Upon completion of this tutorial, you should be able to:
When multiple nodes share a single communication channel (e.g., a wireless medium, a coaxial cable bus), they need a multiple access protocol to coordinate transmissions and avoid collisions. This tutorial covers the major categories of multiple access protocols: channel partitioning (TDMA, FDMA, CDMA), random access (ALOHA, CSMA, CSMA/CD, CSMA/CA), and taking-turns (polling, token passing). We analyze their performance, trade-offs, and real-world applications, with a focus on Ethernet (CSMA/CD) and Wi-Fi (CSMA/CA). Advanced topics such as hidden terminals and exposed terminals are also discussed.
In a broadcast network, multiple nodes share a single communication channel. If two or more nodes transmit simultaneously, their signals overlap, causing a collision and rendering the frames unreadable. The multiple access protocol defines how nodes share the channel to minimize collisions and efficiently utilize the bandwidth.
Key performance metrics:
Protocols are classified into three categories: channel partitioning, random access, and taking-turns.
TDMA divides time into slots and assigns each node a dedicated time slot. Nodes transmit only during their slot, thus no collisions occur. However, if a node has no data to send, its slot is wasted, leading to inefficiency under light load. TDMA is used in cellular networks (GSM) and satellite systems.
FDMA divides the available frequency spectrum into channels, each assigned to a node. Nodes transmit on their own frequency. Like TDMA, it avoids collisions but wastes spectrum if a node is idle. FDMA is used in radio and TV broadcasting, and early cellular systems.
CDMA assigns each node a unique code (spreading sequence). Nodes transmit simultaneously on the same frequency; the receiver can recover the original signal by correlating with the code. CDMA is used in 3G cellular and GPS. It provides resistance to interference and does not require time/frequency coordination, but is more complex.
| Protocol | Resource Partition | Advantages | Disadvantages |
|---|---|---|---|
| TDMA | Time slots | No collisions, predictable | Wasted slots, synchronization needed |
| FDMA | Frequency bands | No collisions, continuous | Wasted bandwidth, hardware complexity |
| CDMA | Code sequences | All nodes simultaneously, robust | Complex, near-far problem |
In random access, nodes transmit whenever they have data, with a mechanism to handle collisions. They are suitable for bursty traffic.
In Pure ALOHA, a node transmits immediately when it has a frame. If a collision occurs, the node waits a random time and retransmits. The maximum throughput is about 18.4% (1/(2e)) of the channel capacity. This was used in early satellite networks.
Slotted ALOHA divides time into slots equal to the frame transmission time. Nodes are synchronized and transmit only at the start of a slot. Collisions occur only if two nodes transmit in the same slot. Maximum throughput is about 36.8% (1/e). This improves on pure ALOHA by reducing the vulnerable period.
CSMA requires nodes to sense the channel before transmitting. If the channel is busy, they defer. This reduces collisions. Variants:
Performance depends on propagation delay: if the channel is sensed idle but a transmission from another node is already in transit (propagation delay), a collision can still occur.
CSMA/CD adds collision detection: nodes listen while transmitting and abort if they detect a collision, saving bandwidth. This is used in Ethernet (wired). After detecting a collision, nodes send a jamming signal and back off (exponential backoff).
CSMA/CA is used in wireless networks (Wi-Fi) because collision detection is not feasible (the transmitter cannot listen and transmit simultaneously on the same frequency). It uses virtual carrier sensing (RTS/CTS) and explicit acknowledgments to avoid collisions.
These protocols coordinate access by taking turns, avoiding collisions.
Let G be the average number of transmission attempts per frame time (including retransmissions). The vulnerable period is 2T (where T is frame time). The probability that no other transmission occurs during this period is e^{-2G}. Throughput S = G × e^{-2G}. Maximum at G=0.5, giving S_max = 1/(2e) ≈ 0.184.
Vulnerable period is T. Probability of no other transmission in the same slot is e^{-G}. Throughput S = G × e^{-G}. Maximum at G=1, S_max = 1/e ≈ 0.368.
