ALOHA Protocols MCQs: 10 Solved Questions with Step-by-Step Explanations

ALOHA questions can look like formula recall, but the usual mistakes happen one step earlier. A solver uses the pure ALOHA exponent in a slotted question, forgets to express the arrival rate per frame or slot time, or confuses the success probability of one named station with that of any one among n stations.

Eight of the ten questions below are previous-year questions from Coal India, RPSC, TPSC, ISRO, UGC NET and GATE papers; the other two are unattributed practice numericals on the same formulas. Attempt each question before reading its explanation.

ALOHA MCQs: the formula map to use before solving

Let T = L/R, where L is frame length in bits and R is channel rate in bits per second. For aggregate attempt rate lambda in frames per second, G = lambda T.

G is dimensionless offered load in attempts per frame time or slot. S is normalised successful throughput. For successful frames per second, multiply S by 1/T.

For n independent stations transmitting with probability p, P(exactly one transmission) = np(1-p)^(n-1). For one specified host, P(the specified host succeeds) = p(1-p)^(n-1).

Slotted and Pure ALOHA formulas and maximum throughput

Question 1, Coal India 2020

Maximum throughput of slotted ALOHA network is:

A. 18.4%

B. 36.8%

C. 50%

D. 35.8%

Answer: B. 36.8%

Slotted ALOHA has S = Ge^(-G). Differentiation gives the peak at G = 1, so S_max = 1/e = 0.367879, or about 36.8%. Pure ALOHA peaks at 1/(2e), about 18.4%, which explains distractor A.

Question 2, RPSC Programmer Paper 1 2024

In a slotted ALOHA with G-stations attempting to transmit in one slot, the throughput is equal to:

A. G × e⁻ᴳ

B. G × e⁻²ᴳ

C. G × eᴳ

D. 2G × e⁻ᴳ

Answer: A. G × e⁻ᴳ

A successful slot contains exactly one attempt. Under the Poisson model, P(X = 1) = e^(-G)G^1/1! = Ge^(-G). The expression Ge^(-2G) belongs to pure ALOHA because its vulnerable interval spans two frame times. A positive exponent cannot represent collision-limited throughput.

Question 3, TPSC System Analyst 2026

At what value of load G the maximum throughput S of PURE ALOHA occurs ?

A. 0.5

B. 1.0

C. 1.5

D. 2.0

Answer: A. 0.5

For pure ALOHA, S = Ge^(-2G). Therefore, dS/dG = e^(-2G)(1 - 2G). Setting this derivative to zero gives G = 1/2. As a check, S_max = 0.5e^(-1) = 0.18394, but the question asks for G, not S. The same differentiate-then-substitute step gives G = 1 for slotted ALOHA. Both derivations, and the rest of this subtopic's practice questions, sit in the ALOHA Protocols module.

ALOHA offered load and normalised throughput numericals

Question 4, ISRO 2015

A certain population of ALOHA users manages to generate 70 request/sec. If the time is slotted in units of 50 msec, then channel load would be

A. 4.25

B. 3.5

C. 350

D. 450

Answer: B. 3.5

Convert 50 msec = 0.05 s. Then G = lambda T = 70 × 0.05 = 3.5 attempts per slot. The value 350 comes from multiplying before converting to seconds. No throughput formula is needed because only load is requested.

Question 5, UGC NET June 2016

A pure ALOHA Network transmits 200 bit frames using a shared channel with 200 Kbps bandwidth. If the system (all stations put together) produces 500 frames per second, then the throughput of the system is ______.

A. 0.384

B. 0.184

C. 0.286

D. 0.586

Answer: B. 0.184

Frame time is T = 200/200000 = 0.001 s. Thus, G = 500 × 0.001 = 0.5. Since this is pure ALOHA, S = Ge^(-2G) = 0.5e^(-1) = 0.183939, which rounds to 0.184. The slotted formula would give 0.5e^(-0.5) = 0.3033, so it does not fit the protocol named in the stem.

Question 6, UGC NET August 2016

A slotted ALOHA network transmits 200 bits frames using a shared channel with 200 kbps bandwidth. If the system (all stations put together) produces 1000 frames per second, then the throughput of the system is _______.

