CPM (Critical Path Method) and PERT (Program Evaluation and Review Technique) are the two foundational project-scheduling techniques used to plan, monitor, and control construction and engineering projects — from building the network diagram itself, through forward/backward pass and float calculations, time-cost trade-off (crashing), PERT's probabilistic treatment of uncertain durations, to Gantt charts and resource scheduling. Every formula, worked step-by-step example, diagram, and exam-pattern table is included.
After studying this chapter you will be able to:
Prerequisite: Highway Engineering (large-scale road and pavement construction projects rely on the network scheduling techniques developed here to plan and monitor progress). CPM-PERT completes the civil engineering study material series on Mock Test Mitra — from here, head back to Study Material to revisit any subject or start practising topic-wise mock tests.
| Term | Definition |
|---|---|
| Activity | A task or work package that consumes time and/or resources |
| Event (node) | A point in time representing the start or end of one or more activities; consumes no time |
| Predecessor | An activity that must be completed before the current activity can start |
| Successor | An activity that cannot start until the current activity is finished |
| Dummy activity | A fictitious activity with zero duration used in AOA networks to show logical dependency without actual work |
| Network | A directed graph of all activities and their precedence relationships |
| Feature | AOA (Activity on Arrow) | AON (Activity on Node / Precedence Diagram) |
|---|---|---|
| Representation | Activities on arrows; nodes are events | Activities on nodes; arrows show dependencies |
| Dummy activities | Required to avoid ambiguity | Not needed |
| Start and end | Single start node, single end node | Single start node, single end node |
| Lags | Not easily handled | Supports SS, FF, SF, FS with lags |
| Software use | Older; PERT originated here | Modern PM software (MS Project, Primavera) |
AON networks have largely superseded AOA in modern practice specifically because they eliminate the need for dummy activities altogether and natively support the four dependency types (Finish-to-Start, Start-to-Start, Finish-to-Finish, Start-to-Finish) with lags — but AOA remains the standard teaching format for GATE/ESE since the classical PERT technique was originally developed on arrow diagrams.
A valid AOA network follows strict conventions: each activity gets exactly one arrow; whenever two activities would otherwise share the same start and end node (making them indistinguishable), a dummy activity (drawn as a dashed arrow with zero duration and zero resource consumption) is inserted to break the ambiguity; the whole network must reduce to a single start event and single end event; no circular dependencies (loops) are permitted; and every event is uniquely numbered.
One arrow per activity; dummy (dashed, zero duration) needed when two activities share both start and end nodes
Single start event, single end event; no loops; unique event numbering
Given: Activities A and B both start at event 1 and both must finish before activity C can start (i.e., C depends on both A and B), and A and B have no other dependents. If drawn naively, A and B would both go from node 1 to node 2. What is needed?
Solution: Two activities cannot share the same start and end node pair (this makes them indistinguishable when referenced by their node numbers, e.g., both would be called "activity 1-2").
Answer: A dummy activity must be inserted — e.g., draw A as 1→2, B as 1→3, then a dummy 3→2 (zero duration) so C can start from node 2 depending on both A (directly) and B (via the dummy).
Given: Activity list: A (no predecessor), B (no predecessor), C (depends on A), D (depends on A and B). Describe the network logic.
Solution: A and B both start the project independently (from the start event). C depends only on A finishing. D depends on both A and B finishing — requiring either a merge node after both A and B, or a dummy if A and B are drawn to different end nodes.
Answer: The network has a burst from the start event into A and B, with C branching from A's end node, and D requiring a merge (possibly via dummy) of both A's and B's end nodes.
Given: A project needs to model a Start-to-Start (SS) relationship with a 2-day lag between two activities. Which network type handles this natively?
Solution: AOA (Activity on Arrow) networks only support Finish-to-Start relationships directly through the node-and-arrow structure; SS/FF/SF relationships with lags require workarounds or aren't supported.
Answer: AON (Activity on Node / Precedence Diagram Method) natively supports SS, FF, SF, and FS relationships with lags — the correct choice here.
