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g Consider a packet that begins at end system A and travels over three links to a destination end system B. These three links are connected by two routers. Assume we ignore queueing delays and processing delays, and assume that the packet is 100 bytes, the propagation speed on all three links is 2*10^8 m/s, the link rate of three links are 1 Mbps, 2 Mbps, and 5 Mbps, respectively, the length of all three links is 10 km. For these values, what is the end-to-end delay

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batolisis

Answer:

0.0015 seconds

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

[tex]\frac{L}{R1} + \frac{L}{R2} + \frac{L}{R3} + \frac{D1}{S1} + \frac{D2}{S2} + \frac{D3}{S3} + D_{proces}[/tex]

=( 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

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