1. **Problem Statement:** Determine a new network configuration by selecting cables (edges) to retain in the graph connecting districts a, b, c, d, e, f, g, h, i such that all districts remain connected and the total yearly maintenance cost is minimized. 2. **Relevant Concept:** This is a classic Minimum Spanning Tree (MST) problem in graph theory. - MST is a subset of edges connecting all vertices with no cycles and minimum total edge weight. - Common algorithms: Kruskal's or Prim's algorithm. 3. **Given Graph Edges and Weights (in thousands):** - (a,b): 3 - (b,c): 2 - (a,d): 1 - (b,e): 7 - (c,f): 12 - (d,e): 6 - (e,f): 11 - (d,g): 4 - (e,h): 10 - (f,i): 5 - (g,h): 8 - (h,i): 9 4. **Step-by-step MST Construction using Kruskal's Algorithm:** - Sort edges by weight ascending: $$ (a,d)=1, (b,c)=2, (a,b)=3, (d,g)=4, (f,i)=5, (d,e)=6, (b,e)=7, (g,h)=8, (h,i)=9, (e,h)=10, (e,f)=11, (c,f)=12 $$ - Start adding edges without forming cycles: 1. Add (a,d) weight 1 2. Add (b,c) weight 2 3. Add (a,b) weight 3 (connects a,b,c,d) 4. Add (d,g) weight 4 5. Add (f,i) weight 5 6. Add (d,e) weight 6 7. Add (b,e) weight 7 (connects e to the main component) - Now vertices a,b,c,d,e,g,f,i are connected except h. 8. Add (g,h) weight 8 (connects h) - Adding any other edge would create cycles or is unnecessary. 5. **Edges to Retain:** $$ \{(a,d), (b,c), (a,b), (d,g), (f,i), (d,e), (b,e), (g,h)\} $$ 6. **Proof of Minimality:** - The MST includes all vertices with minimum total weight. - Total weight: $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$$ - Any other spanning tree would have equal or greater total weight. 7. **Conclusion:** The cables to retain for minimum maintenance cost while keeping all districts connected are: - a-d - b-c - a-b - d-g - f-i - d-e - b-e - g-h This configuration ensures connectivity and minimal yearly maintenance cost of 36 (thousands).