AP CSP Day 49: Fault Tolerance & Redundancy | Cycle 2

Key Concepts

Calculating fault tolerance precisely requires determining the minimum number of simultaneous link failures that can disconnect any node from the rest of the network. A node with only two connections becomes isolated if both fail, regardless of how many redundant connections exist elsewhere in the network. AP CSP Cycle 2 fault tolerance questions present complex network diagrams and ask students to identify the most vulnerable node or the minimum cut that disconnects two specific nodes. Systematic path-counting from the most constrained nodes outward is the most reliable exam strategy.

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Fault Tolerance: Calculating Minimum Cuts

What Is a Minimum Cut?

A minimum cut is the smallest set of connections whose simultaneous removal disconnects two specific nodes or splits the network into two parts. The minimum cut value equals the maximum number of link-disjoint paths between two nodes.

Finding Vulnerable Pairs

Two nodes are most vulnerable if there are few paths between them that share no links. If all paths between two nodes share one common link, removing that single link disconnects them, making their minimum cut equal to 1.

Common Trap: Counting total connections to a node rather than counting paths between two specific nodes. Fault tolerance is always calculated between a specific pair of nodes, not for the network as a whole.
Exam Tip: For each pair of nodes specified in an AP exam question, enumerate all distinct paths between them. Count how many links would need to fail simultaneously to disconnect them. That number is the fault tolerance level for that pair.
Big Idea 4: Computing Systems & Networks
Cycle 2 • Day 49 Practice • Hard Difficulty
Focus: Fault Tolerance & Redundancy

Practice Question

A network connects 5 devices with these links: A-B, A-C, B-C, B-D, C-D, C-E, D-E. What is the minimum number of connections that must fail before device A can no longer communicate with device E?

Why This Answer?

Device A connects to B and C. If both A-B and A-C fail, device A is completely isolated and cannot reach any other device, including E. Alternatively, if C-E and D-E both fail, device E becomes isolated. Either way, the minimum number of failures needed is 2.

Why Not the Others?

A) Removing any single connection still leaves alternative paths between A and E (for example, if A-B fails, A-C-D-E still works). C) Two failures are sufficient to isolate either A or E. D) Four failures is more than the minimum needed.

Common Mistake
Watch Out!

Students test only one path and conclude that one failure is sufficient. To find the minimum, you must verify that ALL alternative paths are also broken.

AP Exam Tip

To find minimum failures for disconnection, identify the node with the fewest connections — its degree equals the minimum cuts needed to isolate it. Check both endpoints.

Keep Practicing!

Consistent daily practice is the key to AP CSP success.

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