VIPSolutionsGive step-by-step solution with explanation and final answer:5. A manufacturing company produces different variations of a product at different work centers in its plant on a daily basis. Following is a network showing the various work centers in the plant, the daily capacities at cach work center, and the flow of the partially completed products between work centers: 2 (2 9 ss Ed 0 [J 6 12 SET 90 30, 65 60 J ol 3 10 90/120 40 65 0 1 7 13 Sel | 8 50 0 7 4 | 60 J 8 14 110 | 0 DARN 5 5 1" 30 0 J i Node 1 represents the point where raw materials enter the process, and node 15 is the ¢ packaging and distribution center. Determine the maximum number of units that can be 1 completed each day and the number of units processed at cach work center.
Question:
Give step-by-step solution with explanation and final answer:
5. A manufacturing company produces different variations of a product at different work
centers in its plant on a daily basis. Following is a network showing the various work
centers in the plant, the daily capacities at cach work center, and the flow of the partially
completed products between work centers:
2 (2 9 ss Ed
0
[J 6 12
SET 90
30, 65 60 J
ol 3 10
90/120 40 65 0
1 7 13 Sel |
8 50 0 7
4 |
60 J
8 14 110 |
0 DARN 5
5 1"
30 0 J
i Node 1 represents the point where raw materials enter the process, and node 15 is the ¢
packaging and distribution center. Determine the maximum number of units that can be
1 completed each day and the number of units processed at cach work center.
This is AI generated Response.
Asked by: Test User 100170
Created at: 2025-05-10 09:03:26
Asked by: Test User 100170
Created at: 2025-05-10 09:03:26
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Let's solve the **maximum flow** problem for the given network using the **Ford-Fulkerson method** (or **Edmonds-Karp algorithm** for clarity), where
- Node **1** is the **source** (raw materials entry)
- Node **15** is the **sink** (packaging/distribution center)
We'll find the **maximum number of units** that can be completed each day and the **flow through each work center**.
---
### Step 1: **List all capacities and network structure**
From the diagram, note the capacity on each edge (from node to node).
### Step 2: **Identify all paths from 1 (source) to 15 (sink)**
Let’s write out a few potential paths from 1 to 15 and their bottleneck capacities:
#### Possible paths:
1. **1 → 2 → 9 → 12 → 15**
- 1-2: 90
- 2-9: 50
- 9-12: 55
- 12-15: 90
- **Bottleneck: 50**
2. **1 → 2 → 6 → 10 → 13 → 15**
- 1-2: 90
- 2-6: 25
- 6-10: 65
- 10-13: 65
- 13-15: 190
- **Bottleneck: 25**
3. **1 → 3 → 7 → 13 → 15**
- 1-3: 120
- 3-7: 20
- 7-13: 50
- 13-15: 190
- **Bottleneck: 20**
4. **1 → 4 → 8 → 14 → 15**
- 1-4: 75
- 4-8: 25
- 8-14: 70
- 14-15: 110
- **Bottleneck: 25**
5. **1 → 5 → 11 → 14 → 15**
- 1-5: 85
- 5-11: 30
- 11-14: 60
- 14-15: 110
- **Bottleneck: 30**
---
### Step 3: **Augment flow along these paths**
Let's augment the flow along the highest capacity paths first, update capacities, and sum up the total flow.
