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4 votes
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If a microcomputer operates at $5$ MHz with an $8$-bit bus and a newer version operates at $20$ MHz with a $32$-bit bus, the maximum speed-up possible approximately will be

  1. $2$
  2. $4$
  3. $8$
  4. $16$
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6 Answers

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7 votes
Best answer
It should be max(20/5, 32/8) = 4. Increasing the bandwidth aids in achieving the maximum speed up. Suppose we had a memory intensive program then increase in bandwidth becomes critical whereas for a CPU intensive process, increase in clock speed becomes critical.
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3 Comments

Thanks
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what does this "critical" mean?

even at these critical situations that u mention , speed up is calculated the same way ?

if not then how is it calculated for both CPU and MEmory intensive process

would be grateful if explained with a small example for both the above cases
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Isn't "bandwidth = frequency x #  of bits transferred in one cycle" here? Why are we taking bandwidth as just data bus size? That means we are calculating Bandwidth as # of bits that can be transferred per clock cycle time instead of # of bits that can be transferred per unit time. And more over here clock cycle time for both the cases are different.

However, I understood the concept of taking max of two parameters.
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6 votes
6 votes

Answer is B).

The second version is the best as compared to first, but the second version has both high bandwidth as well as increased CPU speed. We will not multiply as newer is already superior than first.

so 20/5 = 4 and 32/8 = 4

Hence we will take the maximum of these speedups which is 4.

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2 Comments

why not multiply?
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@kapil bro plz tell me what do u mean here we We "will not multiply" means???  and tell me in which case we will multiply tell me that case also??
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3 votes
3 votes

Old version: 5Mhz processor and 8-bit bus.

New version: 20Mhz processor and 32-bit bus.

Processor speedup: $\frac{20}{5}=4$

Memory access speedup: $\frac{32}{8}=4$

 

Both of them would increase the speed by a factor of 4, but a process either uses CPU or uses the bus (for I/O) at a given time — not both simultaneously. Hence, speedup would be 4, and not 16.

Option B


Analogy: If glasses help you see up to extra 40 feet clearly, using glasses and a hearing-aid would still help you see up to extra 40 feet clearly. :P

 

1 vote
1 vote
Data sent in one cycle by older version (x) =  5*10^6* 8 bit

Datasentinonecyclebynewerversion (y) = 20*10^6* 32 bit

Speed up = y/x = 16

3 Comments

It is a usual mistake :) See which unit we are increasing the speed.. We are increasing the speed of CPU and also increasing the bus width. Both will improve the performance, but won't multiply the performance.
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Then what would be answer sir ??
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it should be max(20/5, 32/8) = 4. Increasing the bandwidth aids in achieving the maximum speed up. Suppose we had a memory intensive program then increase in bandwidth becomes critical whereas for a CPU intensive process, increase in clock speed becomes critical.
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Answer:

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