diff --git a/docs/make.jl b/docs/make.jl
index 0c8e566..7afdd04 100644
--- a/docs/make.jl
+++ b/docs/make.jl
@@ -4,7 +4,7 @@ using CEED
 # Literate for tutorials
 const literate_dir = joinpath(@__DIR__, "..", "tutorials")
 const tutorials_src =
-    ["StaticDesigns.jl", "StaticDesignsFiltration.jl", "GenerativeDesigns.jl"]
+    ["SimpleStatic.jl", "StaticDesigns.jl", "StaticDesignsFiltration.jl", "GenerativeDesigns.jl"]
 const generated_dir = joinpath(@__DIR__, "src", "tutorials/")
 
 # copy tutorials src
@@ -29,6 +29,7 @@ end
 pages = [
     "index.md",
     "Tutorials" => [
+        "tutorials/SimpleStatic.md",
         "tutorials/StaticDesigns.md",
         "tutorials/StaticDesignsFiltration.md",
         "tutorials/GenerativeDesigns.md",
diff --git a/docs/src/tutorials/GenerativeDesigns.jl b/docs/src/tutorials/GenerativeDesigns.jl
index 42078cf..c4cf392 100644
--- a/docs/src/tutorials/GenerativeDesigns.jl
+++ b/docs/src/tutorials/GenerativeDesigns.jl
@@ -249,7 +249,7 @@ designs = efficient_designs(
     6,
     evidence;
     solver,
-    mdp_options = (; max_parallel = 2, costs_tradeoff = [0, 1.0]),
+    mdp_options = (; max_parallel = 2, costs_tradeoff = (0, 1.0)),
     repetitions = 5,
 );
 
