Answer:
2500 kb
Explanation:
Here, we are to calculate the bandwidth delay product
From the question, we are given that
band width = 500 Mbps
The bandwidth-delay product is = 500 x 10^6 x 25 x 10^-3
= 2500 Kbits
Answer:
Option d) B is 1.33 times faster than A
Given:
Clock time, 
No. of cycles per instructions, 
Solution:
Let I be the no. of instructions for the program.
CPU clock cycle, 
= 2.0 I
CPU clock cycle, 
= 1.0 I
Now,
CPU time for each can be calculated as:
CPU time, T = 
Thus B is faster than A
Now,
Performance of B is 1.33 times that of A
Answer: It allows you to locate materials, be aware of your assignments and plan time to get things done.
Hope it helped.
Answer:
b. lw $t4, 4($t0)
c. add $t3, $t5, $t4
Explanation:
Pipeline hazard prevents other instruction from execution while one instruction is already in process. There is pipeline bubbles through which there is break in the structural hazard which preclude data. It helps to stop fetching any new instruction during clock cycle.
