Question

Consider a paging hardware with a TLB. Assume that the entire page table and all the pages are in the physical memory. It takes 10 milliseconds to search the TLB and 80 milliseconds to access the physical memory. If the TLB hit ratio is 0.6, the effective memory access time (in milliseconds) is _____.

Official answer and solution
Official answer
122
From
Official GATE 2014 answer key

Solution

Use the standard effective memory access time formula incorporating TLB hit and miss cases.

  1. Given Parameters:

    • TLB access time: tTLB=10t_{\text{TLB}} = 10 ms
    • Physical memory access time: tmem=80t_{\text{mem}} = 80 ms
    • TLB hit ratio: h=0.6h = 0.6
    • TLB miss ratio: 1−h=0.41 - h = 0.4
  2. Case Analysis:

    • TLB Hit: TLB lookup finds the frame number (1010 ms), then physical memory is accessed once for the target operand (8080 ms).
      Timehit=10+80=90 ms.\text{Time}_{\text{hit}} = 10 + 80 = 90\text{ ms}.
    • TLB Miss: TLB lookup misses (1010 ms), physical memory is accessed for the page table entry (8080 ms), and then physical memory is accessed for the target operand (8080 ms).
      Timemiss=10+80+80=170 ms.\text{Time}_{\text{miss}} = 10 + 80 + 80 = 170\text{ ms}.
  3. Effective Memory Access Time (EMAT):

    EMAT=h×Timehit+(1−h)×Timemiss\text{EMAT} = h \times \text{Time}_{\text{hit}} + (1 - h) \times \text{Time}_{\text{miss}}
    EMAT=0.6×90+0.4×170=54+68=122 ms.\text{EMAT} = 0.6 \times 90 + 0.4 \times 170 = 54 + 68 = 122\text{ ms}.

Alternatively:

EMAT=tTLB+tmem+(1−h)tmem=10+80+0.4×80=90+32=122 ms.\text{EMAT} = t_{\text{TLB}} + t_{\text{mem}} + (1 - h)t_{\text{mem}} = 10 + 80 + 0.4 \times 80 = 90 + 32 = 122\text{ ms}.
Final answer

122

Common trap

Omitting the TLB search time on a TLB miss.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2014 Computer Science and Information Technology (CS) Session 3
Question
43 of 65, printed as CS Q.33
Section
Computer Science and Information Technology

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