Fig. 1.1 — AOA network with a dummy activity: C depends on both A and B; since A ends at node 3 and B ends at node 2, a dummy (dashed, zero duration) connects node 3 to node 2 so C can correctly depend on both.
| Symbol | Name | Definition |
|---|---|---|
| \(E_i\) / \(TE_i\) / EST | Earliest Event Time | Earliest time an event can occur (finish of all preceding activities) |
| \(L_i\) / \(TL_i\) / LST | Latest Event Time | Latest time an event can occur without delaying the project |
| Symbol | Name | Formula |
|---|---|---|
| EST | Earliest Start Time | \(EST(i\text{–}j)=E_i\) |
| EFT | Earliest Finish Time | \(EFT(i\text{–}j)=EST+D=E_i+D\) |
| LST | Latest Start Time | \(LST(i\text{–}j)=LFT-D=L_j-D\) |
| LFT | Latest Finish Time | \(LFT(i\text{–}j)=L_j\) |
The forward pass sweeps through the network from the start event (time zero) toward the end, computing the earliest time each event can occur. Whenever a node has multiple incoming activities (a merge point), its earliest event time must be the maximum of all the incoming paths' earliest finish times — because an event genuinely cannot occur until every activity leading into it has completed, so the slowest incoming path governs. The earliest time of the final node directly gives the overall project duration.
The backward pass sweeps in the opposite direction, from the end event (set equal to the project deadline) back toward the start, computing the latest time each event can occur without delaying the project. Where a node has multiple outgoing activities (a burst point), its latest event time must be the minimum of all the outgoing paths' latest start times — since the event must be ready early enough to satisfy the most demanding (earliest-deadline) successor. A completed backward pass should always recover \(L_1=E_1=0\) at the start node — this is the standard consistency check that the calculation was done correctly.
\( E_1=0 \)
\( E_j = \max_{(i\to j)}[E_i+D(i\text{-}j)] \)
Project duration \(=E_n\)
\( L_n = E_n \)
\( L_i = \min_{(i\to j)}[L_j-D(i\text{-}j)] \)
Check: \( L_1=E_1=0 \)
\( EST=E_i \); \( EFT=E_i+D \)
\( LFT=L_j \); \( LST=L_j-D \)
Given: Node 4 is reached by activity A (from node 2, \(E_2=5\), duration 4) and activity B (from node 3, \(E_3=7\), duration 3). Find \(E_4\).
Solution:
Via A: \(5+4=9\); Via B: \(7+3=10\)
\( E_4 = \max(9,10) = 10 \)
Answer: \(E_4=10\) — governed by the slower path (via B).
Given: Node 2 feeds activity C (to node 4, \(L_4=15\), duration 5) and activity D (to node 5, \(L_5=12\), duration 4). Find \(L_2\).
Solution:
Via C: \(15-5=10\); Via D: \(12-4=8\)
\( L_2 = \min(10,8) = 8 \)
Answer: \(L_2=8\) — governed by the more demanding successor (via D).
Given: Activities: A(1-2, 3 days), B(1-3, 5 days), C(2-4, 4 days), D(3-4, 2 days). Find project duration and \(L_1\) (as a check).
Solution:
Forward: \(E_1=0\); \(E_2=0+3=3\); \(E_3=0+5=5\); \(E_4=\max(3+4,\ 5+2)=\max(7,7)=7\)
Backward: \(L_4=7\); \(L_3=7-2=5\); \(L_2=7-4=3\); \(L_1=\min(3-3,\ 5-5)=\min(0,0)=0\)
Answer: Project duration \(=7\) days; \(L_1=0=E_1\) ✓ (check passes) — note both paths through node 4 are equally critical (two critical paths).
Fig. 2.1 — Forward and backward pass: E (earliest) and L (latest) event times computed for a small network with two paths of equal length (7 days each), giving two critical paths.
| Float Type | Formula | Meaning |
|---|---|---|
| Total Float (TF) | \(TF=L_j-E_i-D(i\text{–}j)=LST-EST=LFT-EFT\) | Maximum delay without delaying project completion |
| Free Float (FF) | \(FF=E_j-E_i-D(i\text{–}j)=E_j-EFT\) | Delay without delaying successor's earliest start |
| Independent Float (IF) | \(IF=E_j-L_i-D(i\text{–}j)\ge0\) (negative \(=\) zero) | Float available regardless of when predecessors finish or successors start |
| Interfering Float | Interfering Float \(=TF-FF\) | Portion of TF that, if used, delays a successor |
The three float measures always satisfy \(IF\le FF\le TF\), reflecting progressively looser assumptions about the surrounding activities: Independent Float assumes the worst case for both neighbours (predecessors finish as late as possible, successors must start as early as possible), Free Float assumes predecessors finish at their earliest, and Total Float makes no allowance for neighbours at all — it's the delay possible if the activity had the entire path's slack to itself. Free Float can always be consumed without affecting any other activity's schedule, whereas Total Float is shared along a path — using it on one activity reduces the float remaining for others sharing that same slack.
