In the network of question 1, activity B has a total float of 1 day but a free float of zero. The practical meaning is that

B can slip one day without delaying the project, but doing so immediately delays its successor D and consumes D’s float

B is critical, since a free float of zero is the definition of a critical activity

B cannot slip at all, since the smaller of the two floats always governs the activity

B and D can each slip one day independently, since each has a total float of one day

Which of the following statements about construction contracts and tendering are correct?

An item-rate contract leaves quantity risk with the client and rate risk with the contractor

A cost-plus-percentage-fee contract gives the contractor an incentive to overspend, so a fixed fee is preferable

An abnormally low tender is a risk rather than a saving, and need not be accepted

An item-rate contract leaves quantity risk with the client and rate risk with the contractor

A cost-plus-percentage-fee contract gives the contractor an incentive to overspend, so a fixed fee is preferable

An abnormally low tender is a risk rather than a saving, and need not be accepted

A lump-sum contract is the right choice when quantities cannot be defined in advance

In the network of question 1 (duration 13 days, critical path A–C–F), activity E is shortened from 7 days to 3 days at considerable expense. The effect on the project duration is that

It is unchanged at 13 days, because E was never on the critical path

It falls to 9 days, since E was the longest activity in the network

It falls by 1 day, in proportion to E’s total float

It falls to 12 days, because shortening any activity shortens the project a little

Construction Management: Contracts, Rate Analysis, Estimation, CPM and PERT

This is the one section of the Civil paper about money, time and risk rather than about materials or forces, and the organising idea is that all three are questions about who bears the uncertainty. A contract is a device for allocating it: an item-rate contract puts quantity risk on the client and rate risk on the contractor, a lump-sum contract puts nearly all of it on the contractor, and a cost-plus contract puts nearly all of it back on the client — which is why the choice of contract is decided by how well the work can be defined in advance, not by which looks cheapest on the day. Network analysis is the same question asked about time. CPM assumes durations are known and asks which activities control the finish date; PERT assumes they are not and asks how confident you can be about a date at all. The single most useful result in the chapter comes out of the second: the critical path is the longest path through the network, not the sum of the durations, so most activities have slack and shortening them achieves nothing — a fact that overturns the intuition that speeding up work speeds up a project. The worked network below has seven activities totalling 29 days of work and finishes in 13, with only three activities critical. Compress any of the other four and the project finishes on exactly the same day.

Projects, tendering and contracts

Contract types, read as an allocation of risk
Type	How the contractor is paid	Who carries the risk	Right when
Item rate (schedule of rates)	A rate per unit of each item, times the quantity actually executed	Quantity risk on the client, rate risk on the contractor	Items are clear but quantities cannot be fixed in advance — most public works
Lump sum	One fixed price for the whole defined work	Almost all on the contractor	The scope is completely defined and unlikely to change
Cost plus (percentage or fixed fee)	Actual cost, plus a margin	Almost all back on the client	Emergency or genuinely undefinable work. A percentage fee rewards overspending, so a fixed fee is preferred
Labour rate / piece work	Labour only; the client supplies materials	Material risk on the client	The client can procure better or wants to control quality of materials

Types of construction project shape all of this before a contract is drafted. A building is repetitive, largely enclosed and dominated by finishes and services; a highway or canal is linear, so its work fronts move rather than stack and its programme is a question of where the machines are; an industrial plant is dominated by procurement lead times rather than by construction; and a dam or a tunnel is dominated by a single risk — the ground — that no amount of planning removes. The right contract, the right programme and the right form of estimate all follow from which of those a project is. Tendering exists to discover a price competitively and to be seen to do so, which is why the procedure is as important as the outcome in public works: notice inviting tender, prequalification where the work needs it, issue of documents, a sealed earnest money deposit with each bid to make a frivolous bid costly, public opening, evaluation and award. The lowest tender is not automatically accepted — an abnormally low bid is a risk rather than a bargain, because a contractor who cannot complete at that price will either claim, slow down or abandon, and all three cost the client more than the difference. Security deposit and retention money then hold performance after award, and the defect liability period keeps a claim open after completion.