diff --git a/docs/src/tutorials/GenerativeDesigns.md b/docs/src/tutorials/GenerativeDesigns.md
index bfcd8ae..57f87ce 100644
--- a/docs/src/tutorials/GenerativeDesigns.md
+++ b/docs/src/tutorials/GenerativeDesigns.md
@@ -287,7 +287,7 @@ designs = efficient_designs(
     6,
     evidence;
     solver,
-    mdp_options = (; max_parallel = 2, costs_tradeoff = [0, 1.0]),
+    mdp_options = (; max_parallel = 2, costs_tradeoff = (0, 1.0)),
     repetitions = 5,
 );
 nothing #hide
diff --git a/docs/src/tutorials/SimpleStatic.jl b/docs/src/tutorials/SimpleStatic.jl
new file mode 100644
index 0000000..5e6b8ae
--- /dev/null
+++ b/docs/src/tutorials/SimpleStatic.jl
@@ -0,0 +1,140 @@
+# # Static Experimental Designs
+
+# Consider the following scenario. There exists a set of experiments, each of which, when performed, yields
+# measurements on one or more observables (features). Each subset of observables (and therefore each subset of experiments)
+# has some "information value", which is intentionally vaguely defined for generality, but for example, may be 
+# a loss function if that subset is used  to train some machine learning model. It is generally the value of acquiring that information. 
+# Finally, each experiment has some monetary cost and execution time to perform the experiment, and
+# the user has some known tradeoff between overall execution time and cost.
+# 
+# CEED.jl provides tools to take these inputs and produce a set of optimal "arrangements" of experiments for each
+# subset of experiments that form a Pareto front along the tradeoff between information gain and total combined cost
+# (monetary and time). This allows informed decisions to be made, for example, regarding how to allocate scarce
+# resources to a set of experiments that attain some acceptable level of information (or, conversely, reduce
+# uncertainty below some level).
+# 
+# The arrangements produced by the tools introduced in this tutorial are called "static" because they implicitly
+# assume that future data will have exactly the information gain of each experiment as the "historical" input.
+# 
+# This tutorial introduces the theoretical framework behind static experimental designs with synthetic data.
+# For examples using real data, please see our other tutorials.
+
+# ## Theoretical Framework
+
+# ### Experiments
+
+# Let $E = \{ e_1, \ldots, e_n\}$ be a set of $n$ experiments (i.e., $|E|=n$). Each experiment $e \in E$ has an
+# associated tuple $(m_{e},t_{e})$, giving the monetary cost and time duration required to perform experiment $e$.
+# 
+# Consider $P(E)$, the power set of experiments (i.e., every possible subset of experiments). Each subset of
+# experiments $S\in P(E)$ has an associated value $v_{S}$, which is the value of the experiments contained in $S$.
+# This may be given by the loss function associated with a prediction task where the information yielded from $S$
+# is used as predictor variables, or some other notion of information value.
+
+# ### Arrangements
+
+# If experiments within a subset $S$ can be performed simultaneously (in parallel), then each $S$ may be arranged 
+# optimally with respect to time. An arrangement $O_{S}$ of $S$ is a partition of the experiments in $S$ such that
+# the size of each partition is not larger than the maximum number of experiments that may be done in parallel.
+# 
+# Let $l$ be the number of partitions, and $o_{i}$ a partition in $O_{S}$. Then the arrangement is a surjection from $S$
+# onto $O_{S}$. If no experiments can be done in parallel, then $l=|S|$. If all experiments are done in parallel, then
+# $l=1$. Other arrangements fall between these extremes.
+
+# ### Optimal Arrangements
+
+# To find the optimal arrangement for each $S$ we need to know the cost of $O_{S}$. The monetary cost of $O_{S}$ is simply
+# the sum of the costs of each experiment:
+# $$m_{O_{S}}=\sum_{e\in S} m_{e}$$
+# The total time required is the sum of the maximum time *of each partition*. This is because while each partition in the 
+# arrangement is done in serial, experiments within partitions are done in parallel.
+# $$t_{O_{S}}=\sum_{i=1}^{l} \text{max} \{ t_{e} e \in o_{i}\}$$
+# Given these costs and a parameter $\lambda$ which controls the tradeoff between monetary cost and time, the combined
+# cost of an arrangement is:
+# $$\lambda m_{O_{S}} + (1-\lambda) t_{O_{S}}$$
+# 
+# For instance, consider the experiments $S = \{e_{1},e_{2},e_{3},e_{4}\}$, with associated costs $(1, 1)$, $(1, 3)$, $(1, 2)$, and $(1, 4)$.
+# If we conduct experiments $e_1$ through $e_4$ in sequence, this would correspond to an arrangement 
+# $O_{S} = (\{ e_1 \}, \{ e_2 \}, \{ e_3 \}, \{ e_4 \})$ with a total cost of $m_{O_{S}} = 4$ and $t_{O_{S}} = 10$.
+# 
+# However, if we decide to conduct $e_1$ in parallel with $e_3$, and $e_2$ with $e_4$, we would obtain an arrangement 
+# $O_{S} = (\{ e_1, e_3 \}, \{ e_2, e_4 \})$ with a total cost of $m_{O_{S}} = 4$, and $t_{O_{S}} = 3 + 4 = 7$.
+# 
+# Continuing our example and assuming a maximum of two parallel experiments, the optimal arrangement is to conduct 
+# $e_1$ in parallel with $e_2$, and $e_3$ with $e_4$. This results in an arrangement $O_{S} = (\{ e_1, e_2 \}, \{ e_3, e_4 \})$ with a total cost of $m_o = 4$ and $t_o = 2 + 4 = 6$.
+# 
+# In fact, it can be readily demonstrated that the optimal arrangement can be found by ordering the experiments in 
+# S in descending order according to their execution times. Consequently, the experiments are grouped sequentially 
+# into partitions whose size equals the maximum number of parallel experiments, except possibly for the final set,
+# if the maximum number of parallel experiments does not divide $S$ evenly.
+
+# ## Synthetic Data Example
+
+# First we load necessary packages.
+
+using CEED, CEED.StaticDesigns
+using Combinatorics: powerset
+using DataFrames
+using POMDPs, POMDPTools, MCTS
+
+# This tutorial presents a synthetic example of using CEED to optimize static experimental design.
+# We consider a situation where there are 3 experiments, and we draw a value of their "loss function"
+# or "entropy" from the uniform distribution on the unit interval for each. 
+# 
+# For each $S\in P(E)$, we simulate the information value ($v_{S}$) of $S$ as the product of
+# the values for each individual experiment.
+# Therefore, because smaller values are better, any subset containing multiple experiments is guaranteed to be 
+# more "valuable" than any component experiment.
+
+experiments = ["e1","e2","e3"];
+experiments_val = Dict([e => rand() for e in experiments]);
+
+experiments_evals = Dict(
+    map(Set.(collect(powerset(experiments)))) do s
+        if length(s) > 0
+            s => prod([experiments_val[i] for i in s])
+        else
+            return s => 1.0
+        end 
+    end
+);
+
+# Better experiments are more costly, both in terms of time and monetary cost. We print
+# the data frame showing each experiment and its associated costs.
+
+experiments_costs = Dict(
+    sort(collect(keys(experiments_val)), by=k->experiments_val[k], rev=true) .=> tuple.(1:3,1:3)
+);
+
+DataFrame(
+    experiment=collect(keys(experiments_costs)),
+    time=getindex.(values(experiments_costs),1),
+    cost=getindex.(values(experiments_costs),2)
+)
+
+# We can plot the experiments ordered by their "loss function".
+
+plot_evals(experiments_evals; f = x->sort(collect(keys(x)), by = k->x[k], rev=true), ylabel = "loss")
+
+# We print the data frame showing each subset of experiments and its overall loss value.
+
+DataFrame(
+    S=collect.(collect(keys(experiments_evals))),
+    value=collect(values(experiments_evals))
+)
+
+# Now we are ready to find the subsets of experiments giving an optimal tradeoff between information
+# value and combined cost (where we use $\lambda=0.5$). CEED exports a function `efficient_designs`
+# which formulates the problem of finding optimal arrangements as a Markov Decision Process and solves
+# optimal arrangements for each subset on the Pareto frontier.
+# 
+# Note that because we set the maximum number of parallel experiments equal to 2, the complete subset
+# of experiments groups the experiments with long execution times together (see plot legend; each group/partition is
+# prefixed with a number).
+
+max_parallel = 2;
+tradeoff = (0.5, 0.5);
+
+designs = efficient_designs(experiments_costs, experiments_evals, max_parallel=max_parallel, tradeoff=tradeoff);
