Mostly finished ex3.

code/3.R

@@ -24,7 +24,7 @@ objective[c(idx_x_12, idx_x_13, idx_x_23, idx_x_24, idx_x_34)] <- c(cost_12, cos
 # (Node 1): x_12 + x_13 = 1
 # (Node 2): x_12 - x_24 - x_23 = 0
 # (Node 3): x_13 + x_23 - x_34 = 0
-# (Node 4): -x_24 -x_34 = -1
+# (Node 4): -x_24 - x_34 = -1
 mat <- matrix(
   0,
   nrow = 5,

typst/appendixes/3.typ

@@ -0,0 +1,6 @@
+#import "/typst/util.typ" as util: indent_par, code_figure
+
+#code_figure(
+  text(size: 0.8em, raw(read("/code/3.R"), lang: "R", block: true)),
+  caption: "Code used for exercise 3",
+)

typst/exercises/2.typ

@@ -6,7 +6,7 @@
 
 #indent_par[We define additional variables to determine the number of circuits for each link, named $y_"ab"$. We also remove the variable $r$, since our objective is now to minimize the overall cost, which only depends on the $y_"ab"$ variables and other constants.]
 
-#indent_par[The following are our updated equations.]
+#indent_par[The following are our updated equations in table 5.]
 
 #figure(
 	pad(1em, table(
@@ -45,7 +45,7 @@
 
 #indent_par[Similarly to the previous exercise, for bifurcated flow, we allow $x_(n_1 n_2 ... n_r)$ to be real numbers, but we must restrict $y_"ab"$ to a positive integer, because we can only have an integer amount of modules.]
 
-#indent_par[This results in the following solution in table 5:]
+#indent_par[This results in the following solution in table 6:]
 
 #figure(
 	pad(1em, table(
@@ -106,3 +106,5 @@
 )
 
 #indent_par[We also reach a network cost of 10000€.]
+
+#pagebreak()

typst/exercises/3.typ

@@ -0,0 +1,52 @@
+#import "/typst/util.typ" as util: indent_par, code_figure
+
+#indent_par[We've resorted to the node-link routing formulation to determine the shortest path.]
+
+#indent_par[Since we're using the same network and flows as in exercise 1, the routes specified in table 1 stay the same as well]
+
+#indent_par[Similarly to the previous 2 exercise, we define the variables $x_"ab"$, where $a -> b$ is a link, as the ratio of traffic passing through that link.]
+
+#indent_par[The following are our equations in table 8.]
+
+#figure(
+	pad(1em, table(
+		columns: (auto, 1fr),
+		align: left + horizon,
+
+		[Description], [Equation],
+		[Node 1], [$x_12 + x_13 = 1$],
+		[Node 2], [$x_12 - x_24 - x_23 = 0$],
+		[Node 3], [$x_13 + x_23 - x_34 = 0$],
+		[Node 4], [$-x_24 - x_34 = -1$],
+	)),
+	kind: table,
+	caption: [Problem equations]
+)
+
+#indent_par[Our objective function is the following in equation 1:]
+
+$ sum_i c_"ab" dot x_"ab" $
+
+#indent_par[Where $c_"ab"$ is the cost of the $a -> b$ link.]
+
+#indent_par[For both of the following cases, the code we have used is presented in the appendix.]
+
+#indent_par[We achieve the following results in table 9:]
+
+#figure(
+	pad(1em, table(
+		columns: (auto, auto),
+		align: left,
+
+		[Variable], [ Value ],
+		[$x_12$], [1],
+		[$x_13$], [0],
+		[$x_23$], [1],
+		[$x_24$], [0],
+		[$x_34$], [1],
+	)),
+	kind: table,
+	caption: [Solution]
+)
+
+#indent_par[The total cost ends end at 3. We conclude that the best solution is the route $1 -> 2 -> 3 -> 4$, with a cost of 3.]

typst/main.typ

@@ -53,14 +53,18 @@
 === 2. Exercise 2
 #include "exercises/2.typ"
 
+=== 3. Exercise 3
+#include "exercises/3.typ"
+
 #pagebreak()
 
 = Appendix
 
 === 1. Exercise 1
-
 #include "appendixes/1.typ"
 
-=== 2. Exercise 2
-
+=== 2. Exercise 2 <appendix2>
 #include "appendixes/2.typ"
+
+=== 3. Exercise 3 <appendix3>
+#include "appendixes/3.typ"