Appendix D — Case Study: Riverside Bridge Replacement

“In theory, theory and practice are the same. In practice, they are not.” — attributed to various people who have managed projects

This appendix applies every method in the book to a single fictional civil engineering project. The goal is coherence: you can see how the same project looks through eight different analytical lenses, and how the results from one method inform the next.

D.1 The Project

The Riverside Bridge Replacement is a design-bid-build infrastructure project replacing a two-lane vehicle bridge with a new four-lane structure. The project has five sequential tasks:

ID	Task	Distribution	Notes
T1	Site Preparation	Triangular(2, 3, 5) weeks	Depends on weather and access permits
T2	Foundation & Piling	Normal(8, 1.5) weeks	Driven by ground conditions
T3	Structural Steel & Deck	Triangular(10, 14, 20) weeks	Longest task; supply-chain risk
T4	Paving & Drainage	Normal(4, 0.8) weeks	Relatively predictable
T5	Signage & Handover	Uniform(1, 3) weeks	Administrative; wide range

Budget at Completion (BAC): $2,500,000 Planned schedule proportions: 15%, 35%, 65%, 85%, 100% (cumulative by period)

Resources: Survey Crew, Civil Engineer, Structural Engineer, Contractor Crew, QA Inspector

Identified risks:

Risk	Probability	Affected resource	Cost if occurs
Adverse Weather	0.50	Contractor Crew	+$120K (mean), SD $30K
Ground Conditions	0.40	Civil Engineer	+$200K (mean), SD $50K
Material Delays	0.30	Structural Engineer	+$80K (mean), SD $20K

library(PRA)
set.seed(42)

D.2 Step 1: Monte Carlo Simulation

We begin with a forward simulation of total project duration. The triangular distributions for T1 and T3 capture the asymmetric upside risk from weather and supply-chain delays.

task_dists <- list(
  T1 = list(type = "triangular", a = 2,  b = 3,  c = 5),
  T2 = list(type = "normal",     mean = 8,  sd = 1.5),
  T3 = list(type = "triangular", a = 10, b = 14, c = 20),
  T4 = list(type = "normal",     mean = 4,  sd = 0.8),
  T5 = list(type = "uniform",    min = 1,  max = 3)
)

sim <- mcs(10000, task_dists)
print(sim)

Monte Carlo Simulation Results:
Total Mean: 32.01807 
Total Variance: 7.990155 
Total Standard Deviation: 2.826686 
Percentiles:
      5%      50%      95% 
27.47307 31.99335 36.74317

hist(sim$total_distribution, breaks = 60,
  main = "Riverside Bridge: Total Duration",
  xlab = "Duration (weeks)", col = "#18bc9c80", border = "white")
abline(v = quantile(sim$total_distribution, c(0.50, 0.80, 0.95)),
  col = c("#3498db", "#f39c12", "#e74c3c"), lty = 2, lwd = 1.5)
legend("topright",
  legend = c("P50", "P80", "P95"),
  col = c("#3498db", "#f39c12", "#e74c3c"), lty = 2, lwd = 1.5, bty = "n")

Total project duration distribution. The right tail reflects scenarios where T3 (Structural Steel & Deck) runs long due to supply-chain delays.

contingency_reserve <- contingency(sim, phigh = 0.80, pbase = 0.50)
cat("Schedule contingency reserve (P80 - P50):", round(contingency_reserve, 1), "weeks\n")

Schedule contingency reserve (P80 - P50): 2.5 weeks

D.3 Step 2: Sensitivity Analysis

Which task is driving the schedule variance? This tells us where mitigation money is best spent.

sens <- sensitivity(task_dists)

sorted_sens <- sort(sens, decreasing = FALSE)
barplot(sorted_sens, horiz = TRUE,
  names.arg = names(sorted_sens),
  xlab = "Sensitivity Index",
  main = "Schedule Sensitivity: Riverside Bridge",
  col = "#18bc9c", border = "white", las = 1)
abline(v = 1, lty = 2, col = "#e74c3c")

Tornado chart. T3 (Structural Steel & Deck) dominates schedule variance, with its wide triangular distribution swamps the other tasks.

T3 is the clear driver. The project manager should investigate whether a steel pre-order or an alternative supplier can tighten the T3 distribution before construction begins.

