Answer:
0.0051825 ms
Explanation:
From the given information:
The clock time is = [tex]\dfrac{1}{clok \ rate}[/tex]
[tex]=\dfrac{1}{6} Ghz[/tex]
[tex]=\dfrac{1}{6}\times 10^9[/tex]
To milliseconds, we have;
= 0.000000625 ms
However, the execution time is the sum total of all instruction types multiplied by the ratio of cycles taken per instruction and cycle time.
Mathematically;
Execution time is: [tex]\Bigg[Sum \ of \ all \Big(instruction \ type \times \dfrac{cycles taken }{instruction} \Big) \Bigg] \times cycle \ time[/tex]
Recall that:
the cycles taken according to the ALU = 400 × 7 =2800
cycles required according to the memory instructions = 300 × 12 = 3600
cycles for branching = 90×21 = 1890
THUS;
Total cycles = 2800 + 3600 + 1890
= 8290 cycles
Finally, execution time = 8290 cycles × 0.000000625 ms
= 0.0051825 ms
