(a) For a small project, the number of masons required is shown. The table also indicates the duration of each activity along…

(a) Resource allocation, unlimited and limited masons (20 marks)

Network. 1-2 comes first. Then 2-3, 2-5 and 2-4 follow. 3-4 and 3-6 follow 2-3. 4-6 follows 2-4 and 3-4. 5-6 follows 2-5. The project ends when 3-6, 4-6 and 5-6 are done.

CPM. Path lengths to the end are 1-2-5-6 = 4 + 9 + 10 = 23 days, 1-2-3-4-6 = 21, 1-2-3-6 = 18 and 1-2-4-6 = 14. The critical path is 1-2, 2-5, 5-6, 23 days. Floats: 2-3 is 2 days (start 4 to 6), 3-4 is 2, 4-6 is 2 (latest start 15), 3-6 is 5 (latest start 15) and 2-4 is 7.

Unlimited resources (all activities at earliest start). Mason demand by period:

• 0–4 d: 2
• 4–6 d: 3 + 4 + 4 = 11
• 6–10 d: 3 + 4 = 7
• 10–13 d: 3 + 4 + 4 = 11
• 13–18 d: 4 + 2 + 4 = 10
• 18–21 d: 4 + 2 = 6
• 21–23 d: 2

Duration 23 days, peak 11 masons, total 163 mason-days. The histogram is uneven.

Limited resources (levelling within the float). 5-6 (13–23 d) and 2-5 (4–13 d) are critical and fixed. Since 3-6 and 4-6 each last 8 days and must both finish by day 23, they must overlap during days 15–18 with 5-6 also running. The lowest possible limit without delay is therefore 10 masons. A feasible schedule is:

• 1-2: 0–4; 2-4: 4–6; 2-5: 4–13; 2-3: 6–12; 3-4: 12–15
• 3-6: 13–21; 4-6: 15–23; 5-6: 13–23

Demand is 2, 8, 7, 7, 9, 10 and 6 masons over successive periods. The duration stays 23 days with peak 10. With fewer than 10 masons the project overruns.

Unlimited resources: shortest time, simple scheduling, but a high peak, idle labour and costly hiring. Limited resources: a smoother histogram and lower peak, but it consumes float, leaves less flexibility, and may lengthen the project if fewer masons are available.

(b) Set-back distances on the curve (15 marks)

R = 400 m, L = 200 m. The lane is 1.9 m inside the centre line, so the radius of the inner lane = 400 − 1.9 = 398.1 m. The set-back m is measured from the centre line.

(i) S = 100 m < L. α = S/(2 × 398.1) = 0.1256 rad. m = R − 398.1 cos α = 400 − 398.1 × 0.9921 = 5.04 m.

(ii) S = 320 m > L. The sight line extends beyond the curve, so α = L/(2 × 398.1) = 0.2512 rad, and the straight extensions are (S − L)/2 = 60 m on each side. m = R − 398.1 cos α + 60 sin α = 400 − 385.6 + 14.9 = 29.3 m.

(c) Collimation error and corrected readings (15 marks)

At P, midway, the sight lengths are equal (25 m each), so the error cancels. True difference = 2.4 − 1.4 = 1.00 m, so B is 1.00 m above A.

At Q (on BA produced), QA = 10 m and QB = 60 m. Let e be the error in the reading per metre of sight length. Then (2.50 − 1.40) = 1.00 + (10 − 60)e, so 1.10 = 1.00 − 50e and e = −0.002 m per m. The reading is too low by 2 mm per metre, so the line of sight inclines downward.

Readings for a horizontal line of sight from Q:

• A: 2.50 + 0.002 × 10 = 2.52 m
• B: 1.40 + 0.002 × 60 = 1.52 m

Check: 2.52 − 1.52 = 1.00 m.

RL of B = RL of A + 1.00 = 200.00 + 1.00 = 201.00 m.