(a) A contractor intends to bid for erecting a statue at a square in an urban area. The various activities of the entire project…

(a)

The network has two paths: A–D and B–C–D. Normal durations: A–D = 8 + 6 = 14 days; B–C–D = 4 + 8 + 6 = 18 days. Hence normal project duration μ = 18 days, critical path B–C–D. Variance = 4 day², so standard deviation σ = √4 = 2 days. For 99% probability, interpolate between 98.93% (Z = 2.3) and 99.18% (Z = 2.4): Z = 2.3 + ((99.00 − 98.93)/(99.18 − 98.93)) × (2.4 − 2.3) Z = 2.3 + (0.07/0.25) × 0.1 = 2.328 ≈ 2.33. Using T = μ + Zσ: T = 18 + 2.33 × 2 = 22.66 days ≈ 22.7 days. So the contractor can expect completion within about 22.66 days with 99% probability.

For crashing to 17 days, cost slopes: A = (9000 − 6000)/(8 − 5) = 1000 ₹/day B = (4000 − 2000)/(4 − 3) = 2000 ₹/day C = (5500 − 5000)/(8 − 7) = 500 ₹/day D = (7500 − 3000)/(6 − 3) = 1500 ₹/day

The critical path is B–C–D. To reduce project duration from 18 to 17 days, crash the critical activity with least slope, i.e. C, by 1 day. New C duration = 7 days. Direct cost = 6000 + 2000 + 5500 + 3000 = 16,500 ₹. Indirect cost for 17 days = 500 × 17 = 8,500 ₹. Total project cost = 16,500 + 8,500 = 25,000 ₹.

(b)(i)

Required amber time for safe signal design is given by A = t_r + v/(2a) + (W + L)/v, where t_r = perception-reaction time, v = speed, a = comfortable deceleration, W = road width, L = car length. Speed v = 60 kmph = 60 × 5/18 = 50/3 m/s = 16.67 m/s. A = 1.2 + (50/3)/(2 × 3) + (20 + 4)/(50/3) A = 1.2 + 25/9 + 36/25 A = 1.2 + 2.7778 + 1.44 = 5.4178 s ≈ 5.42 s. Provided amber duration = 4.0 s. Since 4.0 s < 5.42 s, the amber period is insufficient. If amber and all-red are considered separately, stopping requirement is 1.2 + 2.7778 = 3.98 s and clearing time is 1.44 s, so total change interval still needs 5.42 s. Therefore the driver’s claim is correct.

(b)(ii)

Subsurface drainage is provided to intercept seepage and lower the water table below the road subgrade. A longitudinal perforated pipe or French drain is laid in a trench along one or both shoulders of the road. The trench is filled with graded gravel/sand filter around the pipe, and a geotextile is wrapped around the filter to prevent clogging. Transverse collector drains are provided at intervals to intercept cross-flow and join the longitudinal drain. The collected water is led by outlet pipes to a side drain, natural stream, or sump. Sketch description: In cross-section, the road crust lies above the lowered water table; on the shoulder, a trench contains a perforated pipe at the bottom, surrounded by gravel and geotextile, with an outlet pipe. In plan, longitudinal drains run parallel to the road and are connected by transverse drains at regular spacing. This draws down the water table and keeps the subgrade dry.

(c)

Train resistance on a BG track is taken from the standard empirical expression R = 0.0016W + 0.00008WV + 0.0000006WV² (tonnes, W in tonnes, V in km/h), where the first term is the rolling resistance (given here as 1.6 kg/t for wagons and 2.0 kg/t for the locomotive), the second is the speed-dependent resistance and the third is wind resistance. Since the speed is specified, the speed and wind terms must be included. At V = 60 km/h: speed term = 0.00008 × 60 = 0.0048 t/t (4.8 kg/t); wind term = 0.0000006 × 60² = 0.00216 t/t (2.16 kg/t). Together = 6.96 kg/t.

Level straight track.

• Locomotive: 100 × (2.0 + 6.96) = 896 kg = 0.896 t
• Each wagon: 80 × (1.6 + 6.96) = 684.8 kg = 0.6848 t
• Hauling capacity = total resistance: 16 = 0.896 + 0.6848n, so n = 15.104/0.6848 = 22.06
• Maximum number of wagons = 22

Gradient 1 in 150 on a 1° curve (same train).

• Train weight W = 100 + 22 × 80 = 1,860 t
• Level-track resistance = 0.896 + 22 × 0.6848 = 15.962 t
• Gradient resistance = W/150 = 1,860/150 = 12.400 t
• Curve resistance (BG) = 0.0004 W D = 0.0004 × 1,860 × 1 = 0.744 t
• Required hauling capacity = 15.962 + 12.400 + 0.744 = 29.11 t ≈ 29.1 tonnes

If only the given rolling resistances are used and the speed terms are ignored, the arithmetic gives n = (16,000 − 200)/128 = 123 wagons and a required hauling capacity of about 86.2 t on the gradient. That result ignores the resistance that rises with speed, which is why a 16 t tractive effort appears to haul nearly 10,000 t; the value of 22 wagons is the realistic one.