use definition lists instead of tables

Lecture Notes/W1 - Basic Inventory Models.md

@@ -18,14 +18,14 @@ A coffee shop needs to decide how much coffee beans to purchase and how frequent
 
 #### Assumption
 
-1. Everything is deterministic
-2. Demand is constant (always the same amount of customers)
-3. Only 1 product (1 type of coffee)
-4. No backorders (coffee bean provider is always available)
-5. Zero load time (supply instantly appears when ordered)
-6. No discounts (coffee is always flat rate)
+- Everything is deterministic
+- Demand is constant (always the same amount of customers)
+- Only 1 product (1 type of coffee)
+- No backorders (coffee bean provider is always available)
+- Zero load time (supply instantly appears when ordered)
+- No discounts (coffee is always flat rate)
 
-#### Input
+### Variables
 
 $d$  
 :   Total annual demand (in this case pounds of coffee needed per year) 
@@ -39,13 +39,10 @@ $c$
 $h$
 :    Holding / Storage cost per item per year                            
 
-
-#### Decision Variables / Output
-
-$Q$
+$Q$ [!badge variant="success" text="Decision Variable"]
 :    Quantity
 
-#### Optimal Order Quantity $Q^*$
+### Optimal Order Quantity $Q^*$
 
 Since orders are fulfilled instantly, we only order a batch of coffee beans when stock goes to 0.
 
@@ -119,18 +116,9 @@ $$
 
 ## The News Vendor Model
 
-### Scenario. Selling Newspapers
-
 A newspaper vendor needs to decide how many newspapers to buy from the news publisher. Newspapers expire in a day. Expired newspapers can be salvaged (returned to vendor) for a lower price.
 
-#### Decision Variable
-
-$B$
-:   The number of newspapers to buy from the supplier
-
----
-
-#### Input & Objective Function
+### Variables
 
 $d$
 :   Discrete R.V., the uncertain demand of newspapers                  
@@ -147,6 +135,9 @@ $c$
 $s$          
 :   Salvage price, sale price for expired newspapers. (e.g. recycling the paper) 
 
+$B$ [!badge variant="success" text="Decision Variable"]
+:   The number of newspapers to buy from the supplier
+
 
 The 3 prices have the following inequality:
 
@@ -292,7 +283,7 @@ s\uparrow\implies p-s\darr \implies\frac{p-c}{p-s}\uarr\implies B^*\uarr
 $$
 
 ##  Buying services
-[!badge variant='warning' text="Key Example."]
+[!badge variant='warning' text="Key Example"]
 
 ### Scenario. Signing a contract
 
@@ -301,24 +292,28 @@ Suppose we want to sign a contract with a lighting company to purchase their mai
 - Contract too much: We get refunded for remaining services at a less price
 - Contract too little: We need to pay more for the extra services on the fly
 
-#### Decision Variable
-
-|Variable| Definition|
-| --- | --- |
-| $B$ | The number of services to buy from the contracting company |
+#### Variables
 
-#### Input & Objective Function
+$D$     
+:   A discrete random variable, the uncertain number of services we actually need 
 
+$c$     
+:   Price of each service on the contract                                         
 
-|Variable| Definition|
-| --- | --- |
-| $D$    | A discrete random variable, the uncertain number of services we actually need |
+$s$     
+:   Refund price                                                                  
+
+$p$     
+:   Price of extra services                                                       
+
+$q(d)$  
+:   PMF of $D$, same as $\Bbb P(D=d)$, in this case $d = 0,1,2,\dots$             
+
+$Q(d)$  
+:   CDF of $D$, same as $\Bbb P(D\leqslant d)$               
 
-| $c$    | Price of each service on the contract                                         |
-| $s$    | Refund price                                                                  |
-| $p$    | Price of extra services                                                       |
-| $q(d)$ | PMF of $D$, same as $\Bbb P(D=d)$, in this case $d = 0,1,2,\dots$             |
-| $Q(d)$ | CDF of $D$, same as $\Bbb P(D\leqslant d)$                                    |
+$B$  [!badge variant="success" text="Decision Variable"]
+:   The number of services to buy from the contracting company 
 
 The objective is to minimize the expected total cost $\text {TC}$ by picking the optimal $B$.
 

Lecture Notes/W2 - Linear Programming.md

@@ -6,7 +6,7 @@ icon: check-circle
 # W2 - Linear Programming
 
 
-!!!secondary
+!!!warning
 We will use $\bold x$ to indicate a vector, $x_i$ to indicate the element of $\bold x$. The $\leqslant, \geqslant$ comparators are element–wise comparison. $\bold 0$ is the zero vector.
 !!!
 
