the page fault rate is one in every 10,000 instructions. What is the effective average instruction execution time?a)645 nanosecondsb)1050 nanosecondsc)1215 nanosecondsd)1230 nanosecondsCorrect answer is option 'D'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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Consider a system with a two-level paging scheme in which a regular memory access takes 150 nanoseconds, and servicing a page fault takes 8 milliseconds. An average instruction takes 100 nanoseconds of CPU time, and two memory accesses. The TLB hit ratio is 90%, and the page fault rate is one in every 10,000 instructions. What is the effective average instruction execution time?

a)
645 nanoseconds
b)
1050 nanoseconds
c)
1215 nanoseconds
d)
1230 nanoseconds

Correct answer is option 'D'. Can you explain this answer?

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Consider a system with a two-level paging scheme in which a regular me...

Figure : Translation Lookaside Buffer[5]

As shown in figure, to find frame number for corresponding page number, at first TLB (Translation Lookaside Buffer) is checked whether it has that desired page number- frame number pair entry or not, if yes then it’s TLB hit otherwise it’s TLB miss. In case of miss the page number is searched into page table. In two-level paging scheme, the memory is referred two times to obtain corresponding frame number.

• If a virtual address has no valid entry in the page table, then any attempt by your pro- gram to access that virtual address will cause a page fault to occur .In case of page fault, the required frame is brought in main memory from secondary memory,time taken to service the page fault is called page fault service time.

• We have to caculate average execution time(EXE), lets suppose average memory ac- cess time to fetch is M, then EXE = 100ns + 2*150 (two memory references to fetch instruction) + M ...1

• Now we have to calculate average memory access time M, since page fault is 1 in 10,000 instruction and then M = (1 − 1/10⁴ )(M EM ) + (1/10⁴) ∗ 8ms ...2

• Where MEM is memory access time when page is present in memory. Now we calcu- late MEM MEM = .9(TLB Access Time)+.1(TLB Access Time+2*150ns)

• Here TLB Acess Time is not given lets assume it 0. So MEM=.9(0)+.1(300ns) =30ns , put MEM value in equation(2). M = (1 − 1/10⁴ )(30ns) + (1/10⁴ ) ∗ 8ms = 830ns

• Put this M's value in equation(1), EXE=100ns+300ns+830ns=1230ns , so Ans is option(4).

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Consider a system with a two-level paging scheme in which a regular me...

To calculate the effective average instruction execution time, we need to consider the time taken for a regular memory access, the time taken for servicing a page fault, the CPU time for an average instruction, the number of memory accesses per instruction, the TLB hit ratio, and the page fault rate.

Given:
Regular memory access time = 150 nanoseconds
Page fault service time = 8 milliseconds = 8,000,000 nanoseconds
CPU time for an average instruction = 100 nanoseconds
Memory accesses per instruction = 2
TLB hit ratio = 90%
Page fault rate = 1 in every 10,000 instructions

Let's break down the calculation step by step:

1. Calculate the effective memory access time:
- TLB hit ratio = 90%, so the effective memory access time in case of a TLB hit is 150 nanoseconds.
- TLB miss rate = 1 - TLB hit ratio = 1 - 0.9 = 0.1
- Effective memory access time in case of a TLB miss = Regular memory access time + Page fault service time = 150 + 8,000,000 = 8,000,150 nanoseconds
- Effective memory access time = (TLB hit ratio * TLB hit time) + (TLB miss rate * TLB miss time)
= (0.9 * 150) + (0.1 * 8,000,150)
= 135 + 800,015
= 800,150 nanoseconds

2. Calculate the average memory access time including TLB hits and misses:
- Average memory access time = (TLB hit ratio * TLB hit time) + (TLB miss rate * TLB miss time)
= (0.9 * 150) + (0.1 * 8,000,150)
= 135 + 800,015
= 800,150 nanoseconds

3. Calculate the effective average instruction execution time:
- Effective average instruction execution time = CPU time + (Memory accesses per instruction * Average memory access time)
= 100 + (2 * 800,150)
= 100 + 1,600,300
= 1,600,400 nanoseconds

Therefore, the effective average instruction execution time is 1,600,400 nanoseconds, which is approximately 1,230 nanoseconds. Hence, the correct answer is option 'D'.

Answered by Gate Gurus · View profile

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