Question

Let \(p \leq 1\) be the fraction of machines in a network that are big-endian; the remaining \(1-p\) fraction are little-endian. Suppose we choose two machines at random and send an int from one to the other. Give the average number of byte-order conversions needed for both big-endian network byte order and receiver-makes-right, for \(p=0.1, p=0.5\), and \(p=0.9\). Hint: The probability that both endpoints are big-endian is \(p^{2} ;\) the probability that the two endpoints use different byte orders is \(2 p(1-p)\).

Short answer

For big-endian network byte order and receiver-makes-right: E=0.18 for p=0.1 and p=0.9; E=0.5 for p=0.5.

Step-by-step solution

Understanding Probabilities

First, let's identify the probabilities given: 1. The probability that both machines are big-endian is given by: \[ P(\text{B and B}) = p^2 \] 2. The probability that both machines are little-endian is: \[ P(\text{L and L}) = (1-p)^2 \] 3. The probability that one machine is big-endian and the other is little-endian (or vice versa) is given by: \[ P(\text{B and L or L and B}) = 2p(1-p) \]

Calculate Conversions for Big-endian Network Byte Order

In a big-endian network byte order, only conversions between a big-endian machine and a little-endian machine need to count. The average number of conversions can be calculated as follows: \[ E_{\text{big-endian network}} = 2p(1-p) \]

Calculate Conversions for Receiver-makes-right

In a receiver-makes-right method, the data is converted to the byte order of the receiving machine. Therefore, each pair of differing byte orders needs conversion, considering the total probability of different byte orders: \[ E_{\text{receiver-makes-right}} = 2p(1-p) \]

Apply the Given Values of p

Now, compute the average number of conversions for each given value of p:For \( p = 0.1 \): \[ E = 2(0.1)(1-0.1) = 0.18 \] For \( p = 0.5 \): \[ E = 2(0.5)(1-0.5) = 0.5 \] For \( p = 0.9 \): \[ E = 2(0.9)(1-0.9) = 0.18 \]

Key concepts

big-endian

Big-endian is a method of storing or transmitting data where the most significant byte is stored or transmitted first. This format is commonly used in network protocols to ensure consistency. For example, if we have a 32-bit data representation of the number 0x12345678, in big-endian format, it would be stored as follows:
• Byte 1: 0x12
• Byte 2: 0x34
• Byte 3: 0x56
• Byte 4: 0x78

This makes it easier to read data values without needing additional conversion steps, and many Internet protocols, such as IP and TCP, use big-endian format for this reason. Big-endian helps in preserving the natural ordering of numeric values which is intuitive for humans and certain algorithms.

little-endian

Little-endian is a method of storing or transmitting data where the least significant byte is stored or transmitted first. This format is often used in computer architecture, particularly in x86 processors. For the same 32-bit representation of the number 0x12345678, little-endian format would be stored as:
• Byte 1: 0x78
• Byte 2: 0x56
• Byte 3: 0x34
• Byte 4: 0x12

Little-endian is useful in tasks such as multi-byte addition, where starting from the least significant byte can be more efficient. However, it can lead to complications when interfacing with networks or systems that use big-endian format.

probability calculations

Probability calculations are essential in determining the likelihood of different byte-order configurations in network communication. The exercise includes key probabilities you need to know:
1. The probability that both machines are big-endian is given by: \( P(\text{B and B}) = p^2 \)
2. The probability that both machines are little-endian is: \( P(\text{L and L}) = (1-p)^2 \)
3. The probability that one machine is big-endian and the other is little-endian (or vice versa) is: \( P(\text{B and L or L and B}) = 2p(1-p) \)

These calculations help us understand scenarios involving communication between machines with different byte orders. By knowing the probabilities, we can calculate the expected number of byte-endian conversions needed for different methods, like big-endian network byte order and receiver-makes-right.

network byte order

Network byte order usually means using big-endian byte arrangement for transmitting data over networks. It ensures all communicating devices interpret the bytes in the same order. This consistency prevents misinterpretation of data caused by varying endianness. When data is sent from a big-endian machine to a little-endian machine, or vice versa, byte-order conversions might be required to adhere to network protocol standards.

Let’s compute the average number of byte conversions for big-endian network byte order:
Given by: \( E_{\text{big-endian network}} = 2p(1-p) \)
Here, only conversions between a big-endian machine and a little-endian machine are counted.

receiver-makes-right

Receiver-makes-right is a method where the receiving machine converts the incoming data to its own byte order. This way, the burden of byte-order conversion falls on the receiver:

Given by: \( E_{\text{receiver-makes-right}} = 2p(1-p) \)
Both big-endian to little-endian and little-endian to big-endian conversions are considered. Every time machines with different byte orders communicate, the receiver corrects the byte format. This method can be advantageous as it ensures data is always in the correct format for the receiving system, though it requires the receiver to have conversion logic implemented.