Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of slow start phase is 2 KB and the threshold at the start of first transmission is 24 KB. Assume that 3 duplicate ACK are received during the 5th transmission what is the congestion window size at the end of 10th transmission?
The answer given is 16KB which is correct but getting doubt on how they solve the question.
I think the steps should be:-
1MSS = 2KB.
(in MSS) = 1, 2, 4, 8, 12, 1, 2, 4, 6,7
So after 10th transmission it will be 8 which is 16KB.
And they solved (in KB) = 2,4,8,16, 24(TO then Threshold = 24/2 = 12) 12,13,14,15,16.
This is how they did.
Doubt :- Here it is mentioned that "3 duplicate ack". Now, if we go along TCP Tahoe then we follow what is given in above. But what steps to follow when TCP Reno is asked?
An instruction pipeline consists of following 5 stages:
IF = Instruction Fetch, ID = Instruction Decode, EX = Execute,
MA = Memory Access and WB = Register Write Back
Now consider the following code:
Assume that each stage takes 1 clock cycle for all the instructions. The number of clock cycles are required to execute the code, without operand forwarding over a bypass network ________.
I am getting 12, by doing overlapping between WB and ID stage i.e. in one half of cycle doing WB and in another half doing ID.
Given answer is 14.