subject

Consider the gbn protocol with a sender window size of 4 and a sequence number range of 1024. suppose that at time t, the next in-order packet that the receiver is expecting has a sequence number of "k". assume that the medium does not reorder messages. answer the following questions:

1.what are the possible sets of sequence numbers inside the sender’s window at time t? justify your answer.

2. what are all possible values of the ack field in all possible messages currently propagating back to the sender at time t? justify your answer.

ansver
Answers: 1

Another question on Computers and Technology

question
Computers and Technology, 21.06.2019 18:20
Write the pseudocode for the following: a function called fahrenheittocelsius that accepts a real fahrenheit temperature, performs the conversion, and returns the real celsius temperature. a function called celsiustofahrenheit that accepts a real celsius temperature, performs the conversion, and returns the real fahrenheit temperature. a main module that asks user to enter a fahrenheit temperature, calls the fahrenheittocelsius function, and displays the celsius temperature with a user-friendly message. it then asks the user to enter a celsius temperature, calls the celsiustofahrenheit function, and displays the fahrenheit temperature with a user-friendly message.
Answers: 1
question
Computers and Technology, 21.06.2019 20:50
What are the advantages of google cloud ?
Answers: 2
question
Computers and Technology, 22.06.2019 15:30
What is a costume plot? why is it important to a film or theater production?
Answers: 2
question
Computers and Technology, 22.06.2019 20:00
What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }
Answers: 3
You know the right answer?
Consider the gbn protocol with a sender window size of 4 and a sequence number range of 1024. suppos...
Questions
question
Mathematics, 02.12.2021 18:40
question
English, 02.12.2021 18:40
Questions on the website: 13722361