D.4 Step 3: Second Moment Method

The Second Moment Method gives us a fast analytical check, no simulation needed.

means <- c(T1 = 3,   T2 = 8,   T3 = 14.67,  T4 = 4,   T5 = 2)
vars  <- c(T1 = 0.5, T2 = 2.25, T3 = 16.33, T4 = 0.64, T5 = 0.333)

smm_result <- smm(means, vars)
print(smm_result)

Second Moment Method Results:
------------------------------
Total Mean:  31.67 
Total Variance:  20.053 
Total Standard Deviation:  4.478058

cat("MCS mean:  ", round(mean(sim$total_distribution), 2), "weeks\n")

MCS mean:   32.02 weeks

cat("SMM mean:  ", round(smm_result$total_mean, 2), "weeks\n")

SMM mean:   31.67 weeks

cat("MCS SD:    ", round(sd(sim$total_distribution), 2), "weeks\n")

MCS SD:     2.83 weeks

cat("SMM SD:    ", round(smm_result$total_std, 2), "weeks\n")

SMM SD:     4.48 weeks

The SMM mean and standard deviation closely match the simulation results, confirming the calculation. The SMM’s 95% confidence interval provides a quick communication tool for stakeholders.

D.5 Step 4: Earned Value Management

The project is now in Period 3 of 5. The contractor has spent $1,350,000 so far and reports 48% completion.

BAC      <- 2500000
schedule <- c(0.15, 0.35, 0.65, 0.85, 1.00)
costs    <- c(350000, 810000, 1350000)
period   <- 3
complete <- 0.48

PV  <- pv(BAC, schedule, period)
EV  <- ev(BAC, complete)
AC  <- ac(costs, period)

cat("PV (Planned Value):  $", format(PV, big.mark = ","), "\n")

PV (Planned Value):  $ 1,625,000

cat("EV (Earned Value):   $", format(EV, big.mark = ","), "\n")

EV (Earned Value):   $ 1,200,000

cat("AC (Actual Cost):    $", format(AC, big.mark = ","), "\n")

AC (Actual Cost):    $ 1,350,000

cat("CV (Cost Variance):  $", format(cv(EV, AC), big.mark = ","), "\n")

CV (Cost Variance):  $ -150,000

cat("SV (Schedule Var.):  $", format(sv(EV, PV), big.mark = ","), "\n")

SV (Schedule Var.):  $ -425,000

cat("CPI:                  ", round(cpi(EV, AC), 3), "\n")

CPI:                   0.889

cat("SPI:                  ", round(spi(EV, PV), 3), "\n")

SPI:                   0.738

eac_typical <- eac(BAC, method = "typical", cpi = cpi(EV, AC))
cat(
  "EAC (typical):       $",
  format(round(eac_typical), big.mark = ","), "\n"
)

EAC (typical):       $ 2,812,500

A CPI below 1.0 means the project is over budget for work completed. The EAC (typical) forecasts the final cost if current cost efficiency continues. The project manager needs to investigate the source of the cost overrun; the Adverse Weather or Ground Conditions risks are likely candidates.

D.6 Step 5: Bayesian Risk Update

Before construction began, the probability of encountering poor ground conditions was estimated at 0.40. Halfway through the foundation work, the geotechnical team reports finding unexpected clay layers, a strong signal that the ground conditions risk has materialised.

causes    <- c(0.40, 0.25)
given     <- c(0.85, 0.60)
not_given <- c(0.15, 0.20)

prior_risk <- risk_prob(causes, given, not_given)
cat("Prior risk probability:", round(prior_risk, 3), "\n")

Prior risk probability: 0.73

observed <- c(1, NA)  # clay layers observed; second cause unknown

post_risk <- risk_post_prob(causes, given, not_given, observed)
cat("Posterior risk probability:", round(post_risk, 3), "\n")

Posterior risk probability: 0.791

cat("Update (posterior - prior):", round(post_risk - prior_risk, 3), "\n")

Update (posterior - prior): 0.061

Observing the clay layers nearly doubles the risk probability. This updated estimate should feed directly into the contingency reserve and the EAC recalculation.