Rate analysis and estimation

Rate analysis builds the rate for one unit of an item from first principles — the quantities of each material, the labour of each trade, the plant, then overheads and profit as percentages — and it is written against standard specifications — the codified description of what the item is and how it must be executed, which is what makes two contractors’ rates for the same item comparable at all. The reason a rate is built rather than looked up is that it is local and current: the same cubic metre of concrete costs differently in two districts because the aggregate haul differs. Cost estimation then multiplies those rates by quantities, and it comes in stages of rising confidence: a plinth-area or cube-rate estimate at the sketch stage, worth a percentage rather than a figure; a detailed estimate once drawings exist, item by item from measured quantities; and a revised estimate when the scope has moved far enough that the detailed one is no longer a description of the work. Project planning then sequences what has been estimated, which is what the rest of this chapter is about. The quantities themselves come from measurement. The two standard methods of measuring wall lengths look different and are not.

🧠 Centre-line and long-wall–short-wall give exactly the same answer

Take a room 6 m × 4 m inside with walls 0.4 m thick. The outside dimensions are 6.8 × 4.8 and the centre-line dimensions 6.4 × 4.4. Centre-line method: total length = 2(6.4 + 4.4) = 21.60 m. Long-wall–short-wall method: the long walls are measured out to out, 2 × 6.8 = 13.60 m, and the short walls in to in, 2 × 4.0 = 8.00 m, giving 21.60 m. The two agree exactly, and they must — they are two bookkeeping routes to the same physical volume of masonry, which at 3 m high is 21.60 × 0.4 × 3 = 25.92 m³. That identity is the check to apply to any answer in this topic; if the two methods disagree, one of them has been set out wrongly. Where the methods genuinely differ is in which is less error-prone for a given plan. The centre-line method is quicker and is the natural choice for a building of uniform wall thickness — but it needs a deduction of t/2 at every junction where a cross wall meets a main wall, because the centre lines cross and the corner would otherwise be counted twice. Add one internal cross wall to the room above: its centre-line length is 4.4 m, so the naive total is 26.0 m, and the correct total is 26.0 − 2 × (0.4/2) = 25.60 m. One junction costs 0.4 m — small, and it multiplies with every internal wall, which is why the long-wall–short-wall method is preferred for a plan with many cross walls or with walls of differing thickness, where the deductions become the whole difficulty.

CPM: the longest path, and the two kinds of float

A network is a set of activities with precedence relations, and CPM asks one question of it: which activities determine the finish date? The method is two passes. The forward pass computes, for each activity, the earliest start — the latest of the earliest finishes of everything that must precede it — and hence the earliest finish; the largest earliest finish is the project duration. The backward pass then computes the latest finish — the earliest of the latest starts of everything that follows — and hence the latest start. Total float is LS − ES, and the activities with zero float form the critical path.

A worked network: A(4), B(6), C(5) after A, D(2) after B, E(7) after A, F(4) after C and D, G(1) after E
Activity	Duration	ES / EF	LS / LF	Total float	Free float
A	4	0 / 4	0 / 4	0 — critical	0
B	6	0 / 6	1 / 7	1	0
C	5	4 / 9	4 / 9	0 — critical	0
D	2	6 / 8	7 / 9	1	1
E	7	4 / 11	5 / 12	1	0
F	4	9 / 13	9 / 13	0 — critical	0
G	1	11 / 12	12 / 13	1	1

🎯 29 days of work, 13 days of project, three activities that matter

The seven activities total 29 days of work and the project finishes in 13, because work that can run in parallel does. The critical path is A–C–F, 4 + 5 + 4 = 13, and it is the longest path through the network — which is the definition, and the opposite of the intuition that the critical path is the busiest or the most expensive. Three consequences follow directly and each is examined. Shortening a non-critical activity achieves nothing. Compress E from 7 days to 3 and the project still finishes on day 13, because E was never controlling it. Effort spent there is wasted, and this is the single most common management error the technique exists to prevent. Crashing the critical path moves the bottleneck. Shorten C by 2 days and A–C–F becomes 11, at which point A–E–G (4 + 7 + 1 = 12) is longer and becomes critical — so crashing is iterative: recompute the network after every change, because the path you were compressing may no longer be the one that matters. And float is not free capacity. Total float is how long an activity can slip without delaying the project; free float is how long it can slip without delaying its own successor. B in this network has a total float of 1 but a free float of 0 — delay B by a day and the project still finishes on time, but D is pushed and its float is consumed. Two activities each using their total float can therefore combine to delay the project, which is why float belonging to a chain is shared rather than owned, and why a schedule tracked on total float alone will look healthy right up to the point where it is not.

