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Suppose a shared medium \(\mathrm{M}\) offers to hosts \(\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{N}\) in round-robin fashion an opportunity to transmit one packet; hosts that have nothing to send immediately relinquish M. How does this differ from STDM? How does network utilization of this scheme compare with STDM?

Short Answer

Expert verified
Round-robin reallocates turns by bypassing idle hosts, increasing efficiency, unlike STDM that assigns fixed slots regardless of data presence.

Step by step solution

01

- Understand the Round-Robin Scheme

The round-robin scheme gives each host an opportunity to transmit a packet in a cyclic order. If a host has no packet to send, it forfeits its turn.
02

- Define STDM

Synchronous Time-Division Multiplexing (STDM) divides the time into fixed intervals where each host gets a dedicated time slot to transmit a packet, regardless of whether it has data to send or not.
03

- Compare Round-Robin and STDM

In STDM, time slots are allotted whether a host has data to send or not, which can lead to wasted bandwidth. In the round-robin scheme, if a host has no data to transmit, the opportunity is passed to the next host, reducing idle periods.
04

- Analyze Network Utilization

Round-robin method is more efficient in terms of bandwidth utilization because it does not allocate resources to idle hosts. STDM can result in poor utilization if many hosts don't have data to send during their time slots.

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Key Concepts

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

Round-Robin Scheme
The Round-Robin Scheme ensures that each host connected to the shared medium gets a fair chance to transmit data. In this system, the hosts take turns in a cyclic order. If one host doesn't have any data to send, it simply skips its turn and the next host is given the opportunity. This process continues until all hosts have had their chance. Since a host's unused slot can be immediately taken by another, there is a reduction in idle time. This scheme is dynamic and adaptable to current network traffic, making it efficient and reducing delays. By forfeiting the turn when no data is available to send, the system maintains high responsiveness and low latency.
Synchronous Time-Division Multiplexing (STDM)
Synchronous Time-Division Multiplexing (STDM) divides available bandwidth into fixed time slots, each allocated to a specific host. Regardless of whether the host has data to send, these slots remain reserved. One advantage of STDM is its predictability and simplicity in implementation. Each host knows its exact time to transmit, which can simplify timing and synchronization. However, the major downside is inefficiency in bandwidth utilization. If a host doesn’t have data to send during its allocated slot, that time slice goes wasted. This can lead to potential underutilization of the network capacity, particularly in environments where not all hosts are constantly transmitting data.
Network Utilization
Network utilization refers to how effectively the total available network resources are used. In the context of the Round-Robin Scheme versus STDM, Round-Robin typically results in higher network utilization. This is because it dynamically reallocates the transmission opportunity to hosts ready to send data. On the contrary, STDM might lead to wasted bandwidth if some hosts don't utilize their fixed time slots. Dynamic systems like Round-Robin are better at adapting to varying traffic loads, thereby maintaining higher overall efficiency and minimizing unused resources.
Bandwidth Allocation
Bandwidth allocation is the method of distributing available network capacity among various users or applications. In the Round-Robin Scheme, bandwidth is allocated on an as-needed basis. Each host transmits in its turn if it has data, passing the opportunity if not. Synchronous Time-Division Multiplexing (STDM), on the other hand, allocates fixed time slots to each host in a repetitive cycle. This static allocation can lead to inefficient use of bandwidth, as slots are reserved regardless of the actual need. Adaptive systems like Round-Robin are generally more flexible, ensuring that bandwidth is allocated where it’s needed most at any given time.
Transmission Efficiency
Transmission efficiency measures how effectively data is transmitted across the network with minimal waste. The Round-Robin Scheme enhances efficiency by ensuring that only those hosts with data to send are transmitting, thus minimizing idle times. STDM can suffer from lower efficiency because time slots are allocated regardless of whether hosts have data to transmit. If many slots go unused, it results in wasted transmission opportunities, degrading efficiency. Thus, from a transmission efficiency perspective, the adaptability of the Round-Robin Scheme generally outperforms the rigid structure of STDM.