The critical path is the longest path through the network from start to finish, and every activity on it has zero total float (and consequently zero free float and zero independent float too) — meaning any delay on a critical activity delays the entire project. The critical path can be identified either by marking every activity with \(TF=0\) and tracing the connected chain from start to end, or by exhaustively listing every path through the network and comparing total durations. A network is not guaranteed to have only one critical path — multiple critical paths can coexist, all sharing the same (longest) duration, which is a critical consideration when crashing (since shortening only one critical path leaves the others still governing project duration).
\( TF = L_j - E_i - D = LST-EST \)
\( FF = E_j - E_i - D \)
\( IF = E_j - L_i - D \) (\(\ge0\), else 0)
\( IF \le FF \le TF \)
Critical activity: \( TF=0 \)
Given: Activity (2-4) has \(E_2=3\), \(L_4=12\), duration \(D=6\). Find TF.
Solution:
\( TF = 12-3-6 = 3\ \text{days} \)
Answer: \(TF=3\) days — this activity can be delayed up to 3 days without delaying the project.
Given: Activity (3-5) has \(E_3=5\), \(L_3=7\), \(E_5=14\), duration \(D=6\). Find FF and IF.
Solution:
\( FF = E_5-E_3-D = 14-5-6 = 3\ \text{days} \)
\( IF = E_5-L_3-D = 14-7-6 = 1\ \text{day} \)
Answer: \(FF=3\) days, \(IF=1\) day — confirming \(IF\le FF\).
Given: A network has three paths from start to finish with durations 18, 22, and 22 days. Identify the critical path(s) and project duration.
Solution:
The critical path is the longest path — here two paths tie at 22 days, both longer than the 18-day path.
Answer: Project duration \(=22\) days, with TWO critical paths (both 22-day paths); the 18-day path has \(TF=22-18=4\) days of float on its non-shared activities.
Fig. 3.1 — Float types on an activity timeline: Free Float is the smallest (delay without affecting the immediate successor), Total Float extends to the latest finish time, and Independent Float is bounded by the most conservative assumptions about neighbouring activities.
| Parameter | Normal | Crash |
|---|---|---|
| Duration | Normal time \(t_n\) (maximum economical) | Crash time \(t_c\) (minimum possible) |
| Cost | Normal cost \(C_n\) (minimum for normal time) | Crash cost \(C_c\) (maximum for crash time) |
| Relationship | \(t_n>t_c;\ C_n| Crashing reduces duration but increases direct cost | |
The cost slope expresses how much direct cost rises for every unit of time saved by crashing a particular activity — the additional resources (overtime, extra crews, expedited materials) needed to compress that activity's duration. Since the goal is to reduce project duration as cheaply as possible, the activity with the lowest cost slope among the critical activities should always be crashed first.
Crashing proceeds iteratively: identify all currently critical activities (\(TF=0\)), select the one with the lowest cost slope, crash it by one time unit, then recheck the critical path — because crashing one activity can reduce its path's duration enough that a previously non-critical path becomes newly critical (or ties for critical) alongside it. When multiple critical paths exist simultaneously, activities on every critical path must be crashed together in that step (potentially by crashing a single shared activity, if one exists on all paths, or several activities in parallel otherwise) — crashing only one path leaves project duration unchanged since the other critical path(s) still govern.
Direct costs (labour, materials, equipment for the activities themselves) rise as duration is compressed, while indirect costs (site overheads, supervision, equipment rental) fall as duration shortens, since less time on site means less overhead accumulated. The total project cost is the sum of both, and it traces out a U-shaped curve as duration varies — the optimal duration is the point of minimum total cost, found where the marginal cost of crashing one more day just equals the marginal indirect-cost saving from finishing one day sooner. Beyond this point, further crashing costs more than it saves.
