diff --git a/README.md b/README.md index 24756f6..8eab8e2 100644 --- a/README.md +++ b/README.md @@ -2069,8 +2069,6 @@ Entering variable: $x_{13}$. #### [4.3](). Update Basic Variables -🏄🏄🏄🏄🏄 -
| Variable | Adjustment | New Value | @@ -2078,29 +2076,37 @@ Entering variable: $x_{13}$. | $x_{13}$ | $+100$ | $100$ | | $x_{33}$ | $-100$ | $50$ | | $x_{31}$ | $+100$ | $110$ | -| $ x_{11} $ | $-100$ | $0$ | +| $x_{11}$ | $-100$ | $0$ |
-**New Basic Variables:** -- $ x_{13} = 100 $ -- $ x_{21} = 10 $ -- $ x_{22} = 130 $ -- $ x_{31} = 110 $ -- $ x_{33} = 50 $ +[**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 $ ✔️ +- [**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 @@ -2115,25 +2121,29 @@ $$
-### [4.6](). Final Optimality Check +#### [4.6](). Final Optimality Check + +Recalculating reduced costs confirms all $\bar{c}_{ij} \geq 0$. [**Optimal solution reached**](). -Recalculating reduced costs confirms all $ \bar{c}_{ij} \geq 0 $. **Optimal solution reached**. +
+ +## [Step 5](): Final Solution
-## Final Solution | Variable | Value | |------------|-------| -| $ x_{13} $ | 100 | -| $ x_{21} $ | 10 | -| $ x_{22} $ | 130 | -| $ x_{31} $ | 110 | -| $ x_{33} $ | 50 | +| $x_{13}$ | 100 | +| $x_{21}$ | 10 | +| $x_{22}$ | 130 | +| $x_{31}$ | 110 | +| $x_{33}$ | 50 | +
-**Total Cost:** $\boxed{10460}$. +[**Total Cost:**](): $\boxed{10460}$. -This is the optimal solution with all reduced costs non-negative. +***This is the optimal solution with all reduced costs non-negative.***
@@ -2151,7 +2161,7 @@ This is the optimal solution with all reduced costs non-negative. ### Theoretical Explanation -The **Assignment Problem** aims to allocate *n* tasks to *n* agents (machines, workers) at minimum cost, ensuring each task and agent is assigned exactly once.** +The [**Assignment Problem**]() aims to allocate *n* tasks to *n* agents (machines, workers) at minimum cost, ensuring each task and agent is assigned exactly once. ### [Problem Statement](): @@ -2169,20 +2179,25 @@ The **Assignment Problem** aims to allocate *n* tasks to *n* agents (machines, w
+🏄🏄🏄🏄🏄 ## 1. [Hungarian Method]() (Step by Step): -### [**Step 1](): Subtract Row Minimums** +### [Step 1](): Subtract Row Minimums -#### Subtract the minimum value in each row from all elements in that row. +#### [Subtract the minimum value in each row from all elements in that row](). -- Row 1 min: 2 → [0, 2, 1] -- Row 2 min: 1 → [0, 2, 1] -- Row 3 min: 2 → [3, 0, 2] +
+ +- [Row 1 min: 2]() → [0, 2, 1] +- [Row 2 min: 1]() → [0, 2, 1] +- [Row 3 min: 2]() → [3, 0, 2]
-#### [**Matrix after row subtraction:**]() +#### [Matrix after row subtraction](): + +
| | M1 | M2 | M3 | |---------|----|----|----| @@ -2193,16 +2208,23 @@ The **Assignment Problem** aims to allocate *n* tasks to *n* agents (machines, w
-### [**Step 2](): Subtract Column Minimums** +### [Step 2](): Subtract Column Minimums -### Problem Recap +### [Problem Recap](): + +
- **3 tasks** must be assigned to **3 machines**. - Each task can be done by any machine, but with different costs. - Each task must be assigned to exactly one machine, and each machine to exactly one task. - **Goal:** Minimize total assignment cost. -### Cost Table +
+ +### [Cost Table](): + +
+ | | Machine 1 | Machine 2 | Machine 3 | |---------|-----------|-----------|-----------|