+
+plot_front(designs; labels = make_labels(designs), ylabel = "loss")
\ No newline at end of file
diff --git a/docs/src/tutorials/SimpleStatic.md b/docs/src/tutorials/SimpleStatic.md
new file mode 100644
index 0000000..4eccafb
--- /dev/null
+++ b/docs/src/tutorials/SimpleStatic.md
@@ -0,0 +1,158 @@
+```@meta
+EditURL = "SimpleStatic.jl"
+```
+
+# Static Experimental Designs
+
+Consider the following scenario. There exists a set of experiments, each of which, when performed, yields
+measurements on one or more observables (features). Each subset of observables (and therefore each subset of experiments)
+has some "information value", which is intentionally vaguely defined for generality, but for example, may be
+a loss function if that subset is used  to train some machine learning model. It is generally the value of acquiring that information.
+Finally, each experiment has some monetary cost and execution time to perform the experiment, and
+the user has some known tradeoff between overall execution time and cost.
+
+CEED.jl provides tools to take these inputs and produce a set of optimal "arrangements" of experiments for each
+subset of experiments that form a Pareto front along the tradeoff between information gain and total combined cost
+(monetary and time). This allows informed decisions to be made, for example, regarding how to allocate scarce
+resources to a set of experiments that attain some acceptable level of information (or, conversely, reduce
+uncertainty below some level).
+
+The arrangements produced by the tools introduced in this tutorial are called "static" because they implicitly
+assume that future data will have exactly the information gain of each experiment as the "historical" input.
+
+This tutorial introduces the theoretical framework behind static experimental designs with synthetic data.
+For examples using real data, please see our other tutorials.
+
+## Theoretical Framework
+
+### Experiments
+
+Let $E = \{ e_1, \ldots, e_n\}$ be a set of $n$ experiments (i.e., $|E|=n$). Each experiment $e \in E$ has an
+associated tuple $(m_{e},t_{e})$, giving the monetary cost and time duration required to perform experiment $e$.
+
+Consider $P(E)$, the power set of experiments (i.e., every possible subset of experiments). Each subset of
+experiments $S\in P(E)$ has an associated value $v_{S}$, which is the value of the experiments contained in $S$.
+This may be given by the loss function associated with a prediction task where the information yielded from $S$
+is used as predictor variables, or some other notion of information value.
+
+### Arrangements
+
+If experiments within a subset $S$ can be performed simultaneously (in parallel), then each $S$ may be arranged
+optimally with respect to time. An arrangement $O_{S}$ of $S$ is a partition of the experiments in $S$ such that
+the size of each partition is not larger than the maximum number of experiments that may be done in parallel.
+
+Let $l$ be the number of partitions, and $o_{i}$ a partition in $O_{S}$. Then the arrangement is a surjection from $S$
+onto $O_{S}$. If no experiments can be done in parallel, then $l=|S|$. If all experiments are done in parallel, then
+$l=1$. Other arrangements fall between these extremes.
+
+### Optimal Arrangements
+
+To find the optimal arrangement for each $S$ we need to know the cost of $O_{S}$. The monetary cost of $O_{S}$ is simply
+the sum of the costs of each experiment:
+$$m_{O_{S}}=\sum_{e\in S} m_{e}$$
+The total time required is the sum of the maximum time *of each partition*. This is because while each partition in the
+arrangement is done in serial, experiments within partitions are done in parallel.
+$$t_{O_{S}}=\sum_{i=1}^{l} \text{max} \{ t_{e} e \in o_{i}\}$$
+Given these costs and a parameter $\lambda$ which controls the tradeoff between monetary cost and time, the combined
+cost of an arrangement is:
+$$\lambda m_{O_{S}} + (1-\lambda) t_{O_{S}}$$
+
+For instance, consider the experiments $S = \{e_{1},e_{2},e_{3},e_{4}\}$, with associated costs $(1, 1)$, $(1, 3)$, $(1, 2)$, and $(1, 4)$.
+If we conduct experiments $e_1$ through $e_4$ in sequence, this would correspond to an arrangement
+$O_{S} = (\{ e_1 \}, \{ e_2 \}, \{ e_3 \}, \{ e_4 \})$ with a total cost of $m_{O_{S}} = 4$ and $t_{O_{S}} = 10$.
+
+However, if we decide to conduct $e_1$ in parallel with $e_3$, and $e_2$ with $e_4$, we would obtain an arrangement
+$O_{S} = (\{ e_1, e_3 \}, \{ e_2, e_4 \})$ with a total cost of $m_{O_{S}} = 4$, and $t_{O_{S}} = 3 + 4 = 7$.
+
+Continuing our example and assuming a maximum of two parallel experiments, the optimal arrangement is to conduct
+$e_1$ in parallel with $e_2$, and $e_3$ with $e_4$. This results in an arrangement $O_{S} = (\{ e_1, e_2 \}, \{ e_3, e_4 \})$ with a total cost of $m_o = 4$ and $t_o = 2 + 4 = 6$.
+
+In fact, it can be readily demonstrated that the optimal arrangement can be found by ordering the experiments in
+S in descending order according to their execution times. Consequently, the experiments are grouped sequentially
+into partitions whose size equals the maximum number of parallel experiments, except possibly for the final set,
+if the maximum number of parallel experiments does not divide $S$ evenly.
+
+## Synthetic Data Example
+
+First we load necessary packages.
+
+````@example SimpleStatic
+using CEED, CEED.StaticDesigns
+using Combinatorics: powerset
+using DataFrames
+using POMDPs, POMDPTools, MCTS
+````
+
+This tutorial presents a synthetic example of using CEED to optimize static experimental design.
+We consider a situation where there are 3 experiments, and we draw a value of their "loss function"
+or "entropy" from the uniform distribution on the unit interval for each.
+
+For each $S\in P(E)$, we simulate the information value ($v_{S}$) of $S$ as the product of
+the values for each individual experiment.
+Therefore, because smaller values are better, any subset containing multiple experiments is guaranteed to be
+more "valuable" than any component experiment.
+
+````@example SimpleStatic
+experiments = ["e1","e2","e3"];
+experiments_val = Dict([e => rand() for e in experiments]);
+
+experiments_evals = Dict(
+    map(Set.(collect(powerset(experiments)))) do s
+        if length(s) > 0
+            s => prod([experiments_val[i] for i in s])
+        else
+            return s => 1.0
+        end
+    end
+);
+nothing #hide
+````
+
+Better experiments are more costly, both in terms of time and monetary cost. We print
+the data frame showing each experiment and its associated costs.
+
+````@example SimpleStatic
+experiments_costs = Dict(
+    sort(collect(keys(experiments_val)), by=k->experiments_val[k], rev=true) .=> tuple.(1:3,1:3)
+);
+
+DataFrame(
+    experiment=collect(keys(experiments_costs)),
+    time=getindex.(values(experiments_costs),1),
+    cost=getindex.(values(experiments_costs),2)
+)
+````
+
+We can plot the experiments ordered by their "loss function".
+
+````@example SimpleStatic
+plot_evals(experiments_evals; f = x->sort(collect(keys(x)), by = k->x[k], rev=true), ylabel = "loss")
+````
+
+We print the data frame showing each subset of experiments and its overall loss value.
+
+````@example SimpleStatic
+DataFrame(
+    S=collect.(collect(keys(experiments_evals))),
+    value=collect(values(experiments_evals))
+)
+````
+
+Now we are ready to find the subsets of experiments giving an optimal tradeoff between information
+value and combined cost (where we use $\lambda=0.5$). CEED exports a function `efficient_designs`
+which formulates the problem of finding optimal arrangements as a Markov Decision Process and solves
+optimal arrangements for each subset on the Pareto frontier.
+
+Note that because we set the maximum number of parallel experiments equal to 2, the complete subset
+of experiments groups the experiments with long execution times together (see plot legend; each group/partition is
+prefixed with a number).
+
+````@example SimpleStatic
+max_parallel = 2;
+tradeoff = (0.5, 0.5);
+
+designs = efficient_designs(experiments_costs, experiments_evals, max_parallel=max_parallel, tradeoff=tradeoff);
+
+plot_front(designs; labels = make_labels(designs), ylabel = "loss")
+````
+
diff --git a/docs/src/tutorials/StaticDesigns.jl b/docs/src/tutorials/StaticDesigns.jl
index effdcf5..e7a0422 100644
--- a/docs/src/tutorials/StaticDesigns.jl
+++ b/docs/src/tutorials/StaticDesigns.jl
@@ -4,23 +4,21 @@
 
