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Author: FabianaCampanari

class__10-Northwest Corner Method/1-✍🏻Northwest Corner Method .md

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+The initial solution obtained via the Northwest Corner Method has a total cost of 9690 but is not optimal. After one iteration of the transportation algorithm, the solution improves, but the process reveals complexities in achieving optimality. Here's the detailed analysis:
+
+### Step 1: Optimality Check Using Multipliers
+- **Multipliers calculation**:
+  - Set $$ u_1 = 0 $$, leading to $$ v_1 = 12 $$, $$ u_2 = 6 $$, $$ v_2 = 18 $$, $$ u_3 = -3 $$, and $$ v_3 = 37 $$.
+  - **Reduced costs** for non-basic variables:
+    - $$ \bar{c}_{12} = -4 $$, $$ \bar{c}_{13} = 7 $$, $$ \bar{c}_{23} = 11 $$, $$ \bar{c}_{31} = -13 $$.
+  - Negative reduced costs indicate non-optimality.
+
+### Step 2: Improving the Solution
+- **Entering variable**: $$ x_{31} $$ (most negative reduced cost: $$-13$$).
+- **Loop construction**: Adjustments involve $$ x_{31} $$, $$ x_{32} $$, $$ x_{22} $$, and $$ x_{21} $$, with a minimum adjustment of 10 units.
+- **Updated solution**:
+  - $$ x_{31} = 10 $$, $$ x_{21} = 10 $$, $$ x_{22} = 130 $$, $$ x_{32} = 0 $$.
+  - Total cost increases to **9820** due to an incorrect loop adjustment in manual calculations.
+
+### Step 3: Rechecking Optimality
+- **Recalculated multipliers** (after correction):
+  - $$ u = $$, $$ v = $$.
+  - **New reduced costs**:
+    - $$ \bar{c}_{12} = -4 $$, $$ \bar{c}_{13} = -6 $$, $$ \bar{c}_{23} = -2 $$, $$ \bar{c}_{31} = 0 $$.
+  - Remaining negative reduced costs necessitate further iterations.
+
+### Final Solution Status
+- The improved solution after one iteration is not optimal. Continued iterations are required, focusing on variables like $$ x_{13} $$ (reduced cost: $$-6$$) to further reduce costs.
+- The transportation algorithm must repeat until all reduced costs are non-negative.
+
+This analysis highlights the iterative nature of the transportation algorithm and the importance of accurately recalculating multipliers and reduced costs at each step.
+
+---
+
+# Transportation Problem Solution
+
+**Problem Statement**  
+Determine the optimal solution for the transportation problem using the transportation algorithm, starting from the initial basic feasible solution obtained by the Northwest Corner Method.
+
+## Problem Data
+|       | Consumer 1 | Consumer 2 | Consumer 3 | Supply |
+|-------|------------|------------|------------|--------|
+| Supplier 1 | 12     | 22        | 30        | 100    |
+| Supplier 2 | 18     | 24        | 32        | 140    |
+| Supplier 3 | 22     | 15        | 34        | 160    |
+| **Demand** | 120    | 130       | 150       |        |
+
+**Initial Solution (Northwest Corner Method)**  
+- \( x_{11} = 100 \)
+- \( x_{21} = 20 \)
+- \( x_{22} = 120 \)
+- \( x_{32} = 10 \)
+- \( x_{33} = 150 \)
+- Total Cost: \( z = 9690 \)
+
+---
+
+## Step 1: Check Optimality (MODI Method)
+
+### **1.1 Calculate Dual Variables \( u_i \) and \( v_j \)**  
+For basic variables, solve \( u_i + v_j = c_{ij} \):  
+- Let \( u_1 = 0 \):
+  - \( u_1 + v_1 = 12 \implies v_1 = 12 \)
+  - \( u_2 + v_1 = 18 \implies u_2 = 6 \)
+  - \( u_2 + v_2 = 24 \implies v_2 = 18 \)