\( \text{Cost slope} = \dfrac{C_c-C_n}{t_n-t_c} \)
Units: cost per unit time (₹/day)
Total cost \(=\) Direct cost \(+\) Indirect cost
Optimal duration: minimum total project cost
Given: An activity has \(t_n=10\) days (\(C_n=₹40{,}000\)) and \(t_c=6\) days (\(C_c=₹60{,}000\)). Find the cost slope.
Solution:
\( \text{Cost slope} = \dfrac{60{,}000-40{,}000}{10-6} = \dfrac{20{,}000}{4} = ₹5000/\text{day} \)
Answer: \(₹5000\) per day of crashing.
Given: Two critical activities have cost slopes ₹3000/day and ₹4500/day respectively, both with available crash days remaining. Which is crashed first?
Solution: The activity with the lower cost slope achieves the same one-day reduction in project duration at lower additional cost.
Answer: Crash the ₹3000/day activity first — it is cheaper to compress.
Given: Crashing the project by one more day would cost ₹8000 (sum of cost slopes on all currently critical paths), while the indirect cost saving for finishing one day earlier is ₹6000. Should crashing continue?
Solution: The marginal crashing cost (₹8000) exceeds the marginal indirect-cost saving (₹6000), so this additional day of crashing increases total project cost by ₹2000 net.
Answer: No — crashing should stop here; the current duration (before this step) is at or near the optimal (minimum total cost) point.
Fig. 4.1 — Time-cost trade-off: direct cost rises and indirect cost falls as project duration shortens; the total cost curve reaches its minimum at the optimal duration, where further crashing costs more than it saves.
| Feature | CPM | PERT |
|---|---|---|
| Activity durations | Deterministic (single estimate) | Probabilistic (three time estimates) |
| Application | Construction, manufacturing (well-defined tasks) | R&D, new product development (uncertain tasks) |
| Focus | Time–cost trade-off | Probability of completing by a deadline |
| Origin | DuPont Corporation, 1956 | US Navy Polaris missile project, 1958 |
CPM assumes each activity's duration is known with confidence, since construction tasks are typically well-understood and repeatable; PERT was developed for exactly the opposite situation — genuinely novel R&D work (originally, missile development) where a single duration estimate would be unreliable, so PERT instead models each activity's duration as a random variable.
| Estimate | Symbol | Definition |
|---|---|---|
| Optimistic time | \(t_o\) (or \(a\)) | Minimum time if everything goes perfectly; probability \(\approx1\%\) |
| Most likely time | \(t_m\) (or \(m\)) | Time that would occur most often if activity repeated many times; mode of distribution |
| Pessimistic time | \(t_p\) (or \(b\)) | Maximum time if everything goes wrong; probability \(\approx1\%\) |
PERT assumes each activity's duration follows a Beta distribution (bounded between \(t_o\) and \(t_p\), and unimodal at \(t_m\)) — chosen because it can flexibly represent skewed uncertainty (e.g., more room for things to go wrong than right) unlike a symmetric normal distribution. The expected duration weights the most-likely estimate four times as heavily as either extreme, since the mode is considered the single best guess, while the optimistic and pessimistic bounds each get equal but lesser weight. The variance is derived directly from the spread between the two extreme estimates — a wider gap between optimistic and pessimistic times signals greater uncertainty (larger variance) in that activity's actual duration.
Because the Central Limit Theorem justifies treating the sum of many independent random activity durations as approximately normally distributed, the expected project duration is simply the sum of the individual expected activity times along the critical path, and — critically — the project's variance is the sum of the individual variances (never the sum of standard deviations, which is a very common mistake) along the same critical path. The probability of finishing by a target date is then found by converting the gap between the target and the expected duration into a standard normal Z-score, and reading the corresponding probability from the standard normal cumulative distribution.