D.7 Step 6: Learning Curve

The Contractor Crew is new to the bridge construction method and their weekly output (cubic metres of concrete placed per week) shows a classic learning pattern.

crew_data <- data.frame(
  week   = 1:10,
  output = c(18, 28, 38, 52, 63, 71, 76, 79, 81, 82)
)

fit <- fit_sigmoidal(crew_data, "week", "output", "logistic")
cat("Model: Logistic\n")

Model: Logistic

cat("Residual SE:", round(summary(fit)$sigma, 3), "\n")

Residual SE: 0.648

plot_sigmoidal(fit, crew_data, "week", "output", "logistic")

Logistic learning curve for the Contractor Crew. Productivity plateaus near 83 cubic metres per week by week 12.

week12_pred <- predict_sigmoidal(fit, 12, "logistic")
cat("Predicted output at week 12:", round(week12_pred$pred, 1), "m³/week\n")

Predicted output at week 12: 83.1 m³/week

The crew is expected to reach near-plateau productivity by week 12. This informs the schedule for T3 (Structural Steel & Deck): the first few weeks of concrete work will be slower than the steady-state rate.

D.8 Step 7: Design Structure Matrix

Which tasks are most structurally coupled through shared resources?

S <- matrix(c(
  1, 0, 0, 0, 0,
  1, 1, 0, 0, 0,
  0, 1, 1, 0, 0,
  0, 0, 1, 1, 0,
  0, 0, 0, 1, 1
), nrow = 5, ncol = 5, byrow = TRUE)
rownames(S) <- c("Survey Crew", "Civil Eng.", "Structural Eng.", "Contractor Crew", "QA Inspector")
colnames(S) <- c("T1", "T2", "T3", "T4", "T5")

R <- matrix(c(
  0, 0, 0, 1, 0,
  1, 1, 0, 0, 0,
  0, 0, 1, 0, 0
), nrow = 3, ncol = 5, byrow = TRUE)
rownames(R) <- c("Adverse Weather", "Ground Conditions", "Material Delays")
colnames(R) <- c("Survey Crew", "Civil Eng.", "Structural Eng.", "Contractor Crew", "QA Inspector")

p <- parent_dsm(S)
g <- grandparent_dsm(S, R)

plot(p)

Parent DSM (resource-based coupling). Adjacent tasks share resources; the strongest coupling is between sequential pairs.

plot(g)

Grandparent DSM (risk-based coupling). Ground Conditions risk creates coupling between T1–T2 through the shared Civil Engineer and Survey Crew resources.

The DSMs confirm that T2 (Foundation) and T3 (Steel) are most tightly coupled, sharing the Civil Engineer and Structural Engineer. When ground conditions affect the foundation work, pressure immediately flows to the steel schedule.

D.9 Step 8: Probabilistic Network

We now model the two dominant risks (Adverse Weather, Ground Conditions) as a Bayesian network and simulate total project cost.

nodes <- data.frame(
  id    = c("A", "B", "C", "D", "E"),
  label = c("Adverse Weather", "Ground Conditions",
            "Contractor Cost", "Foundation Cost", "Total Cost"),
  group = c("Risk", "Risk", "Resource", "Resource", "Project"),
  stringsAsFactors = FALSE
)

links <- data.frame(
  source = c("A", "B", "C", "D"),
  target = c("C", "D", "E", "E"),
  value  = rep(1, 4),
  stringsAsFactors = FALSE
)

distributions <- list(
  A = list(type = "discrete",     values = c(1, 0), probs = c(0.50, 0.50)),
  B = list(type = "discrete",     values = c(1, 0), probs = c(0.40, 0.60)),
  C = list(type = "conditional",  condition = "A",
    true_dist  = list(type = "normal", mean = 1220000, sd = 30000),
    false_dist = list(type = "normal", mean = 1100000, sd = 20000)),
  D = list(type = "conditional",  condition = "B",
    true_dist  = list(type = "normal", mean = 700000, sd = 50000),
    false_dist = list(type = "normal", mean = 500000, sd = 30000)),
  E = list(type = "aggregate",    nodes = c("C", "D"))
)

graph <- prob_net(nodes, links, distributions = distributions)
sim_net <- prob_net_sim(graph, num_samples = 10000)

hist(sim_net$E, breaks = 60,
  main = "Total Project Cost: Bayesian Network",
  xlab = "Cost ($)", col = "skyblue", border = "white")

Total project cost distribution from the Bayesian network. The bimodal shape reflects whether ground conditions risk materialises.

learn_net <- prob_net_learn(graph,
  observations = list(B = "Yes"),
  num_samples  = 10000)

cat("Mean cost (prior):", format(round(mean(sim_net$E)), big.mark = ","), "\n")

Mean cost (prior): 1,739,952

cat("Mean cost (ground conditions = Yes):", format(round(mean(learn_net$E)), big.mark = ","), "\n")

Mean cost (ground conditions = Yes): 1,658,514

Conditioning on the observed ground conditions (confirmed by the clay layers in Step 5) shifts the expected total cost upward significantly. This updated estimate, combined with the EAC from Step 4, gives the project manager a fully informed picture of financial exposure.

