Week 8 | Session 4: SC Network Design - Heuristic Approaches (H1 & H2)
Course: Supply Chain Digitization - Module 3: Analytics in SCM
Session Agenda
Section titled “Session Agenda”1. Case - SC Network Design Problem
Section titled “1. Case - SC Network Design Problem”Network Structure: 2-tier network (Manufacturer → Warehouse → Retailer).
- Manufacturers: M1, M2
- Warehouses: W1, W2, W3
- Retailers: R1, R2, R3, R4
Assumptions: Production and handling costs are the same across facilities (simplified). Only transportation costs vary.
Objective: Find the optimal distribution strategy - which manufacturer supplies which warehouse, and which warehouse serves which retailer - to minimize total distribution cost.
Per-unit shipment cost - manufacturer→warehouse (m1, m2) and warehouse→retailer (r1-r4):
| Warehouse | m1 | m2 | r1 | r2 | r3 | r4 |
|---|---|---|---|---|---|---|
| w1 | 1 | 6 | 2 | 1 | 1 | 3 |
| w2 | 4 | 2 | 5 | 4 | 3 | 4 |
| w3 | 3 | 1 | 4 | 3 | 2 | 6 |
| Capacity / Demand | 1,50,000 | 1,20,000 | 45,000 | 35,000 | 68,000 | 70,000 |
2. What is a Heuristic?
Section titled “2. What is a Heuristic?”A rule-based, practical approach to find a good (not necessarily optimal) solution.
- Faster and simpler than full LP optimization.
- Different heuristics can give different costs for the same problem.
- Useful as a starting point.
3. Heuristic 1 - Cheapest Warehouse First
Section titled “3. Heuristic 1 - Cheapest Warehouse First”- Pick the warehouse with lowest total shipping cost to all retailers.
- Route all retailer demand through that warehouse.
- Pick the cheapest manufacturer to supply that warehouse - exhaust its capacity first.
- Fulfill remaining demand from the next manufacturer.
Applying to the Case
Section titled “Applying to the Case”- W1 has the lowest cost to all 4 retailers.
- M1 is cheaper to W1 and supplies first (capacity 1,50,000). Supplies R1, R2, R3 fully, and 2,000 units to R4.
- M2 supplies the balance for R4 (68,000 units).
45,000×2 + 35,000×1 + 68,000×1 + 70,000×3 = 4,03,000M1: 1,50,000×1 + M2: 68,000×6 = 5,58,000Result: Total Cost = ₹9,61,000
4. Heuristic 2 - Cheapest End-to-End Path per Retailer
Section titled “4. Heuristic 2 - Cheapest End-to-End Path per Retailer”- For each retailer individually, evaluate all 6 possible paths (2 Manufacturers × 3 Warehouses).
- Select the path (M → W → Retailer) with the minimum total cost.
- Apply capacity constraints after selecting paths.
For each retailer, the cheapest full manufacturer→warehouse→retailer path is chosen; capacity is then applied:
| Retailer | Cheapest end-to-end path | Units |
|---|---|---|
| R1 | M1 → W1 → R1 | 45,000 |
| R2 | M1 → W1 → R2 | 35,000 |
| R3 | M1 → W1 → R3 | 68,000 |
| R4 | M2 → W2 → R4 (M1 capacity nearly full) | 68,000 + 2,000 |
Applying to the Case
Section titled “Applying to the Case”- R1, R2, R3: Best path is M1 → W1.
- R4: Tie between M1→W1 and M2→W2. Since M1’s capacity is almost full, the balance of 68,000 goes via M2 → W2.
Result: Total Cost = ₹7,57,000 ← better than H1. (H2 routes R4 via W2 instead of W1).
5. Heuristic Comparison & Open Question
Section titled “5. Heuristic Comparison & Open Question”| Heuristic 1 | Heuristic 2 | Optimal (next session) | |
|---|---|---|---|
| Logic | Pick cheapest WH globally, then cheapest Mfr | Find cheapest end-to-end path per retailer | LP / solver |
| Total Cost | ₹9,61,000 | ₹7,57,000 | ? (≤ ₹7,57,000) |
H2 improved cost by ~21% over H1 simply by evaluating full paths. But: Is H2 the best possible? We don’t know yet. The next session will solve this using LP/Solver to find the true optimal.
Session Summary
Section titled “Session Summary”- SC Network Design: 2-tier problem (Mfr → WH → Retailer). Minimize total cost.
- Heuristic 1: Pick cheapest WH for all retailers → assign cheapest Mfr first. Cost: ₹9,61,000.
- Heuristic 2: Evaluate full end-to-end path per retailer → pick min cost path. Cost: ₹7,57,000.
- Key insight: H2 beats H1 because it considers the full Mfr→WH→Retailer path.