PERT, and what a completion date actually means

PERT replaces a single duration with three estimates — optimistic t_o, most likely t_m and pessimistic t_p — and treats the duration as a beta-distributed random variable with t_e = (t_o + 4t_m + t_p)/6 and variance ((t_p − t_o)/6)². The weighting is what makes it useful: the most likely value carries four-sixths of the estimate, so the mean sits near it but is dragged towards whichever tail is longer, and the variance depends only on the spread t_p − t_o — not on where t_m sits within it. An activity with t_o = 4, t_m = 6, t_p = 14 has t_e = 7.0 and a standard deviation of 1.667: the mean is a full day beyond the most likely value, because the pessimistic tail is four times as long as the optimistic one.

⚠️ A PERT date is a 50% date, and variances add while standard deviations do not

Take a critical path of three activities with t_e of 7, 5 and 10 and variances 2.778, 0.444 and 4.000. The expected duration is 22.0 days and the path variance is the sum of the variances, 7.222, so the standard deviation is √7.222 = 2.687 — not the sum of the standard deviations, which would be 4.333 and would overstate the uncertainty by 61%. Variances add for independent random variables; standard deviations never do. Now the point that matters more than the arithmetic: the expected duration is a 50% date. There is a one-in-two chance of finishing after day 22, which is not what a client hears when they are told "22 days". Probabilities are read off the normal distribution with Z = (T_s − T_e)/σ: finishing by day 24 has Z = 0.744, or 77.2%; by day 27, Z = 1.861, or 96.9%; by day 30, 99.9%. The whole value of PERT is in that table — it converts a schedule into a confidence level, so a contract date can be set at 90% or 95% rather than at the coin-flip the expected duration represents. Two honest limitations. The distribution assumption and the (t_o + 4t_m + t_p)/6 weighting are conventions, not derivations, and the answer inherits their arbitrariness. And PERT computes the probability along one path, so it systematically understates the risk of a project with several near-critical paths: any one of them running late delays the finish, and the probability that at least one does is greater than the probability that any given one does. A network with three parallel paths all of expected length 22 has considerably less than a 50% chance of finishing on day 22.

Finally, crashing. Each activity has a normal duration and cost and a crash duration and cost, and the cost slope = (crash cost − normal cost)/(normal time − crash time) is the price of buying one day. An activity normally taking 10 days at ₹50,000 and crashable to 7 days at ₹71,000 has a slope of ₹7,000 per day. The procedure is to crash the critical activity with the lowest slope, one day at a time, recomputing the network after each step because the critical path moves — and to stop when the slope exceeds the indirect cost per day saved, since indirect costs (supervision, plant hire, site establishment) fall with duration while direct costs rise. The optimum duration is where the total is least, and it is generally shorter than the normal duration and longer than the fully crashed one.