CSMA performance depends on the ratio of propagation delay (a) to frame transmission time. For small a, throughput approaches 1. As a increases, collisions become more likely. The maximum throughput for 1-persistent CSMA with a given a can be derived, but we focus on qualitative understanding.
After a collision, nodes must back off (wait) before retransmitting. Ethernet uses binary exponential backoff: after k collisions, the backoff time is chosen uniformly from [0, 2^k - 1] slots. This reduces collision probability in subsequent attempts.
Ethernet (IEEE 802.3) originally used CSMA/CD on a shared coaxial cable. The protocol:
With switched Ethernet, full-duplex links eliminate collisions, and CSMA/CD is no longer used.
Wi-Fi (802.11) uses CSMA/CA with the following:
Wi-Fi also uses virtual carrier sensing via the Network Allocation Vector (NAV) to indicate channel reservation duration.
In wireless, node A can hear B but not C; node C can hear B but not A. If A and C both transmit to B simultaneously, a collision occurs at B even though both sensed the channel idle. RTS/CTS solves this: A sends RTS to B, B replies with CTS (which C can hear), so C defers.
Node B is transmitting to A; node C is within range of B but not A. C wants to transmit to D, but senses channel busy (due to B's transmission) and defers unnecessarily. This reduces throughput. Solutions exist (e.g., directional antennas, busy tone), but are complex.
| Protocol | Type | Throughput (max) | Collisions? | Use Case |
|---|---|---|---|---|
| Pure ALOHA | Random | 18.4% | Yes | Satellite (early) |
| Slotted ALOHA | Random | 36.8% | Yes | Satellite, some cellular |
| 1-persistent CSMA | Random | ~50-80% (depends) | Yes | Wired LANs (early) |
| CSMA/CD | Random | ~100% (under low load) | Yes (aborted) | Ethernet (shared) |
| CSMA/CA | Random | ~50-70% (with overhead) | Yes (avoidance) | Wi-Fi |
| TDMA | Partitioning | 100% (if all slots used) | No | Cellular, satellite |
| Token Passing | Taking-turns | ~100% (if traffic exists) | No | Token Ring, FDDI |
Satellite links often use TDMA or Slotted ALOHA. TDMA provides guaranteed bandwidth but is inefficient with bursty traffic. Slotted ALOHA is used in some systems (e.g., VSAT) because it handles bursty traffic well, though with lower throughput. Modern systems use dynamic allocation (e.g., DAMA) to combine benefits.
Question 1: What is the maximum throughput of Pure ALOHA?
(b) 1/(2e) ≈ 0.184. Pure ALOHA maximum throughput is 18.4%.
Question 2: Which of the following protocols uses virtual carrier sensing to avoid collisions?
(b) CSMA/CA. Wi-Fi uses virtual carrier sensing via RTS/CTS and NAV.
Question 3: In Slotted ALOHA, the vulnerable period is:
(a) One slot time. Transmission starts only at slot boundaries; collisions only occur if another node transmits in the same slot.
Question 4: What is the main reason CSMA/CD cannot be used in wireless networks?
(b) Collision detection is difficult because the transmitter cannot listen and transmit simultaneously. In wireless, a node cannot detect collisions while transmitting due to the half-duplex nature and signal attenuation.
Question 5: Which of the following is a channel partitioning protocol?
(c) FDMA. FDMA (Frequency Division Multiple Access) partitions the frequency spectrum.
Question 6: In CSMA/CD, after a collision, the backoff algorithm used is:
(b) Binary exponential backoff. Ethernet uses this to reduce collision probability.
Question 7: The hidden terminal problem is a concern in:
(b) Wireless networks. Hidden terminals occur when nodes are out of range of each other but can cause collisions at a common receiver.
Question 8: In pure ALOHA, if G = 0.5, what is the throughput?
(c) 0.5 × e^{-1} ≈ 0.184. S = G × e^{-2G} = 0.5 × e^{-1} ≈ 0.184.
Question 9: Which protocol achieves the highest throughput under low load?