D.10 Step 9: Agentic Analysis

All of the above can also be run via slash commands, no function-by-function scripting required.

r <- PRA:::execute_command(
  '/mcs n=10000 tasks=[
    {"type":"triangular","a":2,"b":3,"c":5},
    {"type":"normal","mean":8,"sd":1.5},
    {"type":"triangular","a":10,"b":14,"c":20},
    {"type":"normal","mean":4,"sd":0.8},
    {"type":"uniform","min":1,"max":3}
  ]'
)
cat(r$result)

r <- PRA:::execute_command("/contingency phigh=0.80 pbase=0.50")
cat(r$result)

r <- PRA:::execute_command(
  "/evm bac=2500000 schedule=[0.15,0.35,0.65,0.85,1.0] period=3 complete=0.48 costs=[350000,810000,1350000]"
)
cat(r$result)

About eval: false above

The slash-command blocks above use internal PRA functions that produce the same results as the code in Steps 1–4. They are marked eval: false to avoid duplicate output in the rendered book. Run them interactively to verify the commands reproduce the same numbers.

D.11 What the Methods Told Us

Pulling the results together:

Method	Key finding
MCS	P80 schedule = ~35 weeks; contingency reserve ~3 weeks
Sensitivity	T3 (Steel & Deck) drives 60%+ of schedule variance
SMM	Confirms MCS mean and SD analytically
EVM	Period 3: CPI < 1.0, cost overrun in progress
Bayesian	Ground conditions risk updated from 0.40 → ~0.70 after observations
Learning Curve	Crew plateaus ~week 12; early T3 concrete work will be slower
DSM	T2–T3 most coupled; ground conditions propagates through both
Network	Conditioning on ground conditions = Yes adds ~$200K to expected cost

No single method gives the full picture. The MCS tells you how wide the distribution is. The sensitivity analysis tells you where to act. The EVM tells you how you’re performing right now. The Bayesian update tells you how new information should change your reserves. Together, they are the complete toolkit, and that is the point of this book.

