Merge pull request #858 from Quantum-Software-Development/FabianaCampanari-patch-1

Update README.md

Author: FabianaCampanari

README.md

@@ -1695,13 +1695,222 @@ $$
 
 <br>
 
-### **Conclusion:** Negative reduced costs $\( x_{12}, x_{31} \)$ indicate the [solution is **not optimal]()**.
+#### ***Conclusion:** Negative reduced costs $\( x_{12}, x_{31} \)$ indicate the [solution is **not optimal]()***.
 
 <br>
 
 ## [Step 2](): Improve the Solution
 
 
+#### [2.1](). Select Entering Variable
+
+Most negative reduced cost: $\bar{c}_{31} = -13$.  
+
+**Entering variable:** $x_{31}$.
+
+
+<br>
+
+
+#### [2.2](). Construct the Closed Loop
+
+<br>
+
+- **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: $\min(10, 20) = 10$.
+
+<br>
+
+#### [2.3](). Update Basic Variables**  
+
+<br>
+
+| Variable     | Adjustment | New Value |
+|--------------|------------|-----------|
+| $ x_{31} $   | $+10$      | $10$      |
+| $ x_{32} $   | $-10$      | $0$       |
+| $ x_{22} $   | $+10$      | $130$     |
+| $ x_{21} $   | $-10$      | $10$      |
+
+<br>
+
+#### [**New Basic Variables:**]()  
+
+- $ x_{11} = 100 $  
+- $ x_{21} = 10 $  
+- $ x_{22} = 130 $  
+- $ x_{31} = 10 $  
+- $ x_{33} = 150 $
+
+<br>
+
+#### [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: $150$ ✔️
+
+
+ <br>
+
+#### [2.5](). Calculate New Total Cost**  
+
+<br>
+
+$$
+\begin{align*}
+z &= (12 \times 100) + (18 \times 10) + (24 \times 130) + (22 \times 10) + (34 \times 150) \\
+  &= 1200 + 180 + 3120 + 220 + 5100 \\
+  &= \boxed{9820}
+\end{align*}
+$$
+
+<br>
+
+## [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 $
+
+<br>
+
+### **Result:**  
+
+<br>
+
+$$
+\begin{align*}
+u_1 &= 0, \quad u_2 = 6, \quad u_3 = 10 \\
+v_1 &= 12, \quad v_2 = 18, \quad v_3 = 24 \\
+\end{align*}
+$$
+
+<br>
+
+### **3.2 Compute Reduced Costs Again**  
+
+<br>
+
+| 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 $       | $13$   |
+
+<br>
+
+**Conclusion:** Negative reduced costs ($ x_{12}, x_{13}, x_{23} $) mean the solution is **still not optimal**. Further iterations are required.
+
+<br>
+
+## Final Iteration (Optimal Solution)
+
+### [4.1](). Select Entering Variable**  
+
+Most negative reduced cost: $\bar{c}_{13} = -6$.  
+
+**Entering variable:** $x_{13}$.
+
+<br>
+
+#### [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 $.
+ 
+<br>
+
+#### [4.3](). Update Basic Variables
+
+<br>
+
+| 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
+
+$$
+\begin{align*}
+z &= (22 \times 10) + (24 \times 130) + (30 \times 100) + (22 \times 110) + (34 \times 50) \\
+  &= 220 + 3120 + 3000 + 2420 + 1700 \\
+  &= \boxed{10460}
+\end{align*}
+$$
+
+
+<br>
+
+### [4.6](). Final Optimality Check
+
+Recalculating reduced costs confirms all $ \bar{c}_{ij} \geq 0 $. **Optimal solution reached**.
+
+<br>
+
+## 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.
+
+
+
 
 ### Under Construction 🚛