Key takeaways

A contract type is a risk allocation: item rate splits it, lump sum loads the contractor, cost plus loads the client.
Centre-line and long-wall–short-wall give identical totals — 21.60 m here; the centre-line method needs t/2 deducted at each junction.
The critical path is the longest path, not the sum of durations: 29 days of work finishes in 13.
Total float ≠ free float. B here has total float 1 and free float 0 — using it delays D.
PERT: t_e = (t_o + 4t_m + t_p)/6, and variances add — σ = √7.222 = 2.687, never the sum of the σ’s.
The expected duration is a 50% date; day 24 is 77.2% and day 27 is 96.9%, which is what PERT is for.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A project has activities A (4 days, no predecessor), B (6, none), C (5, after A), D (2, after B), E (7, after A), F (4, after C and D) and G (1, after E). The project duration in days is ______
Numerical answer — type the value.
Show answer
Answer: 13
13 days. Trace the paths rather than adding the durations. There are three routes from start to finish: A–C–F = 4 + 5 + 4 = 13, B–D–F = 6 + 2 + 4 = 12, and A–E–G = 4 + 7 + 1 = 12. The project duration is the longest of them, so 13 days by way of A–C–F, and that is the critical path. Note the two numbers this puts side by side. The seven activities represent 29 days of work, and the project finishes in 13, because work that can run in parallel does — and note also that the critical path is not the one with the most activities or the largest single duration: E alone is 7 days, the longest activity in the network, and it is not critical. Three consequences follow, and they are the whole point of the technique. Shortening a non-critical activity achieves nothing: compress E from 7 days to 3 and the project still finishes on day 13. Crashing the critical path moves the bottleneck: shorten C by 2 days and A–C–F becomes 11, at which point A–E–G at 12 becomes critical, so the network must be recomputed after every change. And float is not surplus capacity — B has a total float of 1 day but a free float of zero, so using its float pushes D and consumes its float instead. A manager who does not know which activities are critical will spend effort where it cannot help, which is precisely the error CPM exists to prevent.
For a PERT activity with optimistic time 4 days, most likely time 6 days and pessimistic time 14 days, the expected time in days is ______
Numerical answer — type the value.
Show answer
Answer: 7
7 days. t_e = (t_o + 4t_m + t_p)/6 = (4 + 24 + 14)/6 = 42/6 = 7.0. The result is worth dwelling on because it is not 6. The most likely duration is 6 days, and the expected duration is a full day longer, because the pessimistic tail runs 8 days past the mode while the optimistic tail runs only 2 days short of it. The weighting 1 : 4 : 1 keeps the estimate anchored near the most likely value but lets it be dragged towards whichever tail is longer — and construction durations are almost always skewed this way, since there are many ways for work to run late and very few for it to finish early. A planner who schedules on the most likely duration is therefore systematically optimistic, activity by activity, and the error accumulates along a path. The variance is ((t_p − t_o)/6)² = (10/6)² = 2.778, so σ = 1.667 days, and note what the variance does and does not depend on: it uses only the spread t_p − t_o, and is completely indifferent to where t_m sits within it. An activity with estimates 4, 12, 14 has the same variance as this one and a t_e of 11 — the mode moves the mean but not the uncertainty. That is a convention of the model rather than a fact about the world, and it is worth knowing as a limitation: PERT captures how wide an estimate is but not how lopsided, so two activities that feel very different to a planner can be indistinguishable in their contribution to schedule risk.
A critical path consists of three activities whose variances are 2.778, 0.444 and 4.000 (days²), with a total expected duration of 22 days. The standard deviation of the project duration in days is ______ (round to three decimal places)
Numerical answer — type the value.
Show answer
Answer: 2.687
2.687 days. Variances add for independent random variables, standard deviations never do: σ² = 2.778 + 0.444 + 4.000 = 7.222, so σ = √7.222 = 2.687 days. Adding the standard deviations instead gives 1.667 + 0.667 + 2.000 = 4.333, which overstates the uncertainty by 61% and is the single commonest error in PERT problems. Having got σ, use it for what PERT actually exists to deliver — a confidence level rather than a date. The expected duration of 22 days is a 50% date: there is a one-in-two chance of overrunning it, which is not what a client understands from "22 days". Read probabilities from Z = (T_s − T_e)/σ: finishing by day 24 gives Z = 2/2.687 = 0.744, hence 77.2%; by day 27, Z = 1.861, hence 96.9%; by day 30, Z = 2.977, hence 99.9%. So committing to a 95% date here means promising about day 26.4, four and a half days beyond the expected duration — and that gap, not the expected duration, is the number a contract should be built on. Two honest limitations belong with the answer. The beta assumption and the 1 : 4 : 1 weighting are conventions rather than derivations. And this calculation follows one path, so it understates risk when a network has several near-critical paths: any one of them slipping delays the finish, and the chance that at least one does exceeds the chance that a particular one does. A network with three parallel paths all expected to take 22 days has appreciably less than a 50% chance of finishing on day 22.