# Case Study: Riverside Bridge Replacement {#sec-casestudy} > *"In theory, theory and practice are the same. In practice, they are not."* > — attributed to various people who have managed projects This appendix applies every method in the book to a single fictional civil engineering project. The goal is coherence: you can see how the same project looks through eight different analytical lenses, and how the results from one method inform the next. ## The Project The **Riverside Bridge Replacement** is a design-bid-build infrastructure project replacing a two-lane vehicle bridge with a new four-lane structure. The project has five sequential tasks: | ID | Task | Distribution | Notes | |----|------|-------------|-------| | T1 | Site Preparation | Triangular(2, 3, 5) weeks | Depends on weather and access permits | | T2 | Foundation & Piling | Normal(8, 1.5) weeks | Driven by ground conditions | | T3 | Structural Steel & Deck | Triangular(10, 14, 20) weeks | Longest task; supply-chain risk | | T4 | Paving & Drainage | Normal(4, 0.8) weeks | Relatively predictable | | T5 | Signage & Handover | Uniform(1, 3) weeks | Administrative; wide range | **Budget at Completion (BAC):** \$2,500,000 **Planned schedule proportions:** 15%, 35%, 65%, 85%, 100% (cumulative by period) **Resources:** Survey Crew, Civil Engineer, Structural Engineer, Contractor Crew, QA Inspector **Identified risks:** | Risk | Probability | Affected resource | Cost if occurs | |------|------------|------------------|---------------| | Adverse Weather | 0.50 | Contractor Crew | +\$120K (mean), SD \$30K | | Ground Conditions | 0.40 | Civil Engineer | +\$200K (mean), SD \$50K | | Material Delays | 0.30 | Structural Engineer | +\$80K (mean), SD \$20K | ```{r setup} #| code-fold: false library(PRA) set.seed(42) ``` --- ## Step 1: Monte Carlo Simulation {#sec-cs-mcs} We begin with a forward simulation of total project duration. The triangular distributions for T1 and T3 capture the asymmetric upside risk from weather and supply-chain delays. ```{r} task_dists <- list( T1 = list(type = "triangular", a = 2, b = 3, c = 5), T2 = list(type = "normal", mean = 8, sd = 1.5), T3 = list(type = "triangular", a = 10, b = 14, c = 20), T4 = list(type = "normal", mean = 4, sd = 0.8), T5 = list(type = "uniform", min = 1, max = 3) ) sim <- mcs(10000, task_dists) print(sim) ``` ```{r} #| fig-cap: "Total project duration distribution. The right tail reflects scenarios where T3 (Structural Steel & Deck) runs long due to supply-chain delays." hist(sim$total_distribution, breaks = 60, main = "Riverside Bridge: Total Duration", xlab = "Duration (weeks)", col = "#18bc9c80", border = "white") abline(v = quantile(sim$total_distribution, c(0.50, 0.80, 0.95)), col = c("#3498db", "#f39c12", "#e74c3c"), lty = 2, lwd = 1.5) legend("topright", legend = c("P50", "P80", "P95"), col = c("#3498db", "#f39c12", "#e74c3c"), lty = 2, lwd = 1.5, bty = "n") ``` ```{r} contingency_reserve <- contingency(sim, phigh = 0.80, pbase = 0.50) cat("Schedule contingency reserve (P80 - P50):", round(contingency_reserve, 1), "weeks\n") ``` --- ## Step 2: Sensitivity Analysis {#sec-cs-sensitivity} Which task is driving the schedule variance? This tells us where mitigation money is best spent. ```{r} sens <- sensitivity(task_dists) ``` ```{r} #| fig-cap: "Tornado chart. T3 (Structural Steel & Deck) dominates schedule variance, with its wide triangular distribution swamps the other tasks." sorted_sens <- sort(sens, decreasing = FALSE) barplot(sorted_sens, horiz = TRUE, names.arg = names(sorted_sens), xlab = "Sensitivity Index", main = "Schedule Sensitivity: Riverside Bridge", col = "#18bc9c", border = "white", las = 1) abline(v = 1, lty = 2, col = "#e74c3c") ``` T3 is the clear driver. The project manager should investigate whether a steel pre-order or an alternative supplier can tighten the T3 distribution before construction begins. --- ## Step 3: Second Moment Method {#sec-cs-smm} The Second Moment Method gives us a fast analytical check, no simulation needed. ```{r} means <- c(T1 = 3, T2 = 8, T3 = 14.67, T4 = 4, T5 = 2) vars <- c(T1 = 0.5, T2 = 2.25, T3 = 16.33, T4 = 0.64, T5 = 0.333) smm_result <- smm(means, vars) print(smm_result) ``` ```{r} cat("MCS mean: ", round(mean(sim$total_distribution), 2), "weeks\n") cat("SMM mean: ", round(smm_result$total_mean, 2), "weeks\n") cat("MCS SD: ", round(sd(sim$total_distribution), 2), "weeks\n") cat("SMM