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Suppose that one byte in a buffer covered by the Internet checksum algorithm needs to be decremented (e.g., a header hop count field). Give an algorithm to compute the revised checksum without rescanning the entire buffer. Your algorithm should consider whether the byte in question is low order or high order.

Short Answer

Expert verified
Identify affected word, decrement that word, then adjust the checksum accordingly while maintaining 16-bit range using carry-around.

Step by step solution

01

Understand Checksum Adjustment

When adjusting a checksum, the Internet checksum algorithm processes data in 16-bit words. Each byte's adjustment affects part of one of these words.
02

Identify Byte Position

Determine whether the byte to be decremented is in a high-order (the first 8 bits of a 16-bit word) or low-order (the last 8 bits of a 16-bit word) position.
03

Convert Byte Position to Word Position

Identify the word affected by the change. For example, if the byte position is an odd number, it affects the low-order byte of the word, otherwise it affects the high-order byte.
04

Calculate Change in Checksum

If the byte is decreased, the checksum should be increased by 1 to offset this decrease. This requires adjusting the 16-bit word it affects.
05

Implement High-order Adjustment

If the high-order byte is decremented, subtract 1 from the checksum directly since the high-order byte contributes more weight (i.e., \(byte \times 256\)).
06

Implement Low-order Adjustment

If the low-order byte is decremented, subtract 1 from the checksum indirectly. This translates to a simple addition of 1 to the checksum because the effect on the sum is less (i.e., \(byte \times 1\)).
07

Recalculate Overall Checksum

Adjust the checksum properly by adding the result from the word calculation. Ensure this keeps the checksum within the 16-bit range using \(\text{carry-around}\).

Key Concepts

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

Checksum Adjustment
When a byte in a buffer needs to be decremented, the Internet checksum algorithm allows for adjustments without recalculating the entire checksum from scratch. This algorithm processes data in 16-bit words. The checksum adjustment involves determining how the change in one byte affects the overall sum in the buffer.
This process helps maintain data integrity and efficient error detection by ensuring the checksum reflects changes accurately, even when tiny adjustments are necessary.
High-Order Byte
In a 16-bit word, the high-order byte represents the first 8 bits (the leftmost). If the byte to be modified falls into the high-order position, this means it has a more significant impact on the overall checksum because of its higher weighting.
When decrementing a high-order byte, the algorithm must subtract 256 (or adjust heavily) from the checksum since the high-order byte contributes more value, as shown in formula: \[ \text{high-order byte} \times 256 \]. The adjustment ensures the checksum accurately compensates for the change.
Low-Order Byte
The low-order byte of a 16-bit word occupies the last 8 bits (the rightmost). This byte contributes less weight to the checksum calculation compared to the high-order byte. If the byte to be decremented is in the low-order position, the adjustment is simpler.
Decrementing a low-order byte means you subtract 1 from its specific value within the word without affecting the higher weighted part. The algorithm needs to add 1 to the original checksum to balance out the decrement impact accurately.
16-bit Word
The Internet checksum algorithm processes data as 16-bit words, which means each word consists of two adjacent 8-bit bytes. Understanding how these bytes work together is crucial for checksum adjustment calculations.
The positioning determines whether you deal with a high-order or low-order byte. Identifying the correct word containing the byte in question helps to apply the correct adjustment value, ensuring the checksum remains efficient and accurate.
Header Hop Count Field
In the context of networks, a header hop count field indicates the number of hops (or steps) a packet has taken through the network. Adjustments to this field, such as decrementing its value, are part of modifying the overall packet header and hence the checksum.
Proper checksum adjustments ensure that even as hop counts change, the error detection capabilities of the checksum remain intact. Keeping track of changes in the header, like the hop count, is vital to maintaining the integrity and efficiency of network communications.

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Most popular questions from this chapter

Suppose that we attempt to run the sliding window algorithm with SWS = RWS = 3 and with MaxSeqNum \(=5 .\) The Nth packet DATA[ \(N]\) thus actually contains \(N\) mod 5 in its sequence number field. Give an example in which the algorithm becomes confused; that is, a scenario in which the receiver expects DATA[5] and accepts DATA[0]-which has the same transmitted sequence number-in its stead. No packets may arrive out of order. Note this implies MaxSeqNum \(\geq 6\) is necessary as well as sufficient.

Draw a timeline diagram for the sliding window algorithm with SWS = RWS = 4 frames for the following two situations. Assume the receiver sends a duplicate acknowledgement if it does not receive the expected frame. For example, it sends DUPACK[2] when it expects to see FRAME[2] but receives FRAME[3] instead. Also, the receiver sends a cumulative acknowledgment after it receives all the outstanding frames. For example, it sends ACK[5] when it receives the lost frame FRAME[2] after it already received FRAME[3], FRAME[4], and FRAME[5]. Use a timeout interval of about \(2 \times\) RTT. (a) Frame 2 is lost. Retransmission takes place upon timeout (as usual). (b) Frame 2 is lost. Retransmission takes place either upon receipt of the first DUPACK or upon timeout. Does this scheme reduce the transaction time? Note that some end-to-end protocols (e.g., variants of TCP) use a similar scheme for fast retransmission.

Consider an ARQ algorithm running over a \(20-\mathrm{km}\) point-to-point fiber link. (a) Compute the propagation delay for this link, assuming that the speed of light is \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in the fiber. (b) Suggest a suitable timeout value for the ARQ algorithm to use. (c) Why might it still be possible for the ARQ algorithm to time out and retransmit a frame, given this timeout value?

Coaxial cable Ethernet was limited to a maximum of \(500 \mathrm{~m}\) between repeaters, which regenerate the signal to \(100 \%\) of its original amplitude. Along one \(500-\mathrm{m}\) segment, the signal could decay to no less than \(14 \%\) of its original value \((8.5 \mathrm{~dB})\). Along \(1500 \mathrm{~m}\), then, the decay might be \((0.14)^{3}=0.3 \%\). Such a signal, even along \(2500 \mathrm{~m}\), is still strong enough to be read; why then are repeaters required every \(500 \mathrm{~m}\) ? $$ \begin{array}{l|l} \hline \text { Item } & \text { Delay } \\ \hline \text { Coaxial cable } & \text { propagation speed } .77 c \\ \text { Link/drop cable } & \text { propagation speed .65c } \\ \text { Repeaters } & \text { approximately 0.6 } \mu \text { s each } \\ \text { Transceivers } & \text { approximately 0.2 } \mu \text { s each } \\ \hline \end{array} $$

Assuming a framing protocol that uses bit stuffing, show the bit sequence transmitted over the link when the frame contains the following bit sequence: 110101111101011111101011111110 Mark the stuffed bits.

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