Hughes Placement: Sample Questions 8 - 10 of 19
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Question 8
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If the cache access time is 1 nano seconds and main memory access time is 2 nano seconds and hit ratio is 0.9, then calculate the average access time.
EditExplanation
Relation between average access time, hit ratio, cache access time and memory access time is,
Where, Tavg = average access time
h = hit ratio
1 - h = miss ratio
Tc = cache access time
Tm = memory access time
Substituting values in equation (i) ,
Average access time is 1.1 nano seconds.
Question 9
Explanation
8085 is a 8 bit processor, where data lines are 8 bits and address lines are 16 bits.
To address 64 kilo bytes of memory we need to use 16 address lines.
Same way, to address 32 kilo bytes of memory 15 address lines are used. Remaining 1 address line is used to enable the chip.
To address 16 kilo bytes of memory, 14 address lines are used.
To address 8 kilo bytes of memory, 13 address lines are used.
To address 4 kilo bytes of memory, 12 address lines are used.
Question 10
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What is the channel capacity if the bandwidth is 500 Hz and there are 8 signal levels?
EditExplanation
Nyquist criteria give the relation between channel capacity (C) , Bandwidth (B) and signal levels (M) , which is given as,
Substituting the values, B = 500 and M = 8 from the given information in the equation (i) ,
Channel capacity,
Channel capacity will be 3000 bits per second.