#### 1. **Path 1 → 2 → 9 → 12 → 15**
- Bottleneck = 50
- Flow: 50 units
- Update capacities:
- 1-2: 90 - 50 = 40
- 2-9: 50 - 50 = 0
- 9-12: 55 - 50 = 5
- 12-15: 90 - 50 = 40
#### 2. **Path 1 → 2 → 6 → 10 → 13 → 15**
- Bottleneck = 25
- Flow: 25 units
- Update capacities:
- 1-2: 40 - 25 = 15
- 2-6: 25 - 25 = 0
- 6-10: 65 - 25 = 40
- 10-13: 65 - 25 = 40
- 13-15: 190 - 25 = 165
#### 3. **Path 1 → 3 → 7 → 13 → 15**
- Bottleneck = 20
- Flow: 20 units
- Update capacities:
- 1-3: 120 - 20 = 100
- 3-7: 20 - 20 = 0
- 7-13: 50 - 20 = 30
- 13-15: 165 - 20 = 145
#### 4. **Path 1 → 4 → 8 → 14 → 15**
- Bottleneck = 25
- Flow: 25 units
- Update capacities:
- 1-4: 75 - 25 = 50
- 4-8: 25 - 25 = 0
- 8-14: 70 - 25 = 45
- 14-15: 110 - 25 = 85
#### 5. **Path 1 → 5 → 11 → 14 → 15**
- Bottleneck = 30
- Flow: 30 units
- Update capacities:
- 1-5: 85 - 30 = 55
- 5-11: 30 - 30 = 0
- 11-14: 60 - 30 = 30
- 14-15: 85 - 30 = 55
---
### Step 4: **Check for remaining paths with positive capacities**
Look for more augmenting paths until no more paths can be found.
#### 6. **Path 1 → 3 → 7 → 8 → 14 → 15**
- 1-3: 100, 3-7: 0, path blocked (capacity 0).
#### 7. **Path 1 → 4 → 8 → 14 → 15**
- 4-8: 0, path blocked.
#### 8. **Path 1 → 5 → 8 → 14 → 15**
- 5-8: 40
- 8-14: 45
- 14-15: 55
- 1-5: 55
- Bottleneck = 40
- Flow: 40 units
- Update:
- 1-5: 55 - 40 = 15
- 5-8: 40 - 40 = 0
- 8-14: 45 - 40 = 5
- 14-15: 55 - 40 = 15
#### 9. **Path 1 → 5 → 8 → 13 → 15**
- 1-5: 15
- 5-8: 0, path blocked.
#### 10. **Path 1 → 4 → 8 → 13 → 15**
- 4-8: 0, path blocked.
#### 11. **Path 1 → 3 → 7 → 13 → 15**
- 3-7: 0, path blocked.
#### 12. **Path 1 → 2 → 6 → 10 → 13 → 15**
- 2-6: 0, path blocked.
#### 13. **Path 1 → 2 → 9 → 12 → 15**
- 2-9: 0, path blocked.
#### 14. **Path 1 → 5 → 11 → 14 → 15**
- 5-11: 0, path blocked.
---
### Step 5: **Sum up all flows**
Add all the bottleneck flows found:
- 50 (Path 1)
- 25 (Path 2)
- 20 (Path 3)
- 25 (Path 4)
- 30 (Path 5)
- 40 (Path 8)
**Total maximum flow = 50 + 25 + 20 + 25 + 30 + 40 = 190 units per day**
---
### Step 6: **Units processed at each work center**
Let's summarize the flow through each node (excluding the source and sink):
- Node 2: 50 (to 9) + 25 (to 6) = 75 units
- Node 3: 20 (to 7)
- Node 4: 25 (to 8)
- Node 5: 30 (to 11) + 40 (to 8) = 70 units
- Node 6: 25 (to 10)
- Node 7: 20 (to 13)
- Node 8: 25 (to 14) + 40 (to 14) = 65 units
- Node 9: 50 (to 12)
- Node 10: 25 (to 13)
- Node 11: 30 (to 14)
- Node 12: 50 (to 15)
- Node 13: 20 (to 15) + 25 (to 15) = 45 units
- Node 14: 25 (to 15) + 30 (to 15) + 40 (to 15) = 95 units
---
## **Final Answers**
1. **Maximum number of units completed each day:**
**190 units**
2. **Number of units processed at each work center:**
- Node 2: **75**
- Node 3: **20**
- Node 4: **25**
- Node 5: **70**
- Node 6: **25**
- Node 7: **20**
- Node 8: **65**
- Node 9: **50**
- Node 10: **25**
- Node 11: **30**
- Node 12: **50**
- Node 13: **45**
- Node 14: **95**
**Let me know if you'd like to see the flows on each individual edge or a flow diagram!**
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