NATIONAL INSTITUTE OF TECHNOLOGY HAMIRPUR
Himachal Pradesh – 177 005, INDIA
COMPUTER SCIENCE & ENGINEERING
Even SEMESTER, 2020
COMPUTER NETWORKS Assignment
(Numerical Assignment)
By ARVIND SINGH
R. No- 17544
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Computer Networks Assignment from NIT Hamirpur: Window Size and Throughput Calculations, Assignments of Computer Networks
Solutions to numerical questions from a computer networks assignment at nit hamirpur. The questions cover topics such as window size calculation, propagation and transmission delays, and maximum file size representation. The solutions include detailed calculations and explanations.
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NATIONAL INSTITUTE OF TECHNOLOGY HAMIRPUR
Himachal Pradesh – 177 005, INDIA
COMPUTER SCIENCE & ENGINEERING
Even SEMESTER, 2020
COMPUTER NETWORKS Assignment
(Numerical Assignment)
By ARVIND SINGH
R. No- 17544
Numerical ASSIGNMENT
ANS 1.
a) Here we have a window size of N = 4. Suppose the receiver has received packet k-1, and has ACKed that and all other preceding packets.
- If all of these ACK’s have been received by sender, then sender’s window is [k, k+N-1].
- Suppose next that none of the ACKs have been received at the sender.
- In this second case, the sender’s window containsk-1 and the N packets up to and includingk-1. The sender’s window is thus [k-N, k-1].
- By these arguments, the senders window is of size 4 and begins somewhere in the range [k- 4 , k]. b) If the receiver is waiting for packetk, then it has received (and ACKed) packetk-1 and theN- 1 packets before that.
- If none of those NACKs have been yet received by the sender, then ACK messages with values of [k-N, k- 1] may still be propagating back.
- Because the sender has sent packets [k-N, k-1], it must be the case that the sender has already received an ACK fork-N-1.
- Once the receiver has sent an ACK fork-N-1, assuming no that the ack message is not lost resulting in retransmission, it will never send an ACK that is less thatk-N-1. Thus the range of in-flight ACK values can range fromk-N-1 tok-1. Given N=4, the range of in-flight ACK values can range from k - 5 to k - 1. ANS 2. Given- Packet size = 32 bytes Round Trip Time = 80 msec Bandwidth = 128 Kbps Calculating Transmission Delay - Transmission delay (Tt) = Packet size / Bandwidth = 32 bytes / 128 Kbps = (32 x 8 bits) / (128 x 10^3 bits per sec) = 2 msec Calculating Propagation Delay - Propagation delay (Tp) = Round Trip Time / 2 = 80 msec / 2 = 40 msec
hence, N = (10.304 * 10^3)/10 = 1030 stations. ANS 5. a) (1 – p(A))^4 p(A) where, p(A)= probability that A succeeds in a slot p(A)= p(A transmits and B does not and C does not and D does not) = p(A transmits) p(B does not transmit) p(C does not transmit) p(D does not transmit) = p(1 – p) (1 – p)(1-p) = p(1 – p)^3 Hence, p(A succeeds for first time in slot 5) = (1 – p(A))^4 p(A) = (1 – p(1 – p)^3 )^4 p(1 – p)^3 b) p(A succeeds in slot 4) = p(1-p)^3 p(B succeeds in slot 4) = p(1-p)^3 p(C succeeds in slot 4) = p(1-p)^3 p(D succeeds in slot 4) = p(1-p)^3 p(either A or B or C or D succeeds in slot 4) = 4 p(1-p)^3 (because these events are mutually exclusive) c) p(some node succeeds in a slot) = 4 p(1-p)^3 p(no node succeeds in a slot) = 1 - 4 p(1-p)^3 Hence, p(first success occurs in slot 3) = p(no node succeeds in first 2 slots) p(some node succeeds in 3 rdslot) = (1 - 4 p(1-p)^3 )^2 4 p(1-p)^3 ANS 6. Time t Event 0 A and B begin transmission 245 A and B and detect collision 245 + 32 = 277 A and B and finish transmitting jam signal 277 + 245 = 522 A’s last bit arrives at B;A detects an idle channel 522 A starts transmitting B returns to step 523 277 + 512 = 789 B must sense idle channel for 96 bit times before it transmits(KB= 1, back-off time =KB·512) 524 522 + 245 = 767 A’s transmission reaches B Because A’s retransmission reaches before B’s scheduled retransmission time (789 + 96) , B refrains from transmitting while A retransmits. Thus A and B do not collide. Thus the factor 512 appearing in the exponential back-off algorithm is sufficiently large.