\( t_e = \dfrac{t_o+4t_m+t_p}{6} \)
\( \sigma^2 = \left(\dfrac{t_p-t_o}{6}\right)^2 \)
\( \sigma = \dfrac{t_p-t_o}{6} \)
\( T_E = \Sigma t_e \) (along critical path)
\( \sigma_p^2 = \Sigma\sigma^2 \) (sum of variances)
\( Z = \dfrac{T_S-T_E}{\sigma_p} \)
\( P(T\le T_S) = \Phi(Z) \)
Given: \(t_o=4\) days, \(t_m=6\) days, \(t_p=14\) days. Find \(t_e\) and \(\sigma^2\).
Solution:
\( t_e = \dfrac{4+4(6)+14}{6} = \dfrac{4+24+14}{6} = \dfrac{42}{6} = 7\ \text{days} \)
\( \sigma^2 = \left(\dfrac{14-4}{6}\right)^2 = \left(\dfrac{10}{6}\right)^2 \approx 2.78 \)
Answer: \(t_e=7\) days, \(\sigma^2\approx2.78\ \text{days}^2\).
Given: The critical path has three activities with \(t_e\) values 7, 10, 5 days and variances 2.78, 1.0, 0.44. Find \(T_E\) and \(\sigma_p^2\).
Solution:
\( T_E = 7+10+5 = 22\ \text{days} \)
\( \sigma_p^2 = 2.78+1.0+0.44 = 4.22 \)
Answer: \(T_E=22\) days, \(\sigma_p^2=4.22\ \text{days}^2\), so \(\sigma_p=\sqrt{4.22}\approx2.05\) days.
Given: \(T_E=22\) days, \(\sigma_p=2.05\) days, target date \(T_S=25\) days. Find the probability of completing by the target.
Solution:
\( Z = \dfrac{25-22}{2.05} = \dfrac{3}{2.05} \approx 1.46 \)
From the standard normal table, \(\Phi(1.46)\approx0.928\)
Answer: Probability of completing by day 25 \(\approx92.8\%\).
Fig. 5.1 — PERT's Beta distribution for activity duration: bounded between the optimistic and pessimistic estimates, peaked (mode) at the most-likely estimate, with the expected time weighting the mode four times as heavily as either extreme.
A Gantt chart (bar chart) is a horizontal bar graph representing the project schedule, developed by Henry Gantt around 1910–1915 — making it far older than CPM or PERT, yet it remains the most widely used tool for communicating schedules to non-technical stakeholders because of its intuitive visual format.
| Feature | Description |
|---|---|
| Horizontal axis | Time scale (days, weeks, months) |
| Vertical axis | List of activities / work packages |
| Bars | Span from EST to EFT (or LST to LFT); length = duration |
| Float | Often shown as thin lines or hatching extending beyond the solid bar to LFT |
| Milestone | Diamond symbol ◆ marking a key event with zero duration |
| Progress tracking | Bars partially filled to show % complete |
A Gantt chart's core strength is also the source of its core weakness: its simple, visual bar-per-activity format is easy to read at a glance and communicates well with clients and field staff, but that same simplicity means it does not clearly show the logical dependencies between activities, nor does it explicitly identify the critical path — two things a network diagram shows immediately. For large, complex projects (roughly beyond 50 activities), a Gantt chart also becomes visually unwieldy and difficult to update when the underlying logic changes. In practice, the two tools are complementary rather than competing: the CPM/PERT network is used to plan the logic and identify the critical path, while the Gantt chart derived from it is used to communicate and track the schedule day-to-day.
Bar spans EST to EFT (or LST to LFT); length = activity duration
Float shown as a thin line/hatch extending the solid bar out to LFT
Given: An activity has \(EST=5\) days, \(EFT=12\) days, \(LFT=15\) days. Sketch the Gantt bar description.
Solution: The solid bar spans from day 5 to day 12 (duration 7 days); a hatched/thin extension continues from day 12 to day 15, representing the 3 days of Total Float.
Answer: Solid bar: day 5–12; float extension: day 12–15 (3 days float).
Given: Two activities' bars on a Gantt chart appear to overlap in time but the chart doesn't indicate whether one depends on the other. What tool would clarify this?
Solution: A Gantt chart shows only when activities occur, not their logical relationships — dependency information must come from elsewhere.
Answer: A CPM/PERT network diagram (AOA or AON) would explicitly show whether a predecessor-successor relationship exists between the two activities.
