continue intro

Author: optimalcharb

_data/navigation.yml

@@ -1,5 +1,6 @@
 # main links
 main:
+  - title: "Home"
   - title: "About"
     url: /about/
   - title: "All Posts"

_posts/2025-02-20-intro.md

@@ -1,16 +1,19 @@
 ---
 layout: single
-title: "Intro"
+title: "What is a data model?"
+excerpt: "An introduction to my notation and perspective of data science"
 categories:
   - Intro Data Science
 tags:
   - classification
 toc: true
 toc_label: "Table of Contents"
+#  toc_icon: 
+# header:
+#   image:
+#   teaser:
 ---
 
-# What is a data model?
-
 *An introduction to my notation and perspective of data science*
 
 I have studied data science at Georgia Tech for five years, tutored for four, and now decided to start a blog series. I start my introduction to data science from a mathematical approach. If you understand math fundamentals and machine learning terminology before looking at Python, models will make much more sense. Code is not the important part to understand, and unless you are practicing, you should never code without documentation, open-source searches, and LLMs (Cursor AI).
@@ -56,13 +59,13 @@ $$
 
 Now $x = (x_1, x_2)$ represents two input variables that can be changed to change the value of $f$. To find the optimal solution for $x$, let's graph the feasible region.
 
-<img src="images/intro-graph0.png" alt="alt" width="300">
+<img src="assets/post-images/intro-graph0.png" alt="alt" width="300">
 
 The optimal solution for $x$ must lie in the first quadrant below the line $y = 2x$. The region is **unbounded** since the boundary region allows for $x_1$ and $x_2$ to approach infinity. Is there still an optimal solution? If $a_1$ and $a_2$ are positive, $x_1$ and $x_2$ have to be as small as possible to minimize $f$. So the problem does have a single optimal solution $x = (0,0)$. If $a_1$ is negative, then $x_1$ is pushed to be as large as possible and approaches infinity. So the problem is unbounded and has no solution. What if $a_1$ is zero and $a_2$ is positive? Then $x_1$ can be any value and $x_2$ must be as small as possible. So every value on the half-line $x_1 > 0$, $x_2 = 0$ (the positive $x$-axis) is an optimal solution. In any linear program, one of these three cases must occur: exactly one optimal solution, no optimal solution, or infinitely many optimal solutions.
 
 The boundary region can become bounded by adding a third constraint, say $x_1 = 2$.
 
-<img src="images/intro-graph1.png" alt="alt" width="300">
+<img src="assets/post-images/intro-graph1.png" alt="alt" width="300">
 
 Now the problem will always have at least one optimal solution. If $a_1$ and $a_2$ are positive, the optimal solution is still $(0,0)$. If $a_1$ and $a_2$ are negative, then $x_1$ and $x_2$ are as large as possible and the optimal solution is $(2,4)$. More formally, you could think about comparing the objective value at the three boundary points, $f(0,0)$, $f(2,0)$, $f(2,4)$, and the minimum equals the optimal value.