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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.

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

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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.**

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