Scenario: A smart factory is deploying 25 industrial sensors in a partial mesh topology. The network architect must decide on the optimal number of connections to balance fault tolerance against cost.
Given Information:
- 25 sensor nodes
- Connection cost: $120 per wireless link (hardware + configuration)
- Reliability requirement: Network must survive 2 simultaneous node failures
- Budget constraint: $3,600 for connectivity infrastructure
Step 1: Calculate Full Mesh Baseline
# Full mesh connections formula: n(n-1)/2
nodes = 25
full_mesh_connections = (nodes * (nodes - 1)) // 2
= (25 * 24) // 2
= 300 connections
full_mesh_cost = 300 × $120 = $36,000
Conclusion: Full mesh is 10× over budget and unnecessary for this deployment.
Step 2: Determine Minimum Connectivity for Fault Tolerance
For network to survive 2 simultaneous failures with reasonable probability: - Each node needs at least 3 connections (if 2 neighbors fail, 1 remains) - Industry standard: 40-60% connectivity ratio provides good fault tolerance
Step 3: Calculate Partial Mesh Options
# Option A: 40% connectivity
connections_40pct = int(300 × 0.40) = 120 connections
cost_40pct = 120 × $120 = $14,400 (still over budget)
# Option B: 30% connectivity (minimum for 3 connections/node avg)
connections_30pct = int(300 × 0.30) = 90 connections
cost_30pct = 90 × $120 = $10,800 (still over budget)
# Option C: Budget-constrained approach
budget_connections = $3,600 / $120 = 30 connections
connectivity_ratio = 30 / 300 = 0.10 (10%)
avg_connections_per_node = (30 × 2) / 25 = 2.4 connections/node
Step 4: Analyze Feasibility
With only 30 connections (10% connectivity): - Average 2.4 connections per node (below 3-connection minimum for 2-failure tolerance) - Risk: Some nodes may have only 1-2 neighbors, creating vulnerability
Step 5: Recommendation
Hybrid Topology Solution:
- Use tree topology backbone with selective mesh for critical nodes
- 24 connections form tree (each node connects to parent): 24 × $120 = $2,880
- 6 connections add redundancy to critical measurement points: 6 × $120 = $720
- Total: $3,600 (on budget)
- Connectivity ratio: 30/300 = 10% (but strategically placed)
Result: Tree provides basic connectivity to all nodes. Selective mesh adds redundancy where it matters most (nodes near production bottlenecks). Network survives single failure anywhere, and survives dual failure except at critical nodes (acceptable for budget constraint).
Key Insight: Connectivity ratio alone does not determine fault tolerance — link placement matters more than the raw percentage.