Which machine, with its bucket carried on a rigid arm, digs below its own standing level and is the usual choice for trenches and foundation pits?

Backhoe, whose bucket is drawn back toward the machine below ground level

Front shovel, whose bucket is pushed away from the machine to cut into a bank

Backhoe, whose bucket is drawn back toward the machine below ground level

Bulldozer, whose blade pushes loose soil ahead of the tractor

Scraper, which cuts, hauls and spreads the soil in a single pass

A road job needs a clean sand-and-gravel sub-base compacted and, elsewhere, a layer of cohesive clay. Which pairing of roller and soil is most appropriate?

A vibratory roller on the sand-and-gravel and a sheepsfoot roller on the clay

A sheepsfoot roller on both the sand-and-gravel and the clay

A vibratory roller on both the sand-and-gravel and the clay

A sheepsfoot roller on the sand-and-gravel and a vibratory roller on the clay

A vibratory roller on the sand-and-gravel and a sheepsfoot roller on the clay

A 100 m³ (bank) excavation is in soil that swells 25% on digging. Each truck carries 10 m³ of loose soil. The minimum number of truck loads needed to carry it all away is

13, because the loose volume is 125 m³ and 12.5 loads must be rounded up

8, found by dividing 100 m³ by 1.25 and then by the 10 m³ capacity of a truck

10, found by dividing the 100 m³ directly by the 10 m³ truck capacity and ignoring swell

12, found by rounding down the loose volume of 125 m³ divided by the truck capacity

13, because the loose volume is 125 m³ and 12.5 loads must be rounded up

As the working radius of a tower crane is increased, its safe lifting capacity

falls, because the overturning moment, load multiplied by radius, is limited

rises, because a longer jib spreads the weight over a larger area of the slab

falls, because the overturning moment, load multiplied by radius, is limited

stays constant, because the counterweight is moved to compensate for the radius exactly

rises until the jib is horizontal and then falls as the hook descends toward the ground

Which of the following statements about construction equipment are correct?

A backhoe digs below its standing level, so it suits trenches and foundation pits

A dragline suits soft or wet ground where a long reach is needed

Needle (immersion) vibrators are used to consolidate concrete in deep members such as columns and beams

A backhoe digs below its standing level, so it suits trenches and foundation pits

A dragline suits soft or wet ground where a long reach is needed

A grader is the machine chosen for bulk excavation of very large volumes of earth

Needle (immersion) vibrators are used to consolidate concrete in deep members such as columns and beams

What is the main limitation of a bar chart as a planning tool?

It does not show the interdependence of activities, so critical activities and float cannot be read

It cannot show how long each activity takes, because the bars carry no length scale at all

It cannot be drawn against a calendar, so the dates have to be written out beside every single bar

It needs three separate time estimates for every activity before any bar can be drawn

In an activity-on-arrow network, a dummy activity is used

to show a logical dependency or keep event numbers unique, using no time or resources

to represent an activity whose duration is very uncertain and so needs a time margin added

to mark which path is critical so that it stands out clearly in the network diagram

to show a logical dependency or keep event numbers unique, using no time or resources

to add a safety allowance of time at the end of the project to absorb delays

An activity has a total float of 6 days and a free float of 2 days. It is delayed by 4 days. Which statement is correct?

The project is not delayed, but a successor’s earliest start is pushed back by 2 days, using up its float

The project finishes 4 days late, because the delay is larger than the free float of the activity

The project is not delayed, but a successor’s earliest start is pushed back by 2 days, using up its float

Nothing else is affected at all, because the delay is smaller than the total float of 6 days

The project is delayed by 2 days, which is the part of the delay beyond the free float of 2 days

On the critical path, activity P has a cost slope of 800 per day with two days of crashing left, and activity Q has slope 500 but is already at its crash time. Activity R has slope 600 with one day left. Non-critical activity S has slope 200. To shorten the project by one day at least cost, crash

activity R, because it is the cheapest critical activity that can still be crashed

activity S, because it has the lowest cost slope of all the activities in the network

activity Q, because it has the lowest slope among the critical activities listed

activity P, because it has the most crashing time left on the critical path

activity R, because it is the cheapest critical activity that can still be crashed

Which of the following statements about floats in a CPM network are correct?

Every activity on the critical path has zero total float when the project is to finish at its earliest completion

The free float of an activity can never exceed its total float

Every activity on the critical path has zero total float when the project is to finish at its earliest completion

The free float of an activity can never exceed its total float

Delaying an activity by exactly its free float always delays the project completion date

Independent float is never less than free float

Which of the following statements about crashing and the time–cost trade-off are correct?

As a project is crashed, its direct cost rises

As a project is shortened, its indirect cost falls

The optimum duration is the one of least total cost, which may lie between the normal and the crash duration

As a project is crashed, its direct cost rises

As a project is shortened, its indirect cost falls

The optimum duration is the one of least total cost, which may lie between the normal and the crash duration

Crashing a non-critical activity with a low cost slope is the cheapest way to shorten the project

Which statement correctly contrasts PERT with CPM?