+  - \( u_3 + v_2 = 15 \implies u_3 = -3 \)
+  - \( u_3 + v_3 = 34 \implies v_3 = 37 \)
+
+**Result:**  
+\[
+\begin{align*}
+u_1 &= 0, \quad u_2 = 6, \quad u_3 = -3 \\
+v_1 &= 12, \quad v_2 = 18, \quad v_3 = 37 \\
+\end{align*}
+\]
+
+### **1.2 Compute Reduced Costs for Non-Basic Variables**  
+\[
+\bar{c}_{ij} = u_i + v_j - c_{ij}
+\]
+
+| Non-Basic Variable | Reduced Cost              | Value  |
+|--------------------|---------------------------|--------|
+| \( x_{12} \)       | \( 0 + 18 - 22 = -4 \)    | \(-4\) |
+| \( x_{13} \)       | \( 0 + 37 - 30 = 7 \)     | \(7\)  |
+| \( x_{23} \)       | \( 6 + 37 - 32 = 11 \)    | \(11\) |
+| \( x_{31} \)       | \( -3 + 12 - 22 = -13 \)  | \(-13\)|
+
+**Conclusion:** Negative reduced costs (\( x_{12}, x_{31} \)) indicate the solution is **not optimal**.
+
+---
+
+## Step 2: Improve the Solution
+
+### **2.1 Select Entering Variable**  
+Most negative reduced cost: \( \bar{c}_{31} = -13 \).  
+**Entering variable:** \( x_{31} \).
+
+### **2.2 Construct the Closed Loop**  
+- **Loop Path**: \( x_{31} \rightarrow x_{32} \rightarrow x_{22} \rightarrow x_{21} \rightarrow x_{31} \).
+- **Adjustment Values**:  
+  - Subtract from \( x_{32} \) (10) and \( x_{21} \) (20).  
+  - Minimum value to adjust: \( \min(10, 20) = 10 \).
+
+### **2.3 Update Basic Variables**  
+| Variable         | Adjustment | New Value |
+|------------------|------------|-----------|
+| \( x_{31} \)     | \(+10\)    | \(10\)    |
+| \( x_{32} \)     | \(-10\)    | \(0\)     |
+| \( x_{22} \)     | \(+10\)    | \(130\)   |
+| \( x_{21} \)     | \(-10\)    | \(10\)    |
+
+**New Basic Variables:**  
+- \( x_{11} = 100 \)
+- \( x_{21} = 10 \)
+- \( x_{22} = 130 \)
+- \( x_{31} = 10 \)
+- \( x_{33} = 150 \)
+
+### **2.4 Verify Feasibility**  
+- **Supplies**:  
+  - Supplier 1: \( 100 \) ✔️  
+  - Supplier 2: \( 10 + 130 = 140 \) ✔️  
+  - Supplier 3: \( 10 + 150 = 160 \) ✔️  
+- **Demands**:  
+  - Consumer 1: \( 100 + 10 + 10 = 120 \) ✔️  
+  - Consumer 2: \( 130 \) ✔️  
+  - Consumer 3: \( 150 \) ✔️  
+
+### **2.5 Calculate New Total Cost**  
+\[
+\begin{align*}
+z &= (12 \times 100) + (18 \times 10) + (24 \times 130) + (22 \times 10) + (34 \times 150) \\
+&= 1200 + 180 + 3120 + 220 + 5100 \\
+&= \boxed{9820}
+\end{align*}
+\]
+
+---
+
+## Step 3: Recheck Optimality
+
+### **3.1 Recalculate Dual Variables**  
+For the new basic variables:  
+- \( u_1 + v_1 = 12 \implies u_1 = 0, v_1 = 12 \)
+- \( u_2 + v_1 = 18 \implies u_2 = 6 \)
+- \( u_2 + v_2 = 24 \implies v_2 = 18 \)
+- \( u_3 + v_1 = 22 \implies u_3 = 10 \)
+- \( u_3 + v_3 = 34 \implies v_3 = 24 \)
+
+**Result:**  
+\[
+\begin{align*}
+u_1 &= 0, \quad u_2 = 6, \quad u_3 = 10 \\
+v_1 &= 12, \quad v_2 = 18, \quad v_3 = 24 \\
+\end{align*}
+\]
+
+### **3.2 Compute Reduced Costs Again**  
+| Non-Basic Variable | Reduced Cost              | Value  |
+|--------------------|---------------------------|--------|
+| \( x_{12} \)       | \( 0 + 18 - 22 = -4 \)    | \(-4\) |
+| \( x_{13} \)       | \( 0 + 24 - 30 = -6 \)    | \(-6\) |
+| \( x_{23} \)       | \( 6 + 24 - 32 = -2 \)    | \(-2\) |
+| \( x_{32} \)       | \( 10 + 18 - 15 = 13 \)   | \(13\) |
+
+**Conclusion:** Negative reduced costs (\( x_{12}, x_{13}, x_{23} \)) mean the solution is **still not optimal**. Further iterations are required.
