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Let's analyze the provided activity network step by step to answer the questions.

### a) **Determine the earliest start time of activity G**.

The earliest start time (EST) of an activity is determined by the completion time of its predecessor activities. Activity G depends on activities C and E. To find the EST of G, we need to calculate the earliest finish time of its predecessors.

1. **Start to A**: \(6\) units.
2. **Start to B**: \(4\) units.

From these:
- For **C** (dependent on A): Earliest finish time of A + time of C = \(6 + 2 = 8\).
- For **D** (dependent on B): Earliest finish time of B + time of D = \(4 + 2 = 6\).
- For **E** (dependent on D): Earliest finish time of D + time of E = \(6 + 3 = 9\).

Now for **G**:
- The earliest start of G is dependent on both C and E, so it's the larger of the two earliest finish times:
  - From C: \(8\)
  - From E: \(9\)

So, the earliest start time for **G** is \(9\).

### b) **Determine the latest start time of activity F**.

To calculate the latest start time (LST) of activity F, we must first know the total project completion time, which will come from part (e). For now, let's tentatively assume that the total completion time is the earliest finish of the Finish node, which depends on F, G, and H.

1. **Earliest finish of G**: \(9 + 6 = 15\)
2. **Earliest finish of F**: \(6 + 4 = 10\)
3. **Earliest finish of H**: \(9 + 4 = 13\)

Thus, the minimum completion time for the project is \(15\) (coming from G's path).

The latest finish time for F, given that the project must be done by time \(15\), is:
\[
\text{LST of F} = \text{Latest finish time of F} - \text{duration of F} = 15 - 4 = 11.
\]

### c) **Determine the float time for activity C**.

Float (or slack) time is the amount of time an activity can be delayed without affecting the overall project completion time. The formula for float is:

\[
\text{Float} = \text{Latest start time} - \text{Earliest start time}.
\]

For **C**:
- Earliest start time = \(6\) (from A),
- Latest start time = \(9 - 2 = 7\) (since the latest start of G is 9, and C takes 2 units of time).

Thus, the float time for **C** is:
\[
7 - 6 = 1 \text{ unit}.
\]

### d) **Write down the critical path**.

The critical path is the sequence of activities that determines the minimum project duration. It's the longest path through the network where float time is zero.

From the network, we can see:
1. Start → B → D → E → G → Finish
2. Duration: \(4 + 2 + 3 + 6 = 15\).

Thus, the **critical path** is **B → D → E → G**, and the total duration is 15 units.

### e) **Determine the minimum completion time for this project**.

As determined from the critical path analysis, the minimum completion time for the project is **15 units**.

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Let me know if you'd like more detailed calculations or have questions!

Here are 5 related questions for further practice:
1. How would the earliest start time change if activity E took 1 more unit of time?
2. What is the total float for non-critical path activities?
3. Can an activity have a negative float time?
4. If the project deadline was 14 units, which activities would need to be expedited?
5. How does delaying activity B impact the overall project completion?

### Tip:
In critical path analysis, focus on identifying the sequence of tasks that cannot afford delays—this sequence will give the most accurate project time estimates.

Consider the activity network below with activities A, B, C, D, E, F, G, and H. Determine the earliest start time of activity G, the latest start time of activity F, the float time for activity C, the critical path, and the minimum completion time for the project.

Explore how to analyze an activity network to find the earliest start time of activities, the latest start time, float, and the critical path. This example uses activities A-H and provides a step-by-step solution for determining the minimum completion time for the project using the Critical Path Method (CPM).

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