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Official PYQ · 1990Q7-bMemory Management and Virtual MemoryNAT1 mark

GATE CS 1990 Q7-b — Memory Management and Virtual Memory

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In a two-level virtual memory, the memory access time for main memory, tM=10−8t_{M}=10^{-8} sec, and the memory access time for the secondary memory, tD=10−3t_D=10^{-3} sec. What must be the hit ratio, HH such that the access efficiency is within 8080 percent of its maximum value?
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Answer: 99.99

The average access time (TavgT_{\text{avg}}) for a two-level memory system is given by:

Tavg=tM+(1−H)tDT_{\text{avg}} = t_M + (1 - H)t_D

where:

  • tM=10−8 sect_M = 10^{-8}\text{ sec} (Main memory access time)
  • tD=10−3 sect_D = 10^{-3}\text{ sec} (Secondary memory access time)
  • HH is the hit ratio

The access efficiency (η\eta) is defined as the ratio of the main memory access time to the average access time:

η=tMTavg=tMtM+(1−H)tD\eta = \frac{t_M}{T_{\text{avg}}} = \frac{t_M}{t_M + (1 - H)t_D}

For the efficiency to be within 80%80\% (or 0.80.8) of its maximum value (which is 11):

tMtM+(1−H)tD≥0.8\frac{t_M}{t_M + (1 - H)t_D} \ge 0.8

Substitute the given values into the inequality:

10−810−8+(1−H)10−3≥0.8\frac{10^{-8}}{10^{-8} + (1 - H)10^{-3}} \ge 0.8 10−8≥0.8×10−8+0.8×(1−H)10−310^{-8} \ge 0.8 \times 10^{-8} + 0.8 \times (1 - H)10^{-3} 0.2×10−8≥0.8×(1−H)10−30.2 \times 10^{-8} \ge 0.8 \times (1 - H)10^{-3} 1−H≤0.2×10−80.8×10−31 - H \le \frac{0.2 \times 10^{-8}}{0.8 \times 10^{-3}} 1−H≤0.25×10−5=2.5×10−61 - H \le 0.25 \times 10^{-5} = 2.5 \times 10^{-6} H≥1−0.0000025=0.9999975H \ge 1 - 0.0000025 = 0.9999975

Expressing HH as a percentage:

H≥99.99975%H \ge 99.99975\%

Thus, the hit ratio must be at least 99.99%99.99\% (or more precisely, 99.99975%99.99975\%).

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