(d) CSMA/CD. Under low load, collisions are rare, and CSMA/CD allows immediate transmission with detection, achieving near 100% throughput.
Question 10: The Network Allocation Vector (NAV) in Wi-Fi is used for:
(b) Virtual carrier sensing. NAV indicates the duration of a channel reservation, allowing nodes to defer even if they cannot sense the carrier.
Question 11: Which of the following is a taking-turns protocol?
(b) Polling. Polling is a taking-turns protocol where a master node grants permission.
Question 12: In Slotted ALOHA, the optimal offer load G that maximizes throughput is:
(b) 1. Throughput S = G e^{-G}, derivative gives G=1.
Question 13: Which problem does the RTS/CTS mechanism primarily address?
(b) Hidden terminal. RTS/CTS reserves the channel and makes other nodes aware of the ongoing transmission, mitigating hidden terminal collisions.
Question 14: What is the vulnerable period for Pure ALOHA?
(b) Two frame times. Because collisions can occur if another transmission starts within one frame time before or after.
Question 15: In Ethernet (CSMA/CD), the minimum frame size is determined by:
(a) The maximum propagation delay and the need to detect collisions. A frame must be long enough to be transmitted while the sender can still detect a collision (i.e., before the first bit reaches the farthest node).
Question 16: Which multiple access protocol is most suitable for real-time, constant-bit-rate traffic?
(c) TDMA. TDMA provides guaranteed time slots and low jitter, ideal for constant-bit-rate traffic like voice.
Question 17: The exposed terminal problem occurs when:
(a) A node can hear a transmission but is not in the same collision domain. The node defers unnecessarily, reducing throughput.
Question 18: In 1-persistent CSMA, if the channel is busy, the node:
(b) Waits for the channel to become idle, then transmits with probability 1. That is the "1-persistent" behavior.
Question 19: Which protocol uses a token to control access?
(b) Token Ring. Token Ring uses a token-passing protocol.
Question 20: What is the primary advantage of CDMA over TDMA and FDMA?
(b) No need for frequency or time coordination. CDMA allows all nodes to transmit simultaneously using different codes, simplifying network planning.
Question 21: In CSMA/CA, the DIFS (Distributed InterFrame Space) is used to:
(c) Ensure the channel is idle before transmission. DIFS is a period of idle time before a node can attempt to transmit.
Question 22: Which of the following is true about Slotted ALOHA compared to Pure ALOHA?
(b) It requires synchronization. Slotted ALOHA needs all nodes to be synchronized to slot boundaries.
Question 23: What is the effect of increasing propagation delay on CSMA performance?
(b) Decreases throughput because the probability of collisions increases. Longer propagation delay means nodes may sense the channel idle while a transmission is already in progress.
Question 24: In a token ring network, if the token is lost, how is it recovered?
(b) A designated node generates a new token. Token Ring uses a monitor node to detect token loss and regenerate it.
Question 25: What is the role of the jamming signal in CSMA/CD?
(b) To ensure all nodes detect a collision. The jamming signal propagates to ensure all stations are aware of the collision.
Question 26: Which protocol is commonly used in Wi-Fi for contention-based access?
(b) CSMA/CA. Wi-Fi uses CSMA/CA for distributed channel access.
Question 27: In pure ALOHA, if the total offered load G is 0.5, what is the probability that a given transmission is successful?
(a) e^{-1} ≈ 0.368. The probability of no other transmission in the vulnerable period (2T) is e^{-2G} = e^{-1} ≈ 0.368.
Question 28: Which access method is used by Ethernet in full-duplex mode?
(c) No access control needed (no collisions). In full-duplex switched Ethernet, each link is point-to-point, so collisions cannot occur.
Question 29: The near-far problem is associated with which multiple access technique?
(c) CDMA. In CDMA, a nearby transmitter may overpower a distant one, making it difficult to decode the distant signal.
Question 30: What is the main advantage of taking-turns protocols over random access protocols?
(c) No collisions and deterministic access. Taking-turns protocols eliminate collisions and provide guaranteed access times.