 # ## Theoretical Framework
 
-# Let us consider a set of $n$ experiments $E = \{ e_1, \ldots, e_n\}$.
-
-# For each subset $S \subseteq E$ of experiments, we denote by $v_S$ the value of information acquired from conducting experiments in $S$.
+# Let us consider a set of $n$ experiments $E = \{ e_1, \ldots, e_n\}$. For each subset $S \subseteq E$ of experiments, we denote by $v_S$ the value of information acquired from conducting experiments in $S$.
 
 # In the cost-sensitive setting of CEED, conducting an experiment $e$ incurs a cost $(m_e, t_e)$. Generally, this cost is specified in terms of monetary cost and execution time of the experiment.
 
-# To compute the cost associated with carrying out a set of experiments $S$, we first need to introduce the notion of an arrangement $o$ of the experiments $S$. An arrangement is modeled as a sequence of mutually disjoint subsets of $S$. In other words, $o = (o_1, \ldots, o_l)$ for a given $l\in\mathbb N$, where $\bigcup_{i=1}^l o_i = S$ and $o_i \cap o_j = \emptyset$ for each $1\leq i < j \leq l$.
+# To compute the cost associated with carrying out a set of experiments $S$, we first need to introduce the notion of an arrangement $o$ of the experiments $S$. An arrangement is a partition of $S$. In other words, $o = (o_1, \ldots, o_l)$ for $l \leq |S|$, where $\bigcup_{i=1}^l o_i = S$ and $o_i \cap o_j = \emptyset$ for each $1\leq i < j \leq l$.
 
-# Given a subset $S$ of experiments and their arrangement $o$, the total monetary cost and execution time of the experimental design is given as $m_o = \sum_{e\in S} m_e$ and $t_o = \sum_{i=1}^l \max \{ t_e : e\in o_i\}$, respectively.
+# Given a subset $S$ of experiments and their arrangement $o$, the total monetary cost and total execution time of the experimental design is given as $m_o = \sum_{e\in S} m_e$ and $t_o = \sum_{i=1}^l \max \{ t_e : e\in o_i\}$, respectively.
 
 # For instance, consider the experiments $e_1,\, e_2,\, e_3$, and $e_4$ with associated costs $(1, 1)$, $(1, 3)$, $(1, 2)$, and $(1, 4)$. If we conduct experiments $e_1$ through $e_4$ in sequence, this would correspond to an arrangement $o = (\{ e_1 \}, \{ e_2 \}, \{ e_3 \}, \{ e_4 \})$ with a total cost of $m_o = 4$ and $t_o = 10$.
 
 # However, if we decide to conduct $e_1$ in parallel with $e_3$, and $e_2$ with $e_4$, we would obtain an arrangement $o = (\{ e_1, e_3 \}, \{ e_2, e_4 \})$ with a total cost of $m_o = 4$, and $t_o = 3 + 4 = 7$.
 