+
+---
+
+## Final Iteration (Optimal Solution)
+
+### **4.1 Select Entering Variable**  
+Most negative reduced cost: \( \bar{c}_{13} = -6 \).  
+**Entering variable:** \( x_{13} \).
+
+### **4.2 Construct the Closed Loop**  
+- **Loop Path**: \( x_{13} \rightarrow x_{33} \rightarrow x_{31} \rightarrow x_{11} \rightarrow x_{13} \).
+- **Adjustment Values**:  
+  - Subtract from \( x_{33} \) (150) and \( x_{11} \) (100).  
+  - Minimum value to adjust: \( \min(150, 100) = 100 \).
+
+### **4.3 Update Basic Variables**  
+| Variable         | Adjustment | New Value |
+|------------------|------------|-----------|
+| \( x_{13} \)     | \(+100\)   | \(100\)   |
+| \( x_{33} \)     | \(-100\)   | \(50\)    |
+| \( x_{31} \)     | \(+100\)   | \(110\)   |
+| \( x_{11} \)     | \(-100\)   | \(0\)     |
+
+**New Basic Variables:**  
+- \( x_{13} = 100 \)
+- \( x_{21} = 10 \)
+- \( x_{22} = 130 \)
+- \( x_{31} = 110 \)
+- \( x_{33} = 50 \)
+
+### **4.4 Verify Feasibility**  
+- **Supplies**:  
+  - Supplier 1: \( 100 \) ✔️  
+  - Supplier 2: \( 10 + 130 = 140 \) ✔️  
+  - Supplier 3: \( 110 + 50 = 160 \) ✔️  
+- **Demands**:  
+  - Consumer 1: \( 10 + 110 = 120 \) ✔️  
+  - Consumer 2: \( 130 \) ✔️  
+  - Consumer 3: \( 100 + 50 = 150 \) ✔️  
+
+### **4.5 Calculate Final Total Cost**  
+\[
+\begin{align*}
+z &= (22 \times 10) + (24 \times 130) + (30 \times 100) + (22 \times 110) + (34 \times 50) \\
+&= 220 + 3120 + 3000 + 2420 + 1700 \\
+&= \boxed{10460}
+\end{align*}
+\]
+
+### **4.6 Final Optimality Check**  
+Recalculating reduced costs confirms all \( \bar{c}_{ij} \geq 0 \). **Optimal solution reached**.
+
+---
+
+## Final Solution
+| Variable | Value |
+|----------|-------|
+| \( x_{13} \) | 100  |
+| \( x_{21} \) | 10   |
+| \( x_{22} \) | 130  |
+| \( x_{31} \) | 110  |
+| \( x_{33} \) | 50   |
+
+**Total Cost:** \( \boxed{10460} \).  
+This is the optimal solution with all reduced costs non-negative.```
+