Exercise 1: Throughput of Pure ALOHA
For a pure ALOHA network, the offered load G is 0.8. Calculate the throughput S. If G is increased to 1.2, what happens to S?
S = G × e^{-2G}.
For G=0.8: S = 0.8 × e^{-1.6} = 0.8 × 0.2019 = 0.1615 (16.15%).
For G=1.2: S = 1.2 × e^{-2.4} = 1.2 × 0.0907 = 0.1088 (10.88%).
Throughput decreases because the collision probability increases with higher offered load.
Exercise 2: Slotted ALOHA Optimal G
Prove that the maximum throughput of Slotted ALOHA occurs at G=1.
S(G) = G e^{-G}. Derivative: S'(G) = e^{-G} - G e^{-G} = e^{-G}(1 - G). Set S'=0 ⇒ 1 - G = 0 ⇒ G=1.
Second derivative: S''(G) = -e^{-G}(1-G) - e^{-G} = e^{-G}(G-2). At G=1, S''(1) = e^{-1}(-1) < 0, so it's a maximum.
Exercise 3: CSMA/CD Collision Detection
In a 10 Mbps Ethernet with a maximum cable length of 2.5 km and propagation speed of 2×10^8 m/s, what is the minimum frame size required for collision detection?
Propagation delay one way = distance / speed = 2500 / (2×10^8) = 1.25×10^{-5} s = 12.5 µs.
Round-trip time = 2 × 12.5 = 25 µs.
At 10 Mbps, bits transmitted in 25 µs = 10×10^6 × 25×10^{-6} = 250 bits.
Thus, minimum frame size = 250 bits (including headers). Ethernet uses 512 bits (64 bytes) to be safe.
Exercise 4: Hidden Terminal Mitigation
Describe how the RTS/CTS handshake works to solve the hidden terminal problem. Include the sequence of frames and how they affect other nodes.
1. Node A sends an RTS (Request to Send) frame to node B, indicating the duration of the upcoming data transmission.
2. Node B responds with a CTS (Clear to Send) frame, also containing the duration. CTS is heard by all nodes within range of B, including hidden nodes that could not hear A.
3. Other nodes (like C) that hear the CTS set their NAV (Network Allocation Vector) for the duration, deferring transmission.
4. A then sends the data frame to B.
5. B acknowledges with an ACK.
This ensures that hidden nodes defer during the data transmission.
Exercise 5: Exposed Terminal Example
Illustrate the exposed terminal problem with a diagram and explain why it reduces throughput.
Nodes: A, B, C, D in a line. A is transmitting to B. C can hear B's transmission (so senses busy) but C wants to transmit to D. Since C is not in B's transmission range (or not causing interference at B), C could actually transmit to D without causing collisions. However, C defers because it senses the channel busy, wasting throughput.
This reduces overall efficiency because C unnecessarily waits.
Exercise 6: TDMA vs. Random Access
A network has 10 nodes, each generating traffic with probability 0.1 per slot (on average). Compare the efficiency of Slotted ALOHA and TDMA with 10 slots per frame.
With TDMA, each node gets 1 slot per frame. The channel is fully utilized only if all nodes have data; otherwise some slots are wasted. Efficiency = (number of nodes with data) / 10 on average. Expected number of busy nodes = 10 × 0.1 = 1, so efficiency ≈ 1/10 = 10% (plus overhead).
With Slotted ALOHA, G = total attempts per slot = 10 × 0.1 = 1. Throughput S = 1 × e^{-1} = 0.368 = 36.8%, much higher. So random access is more efficient for bursty traffic.
Exercise 7: CSMA/CA Backoff
In Wi-Fi, after a collision, the backoff timer is chosen from the range [0, CW-1] where CW (contention window) starts at 15 and doubles up to 1023. If a node experiences 4 collisions, what is the range of backoff slots?
After 1st collision: CW=31, range [0,30].
After 2nd: CW=63, range [0,62].
After 3rd: CW=127, range [0,126].
After 4th: CW=255, range [0,254].
The node picks a random integer in that range, multiplied by the slot time.