-# Given the constraint on the maximum number of parallel experiments, we devise an arrangement $o$ of experiments $S$ such that, for a fixed tradeoff between monetary cost and execution time, the expected combined cost $c_{(o, \lambda)} = \lambda m_o + (1-\lambda) t_o$ is minimized (i.e., the execution time is minimized).
+# Given the constraint on the maximum number of parallel experiments, we devise an arrangement $o$ of experiments $S$ such that, for a fixed tradeoff between monetary cost and execution time, the expected combined cost $c_{s} = \lambda m_o + (1-\lambda) t_o$ is minimized (i.e., the execution time is minimized).
 
-# In fact, it can be readily demonstrated that the optimal arrangement can be found by ordering the experiments in set $S$ in descending order according to their execution times. Consequently, the experiments are grouped sequentially into sets whose size equals the maximum number of parallel experiments, except possibly for the final set.
+# Because $t_{o}$ is the sum of the maximum times of each partition of $S$, to minimize $t_{o}$, experiments with long execution times should be grouped together. Therefore, the optimal arrangement is one such that each partition's size equals the maximum number of parallel experiments, except possibly for the final partition, and that assignment into partitions is done in order of execution time.
 
 # Continuing our example and assuming a maximum of two parallel experiments, the optimal arrangement is to conduct $e_1$ in parallel with $e_2$, and $e_3$ with $e_4$. This results in an arrangement $o = (\{ e_1, e_2 \}, \{ e_3, e_4 \})$ with a total cost of $m_o = 4$ and $t_o = 2 + 4 = 6$.
 
@@ -112,8 +110,8 @@ perf_eval = evaluate_experiments(
     measure = LogLoss(),
 )
 
-# We plot performance measures evaluated for subsets of experiments.
-plot_evals(perf_eval; ylabel = "logloss")
+# We plot performance measures evaluated for subsets of experiments, sorted by performance measure.
+plot_evals(perf_eval; f = x->sort(collect(keys(x)), by = k->x[k], rev=true), ylabel = "logloss")
 
 # ## Cost-Efficient Designs
 
@@ -134,7 +132,7 @@ plot_front(designs; labels = make_labels(designs), ylabel = "logloss")
 
 # ### Parallel Experiments
 
-# We may exploit parallelism in the experimental arrangement. To that end, we first specify the monetary cost and execution time for each experiment, respectively.
+# The previous example assumed that experiments had to be run sequentially. We can see how the optimal arrangement changes if we assume multiple experiments can be run in parallel. To that end, we first specify the monetary cost and execution time for each experiment, respectively.
 
 experiments_costs = Dict(
     ## experiment => (monetary cost, execution time) => features
diff --git a/docs/src/tutorials/StaticDesigns.md b/docs/src/tutorials/StaticDesigns.md
index 3e209b4..918970f 100644
--- a/docs/src/tutorials/StaticDesigns.md
+++ b/docs/src/tutorials/StaticDesigns.md
@@ -8,23 +8,21 @@ Consider a situation where a group of patients is tested for a specific disease.
 
 ## Theoretical Framework
 
-Let us consider a set of $n$ experiments $E = \{ e_1, \ldots, e_n\}$.
-
-For each subset $S \subseteq E$ of experiments, we denote by $v_S$ the value of information acquired from conducting experiments in $S$.
+Let us consider a set of $n$ experiments $E = \{ e_1, \ldots, e_n\}$. For each subset $S \subseteq E$ of experiments, we denote by $v_S$ the value of information acquired from conducting experiments in $S$.
 
 In the cost-sensitive setting of CEED, conducting an experiment $e$ incurs a cost $(m_e, t_e)$. Generally, this cost is specified in terms of monetary cost and execution time of the experiment.
 
-To compute the cost associated with carrying out a set of experiments $S$, we first need to introduce the notion of an arrangement $o$ of the experiments $S$. An arrangement is modeled as a sequence of mutually disjoint subsets of $S$. In other words, $o = (o_1, \ldots, o_l)$ for a given $l\in\mathbb N$, where $\bigcup_{i=1}^l o_i = S$ and $o_i \cap o_j = \emptyset$ for each $1\leq i < j \leq l$.
+To compute the cost associated with carrying out a set of experiments $S$, we first need to introduce the notion of an arrangement $o$ of the experiments $S$. An arrangement is a partition of $S$. In other words, $o = (o_1, \ldots, o_l)$ for $l \leq |S|$, where $\bigcup_{i=1}^l o_i = S$ and $o_i \cap o_j = \emptyset$ for each $1\leq i < j \leq l$.
 
-Given a subset $S$ of experiments and their arrangement $o$, the total monetary cost and execution time of the experimental design is given as $m_o = \sum_{e\in S} m_e$ and $t_o = \sum_{i=1}^l \max \{ t_e : e\in o_i\}$, respectively.
+Given a subset $S$ of experiments and their arrangement $o$, the total monetary cost and total execution time of the experimental design is given as $m_o = \sum_{e\in S} m_e$ and $t_o = \sum_{i=1}^l \max \{ t_e : e\in o_i\}$, respectively.
 
 For instance, consider the experiments $e_1,\, e_2,\, e_3$, and $e_4$ with associated costs $(1, 1)$, $(1, 3)$, $(1, 2)$, and $(1, 4)$. If we conduct experiments $e_1$ through $e_4$ in sequence, this would correspond to an arrangement $o = (\{ e_1 \}, \{ e_2 \}, \{ e_3 \}, \{ e_4 \})$ with a total cost of $m_o = 4$ and $t_o = 10$.
 
