An inland freight transport authority is evaluating the connectivity of a regional railway… — JAMB UTME

An inland freight transport authority is evaluating the connectivity of a regional railway network serving heavy industrial nodes in West Africa. In the baseline phase, the connected planar network ( $p = 1$ ) consists of $8$ urban freight terminals (vertices) linked by $11$ direct rail corridors (edges). Following an expansion project, $2$ new freight terminals and $4$ additional rail corridors are integrated into the existing system. What is the net change in the topological Alpha index ( $\alpha$ ) of the network as a result of this expansion?

A
An increase of $\frac{2}{33}$
B
An increase of $\frac{1}{72}$
An increase of $\frac{2}{55}$ Answer
D
An increase of $\frac{1}{8}$

Answer

The net change in the topological Alpha index of the network is an increase of

\frac{2}{55}

The correct response accurately applies the Alpha index formula for planar graphs: $\alpha = \frac{e - v + p}{2v - 5}$. For the baseline phase, $\alpha_1 = \frac{11 - 8 + 1}{2(8) - 5} = \frac{4}{11}$. For the expanded phase with 10 vertices and 15 edges, $\alpha_2 = \frac{15 - 10 + 1}{2(10) - 5} = \frac{6}{15} = \frac{2}{5}$. The difference $\frac{2}{5} - \frac{4}{11} = \frac{2}{55}$ represents the exact increase in topological network connectivity.

Step-by-Step Solution

Identify the formula for the Alpha index ($\alpha$) of a planar transport network

The formula is $\alpha = \frac{e - v + p}{2v - 5}$, where $e$ is edges, $v$ is vertices, and $p$ is the number of non-connected subgraphs ($p = 1$ for a connected graph).

The Alpha index measures the degree of connectivity by comparing the actual number of fundamental circuits (cyclomatic number) to the maximum possible number of circuits in a planar graph.

Calculate the baseline Alpha index ($\alpha_1$)

For $v_1 = 8$ and $e_1 = 11$, $\alpha_1 = \frac{11 - 8 + 1}{2(8) - 5} = \frac{4}{11}$.

Substitute the initial numbers of terminals and rail corridors into the formula.

Calculate the post-expansion Alpha index ($\alpha_2$)

After adding 2 terminals and 4 corridors, $v_2 = 10$ and $e_2 = 15$. Thus, $\alpha_2 = \frac{15 - 10 + 1}{2(10) - 5} = \frac{6}{15} = \frac{2}{5}$.

Update total vertices and total edges to compute the expanded network's connectivity ratio.

Determine the net change ($\alpha_2 - \alpha_1$)

$\alpha_2 - \alpha_1 = \frac{2}{5} - \frac{4}{11} = \frac{22 - 20}{55} = \frac{2}{55}$.

Subtract the baseline Alpha index from the post-expansion Alpha index.

Key Concept

Alpha Index of Network Connectivity

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