(a) The network of a certain project is shown in the figure given below with the estimated time duration (in days) of various activities. Determine the following :
(i) Earliest event time
(ii) Latest event time
(iii) Earliest start and finish time of

Question

(a) The network of a certain project is shown in the figure given below with the estimated time duration (in days) of various activities. Determine the following :
(i) Earliest event time
(ii) Latest event time
(iii) Earliest start and finish time of each activity
(iv) Latest start and finish time of each activity
(v) Total float of each activity
(vi) Critical path of the network
(Assume the scheduled completion time of the project equal to the earliest event time of the last event)
20

(b) Prepare a detailed analysis of rate (₹ per sq. m) for a 25 mm thick cement concrete (1 : 2 : 4) floor with cement finishing coat. 15

(c) Sketch a typical diamond crossing and label all its components. Design a diamond crossing between two B.G. tracks crossing each other at an angle of 1 in 8·5. 15

UPSC Answer Check · Accepted Answer

(a)(i) Earliest event times (forward pass):
E1 = 0
E2 = 0 + 6 = 6 days
E3 = 0 + 8 = 8 days
E4 = 8 + 7 = 15 days
E5 = max(6 + 7, 15 + 6) = max(13, 21) = 21 days
E6 = 8 + 10 = 18 days
E7 = max(15 + 9, 21 + 8, 18 + 9) = max(24, 29, 27) = 29 days
E8 = 29 + 6 = 35 days
Project duration = 35 days.

(a)(ii) Latest event times (backward pass), with L8 = E8 = 35 days:
L7 = 35 − 6 = 29 days
L6 = 29 − 9 = 20 days
L5 = 29 − 8 = 21 days
L4 = min(21 − 6, 29 − 9) = min(15, 20) = 15 days
L3 = min(15 − 7, 20 − 10) = min(8, 10) = 8 days
L2 = 21 − 7 = 14 days
L1 = min(14 − 6, 8 − 8) = min(8, 0) = 0 day.

(a)(iii) Earliest start time EST = E_i and earliest finish time EFT = E_i + d:
1–2: EST = 0, EFT = 6 days
1–3: EST = 0, EFT = 8 days
2–5: EST = 6, EFT = 13 days
3–4: EST = 8, EFT = 15 days
3–6: EST = 8, EFT = 18 days
4–5: EST = 15, EFT = 21 days
4–7: EST = 15, EFT = 24 days
5–7: EST = 21, EFT = 29 days
6–7: EST = 18, EFT = 27 days
7–8: EST = 29, EFT = 35 days.

(a)(iv) Latest start time LST = L_j − d and latest finish time LFT = L_j:
1–2: LST = 8, LFT = 14 days
1–3: LST = 0, LFT = 8 days
2–5: LST = 14, LFT = 21 days
3–4: LST = 8, LFT = 15 days
3–6: LST = 10, LFT = 20 days
4–5: LST = 15, LFT = 21 days
4–7: LST = 20, LFT = 29 days
5–7: LST = 21, LFT = 29 days
6–7: LST = 20, LFT = 29 days
7–8: LST = 29, LFT = 35 days.

(a)(v) Total float TF = LST − EST = L_j − E_i − d:
1–2: 8 days
1–3: 0 day
2–5: 8 days
3–4: 0 day
3–6: 2 days
4–5: 0 day
4–7: 5 days
5–7: 0 day
6–7: 2 days
7–8: 0 day.

(a)(vi) Activities with zero total float form the critical path:
1–3–4–5–7–8, duration = 8 + 7 + 6 + 8 + 6 = 35 days.
Hence scheduled completion time = 35 days.

(b) Assumed rates: cement ₹400/bag of 50 kg, sand ₹1600/m³, 12.5 mm coarse aggregate ₹1500/m³, mason ₹800/day, bhisti ₹650/day, coolie ₹550/day. Finishing coat assumed 6 mm thick cement mortar 1:3. For 1 m² floor:
Cement concrete volume = 0.025 m³. Dry volume = 1.54 × 0.025 = 0.0385 m³.
For 1:2:4, sum = 7.
Cement = (1/7) × 0.0385 = 0.0055 m³ = 0.0055 × 1440 = 7.92 kg = 0.158 bag.
Sand = (2/7) × 0.0385 = 0.0110 m³.
Coarse aggregate = (4/7) × 0.0385 = 0.0220 m³.
Finishing coat: wet volume = 0.006 m³. Dry volume = 1.33 × 0.006 = 0.00798 m³.
Cement = (1/4) × 0.00798 = 0.001995 m³ = 2.873 kg = 0.0575 bag.
Sand = (3/4) × 0.00798 = 0.005985 m³.
Total cement = 0.2155 bag ≈ 0.22 bag. Total sand = 0.0170 m³. Aggregate = 0.0220 m³.
Material cost: cement 0.22 × 400 = ₹88.00; sand 0.017 × 1600 = ₹27.20; aggregate 0.022 × 1500 = ₹33.00; water/curing/misc = ₹8.00. Total materials = ₹156.20.
Labour: mason 0.05 day × 800 = ₹40.00; coolie/beldar 0.15 day × 550 = ₹82.50; bhisti 0.025 day × 650 = ₹16.25. Total labour = ₹138.75.
Subtotal = 156.20 + 138.75 = ₹294.95.
Tools and plant @2% = ₹5.90.
Overhead and contractor profit @10% = ₹30.09.
Total rate = 294.95 + 5.90 + 30.09 = ₹330.94 ≈ ₹331 per m², excluding GST.
Validity: if finishing coat thickness or mix changes, adjust cement and sand proportionately.

(c) Typical diamond crossing: two straight tracks cross at a small angle. The four rail intersections form two acute crossings and two obtuse crossings. Sleepers are laid perpendicular to the bisector of the crossing angle and are common to both tracks. Labels: acute crossing/V-crossing, obtuse crossing, wing rail, check rail, point rail, splice rail, nose, common sleepers, distance blocks, gauge tie plates, ballast. The acute crossings lie at the ends of the long diagonal; the obtuse crossings lie on the short diagonal.

Design data: B.G. gauge G = 1.676 m. Crossing angle = 1 in 8.5, so cot α = 8.5.
α = tan⁻¹(1/8.5) = 6°42′35″ ≈ 6.71°.
Obtuse crossing angle = 180° − α = 173°17′25″.
Half angle α/2 = 3°21′17″.
Length of diamond L = G cot(α/2) = 1.676 × cot(3°21′17″).
cot(3°21′17″) ≈ 17.06, so L = 1.676 × 17.06 = 28.60 m.
Short diagonal between obtuse crossings = G sec(α/2) = 1.676 × 1.0017 = 1.68 m.
Longest sleeper at acute crossing = 2G sec(α/2) = 2 × 1.676 × 1.0017 = 3.36 m.
Assume sleeper spacing = 0.60 m. Number of sleepers = L/0.60 + 1 = 28.60/0.60 + 1 ≈ 49.
Thus the diamond crossing requires two acute crossings, two obtuse crossings, common sleepers up to 3.36 m long, and about 49 sleepers at 0.60 m spacing.

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