A room measures 6 m × 4 m internally with walls 0.4 m thick and 3 m high. Using the centre-line method, the volume of masonry in the four walls in m³ is ______ (round to two decimal places, ignoring openings)
Numerical answer — type the value.
Show answer
Answer: 25.92
25.92 m³. The centre-line dimensions are the internal dimensions plus one wall thickness: (6 + 0.4) × (4 + 0.4) = 6.4 × 4.4, so the centre-line length is 2(6.4 + 4.4) = 21.60 m, and the volume is 21.60 × 0.4 × 3 = 25.92 m³. Now verify it by the long-wall–short-wall method, which is the check that makes the answer defensible. Long walls are measured out to out, 2 × (6 + 2 × 0.4) = 2 × 6.8 = 13.60 m; short walls in to in, 2 × 4.0 = 8.00 m; total 21.60 m — exactly the same. It has to be: the two are alternative bookkeeping routes to one physical volume of masonry, and if they disagree, one of them has been set out wrongly. That identity is the single most useful check in this topic. Where the methods differ is in which is safer to apply. The centre-line method is quicker for a building of uniform wall thickness, but it requires a deduction of t/2 at every junction where a cross wall meets a main wall, because the centre lines cross there and the corner is otherwise counted twice. Add one internal cross wall to this room: its centre-line length is 4.4 m, so the naive total is 26.0 m and the correct total is 26.0 − 2 × 0.2 = 25.60 m. One junction costs 0.4 m, and the deductions multiply with every internal wall — which is why the long-wall–short-wall method is preferred for plans with many cross walls or with walls of differing thickness, where getting the deductions right becomes the whole difficulty. Note finally that a real estimate deducts openings (doors, windows) from this figure and adds the lintels and sills separately; the question excluded them, and a real bill of quantities would not.
In the network of question 1, activity B has a total float of 1 day but a free float of zero. The practical meaning is that
1. B can slip one day without delaying the project, but doing so immediately delays its successor D and consumes D’s float
2. B is critical, since a free float of zero is the definition of a critical activity
3. B cannot slip at all, since the smaller of the two floats always governs the activity
4. B and D can each slip one day independently, since each has a total float of one day
Show answer
Answer: A — B can slip one day without delaying the project, but doing so immediately delays its successor D and consumes D’s float
B can slip a day without delaying the project, but D pays for it. The two floats answer two different questions and both are needed. Total float = LS − ES is how long an activity can be delayed without delaying the project; for B that is 1 − 0 = 1 day. Free float is how long it can be delayed without delaying its immediate successor, computed as the earliest start of the successor minus the earliest finish of the activity; for B that is ES(D) − EF(B) = 6 − 6 = 0. So B genuinely has a day of slack as far as the finish date is concerned, and none at all as far as D is concerned — delay B by a day and D starts a day later, using up the one day of float that D had. Option (d) is precisely the error this distinction exists to prevent: the float of 1 day belongs to the chain B–D, not to each of them separately, so B and D cannot each take a day. Two activities in series each "using their float" is how a schedule that looked healthy on total float alone turns out to be late — a project can be tracked activity by activity, every activity within its total float, and still miss its date. That is why planning software reports both, why free float is the figure a site manager can act on without consulting anyone downstream, and why the safe rule when reallocating resources is to move them only into activities with genuine free float. Option (b) is wrong: a critical activity has zero total float; B has one day of it and is not critical. Option (c) inverts the relationship — free float is always less than or equal to total float, and it constrains the successor rather than the project.
An activity normally takes 10 days and costs ₹50,000. Crashed, it takes 7 days and costs ₹71,000. Its cost slope in rupees per day is ______
Numerical answer — type the value.
Show answer
Answer: 7000
₹7,000 per day. cost slope = (crash cost − normal cost)/(normal duration − crash duration) = (71,000 − 50,000)/(10 − 7) = 21,000/3 = ₹7,000 per day. The slope is the price of buying one day on that activity, and it is what makes the crashing procedure mechanical rather than a matter of judgement: crash the critical activity with the lowest slope, one day at a time, and recompute the network after every step because the critical path moves as soon as the path you were shortening is no longer the longest. Two constraints govern when to stop, and both are examined. The first is the crash limit — no activity can be shortened past its crash duration, whatever anyone is willing to pay, because at some point the work is limited by physics and not by resources. The second is the comparison that decides the answer: crashing raises direct costs while indirect costs — supervision, plant hire, site establishment, financing — fall with the duration, so it is worth buying a day only while the slope of the cheapest available critical activity is less than the indirect cost saved per day. The optimum duration is where the total of the two is least, and it is generally shorter than the normal duration (because early days are cheap to buy) and longer than the fully crashed one (because the last days are ruinously expensive). Note that the slope is assumed linear between the two points, which is a convenience rather than a truth — the real cost of compression accelerates, so the linear model understates the cost of the last day bought. And an activity’s slope is irrelevant while it is not critical: paying ₹7,000 to shorten an activity with float buys nothing at all.
