Computers and Technology, 03.06.2021 04:40 imoreno8412

Consider a packet of length 2000 bits which begins at end system A and travels over five links to a destination end system. These five links are connected by four packet switches. The propagation speed on all five links is 3*108 m/s. The transmission rate of all five links is 2 Mbps. For each packet switch, the processing delay is 1 msec. The length of the first link is 8,000 km, the length of the second link is 4,000 km, the length of the third link is 2,000 km and the length of the fourth and fifth link is 1000 km. Assuming no queuing delay, what is the end-to-end delay

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Answer from: donavancoulter712

0.095 sec

Explanation:

From the information given:

Packet size = 2000 bytes

The propagation speed on both the links = $3.0*10^8 \ m/s$

The transmission rates of all three links = 2 Mbps

Packet switch processing delay = 1 msec

The length of the 1st link = 8000 km

The length of the 2nd link = 4000 km

The length of the 3rd link = 2000 km

The length of the 4th link = 1000 km

The length of the 5th link = 1000 km

The 1st end system that needs to transmit the packet onto the1st link = L/R_1

$= \dfrac{2000*8}{2*10^6}$

= 0.008 sec

The packet propagates over the 1st link in d1/s1 is;

$= \dfrac{8000*10^3}{3.0 \times 10^8}$

= 0.027 sec

The packet switch generates a delay of $d_{proc}= 1msec$ , after receiving the whole packet

The 1st end system that needs to transmit the packet onto the 2nd link = L/R_2

$= \dfrac{2000*8}{2*10^6}$

= 0.008 sec

The packet propagates over the 2nd link in d2/s2 is;

$= \dfrac{4000*10^3}{3.0 \times 10^8}$

= 0.013 sec

Again, the packet switch generates a delay of $d_{proc}= 1msec$ , after receiving the whole packet

The 1st end system that needs to transmit the packet onto the 3rd link = L/R_3

$= \dfrac{2000*8}{2*10^6}$

= 0.008 sec

The packet propagates over the 3rd link in d3/s3 is;

$= \dfrac{2000*10^3}{3.0 \times 10^8}$

= 0.007 sec

Again, the packet switch generates a delay of $d_{proc}= 1msec$ , after receiving the whole packet

The 1st end system that needs to transmit the packet onto the 4th link = L/R_4

$= \dfrac{2000*8}{2*10^6}$

= 0.008 sec

The packet propagates over the 4th link in d4/s4 is;

$= \dfrac{1000*10^3}{3.0 \times 10^8}$

= 0.003 sec

Again, the packet switch generates a delay of $d_{proc}= 1msec$ , after receiving the whole packet

The 1st end system that needs to transmit the packet onto the 5th link = L/R_5

$= \dfrac{2000*8}{2*10^6}$

= 0.008 sec

The packet propagates over the 5th link in d5/s5 is;

$= \dfrac{1000*10^3}{3.0 \times 10^8}$

= 0.003 sec

The end-to-end delay = $L/R_1+L/R_2+L/R_3+L/R_4+L/R_5 + d_1/s_1+d_2/s_2+d_3/s_3+d_4/s_4+d_5/s_5+d_{proc}+d_{proc}$

=0.008+0.008+0.008+0.008+0.008+0.027+0.013+0.007+0.003+0.003+0.001+0.001

= 0.095 sec

Hence, the end to end delay = 0.095 sec

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Consider a packet of length 2000 bits which begins at end system A and travels over five links to a...