Maximum channel utilization in case of pure ALOHA is:

Question

Accepted Answer

The correct option is (2):

The maximum channel utilization (throughput) for **Pure ALOHA** is a fundamental concept in computer networking, specifically in the study of random access protocols. Pure ALOHA allows stations to transmit data frames whenever they have them, leading to a high probability of collisions. The maximum theoretical throughput is calculated using the formula $S = G \cdot e^{-2G}$, where $S$ is the throughput and $G$ is the offered channel load. To find the maximum value, we take the derivative of $S$ with respect to $G$ and set it to zero, which yields the optimal load $G\_{max}=0.5$. Substituting $G=0.5$ back into the formula gives $S\_{max} = 0.5 \cdot e^{-1}$, which is approximately $0.5 \cdot 0.3679 \approx 0.1839$. Therefore, the maximum channel utilization is about **$18.39%$**, making **$18%$** the closest and correct choice.

The incorrect option is (1):
The value **$39%$** (or more precisely $36.78%$ or $1/e$) represents the maximum channel utilization for the **Slotted ALOHA** protocol, not the Pure ALOHA protocol described in the question. Slotted ALOHA improves upon Pure ALOHA by dividing time into discrete slots and forcing stations to start transmission only at the beginning of a slot. This synchronization halves the vulnerable period for a collision compared to Pure ALOHA. The formula for Slotted ALOHA's throughput is $S = G \cdot e^{-G}$, which maximizes at an offered load of $G=1$. This yields a maximum utilization of approximately $37%$ to $39%$. Since the question specifically asks for *pure* ALOHA, this value is factually incorrect for the stated protocol.

The incorrect option is (3):
The option **$27%$** is incorrect because it does not correspond to the calculated theoretical maximum channel utilization for the Pure ALOHA protocol. The maximum efficiency for Pure ALOHA is mathematically proven to be fixed at $18.39%$ ($1/2e$). While $27%$ might represent the channel's utilization at a specific, suboptimal offered load ($G$) that is higher than $0.5$, it is not the absolute theoretical maximum throughput possible for the protocol. When answering an exam question like this, you must recall the maximum theoretical value, not a random point on the throughput curve.

The incorrect option is (4):
The option **$54%$** is significantly higher than the theoretical maximum for any version of the ALOHA protocol. The fundamental flaw in the ALOHA family of protocols is the lack of sensing before transmission, which inherently limits their efficiency due to a high rate of packet collisions. A utilization rate of $54%$ represents a level of efficiency that is typically achieved by more complex, controlled protocols, such as Carrier Sense Multiple Access with Collision Detection (**CSMA/CD**) used in Ethernet networks. Therefore, $54%$ is far too optimistic for a basic random access method like Pure ALOHA.