$Packet Size = 500 bytes = 500\times8=4000bits$
$A \rightarrow B\;\; $
$Bandwidth = 4\times 10^6bits $ $Distance = 3000km$
$B\rightarrow C\;\; $
$Bandwidth = 2\times 10^6bits $ $Distance = 6000km$
Compute :
Transmission Time = Packet Size/ Bandwidth .
Propogation Time = Distance / Speed .
$T_{t(A\rightarrow B)} = 4000 bits/ 4\times10^6bits = 0.001s = 1ms$
$T_{p(A\rightarrow B)}= 3000 /3\times10^5 = 0.01s = 10ms$
$T_{t(B\rightarrow C)} = 4000 bits/ 2\times10^6bits = 0.002s = 2ms$
$T_{p(B\rightarrow C)}= 6000 /3\times10^5 = 0.02s = 20ms$
Theory is that the sender sends the 2nd packet immediately after sending 1st packet , So at reciever end the last packet will reach taking additional time of its Transmission i.e $1ms $ in $A\rightarrow B$ and $2ms $$B\rightarrow C$
$A\rightarrow B$
$ Pkt_1 \;\text{is sent at time} = 1ms , \text{it reaches B at time}= 11ms \;\;(1+ 10)$
$ Pkt_2 \;\text{was sent immediately after } Pkt_2 = 2ms , \text{and it reaches B at time}= 12ms \;\;(2+10)$
$B\rightarrow C$
$Pkt_1 \;\text{sent to C at t} = 13ms (11+2 ),\text{ it reaches C at time} = 33ms (13+20)$
$Pkt_2 \;\text{sent to C at t} = 15ms (13+2 ),\text{ it reaches C at time} = \bf35ms (15+20)$
$Total\; End \; to \; End\; Delay = 35ms\; $