Given: A project has a key deliverable ("Foundation Approved") that marks a point in time with no associated work duration. How is this shown on a Gantt chart?
Solution: Since it has zero duration, it cannot be drawn as a bar in the usual sense.
Answer: It is shown as a milestone — a diamond (◆) symbol placed at the appropriate date on the chart.
Fig. 6.1 — Gantt chart: horizontal bars represent each activity's scheduled duration, with a lighter extension showing available float, and a diamond marking a zero-duration milestone.
| Technique | Objective | Constraint |
|---|---|---|
| Resource levelling | Minimise fluctuations in resource usage; smooth the demand profile | Project duration fixed; resources unlimited but peak usage minimised |
| Resource allocation (limited) | Complete project with available resources; may extend duration | Resource limit fixed; project duration may extend |
These two techniques represent opposite trade-offs: levelling holds the schedule (project duration) fixed and adjusts resource deployment within that constraint, whereas limited resource allocation holds the resource ceiling fixed and allows the schedule to stretch if the available workforce or equipment can't handle the theoretical peak demand.
Levelling begins from the earliest-start schedule, plotting a resource histogram (resource units required versus time) to visualise demand peaks and troughs. The key principle is that only non-critical activities may be shifted — shifting them later (within their available float) smooths the resource profile without delaying the project, whereas shifting a critical activity has no available float to absorb the shift and directly extends the project duration. Levelling quality is often measured by minimising the sum of squared resource usage across periods, since squaring penalises sharp peaks much more heavily than a simple sum would, favouring a genuinely smoother profile over one with occasional extreme spikes.
When resources are limited and multiple activities compete for the same resource simultaneously, priority must be assigned by some rule — common heuristics include scheduling the activity with the least total float first (protecting the most schedule-critical work), the shortest-duration activity first (freeing up the resource quickly for others), or the earliest latest-finish-time first. Resource-constrained project scheduling is known to be an NP-hard combinatorial problem in general, meaning no single algorithm guarantees the mathematically optimal schedule efficiently for large projects — these heuristic priority rules instead give practical, near-optimal solutions used throughout the industry.
Minimise \( \Sigma(r_i^2\times d_i) \)
\(r_i\)=resource units in period \(i\); \(d_i\)=duration of period \(i\)
Shift amount \(\le TF\) of the (non-critical) activity being shifted
Given: Activity X has Total Float \(TF=4\) days and is scheduled at its earliest start, causing a resource peak. Can it be shifted to smooth the peak?
Solution: Since X is non-critical (TF > 0), it has room to move without delaying the project.
Answer: Yes — X can be shifted up to 4 days later (its Total Float) to help smooth the resource histogram, as long as the shift doesn't create a new conflict elsewhere.
Given: Activity Y is on the critical path (\(TF=0\)) and also contributes to a resource peak. Can it be shifted?
Solution: A critical activity has zero float, so any shift directly delays the project's completion date.
Answer: No — Y cannot be shifted without extending project duration; the resource peak it contributes to must be resolved by other means (e.g., shifting a different, non-critical activity, or adding resources).
Given: Two activities compete for one crew on the same day: Activity P (TF=2 days, duration 5 days) and Activity Q (TF=0 days, duration 3 days). Which gets priority under the "minimum total float first" rule?
Solution: The rule prioritises the activity with the least float — Q has TF=0 (critical), while P has TF=2 (some slack available).
Answer: Activity Q gets priority (it is critical and cannot be delayed); Activity P, having float, can wait if the crew is unavailable.
Fig. 7.1 — Resource levelling: shifting non-critical activities within their available float smooths a sharp resource-demand peak into a more even profile, without changing the overall project duration.
An estimate is the anticipated cost of a work computed before execution, from drawings and rates. Its accuracy grows as the design matures:
| Type | Basis / Use |
|---|---|
| Preliminary / approximate | Rough cost for a decision, from plinth area or cubic content (per m² / per m³ rate) |
| Plinth-area estimate | Plinth area × plinth-area rate for that class of building |
| Cubic-content estimate | Plinth area × height × unit cubic rate (more reliable than plinth area) |
| Detailed estimate | Quantities of every item measured from drawings, multiplied by analysed rates — the accurate, tender-grade estimate |
| Revised / supplementary | Prepared when the sanctioned cost is exceeded, or extra work is added |
For a building, the quantity of each item (earthwork, foundation concrete, masonry) is obtained by two standard methods:
Earthwork for roads, canals and embankments is computed from cross-sectional areas taken at a regular interval \(d\) using the mid-section, trapezoidal (average end area), or prismoidal (Simpson's) rule — the prismoidal rule is the most accurate but needs an odd number of sections. A bar bending schedule (BBS) lists the shape, size, number and cut length of every reinforcement bar, giving the steel weight (via \(\text{weight} = d^2/162\ \text{kg/m}\) for a bar of diameter \(d\) mm).