 However, if we decide to conduct $e_1$ in parallel with $e_3$, and $e_2$ with $e_4$, we would obtain an arrangement $o = (\{ e_1, e_3 \}, \{ e_2, e_4 \})$ with a total cost of $m_o = 4$, and $t_o = 3 + 4 = 7$.
 
-Given the constraint on the maximum number of parallel experiments, we devise an arrangement $o$ of experiments $S$ such that, for a fixed tradeoff between monetary cost and execution time, the expected combined cost $c_{(o, \lambda)} = \lambda m_o + (1-\lambda) t_o$ is minimized (i.e., the execution time is minimized).
+Given the constraint on the maximum number of parallel experiments, we devise an arrangement $o$ of experiments $S$ such that, for a fixed tradeoff between monetary cost and execution time, the expected combined cost $c_{s} = \lambda m_o + (1-\lambda) t_o$ is minimized (i.e., the execution time is minimized).
 
-In fact, it can be readily demonstrated that the optimal arrangement can be found by ordering the experiments in set $S$ in descending order according to their execution times. Consequently, the experiments are grouped sequentially into sets whose size equals the maximum number of parallel experiments, except possibly for the final set.
+Because $t_{o}$ is the sum of the maximum times of each partition of $S$, to minimize $t_{o}$, experiments with long execution times should be grouped together. Therefore, the optimal arrangement is one such that each partition's size equals the maximum number of parallel experiments, except possibly for the final partition, and that assignment into partitions is done in order of execution time.
 
 Continuing our example and assuming a maximum of two parallel experiments, the optimal arrangement is to conduct $e_1$ in parallel with $e_2$, and $e_3$ with $e_4$. This results in an arrangement $o = (\{ e_1, e_2 \}, \{ e_3, e_4 \})$ with a total cost of $m_o = 4$ and $t_o = 2 + 4 = 6$.
 
@@ -138,10 +136,10 @@ perf_eval = evaluate_experiments(
 )
 ````
 
-We plot performance measures evaluated for subsets of experiments.
+We plot performance measures evaluated for subsets of experiments, sorted by performance measure.
 
 ````@example StaticDesigns
-plot_evals(perf_eval; ylabel = "logloss")
+plot_evals(perf_eval; f = x->sort(collect(keys(x)), by = k->x[k], rev=true), ylabel = "logloss")
 ````
 
 ## Cost-Efficient Designs
@@ -170,7 +168,7 @@ plot_front(designs; labels = make_labels(designs), ylabel = "logloss")
 
 ### Parallel Experiments
 
-We may exploit parallelism in the experimental arrangement. To that end, we first specify the monetary cost and execution time for each experiment, respectively.
+The previous example assumed that experiments had to be run sequentially. We can see how the optimal arrangement changes if we assume multiple experiments can be run in parallel. To that end, we first specify the monetary cost and execution time for each experiment, respectively.
 
 ````@example StaticDesigns
 experiments_costs = Dict(
diff --git a/docs/src/tutorials/StaticDesignsFiltration.jl b/docs/src/tutorials/StaticDesignsFiltration.jl
index db42586..8ea1177 100644
--- a/docs/src/tutorials/StaticDesignsFiltration.jl
+++ b/docs/src/tutorials/StaticDesignsFiltration.jl
@@ -18,7 +18,7 @@
 
 # To compute the cost associated with carrying out a set of experiments $S$, we first need to introduce the notion of an arrangement $o$ of the experiments $S$. An arrangement is modeled as a sequence of mutually disjoint subsets of $S$. In other words, $o = (o_1, \ldots, o_l)$ for a given $l\in\mathbb N$, where $\bigcup_{i=1}^l o_i = S$ and $o_i \cap o_j = \emptyset$ for each $1\leq i < j \leq l$.
 
-# Given a subset $S$ of experiments and their arrangement $o$, the total (discounted) monetary cost and execution time of the experimental design is given as $m_o = \sum{i=1}^l r_{S_{i-1}}\sum_{e\in o_i} m_e$ and $t_o = \sum_{i=1}^l \max \{ t_e : e\in o_i\}$, respectively. Importantly, the factor $r_{S_{i-1}}$ models the fact that a set of entities may have dropped out in the previous experiments, hence saving the resources on running the subsequent experiments.
+# Given a subset $S$ of experiments and their arrangement $o$, the total (discounted) monetary cost and execution time of the experimental design is given as $m_o = \sum_{i=1}^{l} r_{S_{i-1}}\sum_{e\in o_i} m_e$ and $t_o = \sum_{i=1}^{l} \max \{ t_e : e\in o_i\}$, respectively. Importantly, the factor $r_{S_{i-1}}$ models the fact that a set of entities may have dropped out in the previous experiments, hence saving the resources on running the subsequent experiments.
 