SD: ", round(smm_result$total_std, 2), "weeks\n") ``` The SMM mean and standard deviation closely match the simulation results, confirming the calculation. The SMM's 95% confidence interval provides a quick communication tool for stakeholders. --- ## Step 4: Earned Value Management {#sec-cs-evm} The project is now in Period 3 of 5. The contractor has spent \$1,350,000 so far and reports 48% completion. ```{r} BAC <- 2500000 schedule <- c(0.15, 0.35, 0.65, 0.85, 1.00) costs <- c(350000, 810000, 1350000) period <- 3 complete <- 0.48 PV <- pv(BAC, schedule, period) EV <- ev(BAC, complete) AC <- ac(costs, period) cat("PV (Planned Value): $", format(PV, big.mark = ","), "\n") cat("EV (Earned Value): $", format(EV, big.mark = ","), "\n") cat("AC (Actual Cost): $", format(AC, big.mark = ","), "\n") cat("CV (Cost Variance): $", format(cv(EV, AC), big.mark = ","), "\n") cat("SV (Schedule Var.): $", format(sv(EV, PV), big.mark = ","), "\n") cat("CPI: ", round(cpi(EV, AC), 3), "\n") cat("SPI: ", round(spi(EV, PV), 3), "\n") eac_typical <- eac(BAC, method = "typical", cpi = cpi(EV, AC)) cat( "EAC (typical): $", format(round(eac_typical), big.mark = ","), "\n" ) ``` A CPI below 1.0 means the project is over budget for work completed. The EAC (typical) forecasts the final cost if current cost efficiency continues. The project manager needs to investigate the source of the cost overrun; the Adverse Weather or Ground Conditions risks are likely candidates. --- ## Step 5: Bayesian Risk Update {#sec-cs-bayes} Before construction began, the probability of encountering poor ground conditions was estimated at 0.40. Halfway through the foundation work, the geotechnical team reports finding unexpected clay layers, a strong signal that the ground conditions risk has materialised. ```{r} causes <- c(0.40, 0.25) given <- c(0.85, 0.60) not_given <- c(0.15, 0.20) prior_risk <- risk_prob(causes, given, not_given) cat("Prior risk probability:", round(prior_risk, 3), "\n") ``` ```{r} observed <- c(1, NA) # clay layers observed; second cause unknown post_risk <- risk_post_prob(causes, given, not_given, observed) cat("Posterior risk probability:", round(post_risk, 3), "\n") cat("Update (posterior - prior):", round(post_risk - prior_risk, 3), "\n") ``` Observing the clay layers nearly doubles the risk probability. This updated estimate should feed directly into the contingency reserve and the EAC recalculation. --- ## Step 6: Learning Curve {#sec-cs-sigmoidal} The Contractor Crew is new to the bridge construction method and their weekly output (cubic metres of concrete placed per week) shows a classic learning pattern. ```{r} crew_data <- data.frame( week = 1:10, output = c(18, 28, 38, 52, 63, 71, 76, 79, 81, 82) ) fit <- fit_sigmoidal(crew_data, "week", "output", "logistic") cat("Model: Logistic\n") cat("Residual SE:", round(summary(fit)$sigma, 3), "\n") ``` ```{r} #| fig-cap: "Logistic learning curve for the Contractor Crew. Productivity plateaus near 83 cubic metres per week by week 12." plot_sigmoidal(fit, crew_data, "week", "output", "logistic") ``` ```{r} week12_pred <- predict_sigmoidal(fit, 12, "logistic") cat("Predicted output at week 12:", round(week12_pred$pred, 1), "m³/week\n") ``` The crew is expected to reach near-plateau productivity by week 12. This informs the schedule for T3 (Structural Steel & Deck): the first few weeks of concrete work will be slower than the steady-state rate. --- ## Step 7: Design Structure Matrix {#sec-cs-dsm} Which tasks are most structurally coupled through shared resources? ```{r} S <- matrix(c( 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1 ), nrow = 5, ncol = 5, byrow = TRUE) rownames(S) <- c("Survey Crew", "Civil Eng.", "Structural Eng.", "Contractor Crew", "QA Inspector") colnames(S) <- c("T1", "T2", "T3", "T4", "T5") R <- matrix(c( 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0 ), nrow = 3, ncol = 5, byrow = TRUE) rownames(R) <- c("Adverse Weather", "Ground Conditions", "Material Delays") colnames(R) <- c("Survey Crew", "Civil Eng.", "Structural Eng.", "Contractor Crew", "QA Inspector") p <- parent_dsm(S) g <- grandparent_dsm(S, R) ``` ```{r} #| fig-cap: "Parent DSM (resource-based coupling). Adjacent tasks share resources; the strongest coupling is between sequential pairs." plot(p) ``` ```{r} #| fig-cap: "Grandparent DSM (risk-based coupling). Ground Conditions risk creates coupling between T1–T2 through the shared