Rate analysis builds up the unit rate of an item by adding the cost of materials, labour and machinery for a unit quantity, plus water charges, contractor's profit (~10%) and overheads. It is the bridge between measured quantities and money, and is what makes a detailed estimate a priced tender.
Once an estimate is priced, work is awarded and executed through a management framework:
Long wall (out-to-out) \(= L_{CL} + 1\times b\)
Short wall (in-to-in) \(= L_{CL} - 1\times b\)
\( V = d\left[\dfrac{A_1+A_n}{2} + (A_2+A_3+\dots+A_{n-1})\right] \)
\( V = \dfrac{d}{3}\big[(A_1+A_n) + 4(A_2+A_4+\dots) + 2(A_3+A_5+\dots)\big] \)
Needs an odd number of sections
\( V = A_m \times L \) (\(A_m\) = area at mid-section)
\( w = \dfrac{d^2}{162}\ \text{kg/m} \) (\(d\) in mm)
Given: A room with centre-line lengths 5 m (long walls, two of them) and 4 m (short walls, two of them); wall breadth at foundation \(b=0.4\ \text{m}\). Find the lengths to use for the two pairs.
Solution: Long wall (out-to-out) \(= 5 + 0.4 = 5.4\ \text{m}\) each; short wall (in-to-in) \(= 4 - 0.4 = 3.6\ \text{m}\) each.
Answer: 2 long walls of 5.4 m and 2 short walls of 3.6 m.
Given: Cross-sectional areas at 30 m intervals along an embankment: 20, 40, 60, 50, 30 m² (5 sections). Find the volume by Simpson's rule.
Solution: \( V = \dfrac{30}{3}\big[(20+30) + 4(40+50) + 2(60)\big] = 10\big[50 + 360 + 120\big] = 10\times530 = 5300\ \text{m}^3 \)
Answer: Volume ≈ 5300 m³.
Given: 40 bars of 12 mm diameter, each 6 m long. Find the total steel weight.
Solution: Weight per metre \(= \dfrac{12^2}{162} = \dfrac{144}{162} = 0.889\ \text{kg/m}\). Total length \(= 40\times6 = 240\ \text{m}\). Weight \(= 240\times0.889 \approx 213\ \text{kg}\).
Answer: ≈ 213 kg of steel.
Cost is the money spent to produce a property; value is its present worth in the market. Common value terms:
| Term | Meaning |
|---|---|
| Market value | Price obtainable in the open market at a given time |
| Book value | Cost less total depreciation to date (shown in the books) |
| Scrap value | Value of the dismantled materials at the end of life (≈ 10% of cost), less demolition cost |
| Salvage value | Value at end of utility period without being dismantled (useful to another user) |
| Assessed value | Value fixed by an authority for levying tax |
| Distress / forced value | Lower value realised under a forced or urgent sale |
Depreciation is the gradual loss in value of a property due to age, wear, and decay. Obsolescence is loss of value due to the asset going out of date (better designs, changed fashion, new regulations) even while physically sound. Common methods to compute depreciation:
A sinking fund is an amount set aside annually (and invested at compound interest) so that a required sum — the cost of replacement, less scrap value — is accumulated by the end of the asset's life. Years' purchase (YP) is the capital sum whose interest equals one unit of annual net income; for a perpetual income \(YP = 1/i\). The capitalised value of a property = net annual income × YP.
