1. Suppose packets are transmitted at rate rA into the network. What is the end-

Question

1. Suppose packets are transmitted at rate rA into the network. What is the end-to-end delay of the ith packet (i.e., from the time when the sender starts transmitting its first bit to the time when its last bit is received by the receiver)? Provide formula(s) and a few sentences of explanation.

2. What is the maximum rate at which the sender injects packets into the network while still avoiding queueing (so that all packets see the same end-to-end delay)?

Consider two lossless links, A and B, which can respectively support throughput rates r bps and rg bps, each link has propagation delay d. A sender reaches a receiver by transmitting through the chain of links A -> B, sending packets of size L bits. The sender knows only the rate ra that link A is capable of supporting, and thus sends its data into the link at rate r, beginning to transmit each packet immediately after it completes transmitting the previous one. A router, sitting between links A and B, can buffer an infinite amount of data (i.e., no loss) if needed, and requires time t to process each packet of size L bits that proceeds through it. Router Link A Link B sender receiver

Explanation / Answer

End to End delay or one-way-delay refers to the time taken for a packet to be transmitted across a network from source to destination(IP network)

monitoring and differs from RTT(round trip time) in that only path in the one direction from source to destination is measured.

common calculations are:

1.Transmission delay.

2.Propogation delay.

3.Processing delay and queuing delay.

transmission can be calculated as: length/bandwidth.

propogation can be calculated as distance/propogation speed.

efficiency can be calculated as: transmission/(2*transmission)+propogation.

RTT=transmissiondelay+propogationDelay.

Time necessary to send one packet formula: (H+(D/P))/C

H headersize

D bitsize

P numberof packets

C capacity

N number of links(no of routers+1)

total time necessary to send (p+2) packets will be end to end (p+2)*(H+(D/P))/C.

maximun size is: min max

interframe 12bytes 12bytes

MAC 8bytes 8bytes

MAC Destination 6bytes 6bytes

MAC source 6bytes 6bytes

MAC Type/length 2bytes 2bytes

payload 46bytes 1500bytes

checksequence 4 4

total 84bytes 1538bytes

(1,000,000,000b/s)/84*8=1488096(max) 1byte=8bits so 84*8

(1,000,000,000b/s)/1538*8=81274(min)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.