QCAA General Mathematics Networks and decision mathematics 1

A float of 0 indicates a critical activity where the earliest and latest start times are identical. For activity W, these times differ.

This value represents the duration of activity W (1 minute), not the available float time.

Float time is calculated as the difference between the Latest Starting Time (LST) and the Earliest Starting Time (EST). For activity W, $4 - 0 = 4$ minutes.

This value corresponds to the start times for activity Z, rather than the calculated float for activity W.

Activity Q depends on P (duration 3), so its Earliest Start Time (EST) is 3. Since Q is a critical activity, it has zero float, meaning its Latest Start Time (LST) must equal its EST ($3$).

The EST is correct, but the LST is wrong. Because Q is a critical activity, there is no slack, so the Latest Start Time cannot be later than the Earliest Start Time.

This incorrectly identifies the EST as 6. Since P starts at 0 and takes 3 units of time, Q can begin at time 3, not 6.

Both values are incorrect. The EST is determined by the completion of P (time 3), and since Q is critical, the LST must also be 3.

By definition, a tree is an acyclic connected graph, meaning it cannot contain any loops or cycles.

A fundamental property of any spanning tree is that it connects all vertices without forming any cycles; if a cycle existed, an edge could be removed to reduce weight while maintaining connectivity.

While a network with distinct edge weights has a unique MST, networks with duplicate edge weights can have multiple different spanning trees that share the same minimum total weight.

A tree with $n$ vertices always has exactly $n-1$ edges, so the number of vertices is always one greater than the number of edges.

This time implies a project duration of 90 minutes. However, the sum of the critical path activities (PRSV) is $38 + 32 + 34 + 26 = 130$ minutes.

The earliest completion time is determined by the total duration of the critical path PRSV. Summing the durations gives $38 + 32 + 34 + 26 = 130$ minutes (2 hours and 10 minutes). Adding this to the 11:00 am start time results in 1:10 pm.

This time implies a duration of 150 minutes. The correct duration based on the critical path is 130 minutes.

This time implies a duration of 190 minutes (3 hours and 10 minutes), which is one hour longer than the actual critical path duration of 130 minutes.