23-Ind-A6 Systems Simulation · December 2018

Question 7 of 8: 2² Factorial Screening Experiment for Bus Loading

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 7 (Part C.2): 2² Factorial Screening Experiment for Bus Loading (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Partial $2^2$ design (Runs 1–4) and observed average wait time to load a patrol, 3 replications per run (the exam's "5 replications" preamble and the 3-column output table are reconciled by working directly from the 3 printed replicate values per run):

Find. (a) the completed design matrix (AB column) with sign interpretation; (b) ANOVA table, which factor(s) are significant, and the recommended settings.

Approach. Fill in the interaction column as the row-wise product $AB=A\times B$; use the $2^k$-factorial contrast method to get $SS_A$, $SS_B$, $SS_{AB}$ from the run totals, take $SS_{error}=SST-SS_A-SS_B-SS_{AB}$, and compare each mean square against $MS_{error}$.

(a) Completed design matrix and sign interpretation.

$-1$ means the factor is at its low (not-implemented) setting — small buses for A, existing/far-destinations-first loading for B; $+1$ means it is implemented — large buses for A, near-destinations-first loading for B. The $AB$ column is the row-wise product of the $A$ and $B$ signs, not an independently-set factor: it is $+1$ whenever $A$ and $B$ are at the same level and $-1$ whenever they differ. The main effect of $A$ is the average response at $A=+1$ minus the average response at $A=-1$, averaged fairly over both levels of $B$ (each level of $B$ appears once at $A=+1$ and once at $A=-1$, which is what makes the design orthogonal); similarly for $B$. The interaction effect $AB$ measures whether $A$'s effect on wait time depends on the level of $B$ — a large $AB$ contrast means the two factors do not act additively (the effect of switching bus size is different depending on which loading policy is active).

(b) ANOVA.

1. Run totals and contrasts. Totals $T_i=\sum(\text{3 reps})$: $T_1=120,\ T_2=272,\ T_3=182,\ T_4=331$. $$C_A = -T_1+T_2-T_3+T_4 = \boxed{301},\quad C_B = -T_1-T_2+T_3+T_4 = \boxed{121},\quad C_{AB} = T_1-T_2-T_3+T_4 = \boxed{-3}.$$
2. Sums of squares ($N=4$ runs, $n=3$ reps, denominator $N\!\cdot\!n=12$): $$SS_A = \frac{301^2}{12} = \boxed{7550.08},\qquad SS_B = \frac{121^2}{12} = \boxed{1220.08},\qquad SS_{AB} = \frac{(-3)^2}{12} = \boxed{0.75}.$$ $$SS_{error} = SST - SS_A - SS_B - SS_{AB} = 8829 - 7550.08 - 1220.08 - 0.75 = \boxed{58.08},\qquad df_{error}=N(n-1)=8.$$ (A direct recomputation of $SST$ from the 12 raw replicate values gives 8828.92, confirming the exam's stated $SST=8829$ to rounding.)
3. Mean squares, $F$-ratios, and decision. $MS_{error}=58.08/8=7.26$; $F_{0.05,1,8}=5.32$:

Both main effects are overwhelmingly significant ($F_A=1039.9$ and $F_B=168.0$, both far above $F_{crit}=5.32$); the interaction is not ($F_{AB}=0.10 \ll 5.32$). The effect sizes confirm the direction: $\text{effect}_A=C_A/(2n)=+50.17$ min and $\text{effect}_B=C_B/(2n)=+20.17$ min — turning either factor ON (implementing large buses, or near-destinations-first loading) increases average wait time to load a patrol, dramatically so for bus size.

Recommended settings. Because $AB$ is not significant, the two factors act additively and can be set independently — there is no need to consider them jointly. Since both main effects are positive (implementing either factor makes wait time worse), the wait-minimizing recommendation is to keep both factors at their low (not-implemented) setting: continue with the smaller buses and the existing loading policy, which matches the observed lowest average wait time (Run 1, 40.00 min) and is consistent with both factors' effects being additive rather than interacting.