A network designer connects 5 routers so that each ordered pair of routers has a one-way (simplex) point-to-point line. How many simplex lines are required for full pairwise reachability?

Introduction / Context:
Designing interconnections between routers depends on link directionality. With duplex links, a single line supports both directions. With simplex links, each direction needs its own separate line. Counting the number of links required clarifies physical cost and scalability.

Given Data / Assumptions:

• There are n = 5 routers.
• Links are point-to-point and simplex (supporting traffic in only one direction).
• Goal: every ordered pair (A → B) has a dedicated path (complete directed connectivity using direct links).

Concept / Approach:
In a complete mesh with duplex lines, the count is n*(n-1)/2 undirected links. For simplex lines, each undirected pair requires two oppositely directed lines. Therefore, the required count doubles: n*(n-1). For n = 5, that equals 5*4 = 20.

Step-by-Step Solution:

Verification / Alternative check:
General formula: simplex lines = n*(n-1). Substitute n = 5 → 20. Cross-check by enumerating pairs (A,B) where A ≠ B; there are 5*4 such ordered pairs.

Why Other Options Are Wrong:

• 5: corresponds to a ring, not a full mesh.
• 10: equals a full mesh if lines were duplex, not simplex.
• 32: exceeds n*(n-1) for n = 5; not applicable.
• None: incorrect because 20 is exact.

Common Pitfalls:
Forgetting to double the pair count for simplex; confusing point-to-point with shared media; assuming multi-hop routing removes the need for links (the question asks for direct lines between every ordered pair).

Final Answer:
20