Finished.

typst/appendixes/2.typ

@@ -4,3 +4,5 @@
   text(size: 0.8em, raw(read("/code/2.R"), lang: "R", block: true)),
   caption: "Code used for exercise 2",
 )
+
+#pagebreak()

typst/appendixes/4.typ

@@ -6,6 +6,8 @@
   caption: [Implementation of `GreedyRandomized`],
 )
 
+#pagebreak()
+
 ==== c.
 #code_figure(
   text(size: 0.8em, raw(read("/code/4c.R"), lang: "R", block: true)),

typst/appendixes/5.typ

@@ -4,3 +4,5 @@
   text(size: 0.8em, raw(read("/code/5.R"), lang: "R", block: true)),
   caption: [Code used for exercise 5],
 )
+
+#pagebreak()

typst/exercises/1.typ

@@ -56,7 +56,7 @@
 
 #indent_par[In addition, our overall objective is to minimize $r$.]
 
-#indent_par[For both of the following cases, the code we have used is presented in the appendix.]
+#indent_par[For both of the following cases, the code we have used is presented in the appendix in code 1.]
 
 ==== a. Bifurcated routing
 

typst/exercises/2.typ

@@ -2,7 +2,7 @@
 
 #indent_par[In order to optimize the maximum link load, we've used the `lpSolve` package from `R`.]
 
-#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[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[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.]
 
@@ -37,7 +37,7 @@
 
 #indent_par[And finally, our overall objective is to minimize $"price" dot (y_12 + y_13 + y_23 + y_24 + y_34)$, where $"price"$ is the price for each module.]
 
-#indent_par[For both of the following cases, the code we have used is presented in the appendix.]
+#indent_par[For both of the following cases, the code we have used is presented in the appendix in code 2.]
 
 #pagebreak()
 

typst/exercises/3.typ

@@ -2,7 +2,7 @@
 
 #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[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.]
 
@@ -29,7 +29,7 @@ $ sum_i c_"ab" dot x_"ab" $
 
 #indent_par[Where $c_"ab"$ is the cost of the $a -> b$ link.]
 
-#indent_par[The code we have used to calculate this is presented in the appendix.]
+#indent_par[The code we have used to calculate this is presented in the appendix in code 3.]
 
 #indent_par[We achieve the following results in table 9:]
 
@@ -49,6 +49,6 @@ $ sum_i c_"ab" dot x_"ab" $
 	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.]
+#indent_par[The total cost ends up at 3. We conclude that the best solution is the route $1 -> 2 -> 3 -> 4$, with a cost of 3.]
 
 #pagebreak()

typst/exercises/4.typ

@@ -32,7 +32,7 @@
 
 #indent_par[In detail, we store a matrix `λ` with the installed traffic through each node, so we may compute the link delays using it. We update this matrix using delays calculated from `Tr` matrix and the speed of light.]
 
-#indent_par[This function is present in the appendix.]
+#indent_par[This function is present in the appendix in code 4.]
 
 #indent_par[After running the provided script `lsrouteB.R`, we obtained the following results for a kilometric solution in table 11:]
 
@@ -58,7 +58,7 @@
 
 #indent_par[In detail, we normalize the matrix to the $[0.0, 1.0]$ interval and added it to the `Mu` matrix using `Mu = (1 + extra_capacity) * 1e9 / 8e3`. This makes it so that, as specified, we don't remove any capacity and only add it where necessary. However, this lead to a maximum link load of higher than 70%. To fix this, we multiplied the original by the lowest factor that would yield a < 70% maximum link load.]
 
-#indent_par[The code used for this, which includes the extra capacity, is present in the appendix.]
+#indent_par[The code used for this, which includes the extra capacity, is present in the appendix in code 5.]
 
 #indent_par[After running our solution, we obtained the following solution in table 12:]
 

typst/exercises/5.typ

@@ -6,10 +6,10 @@
 #indent_par[First, we declared the following variables:]
 
 - Number of servers, $z_i$, in node $i$.
-- Replica, $g_i^j$, has node $i$ be served by node $j$
+- Replica, $g_i^j$, has node $i$ be served by node $j$.
 
 
-#indent_par[Afterwards, we can construct the following equations:]
+#indent_par[Afterwards, we can construct the following equations, in table 13:]
 
 #figure(
 	pad(1em, table(
@@ -31,7 +31,7 @@ $ sum_i,j g_i^j dot C_i^j $
 
 #indent_par[Where $C_i^j$ is the delay between nodes $i$ and $j$.]
 
-#indent_par[The code used to implement this is present in the appendix.]
+#indent_par[The code used to implement this is present in the appendix in code 6.]
 
 #pagebreak()
 

typst/exercises/6.typ

@@ -7,7 +7,7 @@
 	caption: [Network diagram]
 )
 
-#indent_par[The code used to implement this is present in the appendix.]
+#indent_par[The code used to implement this is present in the appendix in code 7.]
 
 #indent_par[From this, we can see that 12 is the optimal number of circuits to maximize the net revenue.]
 

typst/util.typ

@@ -1,4 +1,4 @@
-#let title = "Lab report 3 - TODO"
+#let title = "Lab report 3 - Optimization"
 #let degree = "Master's Programme in Communication Networks Engineering"
 #let course = "Performance Evaluation and Dimensioning of Networks and Systems"