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Contact [7]
1 year ago
12

10. (P37) Compare GBN, SR, and TCP (no delayed ACK). Assume that the timeout values for all three protocols are sufficiently lon

g such that 5 consecutive data segments and their corresponding ACKs can be received (if not lost in the channel) by the receiving host (Host B) and the sending host (Host A) respectively. Suppose Host A sends 5 data segments to Host B, and the 2nd segment (sent from A) is lost. In the end, all 5 data segments have been correctly received by Host B. a. How many segments has Host A sent in total and how many ACKs has Host B sent in total? What are their sequence numbers? Answer this question for all three protocols. b. If the timeout values for all three protocol are much longer than 5 RTT, then which protocol successfully delivers all five data segments in shortest time interval?
Computers and Technology
1 answer:
MissTica1 year ago
8 0

Answer:

see explaination

Explanation:

For GBN:

A sends will 9 segments in total. These are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2, 3, 4, and 5.

B will send 8 ACKs. They are 4 ACKS with sequence number 1, and 4 ACKS withsequence numbers 2, 3, 4, and 5.

For SR:

A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2. B sends 5 ACKs. They are 4 ACKS with sequence number 1, 3, 4, 5. And there is oneACK with sequence number 2.

For TCP:

A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re-sent segments 2. B sends 5 ACKs. They are 4 ACKS with sequence number 2. There is one ACK withsequence numbers 6.

Let it be observed that TCP always send an ACK with expected sequencenumber.

b). TCP. This is because TCP uses fast retransmit without waiting until time out.

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Answer:

Following are the method definition to this question:

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2 years ago
A contact list is a place where you can store a specific contact with other associated information such as a phone number, email
NikAS [45]

Answer:

C++.

Explanation:

<em>Code snippet.</em>

#include <map>

#include <iterator>

cin<<N;

cout<<endl;

/////////////////////////////////////////////////

map<string, string> contacts;

string name, number;

for (int i = 0; i < N; i++) {

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   cout<<endl;

   contacts.insert(pair<string, string> (name, number));

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/////////////////////////////////////////////////////////////////////

map<string, string>::iterator it = contacts.begin();

while (it != contacts.end())  {

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/////////////////////////////////////////////////////////////////////////////////////////////////////////

I have used a C++ data structure or collection called Maps for the solution to the question.

Maps is part of STL in C++. It stores key value pairs as an element. And is perfect for the task at hand.

8 0
2 years ago
You're the sole IT employee at your company, and you don't know how many users or computers are in your organization. Uh oh! Wha
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Answer

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<u></u>

<u>Definition</u>

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4 0
2 years ago
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s344n2d4d5 [400]

Answer:

def words_in_both(a, b):

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Explanation:

Output:

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5 0
2 years ago
Read 2 more answers
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