(a) For a small project, the number of masons required is shown. The table also indicates the duration of each activity along with the masons required. Carry out the resource allocation with unlimited and limited number of resources. Indicate the adv

Question

(a) For a small project, the number of masons required is shown. The table also indicates the duration of each activity along with the masons required. Carry out the resource allocation with unlimited and limited number of resources. Indicate the advantages and disadvantages of both. 20 marks (b) There is a horizontal curve of radius 400 m and length 200 m on a highway. Compute the set-back distances required from the centre line on the inner side of the curve so as to provide for : (i) stopping sight distance of 100 m (ii) safe overtaking sight distance of 320 m The distance between the centre lines of the road and the inner lane is 1·9 m. 15 marks (c) A dumpy level was set up at P exactly between A and B, 50 m apart. The readings on staff held at A and B were 2·4 m and 1·4 m respectively. The instrument was then shifted and set up at Q on the line BA produced at 10 m from A. The readings on staff held at A and B were respectively 2·50 m and 1·4 m. Determine the staff reading on A and B to give a horizontal line of sight. Determine R.L. of B if that of A is 200·00 m. 15 marks

UPSC Answer Check · Accepted Answer

(a) Use CPM for the unlimited-resource case. Earliest path lengths to node 6 are:
1-2-5-6 = 4 + 9 + 10 = 23 d
1-2-3-6 = 4 + 6 + 8 = 18 d
1-2-3-4-6 = 4 + 6 + 3 + 8 = 21 d
1-2-4-6 = 4 + 2 + 8 = 14 d

Critical path is 1-2, 2-5, 5-6, so unlimited-resource duration = 23 d. Latest starts: 2-3 = 6 d, 3-4 = 12 d, 3-6 = 15 d, 4-6 = 15 d, 2-4 = 13 d.

Earliest-start mason demand:
0-4 d: 2
4-6 d: 3 + 4 + 4 = 11
6-10 d: 3 + 4 = 7
10-13 d: 4 + 3 + 4 = 11
13-18 d: 4 + 4 + 2 = 10
18-21 d: 4 + 2 = 6
21-23 d: 2

Peak = 11 masons; total = 163 mason-days.

For the limited case, take the minimum feasible availability that does not extend the 23 d duration, i.e. 10 masons. Since 5-6 is fixed at 13-23 d, 4-6 cannot start after 15 d, and 3-6 cannot finish before 18 d, 4-6 and 3-6 must overlap while 5-6 is active; hence a 10-mason peak is unavoidable. A valid 10-mason schedule is:
1-2: 0-4 d; 2-5: 4-13 d; 5-6: 13-23 d
2-4: 4-6 d; 2-3: 6-12 d; 3-4: 12-15 d
3-6: 13-21 d; 4-6: 15-23 d

Demand:
0-4: 2; 4-6: 8; 6-12: 7; 12-13: 7; 13-15: 9; 15-21: 10; 21-23: 6.

Thus unlimited: 23 d, peak 11 masons. Limited (10 masons): 23 d, peak 10 masons. Unlimited gives shortest duration but high peak and higher labour cost. Limited/levelled gives smoother demand and lower peak, but consumes float, reduces flexibility, and may extend duration if availability is lower.

(b) Use the horizontal-curve sight-distance setback formula. R = 400 m, L = 200 m, e = 1.9 m, inner-lane radius r = R - e = 398.1 m.

(i) S = 100 m < L.
E = R - r cos(S/(2R))
= 400 - 398.1 cos(100/800)
= 400 - 398.1 cos(0.125 rad)
= 5.006 m ≈ 5.01 m

(ii) S = 320 m > L, so the sight line extends on to the tangents. Curve angle Δ = L/R = 0.5 rad, θ = Δ/2 = 0.25 rad, tangent extension each side x = (S - L)/2 = 60 m.
E = R - r cosθ + x sinθ
= 400 - 398.1 cos(0.25) + 60 sin(0.25)
= 400 - 385.724 + 14.844
= 29.12 m

(c) Use the collimation-error method. Let observed reading = true horizontal reading + k d, where d is distance from the instrument and k is collimation error per metre.

At P, PA = PB = 25 m, so errors cancel. True staff-reading difference = 2.4 - 1.4 = 1.0 m.

At Q, Q is on BA produced beyond A, so QA = 10 m and QB = 60 m. Observed difference = 2.50 - 1.40 = 1.10 m.
1.10 = 1.00 + k(10 - 60) = 1.00 - 50k
k = -0.002 m/m.

Corrected horizontal readings:
A: 2.50 - k(10) = 2.50 + 0.02 = 2.52 m
B: 1.40 - k(60) = 1.40 + 0.12 = 1.52 m
Check: 2.52 - 1.52 = 1.00 m. Curvature and refraction are neglected for these short distances.

Since reading_A - reading_B = RL_B - RL_A, RL_B = 200.00 + 1.00 = 201.00 m.

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