Answer:
I think it would be unfulfilled
Answer:
Check the explanation
Explanation:
CPI means Clock cycle per Instruction
given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M
Execution time is given by following Formula.
Execution Time(CPU time) = CPI*Instruction Count * clock time = 
a)
for system A CPU time is 1.3 * 100, 000 600 106
= 216.67 micro sec.
b)
for system B CPU time is 
= 333.33 micro sec
c) Since the system B is slower than system A, So the system A executes the given program in less time
Hence take CPU execution time of system B as CPU time of System A.
therefore
216.67 micro = =
Instructions = 216.67*750/2.5
= 65001
hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A
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Answer:
Cellular phone technology works on a system of geographically separated zones
called
B. Cells
Answer:
The answer is below
Explanation:
The amount of power dissipated by a processor is given by the formula:
P = fCV²
Where f = clock rate, C = capacitance and V = Voltage
For the old version of processor with a clock rate of f, capacitance C and voltage of V, the power dissipated is:
P(old) = fCV²
For the new version of processor with a clock rate of 20% increase = (100% + 20%)f = 1.2f, capacitance is the same = C and voltage of 20% increase = 1.2V, the power dissipated is:
P(new) = 1.2f × C × (1.2V)² = 1.2f × C × 1.44V² =1.728fCV² = 1.728 × Power dissipated by old processor
Hence, the new processor is 1.728 times (72.8% more) the power of the old processor