Answer:
0.0015 seconds
Step-by-step explanation:
Given Data:
links given = 3 we assume them to be ( 1,2,3)
packet ( L ) = 100 bytes ( 800 bits )
propagation speed ( S1 ) = 2*10^8 m/s
link rates ( R1 , R2, R3 ) = 1 Mbps , 2 Mbps, 5 Mbps
length of links ( d ) = 10 km
note:
d1 = d2 = d3
S1 = S2 = S3
To calculate the end-to-end delay we employ this relationship

=( 800 / 1 * 10^6) + ( 800 / 2*10^6) + ( 800 / 5*10^6 ) + (( 10 * 10^3 / 2*10^8))*3 + 0
= 0.0008 + 0.0004 + 0.00016 + 0.00015
= 0.0015 secs