@@ -112,7 +112,7 @@ $$
 
 |||
 
-!!! **Def.** Standard Form
+!!!info **Def.** Standard Form
 
 $$
 \begin{aligned}

Lecture Notes/W3 - Network Flow Pt 1.md

@@ -9,39 +9,37 @@ icon: check-circle
 
 #### Scenario. Sending supplies to stores
 
-Suppose you own a clothing business and owns both factories and stores.
+Suppose you own a clothing business and owns both factories and stores. Let $\bigcirc$ be the factories (supply nodes) and $\triangle$ be the stores (demand node). The supply system looks like a directed graph:
 
-Let $\bigcirc$ be the factories (supply nodes) and $\triangle$ be the stores (demand node).
-
-The supply system looks like a directed graph:
-
-![](/assets/Screenshot_2023-10-14_at_00.00.58.png){class="image-m"}
-
-The placements are arbitrary
+![The placements are arbitrary](/assets/Screenshot_2023-10-14_at_00.00.58.png){class="image-m"}
 
 There’s also a unit cost to ship cloths from $i$ to $j$.
 
 $$
 {\Large\text{\textcircled{$\normalsize i$}}}\xrightarrow{\normalsize c_{ij}}{\Large\triangle}\!\!\!\!\!\!j
 $$
 
-#### Inputs
+### Variables
+
+$I$  
+:   Set of supply nodes 
 
-|Variable| Definition|
-| --- | --- |
-| $I$ | Set of supply nodes |
-| $J$ | Set of demand nodes |
-| $s_i$ | Supply capacity of node $i$ |
-| $d_j$ | Demand at node $j$ |
-| $c_{ij}$ | Unit transportation cost to go from $i$ to $j$ |
+$J$  
+:   Set of demand nodes 
 
-#### Decision Variables
+$s_i$  
+:   Supply capacity of node $i$ 
 
-|Variable| Definition|
-| --- | --- |
-| $x_{ij}$ | Amount of items to ship from supply node $i$ to demand node $j$. $\forall i\in I, j\in J$ |
+$d_j$  
+:   Demand at node $j$ 
 
-#### Objective
+$c_{ij}$
+:   Unit transportation cost to go from $i$ to $j$ 
+
+$x_{ij}$ [!badge variant="success" text="Decision Variable"]
+:   Amount of items to ship from supply node $i$ to demand node $j$. $\forall i\in I, j\in J$ 
+
+### Objective
 
 We want to minimize the total shipping cost.
 
@@ -56,7 +54,7 @@ x_{ij}&\geqslant 0
 \end{aligned}
 $$
 
-!!!primary **Proposition.** Integer Constraints ⇒ Integer Solutions, Existence
+!!!secondary **Proposition.** Integer Constraints ⇒ Integer Solutions, Existence
 
 1. If $d_j, s_i\in \Z$, the solution are also integers.
 2. If the total demand is less than total supply, the the problem is feasible.
@@ -78,8 +76,7 @@ Define $x_{ij}$ to be the **flow** on the arc $i\to j$.
 Each node $i$ has the flow conservation constraint
 
 
-!!!
-**Def.** Flow Conservation
+!!!info **Def.** Flow Conservation
 
 For each node $i$,
 
@@ -91,7 +88,7 @@ $$
 $$
 !!!
 
-#### Objective
+### Objective
 
 Minimize the total cost. $c_{ij}$ is the unit cost to go from $i\to j$. 
 
@@ -106,7 +103,7 @@ $$
 l_{ij}\leqslant x_{ij}\leqslant u_{ij}
 $$
 
-### Special Case: Transportation Problem
+## Special Case: Transportation Problem
 
 Continue from above, define the vertices to be:
 
@@ -139,17 +136,20 @@ A ={}&\{i\to j: i\text{ is supply node}, j\text{ is demand node}\}\cup{}\\
 \end{aligned}
 $$
 
-#### Decision Variables
+### Decision Variables
+
+$y_i$ and $z_j$ are new for this model.
 
-$y_i$ and $z_j$ are new
+$x_{ij}$
+:   Amount of items to ship from supply node $i$ to demand node $j$. 
+
+$y_i$  
+:   The flow from $u$ to $i$ for each supply node $i$ 
 
-|Variable| Definition|
-| --- | --- |
-| $x_{ij}$ | Amount of items to ship from supply node $i$ to demand node $j$. |
-| $y_i$ | The flow from $u$ to $i$ for each supply node $i$ |
-| $z_j$ | The from from $j$ to $u$ for each demand node $j$ |
+$z_j$  
+:   The from from $j$ to $u$ for each demand node $j$ 
 
-#### Objective
+### Objective
 
 Minimize the total cost.
 
@@ -159,7 +159,7 @@ $$
 
 Note that the under-braced parts are 0, because we only use $y_i, z_j$ for flow conservation. There’s no unit cost associated with arcs from or into $u$.
 
-#### Constraints
+### Constraints
 
 For each supply node $i$, total flow out should equal to total flow in.
 

