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

| Activity | Activity Name | Remark |
|----------|---------------|--------|
| A | Make statue | Starti

Question

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

| Activity | Activity Name | Remark |
|----------|---------------|--------|
| A | Make statue | Starting activity |
| B | Lay foundation | Starting activity |
| C | Construct platform | Follows B (starts after laying foundation) |
| D | Erect statue | Finishing activity and follows A and C |

The project is expected to take 18 days to complete having a variance of 4 days. Determine in how many days the contractor would expect the project to be completed with a probability of 99%. For probability of 98.93%, the corresponding normal deviate Z value is +2.3 and for probability of 99.18%, Z value is +2.4. Further, if the contractor intends to complete the project in 17 days by crashing the activities, determine how much total project cost would the contractor expect. The normal and crash duration, and associated cost are given in the table below for various activities :

| Activity | Normal Activity | | Crash Activity | |
|----------|-----------------|---|----------------|---|
| | Duration (in days) | Cost (in ₹) | Duration (in days) | Cost (in ₹) |
| A | 8 | 6,000 | 5 | 9,000 |
| B | 4 | 2,000 | 3 | 4,000 |
| C | 8 | 5,000 | 7 | 5,500 |
| D | 6 | 3,000 | 3 | 7,500 |

For the entire project, the indirect cost is ₹ 500 per day. (20 marks)

(b) (i) A taxi driver was fined for crossing the traffic signal at right-angled road intersection. He claimed that the signal was faultily designed and the duration of amber light is not sufficient. Using the following data, verify the correctness of the driver's claim :

Road width at intersection = 20 m
Speed limit at road = 60 kmph
Amber light duration = 4.0 s
Comfortable deceleration = 3.0 m/s²
Car length = 4.0 m
Perception reaction time = 1.2 s (10 marks)

(ii) Explain with sketches how the subsurface drainage system is provided to lower the water table in road. (5 marks)

(c)

Calculate the maximum number of wagons of weight 80 tonnes each that can be pulled by a locomotive having hauling capacity of 16 tonnes. The weight of the locomotive is 100 tonnes and the train has to run at a speed of 60 kmph on a straight level BG track. Assume rolling resistance of wagon and locomotive as 1.6 kg/tonne and 2.0 kg/tonne respectively.

Also, calculate the hauling capacity of the locomotive required if the train has to climb a gradient of 1 in 150 in 1° curve. 15 marks

Also, calculate the hauling capacity of the locomotive required if the train has to climb a gradient of 1 in 150 in 1° curve. (15 marks)

UPSC Answer Check · Accepted Answer

(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)
Let n be the number of wagons.
Hauling capacity = 16 tonnes = 16,000 kg.
Resistance of locomotive = 2.0 × 100 = 200 kg.
Resistance of one wagon = 1.6 × 80 = 128 kg.
On level straight track:
16,000 = 200 + n × 128
n = (16,000 − 200)/128 = 15,800/128 = 123.44.
Maximum integer number of wagons = 123.

For the same train:
Wagon weight = 123 × 80 = 9,840 tonnes.
Total train weight = 100 + 9,840 = 9,940 tonnes.
Rolling resistance = 200 + 1.6 × 9,840 = 200 + 15,744 = 15,944 kg.
Gradient resistance for 1 in 150 = (1/150) × 9,940 × 1000 = 66,266.67 kg.
Curve resistance for BG, 1° curve, taking 0.4 kg/tonne per degree = 0.4 × 9,940 = 3,976 kg.
Total required hauling capacity = 15,944 + 66,266.67 + 3,976
= 86,186.67 kg = 86.19 tonnes.
So the locomotive would require a hauling capacity of about 86.19 tonnes to climb the gradient in the 1° curve.

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

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