Exercise 8: Efficiency of Polling
A polling system has 5 stations. The poll frame takes 1 ms, and each station's data frame takes 10 ms. The overhead for polling is 1 ms per station. Calculate the maximum throughput (fraction of time used for data) if all stations always have data.
For each cycle, the master polls each station: 5 polls × 1 ms = 5 ms overhead. Each station sends data: 5 × 10 ms = 50 ms data. Total cycle time = 55 ms. Throughput = data time / cycle time = 50/55 ≈ 0.909 = 90.9%.
Exercise 9: CSMA Collision Probability
In a CSMA network with propagation delay a = 0.01 (relative to frame time), what is the approximate probability that a collision occurs when two nodes start transmissions at the same time? Assume they sense the channel idle.
If both nodes sense idle, they transmit. The collision occurs if the transmission from one node hasn't reached the other before it starts. In the worst case, the second node starts within the propagation delay of the first. The probability depends on the timing. With a small a, collisions are rare. A crude approximation: collision probability roughly a (since if the second node starts within a of the first, it may not yet have detected the carrier). More precise analysis requires modeling.
In this case, a=0.01, so ~1% chance.
Exercise 10: Token Passing Overhead
A token ring network has 10 stations. The token transmission time is negligible, but each station adds 1 bit of latency. The data rate is 100 Mbps. The ring length is 1000 m, propagation speed 2×10^8 m/s. Calculate the total ring latency (propagation + station delays).
Propagation delay = 1000 / (2×10^8) = 5×10^{-6} s = 5 µs.
Station delay = 10 stations × 1 bit = 10 bits. At 100 Mbps, 1 bit = 10 ns, so 10 bits = 100 ns = 0.1 µs.
Total latency = 5 µs + 0.1 µs = 5.1 µs.
Exercise 11: ALOHA vs. Slotted ALOHA
Compare the maximum throughput of pure and slotted ALOHA. By what factor does slotted improve throughput?
Pure ALOHA max = 1/(2e) ≈ 0.184.
Slotted ALOHA max = 1/e ≈ 0.368.
Improvement factor = (1/e) / (1/(2e)) = 2. So slotted ALOHA doubles the maximum throughput.
Exercise 12: CSMA/CD Frame Size
In a 100 Mbps Ethernet with a maximum collision domain diameter of 200 m, what is the minimum frame size? Assume propagation speed 2×10^8 m/s.
Round-trip time = 2 × (200 / 2×10^8) = 2 × 1×10^{-6} = 2×10^{-6} s = 2 µs.
At 100 Mbps, bits in 2 µs = 100×10^6 × 2×10^{-6} = 200 bits.
Minimum frame size = 200 bits (plus some margin). Ethernet uses 512 bits.
Exercise 13: CDMA Example
Suppose two nodes use CDMA with codes: A: (+1 +1 -1 -1), B: (+1 -1 +1 -1). If both transmit data bits dA=1, dB=-1, what is the composite signal and how does the receiver for A recover the data?
Transmitted signals: A: 1 × (+1 +1 -1 -1) = (+1 +1 -1 -1). B: -1 × (+1 -1 +1 -1) = (-1 +1 -1 +1).
Composite = (+1-1, +1+1, -1-1, -1+1) = (0, +2, -2, 0).
Receiver A correlates with its code: dot product = (0×1) + (2×1) + (-2×-1) + (0×-1) = 0 + 2 + 2 + 0 = 4. Divide by code length (4) → data = 1. Correct.
Exercise 14: Backoff Performance
In a CSMA/CD network, after 3 collisions, the backoff range is [0,7]. What is the probability that two nodes with the same backoff range pick the same backoff value?
Backoff values: 0..7 (8 choices). Probability that two nodes pick the same value = 1/8 ≈ 0.125.
If they pick the same, they will collide again on retransmission.
Exercise 15: Comparing Protocols for IoT
In a low-power IoT network with many devices sending small packets infrequently, which multiple access protocol would you recommend? Justify.