\( D = \dfrac{C-S}{n} \) per year
Book value after \(t\) years \(= C - tD\)
Book value \(= C(1-r)^t\)
\(r\) = fixed annual depreciation rate
\( I = \dfrac{S\,i}{(1+i)^n - 1} \)
\(S\) = amount to accumulate, \(i\) = rate, \(n\) = years
Perpetual \( YP = \dfrac{1}{i} \)
Capitalised value \(=\) net annual income \(\times\ YP\)
Given: A building costs ₹20,00,000, has a life of \(n = 50\) years and a scrap value of ₹2,00,000. Find the annual depreciation and the book value after 20 years.
Solution: \( D = \dfrac{20{,}00{,}000 - 2{,}00{,}000}{50} = \dfrac{18{,}00{,}000}{50} = 36{,}000 \) per year. Book value after 20 years \(= 20{,}00{,}000 - 20\times36{,}000 = 12{,}80{,}000\).
Answer: ₹36,000 per year; book value ₹12,80,000 after 20 years.
Given: A property yields a net annual income of ₹1,20,000; the expected rate of interest is \(i = 6\%\). Find its capitalised value.
Solution: \( YP = \dfrac{1}{i} = \dfrac{1}{0.06} = 16.67 \). Capitalised value \(= 1{,}20{,}000 \times 16.67 = 20{,}00{,}000\).
Answer: Capitalised value ≈ ₹20,00,000.
Given: A sum of ₹5,00,000 must be accumulated in \(n = 20\) years at \(i = 5\%\). Find the annual sinking-fund instalment.
Solution: \( I = \dfrac{S\,i}{(1+i)^n - 1} = \dfrac{5{,}00{,}000 \times 0.05}{(1.05)^{20} - 1} = \dfrac{25{,}000}{2.653 - 1} = \dfrac{25{,}000}{1.653} \approx 15{,}125 \).
Answer: ≈ ₹15,125 per year.
\( E_j = \max(E_i+D) \)
\( L_i = \min(L_j-D) \)
\( TF = L_j-E_i-D = LST-EST \)
\( FF = E_j-E_i-D \)
\( IF = E_j-L_i-D\ (\ge0) \)
\( IF\le FF\le TF \); critical: \(TF=0\)
\( \dfrac{C_c-C_n}{t_n-t_c} \)
\( t_e = \dfrac{t_o+4t_m+t_p}{6} \)
\( \sigma^2 = \left(\dfrac{t_p-t_o}{6}\right)^2 \)
\( Z=\dfrac{T_S-T_E}{\sigma_p} \); \(P=\Phi(Z)\)
Minimise \(\Sigma(r_i^2\times d_i)\)
Crash critical activity with min cost slope first
| Topic | GATE Focus | ESE Focus | SSC JE Focus |
|---|---|---|---|
| Project networks | Recognising dummy activity need; AOA construction | Complete network diagrams; AOA vs AON theory | Basic terminology; dummy activity concept |
| Forward/backward pass | Full numerical: E and L for all nodes | Complete tabulation with EST/EFT/LST/LFT | Basic forward pass calculation |
| Float & critical path | TF/FF/IF numericals; multiple critical paths | Complete float table; critical path identification methods | TF calculation; critical path definition |
| Crashing | Cost slope numericals; optimal duration | Complete crashing procedure with multiple paths | Cost slope formula; crash vs normal concept |
| PERT | t_e, variance, Z-score numericals (near-certain) | Complete probabilistic analysis; Beta distribution theory | t_e formula; basic probability concept |
| Gantt charts | Rarely numerical; limitations vs CPM | Theory questions on advantages/limitations | Basic Gantt chart reading |
| Resource scheduling | Levelling vs allocation distinction | Resource histogram; heuristic rules application | Definitions of levelling and allocation |
Q1. Activity (2-5) has \(E_2=8\), \(L_5=20\), duration \(D=7\). Find the Total Float.
Q2. An activity has \(t_n=8\) days (\(C_n=₹20{,}000\)), \(t_c=5\) days (\(C_c=₹35{,}000\)). Find the cost slope.
Q3. An activity has \(t_o=3\), \(t_m=5\), \(t_p=13\) days. Find \(t_e\) and \(\sigma\).
Q4. \(T_E=30\) days, \(\sigma_p=3\) days, target \(T_S=27\) days. Find Z and interpret.
Q5. A network has two paths of 25 and 25 days (tied) and one path of 20 days. How many critical paths exist, and what is the project duration?