Lecture Notes/W4 - Network Flow Pt 2.md

@@ -32,4 +32,3 @@ $$
 \min\sum_{(i\to j)\in A}c_{ij}x_{ij} + \sum_{(i\to j)\in A}d_{ij}y_{ij}
 $$
 
-idk what happened this week

Lecture Notes/W5 - Dynamic Programming Pt 1.md

@@ -48,7 +48,7 @@ We could solve the shortest path problem with a general LP solver, but we have m
 
 The following pseudocode is from my own [algorithm notes](https://www.notion.so/Single-Source-Shortest-Paths-ccab559c3b5c4f018913429bf3b1091c?pvs=21) but it does the same thing as what we did in class.
 
-||| :icon-code: Pseudocode
+|||:icon-code: Pseudocode
 ```c
 function Initialize(G, start):
 	for each vertex v in G:
@@ -74,7 +74,7 @@ function NonNegativeDijkstras(G, start):
 				pred[v] = u
 				queue.UpdatePriority(v, dist[v])
 ```
-||| :icon-rocket: Complexity
+|||:icon-rocket: Complexity
 ```
 // matched to each line
 
@@ -104,7 +104,7 @@ O(log V) for heaps
 
 where `queue.ExtractMin()` grabs the vertex with the shortest distance.
 
-#### EX. In-class practice graph
+=== **Ex.** In-class practice graph
 
 ![Source: Canvas, `Shortest_path_problem_in_class.pdf`](/assets/Screenshot_2023-10-23_at_15.33.56.png){class="image-m"}
 
@@ -129,6 +129,8 @@ Running Dijkstra’s algorithm with $\text{start} = A$ gives us shortest path fr
 │ L            │ ['A', 'E', 'F', 'G', 'K', 'L'] │         61 │  <== Path from A to L
 ╰──────────────┴────────────────────────────────┴────────────╯
 ```
+===
+
 
 ### Optimal Substructure
 
@@ -183,16 +185,16 @@ The graph should also be connected, no isolated subgraphs
 
 We don’t want (1) basically, since there’s complete tour. [Source](https://www.mdpi.com/2075-1680/10/1/19)
 
-#### The ****Dantzig-Fulkerson-Johnson (DFJ) formulation****
+#### The Dantzig-Fulkerson-Johnson (DFJ) formulation
 
 $$
 \forall S\sub V, S\ne \varnothing,\text{ define }\delta ^+(S) =\{i\to j\in A:i\in S, j\notin S\}\\[10pt]
 \sum_{a\in \delta^+(S)}x_a\geqslant 1
 $$
 
-This constraint is what makes TSP an NP-Hard problem, the number of possible $S$’s is the size of the power set of $V$, which is $|{\cal P}(V)|=2^{|V|}$
+This constraint is what makes TSP an NP-Hard problem, the number of possible $S$’s is the size of the power set of $V$, which is ${\cal P}(V)=2^{V}$
 
-#### **The Miller–Tucker–Zemlin (MTZ)** formulation
+#### The Miller–Tucker–Zemlin (MTZ) formulation
 
 For each node $i\in V$, label with $u_i\in\N$ except the starting node.
 
@@ -255,20 +257,27 @@ $$
 
 Every DP problem has the following properties:
 
-|Variable| Definition|
-| --- | --- |
-| $t=1,2,\dots ,T$ | Stages |
-| $s_t$ | State at time $t$ |
-| $v_t(s_t)$ | Value function |
-| $c(s_{t-1}, s_t)$ | Transition cost |
+
+Stages
+:	$t=1,2,\dots ,T$  
+
+State at stage $t$
+:	$s_t$  
+
+Value function 
+:	$v_t(s_t)$  
+
+Transition cost 
+:	$c(s_{t-1}, s_t)$  
 
 $$
 v_t(s_t) = \min/\max\{v_{t-1}(s_{t-1}) + c(s_{t-1}, s_t)\}
 $$
 
-#### Thm. Principle of optimality
+!!!success Theorem. Principle of optimality
 
 If $v_t(s_t)$ is optimal, then all the subproblems $v_{t-1}(s_{t-1})$ are also optimal.
+!!!
 
 ### 0-1 Knapsack
 
@@ -294,11 +303,14 @@ If we don’t have the integer constraint $x_i\in\{0,1\}$, we can just take the
 
 #### DP Formulation
 
-|Element| Corresponding varaible|
-| --- | --- |
-| Stage | The maximum item index $1, 2,\dots,n$. If we are in stage $k$, we only consider items $1,2,\dots, k$. |
-| State | $w$, remaining capacity of the backpack |
-| Value function | $v_k(w)$, max profit given capacity $w$ and items $1,2,\dots k$ |
+Stage  
+:	The maximum item index $1, 2,\dots,n$. If we are in stage $k$, we only consider items $1,2,\dots, k$. 
+
+State  
+:	$w$, remaining capacity of the backpack 
+
+Value function
+:	$v_k(w)$, max profit given capacity $w$ and items $1,2,\dots k$ 
 
 $$
 v_k(w) = \max\left\{\begin{gathered}v_{k-1}(w-a_k) + c_k\\v_{k-1}(w)\end{gathered}\right\}