 # We note that these computations are based on the assumption that monetary cost is associated with the analysis of a single experimental entity, such as a patient. Therefore, the total monetary cost obtained for a specific arrangement is effectively the ['expected'](https://en.wikipedia.org/wiki/Expected_value) monetary cost, adjusted for a single entity. Conversely, we suppose that all entities can be concurrently examined in a specific experiment. As such, the total execution time is equivalent to the longest time until all experiments within an arrangement are finished or all entities have been eliminated (which ocurrs when the filtration rate the experiments conducted so far is zero). Importantly, this distinctly differs from calculating the 'expected lifespan' of an entity in the triage.
 
diff --git a/docs/src/tutorials/StaticDesignsFiltration.md b/docs/src/tutorials/StaticDesignsFiltration.md
index 159f936..d17b6a2 100644
--- a/docs/src/tutorials/StaticDesignsFiltration.md
+++ b/docs/src/tutorials/StaticDesignsFiltration.md
@@ -22,7 +22,7 @@ In the cost-sensitive setting of CEED, conducting an experiment $e$ incurs a cos
 
 To compute the cost associated with carrying out a set of experiments $S$, we first need to introduce the notion of an arrangement $o$ of the experiments $S$. An arrangement is modeled as a sequence of mutually disjoint subsets of $S$. In other words, $o = (o_1, \ldots, o_l)$ for a given $l\in\mathbb N$, where $\bigcup_{i=1}^l o_i = S$ and $o_i \cap o_j = \emptyset$ for each $1\leq i < j \leq l$.
 
-Given a subset $S$ of experiments and their arrangement $o$, the total (discounted) monetary cost and execution time of the experimental design is given as $m_o = \sum{i=1}^l r_{S_{i-1}}\sum_{e\in o_i} m_e$ and $t_o = \sum_{i=1}^l \max \{ t_e : e\in o_i\}$, respectively. Importantly, the factor $r_{S_{i-1}}$ models the fact that a set of entities may have dropped out in the previous experiments, hence saving the resources on running the subsequent experiments.
+Given a subset $S$ of experiments and their arrangement $o$, the total (discounted) monetary cost and execution time of the experimental design is given as $m_o = \sum_{i=1}^{l} r_{S_{i-1}}\sum_{e\in o_i} m_e$ and $t_o = \sum_{i=1}^{l} \max \{ t_e : e\in o_i\}$, respectively. Importantly, the factor $r_{S_{i-1}}$ models the fact that a set of entities may have dropped out in the previous experiments, hence saving the resources on running the subsequent experiments.
 
 We note that these computations are based on the assumption that monetary cost is associated with the analysis of a single experimental entity, such as a patient. Therefore, the total monetary cost obtained for a specific arrangement is effectively the ['expected'](https://en.wikipedia.org/wiki/Expected_value) monetary cost, adjusted for a single entity. Conversely, we suppose that all entities can be concurrently examined in a specific experiment. As such, the total execution time is equivalent to the longest time until all experiments within an arrangement are finished or all entities have been eliminated (which ocurrs when the filtration rate the experiments conducted so far is zero). Importantly, this distinctly differs from calculating the 'expected lifespan' of an entity in the triage.
 
diff --git a/docs/src/tutorials/data/class.csv b/docs/src/tutorials/data/class.csv
new file mode 100644
index 0000000..dda6e44
--- /dev/null
+++ b/docs/src/tutorials/data/class.csv
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diff --git a/tutorials/GenerativeDesigns.jl b/tutorials/GenerativeDesigns.jl
index 42078cf..c4cf392 100644
--- a/tutorials/GenerativeDesigns.jl
+++ b/tutorials/GenerativeDesigns.jl
@@ -249,7 +249,7 @@ designs = efficient_designs(
     6,
     evidence;
     solver,
-    mdp_options = (; max_parallel = 2, costs_tradeoff = [0, 1.0]),
+    mdp_options = (; max_parallel = 2, costs_tradeoff = (0, 1.0)),
     repetitions = 5,
 );
 
