Question

What is the size of the multicast address space? Suppose now that two multicast groups randomly choose a multicast address. What is the probability that they choose the same address? Suppose now that 1,000 multicast groups are ongoing at the same time and choose their multicast group addresses at random. What is the probability that they interfere with each other?

Short answer

The multicast address space has 268,435,456 addresses. The probability of two groups picking the same multicast address is very low, about \(\frac{1}{268,435,456}\). For 1,000 groups, the probability of address conflict is about 0.000186 (or 0.0186%).

Step-by-step solution

Understanding the Multicast Address Space

IPv4 multicast addresses are in the range from 224.0.0.0 to 239.255.255.255. Each octet can have 256 possible values. Since there are 5 sets of Class D addresses available after accounting for the reserved range (224.0.0.0 to 224.0.0.255), calculate the total number by subtracting 256 from 2^28 (all IPv4 addresses), which gives 268,435,456 possible addresses.

Calculating Probability for Two Multicast Groups

The probability that two different groups select the same multicast address is the total number of favorable outcomes (1, i.e., they both pick the same address) over the total possible outcomes (number of multicast addresses). This can be written as \(\frac{1}{268,435,456}\).

Calculating Probability for 1,000 Multicast Groups

For 1,000 groups, use the birthday paradox approximation. The probability that no two groups pick the same address is \(\frac{268,435,456}{268,435,456} \times \frac{268,435,455}{268,435,456} \times \cdots \times \frac{268,434,457}{268,435,456}\). Simplifying gives an approximate expression: \(e^{-\frac{1000^2}{2 \times 268,435,456}}\).

Finding the Probability of Collision

The probability of at least one collision occurring is the complement of the probability that no collision occurs. So, it is \(1 - e^{-\frac{1000^2}{2 \times 268,435,456}}\). Calculate this value to determine the likelihood of address conflict.

Key concepts

IPv4 Multicast Addresses

IPv4 multicast addressing is a method used to deliver information from one sender to multiple receivers simultaneously. This is essential for applications like video conferencing and live streaming, where the same data is sent to many users at once.
IPv4 multicast addresses fall within the range of 224.0.0.0 to 239.255.255.255. This means these addresses are reserved specifically for multicast use and cannot be used for anything else. They are known as Class D addresses in IPv4 because they start with "1110" in their binary form. This range offers a large pool of potential addresses.
However, because the addresses from 224.0.0.0 to 224.0.0.255 are reserved, not all are available for general use. Excluding the reserved range, there are 268,435,456 unique multicast addresses. This count results from the calculation: \(2^{28} - 256\). Here, \(2^{28}\) represents the total possible IPv4 addresses in this range, and 256 accounts for the reserved ones.

• Essential for group communication like IPTV or video conferencing
• Defined by a specific range starting with the binary sequence "1110"
• 268,435,456 unique addresses available for multicast assignments

Birthday Paradox Approximation

The birthday paradox is a fascinating probability concept that helps calculate the likelihood of two or more entities (such as multicast groups) coincidentally sharing the same item (like an IP address).
The classic example is that in a group of 23 people, there's about a 50% chance that two will share the same birthday, despite there being 365 possible birthdays. This counter-intuitive result is due to the rapidly growing number of pairwise comparisons as more people join the group.
In the context of IPv4 multicast addresses, we use the birthday paradox to estimate the probability that 1,000 groups will choose the same address. Each group can select any of the 268,435,456 addresses, making the math pretty tricky without the approximation.
The approximation formula expressed in terms of multicast addresses is:
\[e^{-\frac{n^2}{2m}}\]
Here, \(n\) is the number of groups (in this case, 1,000), and \(m\) is the number of unique multicast addresses. The result gives the probability that none of the groups share an address. By subtracting this from 1, you get the probability of at least one collision.

Address Space Collision Probability

Understanding address collision probabilities is crucial for network design, especially when multiple groups independently select multicast addresses.
An address collision happens when two groups inadvertently pick the same multicast address, causing data intended for one group to be received by another. This affects service quality.
To determine how probable such collisions are, we rely on the birthday paradox formula, which helps us approximate the chance of at least one collision when many entities independently choose from a limited set of options. When 1,000 multicast groups pick addresses, the collision probability can be calculated using:
\[1 - e^{-\frac{1000^2}{2 \times 268,435,456}}\]
This formula shows that even with a large address space, significant use increases the chance of collisions.

• Collisions occur when the same address is chosen by more than one group
• The probability can be calculated using the complement of the "no-collision" probability
• Understanding these probabilities helps in managing and optimizing network resources