Which of the following statements about construction contracts and tendering are correct?
1. An item-rate contract leaves quantity risk with the client and rate risk with the contractor
2. A cost-plus-percentage-fee contract gives the contractor an incentive to overspend, so a fixed fee is preferable
3. An abnormally low tender is a risk rather than a saving, and need not be accepted
4. A lump-sum contract is the right choice when quantities cannot be defined in advance
Show answer
Answer: A — An item-rate contract leaves quantity risk with the client and rate risk with the contractor; B — A cost-plus-percentage-fee contract gives the contractor an incentive to overspend, so a fixed fee is preferable; C — An abnormally low tender is a risk rather than a saving, and need not be accepted
(a), (b) and (c). (a) is the defining feature of the commonest form of public-works contract. The contractor quotes a rate per unit of each item and is paid for the quantity actually executed, so if the excavation turns out to be twice what the drawings suggested, the client pays for twice as much — quantity risk sits with the client. But the rate is fixed at award, so if labour or cement rises, the contractor absorbs it — rate risk sits with them. That split is exactly why the form suits work whose items are clear but whose quantities cannot be pinned down in advance, which describes most earthwork, foundations and repair. (b) is a genuine perverse incentive worth stating plainly: if the fee is a percentage of cost, then every rupee the contractor spends increases their own fee, so the contract rewards exactly the behaviour the client wants to prevent. A fixed fee removes the incentive by decoupling the margin from the spend, and a target-cost arrangement goes further by sharing the saving. (c) is the practice, and the reasoning is worth carrying: a bid materially below the engineer’s estimate usually means the bidder has misread the scope, made an arithmetic error, or intends to recover the shortfall through claims and variations. All three end with the client paying more than the next-lowest bid would have cost, or with an abandoned site — which is why tender conditions reserve the right to reject the lowest bid and why abnormally low bids are investigated rather than simply accepted. (d) is wrong, and it inverts the rule from (a). A lump-sum contract places a fixed price on a fully defined scope, so it is right precisely when quantities are known and unlikely to change. Using it where quantities are uncertain means the contractor must price the risk — inflating the bid — or price it wrongly and then pursue claims for every variation, which is the worst of both arrangements.
In the network of question 1 (duration 13 days, critical path A–C–F), activity E is shortened from 7 days to 3 days at considerable expense. The effect on the project duration is that
1. It is unchanged at 13 days, because E was never on the critical path
2. It falls to 9 days, since E was the longest activity in the network
3. It falls by 1 day, in proportion to E’s total float
4. It falls to 12 days, because shortening any activity shortens the project a little
Show answer
Answer: A — It is unchanged at 13 days, because E was never on the critical path
Unchanged at 13 days. The project duration is the length of the longest path, and E lies on A–E–G, which is 4 + 7 + 1 = 12 days — already shorter than the critical A–C–F at 13. Shortening E to 3 days takes that path to 8 days, which simply makes an already non-critical path even less critical. Nothing about the finish date changes, and the expense buys nothing at all. This is the single most valuable result CPM delivers, and it contradicts a strong intuition. E is the longest activity in the network at 7 days, so it looks like the obvious target for compression — and it is the wrong one, because what governs a project is not the size of its activities but the length of its paths. The corollary matters just as much: only the critical path can be shortened usefully, and even then only until it stops being the longest. Shorten C by 2 days and A–C–F becomes 11, at which point A–E–G at 12 is now the longest and becomes critical — so any further saving requires attention to E after all, and the network must be recomputed after every compression. This is also the case for the distinction between total and free float, and the reason a manager reallocating labour should move it out of activities with genuine free float and into critical ones. Option (b) confuses the longest activity with the longest path. Option (c) misuses float, which measures permissible delay, not achievable saving. Option (d) states the intuition CPM exists to correct: most activities in a real network have slack, and effort spent on them changes nothing.