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Most popular questions from this chapter

One property of addresses is that they are unique; if two nodes had the same address it would be impossible to distinguish between them. What other properties might be useful for network addresses to have? Can you think of any situations in which network (or postal or telephone) addresses might not be unique?

Calculate the total time required to transfer a \(1000-\mathrm{KB}\) file in the following cases, assuming an RTT of \(100 \mathrm{~ms}\), a packet size of \(1 \mathrm{~KB}\) and an initial \(2 \times\) RTT of "handshaking" before data is sent. (a) The bandwidth is \(1.5 \mathrm{Mbps}\), and data packets can be sent continuously. (b) The bandwidth is \(1.5 \mathrm{Mbps}\), but after we finish sending each data packet we must wait one RTT before sending the next. (c) The bandwidth is "infinite," meaning that we take transmit time to be zero, and up to 20 packets can be sent per RTT. (d) The bandwidth is infinite, and during the first RTT we can send one packet \(\left(2^{1-1}\right)\), during the second RTT we can send two packets \(\left(2^{2-1}\right)\), during the third we can send four \(\left(2^{3-1}\right)\), and so on. (A justification for such an exponential increase will be given in Chapter \(6 .)\)

Assume you wish to transfer an \(n\)-byte file along a path composed of the source, destination, seven point-to-point links, and five switches. Suppose each link has a propagation delay of \(2 \mathrm{~ms}\), bandwidth of \(4 \mathrm{Mbps}\), and that the switches support both circuit and packet switching. Thus you can either break the file up into 1-KB packets, or set up a circuit through the switches and send the file as one contiguous bit stream. Suppose that packets have 24 bytes of packet header information and 1000 bytes of payload, that store-and-forward packet processing at each switch incurs a 1 -ms delay after the packet has been completely received, that packets may be sent continuously without waiting for acknowledgments, and that circuit setup requires a 1-KB message to make one round-trip on the path incurring a 1-ms delay at each switch after the message has been completely received. Assume switches introduce no delay to data traversing a circuit. You may also assume that file size is a multiple of 1000 bytes. (a) For what file size \(n\) bytes is the total number of bytes sent across the network less for circuits than for packets? (b) For what file size \(n\) bytes is the total latency incurred before the entire file arrives at the destination less for circuits than for packets? (c) How sensitive are these results to the number of switches along the path? To the bandwidth of the links? To the ratio of packet size to packet header size? (d) How accurate do you think this model of the relative merits of circuits and packets is? Does it ignore important considerations that discredit one or the other approach? If so, what are they?

Consider a simple protocol for transferring files over a link. After some initial negotiation, A sends data packets of size \(1 \mathrm{~KB}\) to B; B then replies with an acknowledgment. A always waits for each ACK before sending the next data packet; this is known as stop-and-wait. Packets that are overdue are presumed lost and are retransmitted. (a) In the absence of any packet losses or duplications, explain why it is not necessary to include any "sequence number" data in the packet headers. (b) Suppose that the link can lose occasional packets, but that packets that do arrive always arrive in the order sent. Is a 2-bit sequence number (that is, \(N\) mod 4) enough for \(\mathrm{A}\) and \(\mathrm{B}\) to detect and resend any lost packets? Is a 1-bit sequence number enough? (c) Now suppose that the link can deliver out of order, and that sometimes a packet can be delivered as much as 1 minute after subsequent packets. How does this change the sequence number requirements?

Consider a closed-loop network (e.g., token ring) with bandwidth \(100 \mathrm{Mbps}\) and propagation speed of \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\). What would the circumference of the loop be to exactly contain one 250 -byte packet, assuming nodes do not introduce delay? What would the circumference be if there was a node every \(100 \mathrm{~m}\), and each node introduced 10 bits of delay?

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