A. 0.268

B. 0.468

C. 0.368

D. 0.568

Answer: C. 0.368

Again, T = 200/200000 = 0.001 s, but now G = 1000 × 0.001 = 1. Slotted ALOHA gives S = Ge^(-G) = 1/e = 0.367879, rounded to 0.368. Since G = 1 is also the peak-load checkpoint, the result must equal the known slotted maximum.

Exactly one transmitter and per-host success probability

Question 7, GATE 2007

There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot?

A. (1-p)⁽ⁿ⁻¹⁾

B. np(1-p)⁽ⁿ⁻¹⁾

C. p(1-p)⁽ⁿ⁻¹⁾

D. 1-(1-p)⁽ⁿ⁻¹⁾

Answer: B. np(1-p)⁽ⁿ⁻¹⁾

Choose the sole transmitter in n ways. That station transmits with probability p, while the other n - 1 stations stay silent with probability (1-p)^(n-1). Thus, the result is np(1-p)^(n-1). Option C applies to one specified station, not any one of n stations.

Question 8

There are 14 stations in a slotted LAN. Each station attempts to transmit with a probability of 0.35 in each time slot. What is the probability that ONLY one station transmits in a given time slot?

A. 0.018

B. 0.019

C. 0.021

D. 0.022

Answer: A. 0.018

Substitute in the exactly-one formula: P = 14 × 0.35 × (1 - 0.35)^13 = 4.9 × 0.65^13 = 0.018116308866, which rounds to 0.018. Keep enough precision: 0.65^13 is about 0.0036975, and rounding it to 0.004 turns the product into 0.0196, which lands on distractor B.

Question 9, GATE Information Technology 2004

Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability p = 0.2 in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support, if each host has to be provided a minimum throughput of 0.16 frames per time slot?

A. 1

B. 2

C. 3

D. 4

Answer: B. 2

Use per-host throughput: 0.2(0.8)^(n-1) >= 0.16, so (0.8)^(n-1) >= 0.8. Check the boundary: n = 2 gives 0.2 × 0.8 = 0.16, while n = 3 gives 0.2 × 0.8^2 = 0.128. Therefore, two is the maximum.

Pure ALOHA vulnerable time and minimum frame size

Question 10

A Pure ALOHA network has a bandwidth of 6.8 × 10⁸ bits/second. It has a vulnerable time of 2.2 µs. What is the size of minimum frame in this network?

A. 762 bits

B. 187 bytes

C. 93.5 bytes

D. 95 bytes

Answer: C. 93.5 bytes

Pure ALOHA's vulnerable time is 2T, so T = 2.2/2 = 1.1 microseconds. Then L = RT = 6.8 × 10^8 × 1.1 × 10^(-6) = 748 bits, and 748/8 = 93.5 bytes. Option B treats the full 2.2 microseconds as one frame time. Options A and D sit within 2% of the right answer (762 bits against 748 bits, and 95 bytes against 93.5 bytes), so this one has to be calculated rather than eyeballed.

ALOHA MCQ answer-review map: diagnose the formula mistake

Four traps deserve a final check:

• Convert milliseconds or microseconds to seconds before multiplying by a per-second rate.

• Do not select Ge^(-2G) merely because a collision is mentioned. First identify pure or slotted ALOHA.

• Include the n choices when the question asks for any one station, not a specified station.

• Keep units attached: 748 is bits, while 748/8 = 93.5 is bytes.

For more numerical practice, try Subnetting MCQs: 12 Solved IP Addressing Questions. For another protocol-comparison set, use TCP and UDP MCQs: 12 Solved Transport Layer Questions.

ALOHA Protocols MCQs: the short version and next practice step

• Pure ALOHA uses Ge^(-2G) and peaks at G = 0.5.

• Slotted ALOHA uses Ge^(-G) and peaks at G = 1.

• Offered load is arrival rate multiplied by frame or slot time.

• Exactly one of n stations transmits successfully with np(1-p)^(n-1).

• Pure ALOHA vulnerable time is twice the frame time.

Redo Questions 4, 7, and 10 without looking. They test the three distinct calculation families: offered load, exactly-one-transmitter probability, and vulnerable time with unit conversion.

KnowledgeGate carries over 20 ALOHA Protocols questions with worked solutions. You can browse the GATE CS Exam preparation category, follow a structured subject path through GATE Guidance by Sanchit Sir, or move to timed mixed practice with the GATE Test Series.