Recommend Slotted ALOHA or CSMA/CA with low overhead. Since traffic is bursty and infrequent, random access avoids the inefficiency of TDMA (wasted slots). Slotted ALOHA is simple and efficient for low load. Alternatively, CSMA/CA can be used with a small contention window to reduce energy consumption. Polling would be inefficient due to overhead.
Homework 1: Derive the throughput formula for Slotted ALOHA (S = G e^{-G}) from first principles. Define all variables and explain the vulnerable period.
Let G be the average number of frame transmission attempts per slot (including retransmissions). The vulnerable period is one slot time. A transmission is successful if no other node transmits in that slot. The number of transmissions in a slot follows a Poisson distribution with mean G. The probability that exactly zero other transmissions occur in the same slot is e^{-G}. The throughput S is the fraction of slots that contain successful transmissions, which is the probability that a transmission attempt is successful times the average number of attempts? More formally: S = G × (probability that a given attempt is successful). The probability that a given attempt is successful is the probability that no other attempts occur in that slot, which is e^{-G}. Thus S = G e^{-G}. Maximum at G=1, S=1/e.
Homework 2: Explain the difference between 1-persistent, non-persistent, and p-persistent CSMA. Under what conditions would each be preferred?
1-persistent is good for low-load, low-latency; non-persistent for high-load; p-persistent for slotted channels like Wi-Fi (though Wi-Fi uses a different backoff).
Homework 3: A network uses Slotted ALOHA with 10 nodes. Each node has a packet to send with probability 0.2 per slot. What is the throughput? If the nodes are synchronized, what is the probability that a given slot has a successful transmission?
G = number of nodes × probability per node = 10 × 0.2 = 2 attempts per slot.
Throughput S = G e^{-G} = 2 e^{-2} = 2 × 0.1353 = 0.2707 (27.1%).
The probability that a slot has a successful transmission is S = 0.2707.
Homework 4: Describe the operation of the Distributed Coordination Function (DCF) in IEEE 802.11. Include the interframe spaces (DIFS, SIFS) and the backoff procedure.
DCF uses CSMA/CA. A node with data to send senses the channel. If idle for DIFS (Distributed InterFrame Space), it transmits. If busy, it waits until idle, then chooses a random backoff timer (slots). The timer decrements while the channel is idle; if the channel becomes busy, it freezes. When timer reaches zero, node transmits. If a collision occurs (no ACK), the node doubles the contention window and retries. SIFS (Short InterFrame Space) is used for ACK and RTS/CTS, having higher priority (shorter time).
Homework 5: Compare and contrast the multiple access protocols used in Ethernet (CSMA/CD) and Wi-Fi (CSMA/CA). Why does Wi-Fi not use CSMA/CD?
Both use carrier sensing before transmission. Ethernet uses CSMA/CD: it detects collisions during transmission and aborts, using exponential backoff. Wi-Fi uses CSMA/CA: it avoids collisions through virtual carrier sensing (RTS/CTS), acknowledgments, and backoff; it cannot detect collisions because the transmitter cannot listen simultaneously. Wi-Fi also has variable channel conditions and hidden terminals. Therefore, CSMA/CD is not feasible in wireless.
Homework 6: In a polling system, how does the overhead affect performance as the number of stations increases? Discuss the trade-off between polling and random access for large numbers of stations.
In polling, overhead increases linearly with number of stations (each poll frame). For large N, the polling cycle becomes long, increasing delay and reducing throughput (fraction of time spent polling). Random access protocols like CSMA/CA have overhead that is relatively independent of N (only collisions cause overhead). However, as N grows, collision probability increases, reducing throughput. For very large N, TDMA or a hybrid may be better. Polling is suitable for small to medium N with steady traffic; random access is better for bursty traffic and large N.
Homework 7: Explain the hidden terminal problem in wireless networks. How does the RTS/CTS mechanism solve it, and what are the drawbacks of using RTS/CTS?
Hidden terminal: A and C cannot hear each other, but both can hear B. They may transmit to B simultaneously, causing collisions. RTS/CTS: A sends RTS to B; B replies with CTS, which is heard by C (since C is within range of B). C defers. Drawbacks: Adds overhead (RTS, CTS frames) and increases latency; may not be efficient for small data frames; still susceptible to collisions if RTS frames collide.
