Question 6 of 7: Line-of-Balance Scheduling for a 10-Floor High-Rise

Question 6: Line-of-Balance Scheduling for a 10-Floor High-Rise (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.

Line-of-Balance vs. CPM. LOB plots each repetitive trade as a single sloped line (or band) across a floor/unit-vs-time chart, where the slope is that trade's production rate (units per day), instead of drawing every floor's activities as separate CPM nodes. Pros: it shows crew continuity directly (a flat, unbroken line means the crew never sits idle between floors), it makes the controlling (slowest) trade visually obvious as the one whose line is least steep, and it scales to buildings with many repeated floors without the network exploding into hundreds of near-identical CPM activities. Cons: it assumes each trade's output rate is steady and known in advance, which fits highly repetitive vertical construction (floor-by-floor fit-out) far better than one-off or highly interdependent scopes; and because it hides the detailed logic between individual tasks on a single floor, it is a poor substitute for CPM where complex, non-repetitive precedence (not just "trade X follows trade Y up the building") drives the schedule.

Given. 10 floors, each needing four sequential tasks A→B→C→D; every task takes 2 days to complete on one floor; the tasks' floor-delivery rates are $r_A=1$, $r_B=2$, $r_C=1$, $r_D=0.5$ floors/day.

Find. The crew count each task needs to sustain its stated rate, the resulting floor-by-floor schedule, and the overall project duration; then sketch the schematic LOB chart.

Approach. A task needs enough crews working in parallel, offset by $1/rate$ days, to keep delivering a new floor every $1/rate$ days despite each floor taking a full 2-day duration: crews $= duration \times rate$. Each floor's start on a task is the later of (i) when the previous task finished that same floor, and (ii) when this task's own crew(s) finished the previous floor.

1. Crews needed per task. $Crews = duration\times rate$: A: $2\times1=2$ crews; B: $2\times2=4$ crews; C: $2\times1=2$ crews; D: $2\times0.5=1$ crew.
2. Task A (first task, no predecessor). Floor $i$ starts at $(i-1)/r_A=(i-1)$ days and finishes 2 days later. Floor 1: day 0→2. Floor 10: day 9→11.
3. Task B (faster than A, so it is always waiting on A). Because $r_B=2 > r_A=1$, B never has to wait on its own crews — it starts each floor the instant A finishes it: $start_B(i)=finish_A(i)=i+1$. Floor 1: day 2→4. Floor 10: day 11→13.
4. Task C (same rate as B's finish-line pace). B's finish times step by 1 day/floor and $1/r_C=1$ day/floor too, so once C locks onto B's heels on floor 1 it stays there: $start_C(i)=i+3$. Floor 1: day 4→6. Floor 10: day 13→15.
5. Task D (the bottleneck — only 1 crew, $1/r_D=2$ days/floor). $D$ can start floor 1 only once $C$ finishes it, at day 6, giving floor 1: day 6→8. From floor 2 on, D's own 2-day/floor pace (its single crew must finish one floor before starting the next) is slower than C keeps supplying floors, so D runs back-to-back: $start_D(i)=2i+4$ for $i\ge1$. Floor 10: $start=24$, $finish=26$. $$\boxed{\text{Project duration} = 26 \text{ days, controlled entirely by Task D's single-crew, 2-day/floor pace}}$$