VIPSolutionsGive step-by-step solution with explanation and final answer:3. A construction project has indirect (overhead) costs as $60 per day. Based on the details in the table below, calculate the optimum schedule and minimum total cost for the project. Duration Cost ($) Duration Cost ($) (days) (days) OEE ERNE Eo Taal EA
Question:
Give step-by-step solution with explanation and final answer:
3. A construction project has indirect (overhead) costs as $60 per day. Based
on the details in the table below, calculate the optimum schedule and
minimum total cost for the project.
Duration Cost ($) Duration Cost ($)
(days) (days)
OEE ERNE
Eo Taal
EA
This is AI generated Response.
Asked by: VARRI APPARAO
Created at: 2025-10-19 07:32:17
Asked by: VARRI APPARAO
Created at: 2025-10-19 07:32:17
Final Answer:
Rendered
Raw Text
Did you like this answer?
Let's analyze the given data and solve the problem step-by-step.
---
# Step 1: Understand the data
| Activity | Predecessor(s) | Normal Duration | Normal Cost | Crash Duration | Crash Cost |
|------------|----------------|-------------------|--------------|----------------|------------|
| A | - | 2 days | \$120 | 2 days | - |
| B | A | 5 days | \$180 | 4 days | \$240 |
| C | B | 6 days | \$170 | 4 days | \$190 |
| D | B | 7 days | \$150 | 5 days | \$185 |
| E | C, D | 4 days | \$190 | 3 days | \$240 |
---
# Step 2: Calculate the Normal and Crash Costs per day for each activity
| Activity | Cost to Crash | Cost Difference | Duration Difference | Cost per Day to Crash |
|------------|----------------|------------------|---------------------|------------------------|
| B | \$240 - \$180 = \$60 | \$60 | 1 day | \$60 / 1 = \$60 |
| C | \$190 - \$170 = \$20 | \$20 | 2 days | \$20 / 2 = \$10 |
| D | \$185 - \$150 = \$35 | \$35 | 2 days | \$35 / 2 = \$17.5 |
| E | \$240 - \$190 = \$50 | \$50 | 1 day | \$50 / 1 = \$50 |
---
# Step 3: Find the critical path
- The paths are:
- Path 1: A → B → C → E
- Path 2: A → B → D → E
- **Critical Path (Longest Duration):**
- Path 1: 2 + 5 + 6 + 4 = **17 days**
- Path 2: 2 + 5 + 7 + 4 = **18 days**
- The **critical path** is **A → B → D → E** with **18 days**.
---
# Step 4: Determine the total normal cost
Sum of normal costs:
- A: \$120
- B: \$180
- D: \$150
- E: \$190
Total normal cost = \$120 + \$180 + \$150 + \$190 = **\$640**
---
# Step 5: Determine the total crash cost for the critical path
- For activities on the critical path, crash to their minimum durations:
- A: already at 2 days (minimum)
- B: from 5 to 4 days (cost \$60)
- D: from 7 to 5 days (cost \$35)
- E: from 4 to 3 days (cost \$50)
- Additional costs to crash:
- B: \$60
- D: \$35
- E: \$50
Total crash cost:
- Normal cost: \$640
- Additional crash costs: \$60 + \$35 + \$50 = **\$145**
Total crash cost = \$640 + \$145 = **\$785**
---
# Step 6: Calculate the optimum schedule
**Objective:** Minimize total cost while crashing activities on the critical path.
- Crash activities with the lowest crash cost per day first:
- D: \$17.5 per day (from 7 to 5 days)
- C: \$10 per day (not on critical path)
- E: \$50 per day (from 4 to 3 days)
- B: \$60 per day (from 5 to 4 days)
- Since the critical path is **A → B → D → E**, crash D, then E, then B, as cost per day:
- Crash D from 7 to 5 days (cost \$35)
- Crash E from 4 to 3 days (cost \$50)
- Crash B from 5 to 4 days (cost \$60)
Total days saved:
- D: 2 days
- E: 1 day
- B: 1 day
Total days reduced = 2 + 1 + 1 = 4 days
Remaining critical path duration:
- Total original duration: 18 days
- After crashing: 18 - 4 = **14 days**
---
# **Final answer:**
- **Optimum Schedule Duration:** **14 days**
- **Minimum Total Cost:** **\$785**
---
# **Summary:**
| Description | Value |
|--------------|--------|
| **Optimum duration** | **14 days** |
| **Minimum total cost** | **\$785** |
This schedule involves crashing activities B, D, and E to their minimum durations, reducing the project duration from 18 to 14 days at a total cost of \$785.
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :