Final answer:
The execution time for computers with clock rates of 1 GHz, 2 GHz, and 2.5 GHz for a program requiring 500,000 clock cycles is calculated using the formula for execution time in computer architecture. As the clock rate increases, the execution time decreases, resulting in faster program execution. However, there are trade-offs between clock rate and performance and increasing clock rates may not always lead to proportional performance improvements.
Step-by-step explanation:
To calculate the execution time of a program, we can use the formula:
Execution time (s) = Total clock cycles / Clock rate (Hz)
For a program requiring 500,000 clock cycles, and computers with clock rates of 1 GHz, 2 GHz, and 2.5 GHz:
A. Applying the formula:
For a computer with a clock rate of 1 GHz:
Execution time = 500,000 / 1,000,000,000 (1 billion)
Execution time = 0.0005 seconds or 0.5 milliseconds
Similarly, for a computer with a clock rate of 2 GHz:
Execution time = 500,000 / 2,000,000,000 (2 billion)
Execution time = 0.00025 seconds or 0.25 milliseconds
And for a computer with a clock rate of 2.5 GHz:
Execution time = 500,000 / 2,500,000,000 (2.5 billion)
Execution time = 0.0002 seconds or 0.2 milliseconds
B. Analyzing the impact of clock rate:
From the above calculations, we can see that as the clock rate increases, the execution time decreases. A higher clock rate means the computer can execute more clock cycles per second, resulting in faster program execution.
C. Discussing trade-offs between clock rate and performance:
While higher clock rates can lead to faster program execution, there are trade-offs to consider. Higher clock rates require more power and generate more heat, which can limit the computer's performance and increase the need for cooling. Additionally, increasing clock rates may not always result in proportional performance improvements, as other factors like memory access and instruction efficiency also affect overall performance.
D. Evaluating speedup achieved with higher clock rates:
To calculate the speedup achieved with higher clock rates, we can compare the execution times. For example, comparing a computer with a clock rate of 2 GHz to a computer with a clock rate of 1 GHz:
Speedup = Execution time of 1 GHz computer / Execution time of 2 GHz computer
Speedup = 0.0005 seconds / 0.00025 seconds
Speedup = 2
This means the 2 GHz computer is twice as fast as the 1 GHz computer.