Homework 8: Derive the maximum throughput of pure ALOHA. Show all steps including the definition of G, the vulnerable period, and the optimization.
Let G = average number of transmission attempts per frame time (including retransmissions). A frame is successful if no other frame is transmitted during the vulnerable period of 2 frame times (the frame itself plus one frame time before). The number of attempts in 2T is Poisson with mean 2G. Probability of zero other attempts = e^{-2G}. Throughput S = G × e^{-2G}. To maximize, differentiate: dS/dG = e^{-2G} - 2G e^{-2G} = e^{-2G}(1 - 2G). Set to 0 ⇒ G=0.5. S_max = 0.5 × e^{-1} = 1/(2e) ≈ 0.184.
Homework 9: What is the "near-far" problem in CDMA, and how is it typically mitigated?
The near-far problem occurs when a mobile station close to the base station transmits at high power, overpowering a distant station's signal, making it hard to decode the distant station. Mitigations include power control (adjusting transmission power so that all signals arrive at roughly equal power), and using orthogonal codes with low cross-correlation.
Homework 10: In a CSMA/CD network, suppose the propagation delay is 5 µs and the transmission rate is 10 Mbps. What is the minimum frame size required for collision detection? If the frame size is smaller, what can happen?
Round-trip delay = 2 × 5 = 10 µs. At 10 Mbps, bits in 10 µs = 10×10^6 × 10×10^{-6} = 100 bits. So minimum frame size = 100 bits (plus jam signal). If smaller, a collision might occur after the sender has finished transmitting, so the sender will not detect it and will assume successful transmission, leading to corrupted data.
Homework 11: Compare the performance of Slotted ALOHA and CSMA under light load and heavy load. Which is more stable?
Under light load: CSMA has very few collisions and high throughput (near 100%). Slotted ALOHA throughput is at most 36.8%. Under heavy load: CSMA suffers from increased collisions (especially if propagation delay is significant), but with backoff it can maintain reasonable throughput. Slotted ALOHA throughput drops as G increases beyond 1 (since S = G e^{-G} decreases). CSMA is more stable because it senses the channel; ALOHA is unstable at high loads.
Homework 12: In token ring networks, how is fairness ensured? What happens if a station has no data to send?
Fairness is ensured because each station gets a turn to transmit when it holds the token. If a station has no data, it simply passes the token to the next station. This guarantees each station gets equal opportunity, and the token circulates continuously.
Homework 13: Explain the concept of "carrier sense" in CSMA. Why is it not sufficient to completely avoid collisions?
Carrier sense means listening to the channel before transmitting to check if it is busy. However, collisions can still occur due to propagation delay: two nodes may both sense the channel idle and transmit simultaneously before their signals reach each other. Also, in wireless, hidden terminals cause collisions even if carrier sense is used locally.
Homework 14: A network uses TDMA with 5 stations. Each station gets a slot of 10 ms in a frame of 50 ms. If station 1 is idle, what happens to its slot? Can the slot be reused by other stations?
In basic TDMA, the slot is wasted; no other station can use it. More advanced dynamic TDMA can allocate idle slots to other stations, but standard TDMA does not. This is a major inefficiency.
Homework 15: In Wi-Fi (CSMA/CA), what is the purpose of the Network Allocation Vector (NAV) and how does it differ from physical carrier sensing?
NAV is a virtual carrier sensing mechanism. It is set by nodes when they hear RTS/CTS frames indicating the duration of a transmission. Nodes defer for that duration even if they cannot sense the physical carrier (e.g., due to being out of range of the sender). Physical carrier sensing detects actual energy on the channel. NAV provides additional protection against hidden terminals.
This tutorial covered the major multiple access protocols used in computer networks:
Understanding these protocols is essential for designing and managing LANs, WLANs, and other shared-media networks. In the next tutorial, we will explore MAC addressing and the Address Resolution Protocol (ARP).