Question

Given three packets of size 1500 bytes each are ready inside Computer 1 to be transmitted onto Link 1. Computer 1 has zero processing delay. Assume each router has zero queuing and zero processing delay. Link 1 is a 10 Mbit/sec link and 5000 km long. Link 2 is 100 Mbits/sec and is 5000 km long. Link 3 is 1000 Mbits/sec and is 5000 km long. Propagation speed is 2 × 10 8 m/s. Assume a router has to receive all of a packet before it can start sending it out along the next link (assume store and forward).

a) What is the total end to end delay for packet number one (time from first bit output on link 1 until last bit received at Computer 2)?

b)How much time goes by between when the last part of packet one is received by router 2 and when the first part of packet two arrives at router 2? (How far apart are the packets in time on link 2?)

c)How much time goes by between when the last part of packet one is received by Computer 2 and when the first part of packet number two arrives at Computer 2? (How far apart are the packets in time on link 3?)

Edit: Regarding the comment, there are 3 total packets each of size 1500 bytes..all of the questions are discussing one or two of the specific packets

Answer 1

Question a)

Here I first calculated the transmission time and propogation time of all links.

DATI (4) Packet Size : C) Prengation Speed: 1500 Byla 1500 X 8 bits 2110 m/s Bandev idth (BW) Transmission Time (TE) lombit/see lux10 bit/sec 1500 x 8 12x16" Bu loxio Sooo km. Gooux 103 m. SoowX10?m0.025 sec Length of Cable (o) Propagation Dolay (To) 2x 108 Linka: Bw 100 mb/sect = 100 x 10 bit/ see . 1500x 8 12x10's 1oox 106 Ta Same as Link 1 Link 3: Во - T = L Bw T 1000 mb/see 1000 x 10 bit/sec 1500 X8 10000 loc Link 1 12x106 S Same as a) End to End delay of packet 1 = (Te at Link 1 + Tp at Link 1) +(It at Link 2 + Tp at Link 2). + at Link 3. + Top at Link 3) (12x1024 +0.02%) +(!2X10 +0:025) + (12X.10€0:025) 3x0025 + 12x10W+ 12x109 + 12x166 12 C 10" tlost 3 x 0.025 + 12x10 (10? + io +1) 12x16x 3xb025 + - 3x0.025 + 0.076332 sec. Be Positive... led With CamScanner

Below here i calculated the timeilne of packet1 and packet2 and we will use this for solving question b) and c).

12x10 C1 (12x10-5) Tt DATE B-2 (2 Computer 1 LI L2 Computer 2 L3 Le Tp = 0.025 Packet 1 Transmission : et turco. Time to ciotol to beog Position of Packet 1 12x10" = 0.0012 LI 090012 0:0 25 20:0262 00262 +12x10.5 10:02632 12 0:02632 +0.025 = 0.65132 R2 0.05132 + 12x10-9= 0.051 332 2 L3 0-051332_+0.102.5 = 0.076.332.1 C2 tog hurit to a who otcout Packet 2 Transmission in Time Position of Packet 2 سے دلی دارم اما در وا را (Packet 2 can start transmission only after Packet 1 is transmitted from CL to LL. It See will start after a 12x1024) [ 12x 10 -4 12x ou + 12x16 " = 0.0024 0.0024+ 0.025E_00274 0.0274 + 12x105 = 0.02752 0-02752 +0.025 = 0.05252 LI RI R2 Be Positive... CSScanned with CamScanner

Question b) and c)

Here i used the above calculation to solve these questions.

DATE b) (As. Packet 1 at R2 0.05.1325- Packet 2 at Ri 0.02 74s. have to only look at first part of we don't need to add transmission time from Rito 12 ) Packet 2 First part of Packet 2 at R2 : 0-027440-025 0.0524 Time difference & 0.05.24-0-05132 So Ioo.00108 see.. ng Packet 1 sat oo 02076332 Packet 2 at R2 : 0105252 (As we have to only look at first part of Packet 2, werdon't need to add transmissian time from R2 to 2. (2) First part of Packet 2 at 12: 0.05252+0 -025 0.07752 Time Difference to 0.0152 - 01076332 0.001188 seco Scanned with CamScanner