Civil Engineer and Survey Crew resources." plot(g) ``` The DSMs confirm that T2 (Foundation) and T3 (Steel) are most tightly coupled, sharing the Civil Engineer and Structural Engineer. When ground conditions affect the foundation work, pressure immediately flows to the steel schedule. --- ## Step 8: Probabilistic Network {#sec-cs-network} We now model the two dominant risks (Adverse Weather, Ground Conditions) as a Bayesian network and simulate total project cost. ```{r} nodes <- data.frame( id = c("A", "B", "C", "D", "E"), label = c("Adverse Weather", "Ground Conditions", "Contractor Cost", "Foundation Cost", "Total Cost"), group = c("Risk", "Risk", "Resource", "Resource", "Project"), stringsAsFactors = FALSE ) links <- data.frame( source = c("A", "B", "C", "D"), target = c("C", "D", "E", "E"), value = rep(1, 4), stringsAsFactors = FALSE ) distributions <- list( A = list(type = "discrete", values = c(1, 0), probs = c(0.50, 0.50)), B = list(type = "discrete", values = c(1, 0), probs = c(0.40, 0.60)), C = list(type = "conditional", condition = "A", true_dist = list(type = "normal", mean = 1220000, sd = 30000), false_dist = list(type = "normal", mean = 1100000, sd = 20000)), D = list(type = "conditional", condition = "B", true_dist = list(type = "normal", mean = 700000, sd = 50000), false_dist = list(type = "normal", mean = 500000, sd = 30000)), E = list(type = "aggregate", nodes = c("C", "D")) ) graph <- prob_net(nodes, links, distributions = distributions) sim_net <- prob_net_sim(graph, num_samples = 10000) ``` ```{r} #| fig-cap: "Total project cost distribution from the Bayesian network. The bimodal shape reflects whether ground conditions risk materialises." hist(sim_net$E, breaks = 60, main = "Total Project Cost: Bayesian Network", xlab = "Cost ($)", col = "skyblue", border = "white") ``` ```{r} learn_net <- prob_net_learn(graph, observations = list(B = "Yes"), num_samples = 10000) cat("Mean cost (prior):", format(round(mean(sim_net$E)), big.mark = ","), "\n") cat("Mean cost (ground conditions = Yes):", format(round(mean(learn_net$E)), big.mark = ","), "\n") ``` Conditioning on the observed ground conditions (confirmed by the clay layers in Step 5) shifts the expected total cost upward significantly. This updated estimate, combined with the EAC from Step 4, gives the project manager a fully informed picture of financial exposure. --- ## Step 9: Agentic Analysis {#sec-cs-agent} All of the above can also be run via slash commands, no function-by-function scripting required. ```{r} #| eval: false r <- PRA:::execute_command( '/mcs n=10000 tasks=[ {"type":"triangular","a":2,"b":3,"c":5}, {"type":"normal","mean":8,"sd":1.5}, {"type":"triangular","a":10,"b":14,"c":20}, {"type":"normal","mean":4,"sd":0.8}, {"type":"uniform","min":1,"max":3} ]' ) cat(r$result) ``` ```{r} #| eval: false r <- PRA:::execute_command("/contingency phigh=0.80 pbase=0.50") cat(r$result) ``` ```{r} #| eval: false r <- PRA:::execute_command( "/evm bac=2500000 schedule=[0.15,0.35,0.65,0.85,1.0] period=3 complete=0.48 costs=[350000,810000,1350000]" ) cat(r$result) ``` ::: {.callout-note} ## About `eval: false` above The slash-command blocks above use internal PRA functions that produce the same results as the code in Steps 1–4. They are marked `eval: false` to avoid duplicate output in the rendered book. Run them interactively to verify the commands reproduce the same numbers. ::: --- ## What the Methods Told Us Pulling the results together: | Method | Key finding | |--------|------------| | MCS | P80 schedule = ~35 weeks; contingency reserve ~3 weeks | | Sensitivity | T3 (Steel & Deck) drives 60%+ of schedule variance | | SMM | Confirms MCS mean and SD analytically | | EVM | Period 3: CPI < 1.0, cost overrun in progress | | Bayesian | Ground conditions risk updated from 0.40 → ~0.70 after observations | | Learning Curve | Crew plateaus ~week 12; early T3 concrete work will be slower | | DSM | T2–T3 most coupled; ground conditions propagates through both | | Network | Conditioning on ground conditions = Yes adds ~\$200K to expected cost | No single method gives the full picture. The MCS tells you how wide the distribution is. The sensitivity analysis tells you where to act. The EVM tells you how you're performing right now. The Bayesian update tells you how new information should change your reserves. Together, they are the complete toolkit, and that is the point of this book.