Can anyone help me with the calculation? I got 0.01 for part 1 and for part 2 99

Question

Can anyone help me with the calculation?

I got 0.01 for part 1 and for part 2 99000 kps

is it correct?

Computer Networking course :)))) Topic: Network delay and throughput

7. [10 pts.] Consider an unreliable transmission path that connects two communication nodes. The links constituting the path may drop packets but do not introduce bit errors. Assume that any given link has bit rate Rb100 Mbps, and that the average packet size is 1000 bits. 1. Compute the probability that a packet gets dropped before reaching its destination knowing that the path to the destination has 10 links, and that the probability a packet gets dropped in any given link is Pd = 10-3 2 Cive an estimate of the aerge thoughut in his cane

Explanation / Answer

Yes, Your answers are correct.

1. Given,

Probability that a packet gets dropped on a link, pd = 10-3

NUmber of links between the source and destination = 10

Therefore,

Probabilty that a packet doesn't reach destination = Probabilty that a packet gets dropped on any of the 10 links

Probabilty that a packet gets dropped on any of the 10 links = 10 * 10-3 = 10-2 = 0.01

Hence,

   Probabilty that a packet doesn't reach destination = 0.01

2. Given,

Bit rate on any link or Bandwidth = 100 Mbps = 100 * 106 bits per second

Average packet size = 1000 bits

Therefore, number of packets that can get transmitted in 1 second = (100 * 106 ) / (1000)

= 1,00,000 packets

But, the probabilty that a packet doesn't reach destination = 0.01

Hence, number of packets that won't reach destination out of 1,00,000 packets = 100000 * 0.01

= 1000 packets

Therefore, 1000 packets won't reach the destination. Hence (1,00,000 - 1000) = 99,000 packets reach the destination.

Hence throughput = number of packets * packet size

= 99,000 * 1000 bits

But 1Kb = 1000 bits

THerefore,

Throughput = 99,000 * 1Kbps = 99000 Kbps

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.