diff --git a/tutorials/SimpleStatic.jl b/tutorials/SimpleStatic.jl
index f060969..7092758 100644
--- a/tutorials/SimpleStatic.jl
+++ b/tutorials/SimpleStatic.jl
@@ -1,19 +1,96 @@
-# # Static Designs
-
-# This tutorial presents a synthetic example of using CEED to optimize static experimental design.
-# We consider a situation where there are 3 experiments, and we draw a value of their "loss function"
-# or "entropy" from the uniform distribution on the unit interval for each. 
-
-# The "evaluation data" for subsets of experiments is the produt of those value for each individual experiment.
-# Therefore, because smaller values are better, the subset containing multiple experiments is guaranteed to be 
-# more "valuable" than any individual experiment.
+# # Static Experimental Designs
+
+# Consider the following scenario. There exists a set of experiments, each of which, when performed, yields
+# measurements on one or more observables (features). Each subset of observables (and therefore each subset of experiments)
+# has some "information value", which is intentionally vaguely defined for generality, but for example, may be 
+# a loss function if that subset is used  to train some machine learning model. It is generally the value of acquiring that information. 
+# Finally, each experiment has some monetary cost and execution time to perform the experiment, and
+# the user has some known tradeoff between overall execution time and cost.
+# 
+# CEED.jl provides tools to take these inputs and produce a set of optimal "arrangements" of experiments for each
+# subset of experiments that form a Pareto front along the tradeoff between information gain and total combined cost
+# (monetary and time). This allows informed decisions to be made, for example, regarding how to allocate scarce
+# resources to a set of experiments that attain some acceptable level of information (or, conversely, reduce
+# uncertainty below some level).
+# 
+# The arrangements produced by the tools introduced in this tutorial are called "static" because they implicitly
+# assume that future data will have exactly the information gain of each experiment as the "historical" input.
+# 
+# This tutorial introduces the theoretical framework behind static experimental designs with synthetic data.
+# For examples using real data, please see our other tutorials.
+
+# ## Theoretical Framework
+
+# ### Experiments
+
+# Let $E = \{ e_1, \ldots, e_n\}$ be a set of $n$ experiments (i.e., $|E|=n$). Each experiment $e \in E$ has an
+# associated tuple $(m_{e},t_{e})$, giving the monetary cost and time duration required to perform experiment $e$.
+
+# ![experiments](assets/static_experiments.png)
+
+# Consider $P(E)$, the power set of experiments (i.e., every possible subset of experiments). Each subset of
+# experiments $S\in P(E)$ has an associated value $v_{S}$, which is the value of the experiments contained in $S$.
+# This may be given by the loss function associated with a prediction task where the information yielded from $S$
+# is used as predictor variables, or some other notion of information value.
+
+# ![experiments](assets/static_powerset.png)
+
+# ### Arrangements
+
+# If experiments within a subset $S$ can be performed simultaneously (in parallel), then each $S$ may be arranged 
+# optimally with respect to time. An arrangement $O_{S}$ of $S$ is a partition of the experiments in $S$ such that
+# the size of each partition is not larger than the maximum number of experiments that may be done in parallel.
+# 
+# Let $l$ be the number of partitions, and $o_{i}$ a partition in $O_{S}$. Then the arrangement is a surjection from $S$
+# onto $O_{S}$. If no experiments can be done in parallel, then $l=|S|$. If all experiments are done in parallel, then
+# $l=1$. Other arrangements fall between these extremes.
+
+# ![experiments](assets/static_arrangement.png)
+
+# ### Optimal Arrangements
+
+# To find the optimal arrangement for each $S$ we need to know the cost of $O_{S}$. The monetary cost of $O_{S}$ is simply
+# the sum of the costs of each experiment:
+# $$m_{O_{S}}=\sum_{e\in S} m_{e}$$
+# The total time required is the sum of the maximum time *of each partition*. This is because while each partition in the 
+# arrangement is done in serial, experiments within partitions are done in parallel.
+# $$t_{O_{S}}=\sum_{i=1}^{l} \text{max} \{ t_{e} e \in o_{i}\}$$
+# Given these costs and a parameter $\lambda$ which controls the tradeoff between monetary cost and time, the combined
+# cost of an arrangement is:
+# $$\lambda m_{O_{S}} + (1-\lambda) t_{O_{S}}$$
+# 
+# For instance, consider the experiments $S = \{e_{1},e_{2},e_{3},e_{4}\}$, with associated costs $(1, 1)$, $(1, 3)$, $(1, 2)$, and $(1, 4)$.
+# If we conduct experiments $e_1$ through $e_4$ in sequence, this would correspond to an arrangement 
+# $O_{S} = (\{ e_1 \}, \{ e_2 \}, \{ e_3 \}, \{ e_4 \})$ with a total cost of $m_{O_{S}} = 4$ and $t_{O_{S}} = 10$.
+# 
+# However, if we decide to conduct $e_1$ in parallel with $e_3$, and $e_2$ with $e_4$, we would obtain an arrangement 
+# $O_{S} = (\{ e_1, e_3 \}, \{ e_2, e_4 \})$ with a total cost of $m_{O_{S}} = 4$, and $t_{O_{S}} = 3 + 4 = 7$.
+# 
+# Continuing our example and assuming a maximum of two parallel experiments, the optimal arrangement is to conduct 
+# $e_1$ in parallel with $e_2$, and $e_3$ with $e_4$. This results in an arrangement $O_{S} = (\{ e_1, e_2 \}, \{ e_3, e_4 \})$ with a total cost of $m_o = 4$ and $t_o = 2 + 4 = 6$.
+# 
+# In fact, it can be readily demonstrated that the optimal arrangement can be found by ordering the experiments in 
+# S in descending order according to their execution times. Consequently, the experiments are grouped sequentially 
+# into partitions whose size equals the maximum number of parallel experiments, except possibly for the final set,
+# if the maximum number of parallel experiments does not divide $S$ evenly.
+
+# ## Synthetic Data Example
+
+# First we load necessary packages.
 
 using CEED, CEED.StaticDesigns
 using Combinatorics: powerset
 using DataFrames
 using POMDPs, POMDPTools, MCTS
 
-# First we mimic the output of `evaluate_experiments`.
+# This tutorial presents a synthetic example of using CEED to optimize static experimental design.
+# We consider a situation where there are 3 experiments, and we draw a value of their "loss function"
+# or "entropy" from the uniform distribution on the unit interval for each. 
+# 
+# For each $S\in P(E)$, we simulate the information value ($v_{S}$) of $S$ as the product of
+# the values for each individual experiment.
+# Therefore, because smaller values are better, any subset containing multiple experiments is guaranteed to be 
+# more "valuable" than any component experiment.
 
 experiments = ["e1","e2","e3"];
 experiments_val = Dict([e => rand() for e in experiments]);
@@ -52,7 +129,14 @@ DataFrame(
     value=collect(values(experiments_evals))
 )
 
-# We use `efficient_designs` to solve the optimal arrangements.
+# Now we are ready to find the subsets of experiments giving an optimal tradeoff between information
+# value and combined cost (where we use $\lambda=0.5$). CEED exports a function `efficient_designs`
+# which formulates the problem of finding optimal arrangements as a Markov Decision Process and solves
+# optimal arrangements for each subset on the Pareto frontier.
+# 
+# Note that because we set the maximum number of parallel experiments equal to 2, the complete subset
+# of experiments groups the experiments with long execution times together (see plot legend; each group/partition is
+# prefixed with a number).
 
 max_parallel = 2;
 tradeoff = (0.5, 0.5);
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