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Suppose an IP packet is fragmented into 10 fragments, each with a \(1 \%\) (independent) probability of loss. To a reasonable approximation, this means there is a \(10 \%\) chance of losing the whole packet due to loss of a fragment. What is the probability of net loss of the whole packet if the packet is transmitted twice, (a) assuming all fragments received must have been part of the same transmission? (b) assuming any given fragment may have been part of either transmission? (c) Explain how use of the Ident field might be applicable here.

Short Answer

Expert verified
a) 1% b) 9.56% c) The Ident field helps differentiate fragments from multiple transmissions.

Step by step solution

01

- Understand the initial conditions

The IP packet has a 10% chance of being lost if any of its 10 fragments is lost. This probability holds for a single transmission.
02

- Probability of successful packet in one transmission

The probability of successfully receiving the entire packet in one transmission can be calculated as: \( P_{\text{success, single}} = 1 - P_{\text{loss, single}} = 1 - 0.1 = 0.9 \)
03

- Case (a) Calculation

For case (a), the packet is transmitted twice, and all fragments in the successful packet must be from the same transmission. The combined probability of losing the entire packet when it is transmitted twice is: \(P_{\text{loss, twice, same transmission}} = P_{\text{loss, single}} \times P_{\text{loss, single}} = 0.1 \times 0.1 = 0.01 (or 1 \%) \)
04

- Case (b) Calculation

For case (b), consider that each fragment can come from either transmission, doubling the chances for each fragment to be successfully received in one of the transmissions. The probability of successfully receiving each fragment is:\( P_{\text{success, fragment}} = 1 - P_{\text{loss, fragment}}^2 = 1 - (0.1 \times 0.1) = 1 - 0.01 = 0.99 \)The probability of the entire packet being successfully received is:\( P_{\text{success, whole packet}} = (0.99)^{10} \)The probability of losing the whole packet is then: \( P_{\text{loss, twice, any transmission}} = 1 - (0.99)^{10} \)
05

- Ident field usage

The Ident field in IP packets helps in reassembling the fragments correctly at the destination. In case of multiple transmissions, the Ident field is crucial because it allows the receiver to distinguish which fragments belong to which transmission, especially when fragments from different transmissions might be mixed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

IP packet loss probability
When dealing with IP packets, fragmentation can affect the probability of the entire packet being lost. Let's break it down. If an IP packet is split into 10 fragments, and each fragment has an independent 1% chance of being lost, the probability of losing at least one fragment (and thus the entire packet) is approximately 10%. This is because the loss of any single fragment means the entire packet becomes useless.
The calculation for the probability of net loss for one transmission, given a 10% individual fragment loss rate, shows how small probabilities can aggregate to influence overall packet loss. When the packet is transmitted twice, there are different scenarios to consider. For instance, if all fragments received must come from the same transmission, the probability of packet loss is compounded. Conversely, if fragments can come from either transmission, the likelihood of total packet loss decreases dramatically.
Packet transmission reliability
The reliability of packet transmission is crucial in network communication. It depends on factors like redundant transmissions and whether fragments from different transmissions can be combined. In our example, the probability of successfully transferring the entire packet in a single transmission is 90%. This is calculated from the complement probability of losing the packet: 1 minus the 10% loss rate.
Now, if the packet is sent twice, and every fragment from the successfully received packet must come from the same transmission, the probability of losing both transmissions is quite low, only 1%. Alternatively, if any given fragment can come from either transmission, then the probability of successfully receiving each fragment increases significantly. Each fragment now has a probability of 99% of being received correctly because it could come from either of the two transmissions. Thus, the overall reliability of the packet dramatically improves when fragments can be combined from multiple transmissions.
Ident field usage
The Ident field in the IP header plays a vital role in ensuring that fragmented packets are correctly reassembled at the destination. It provides a unique identifier for each packet, which is essential when packets are sent more than once. In scenarios involving multiple transmissions, the Ident field helps distinguish between fragments from different transmissions, ensuring that the receiver can reassemble the fragments into the correct packets.
For instance, if you transmit an IP packet twice and fragments from both transmissions converge at the receiver, the Ident field helps sort them, ensuring each fragment is paired with its correct counterpart. This way, the system maintains coherence and reduces packet transmission errors. This is particularly important when considering error recovery and improving packet transmission reliability, as was discussed in our example calculations.

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Most popular questions from this chapter

Read the man page or other documentation for the Unix/Windows utility netstat. Use netstat to display the current IP routing table on your host. Explain the purpose of each entry. What is the practical minimum number of entries?

Suppose IP routers learned about IP networks and subnets the way Ethernet learning bridges learn about hosts: by noting the appearance of new ones and the interface by which they arrive. Compare this with existing distance-vector router learning (a) for a leaf site with a single attachment to the Internet, and (b) for internal use at an organization that did not connect to the Internet. Assume that routers only receive new-network notices from other routers, and that the originating routers receive their IP network information via configuration.

Why do you think IPv4 has fragment reassembly done at the endpoint, rather than at the next router? Why do you think IPv6 abandoned fragmentation entirely? Hint: Think about the differences between IP-layer fragmentation and link-layer fragmentation.

Suppose host \(\mathrm{A}\) is sending to a multicast group; the recipients are leaf nodes of a tree rooted at A with depth \(N\) and with each nonleaf node having \(k\) children; there are thus \(k^{N}\) recipients. (a) How many individual link transmissions are involved if A sends a multicast message to all recipients? (b) How many individual link transmissions are involved if A sends unicast messages to each individual recipient? (c) Suppose A sends to all recipients, but some messages are lost and retransmission is necessary. Unicast retransmissions to what fraction of the recipients is equivalent, in terms of individual link transmissions, to a multicast retransmission to all recipients?

IP hosts that are not designated routers are required to drop packets misaddressed to them, even if they would otherwise be able to forward them correctly. In the absence of this requirement, what would happen if a packet addressed to IP address A were inadvertently broadcast at the link layer? What other justifications for this requirement can you think of?

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