PERT uses three estimates to give the chance of meeting a date; CPM uses one duration and allows time–cost trade-off

PERT uses a single deterministic duration and is chiefly used for crashing a project, while CPM uses three time estimates

PERT is activity-oriented and suited to repetitive work, while CPM is event-oriented and suited to research projects

PERT uses three estimates to give the chance of meeting a date; CPM uses one duration and allows time–cost trade-off

PERT and CPM differ only in drawing style, since both use one duration and give identical outputs for any project

The probability of finishing a project by a given date, calculated in PERT from the critical path alone, tends to be

higher than the true probability, because other near-critical paths can also finish late

lower than the true probability, because the beta distribution always understates the likely time

exactly equal to the true probability whenever the three time estimates are unbiased

unrelated to the true probability, since PERT ignores the variance of the activities

Which of the following statements about PERT time estimates are correct?

The variance of an activity time is ((b − a) ÷ 6)²

The variance of a path is the sum of the variances of its activities, if they are independent

The expected time is the simple average (a + m + b) ÷ 3 of the three estimates

The variance of an activity time is ((b − a) ÷ 6)²

The variance of a path is the sum of the variances of its activities, if they are independent

The standard deviation of a path is the sum of the standard deviations of its activities

Construction Planning and Equipment; Project Management Techniques: Bar Charts, CPM, Crashing and PERT

This chapter answers the first half of Section B1.6 of Part B1 of the Architecture and Planning (AR) paper, “Project Management and Professional Practice”: “Construction planning and equipment; Construction project management techniques e.g. PERT, CPM”. It teaches how a construction job is planned and controlled, what the main earthmoving, concreting and hoisting machines do and how their output is worked out, the bar chart, milestone chart and network diagram, the CPM forward and backward pass with total, free, independent and interfering float, crashing and the time–cost trade-off, and PERT with its three-point estimates and the probability of finishing on time. The numericals are equipment output and fleet size, project duration and floats from a small network, crash cost slopes and net saving, expected time and variance, and probability from a stated normal value.

1. Construction planning: from scope to a controlled programme

Planning decides what is to be done, in what order, with which resources and by when; scheduling puts that plan on a calendar; control compares progress with the plan and corrects the difference. Planning moves from the whole to the part: fix the scope and the end date, break the work into a work breakdown structure (WBS) of packages and activities, estimate each activity’s duration and resources, decide the logical dependencies, draw the schedule, and then monitor it. The order of work is set by the technology (foundations before frame, frame before roof) and by the resources available (one crane cannot serve two tasks at once), not by the drawings alone.

Levels of programme
Level	Horizon	Detail	Used by
Master programme	Whole project, months to years	Major packages and milestones	Client and senior management
Stage or contract programme	A stage or package, weeks to months	Activities, durations, logic	Project manager, contractor
Short-term programme	One to three weeks	Gangs, equipment, deliveries by day	Site engineer, foreman

Resources are labour, materials, equipment, money and time, and a plan is workable only if each is available when needed. Site layout planning places access roads, storage, the crane, mixing or batching area, temporary offices and labour camps so that materials move the shortest distance and safely. Planning is done at three moments: before tendering (method statement and cost), before the contract starts (programme, procurement of long-lead items, resource plan) and during construction (the rolling short-term plan).

⚠️ Planning is not scheduling

A bar chart drawn on a calendar is a schedule, not a plan: it records dates, but not why the activities must follow one another. A plan with sound logic can always be rescheduled when a delay occurs; a schedule drawn without logic can only be redrawn.

2. Construction equipment: earthmoving, concreting and hoisting

Earthmoving and compaction machines in principle
Machine	Working principle	Best used for	Limit
Bulldozer	A blade pushes soil ahead of a tracked tractor; a ripper can tear hard ground	Clearing, short-distance pushing, spreading, rough grading	Uneconomic for long hauls
Scraper	Cuts, loads, hauls and spreads in a single machine	Large-volume earthwork over medium hauls	Needs a firm haul road; poor in rock
Grader	A long blade slung between axles shapes the surface	Fine grading, road-base shaping, side drains	Not a bulk-excavation machine
Backhoe	Bucket on an arm is pulled toward the machine and digs below its standing level	Trenches, foundation pits, basements	Limited reach and depth
Front shovel	Bucket is pushed away from the machine and digs above its standing level	Digging into a bank or face	Poor below its own level
Dragline	A bucket hung on cables is cast out and dragged back	Soft or wet ground, long reach, below the machine	Less accurate in trimming
Clamshell	A two-jaw grab is dropped vertically and closed	Deep narrow shafts and pits, underwater	Slow, low output
Sheepsfoot roller	Projecting feet knead the soil from the bottom of each layer upward	Cohesive clays	Unsuited to clean sands and gravels
Vibratory or smooth-wheel roller	Weight with vibration rearranges the grains into a dense state	Granular soils, sand, gravel, aggregate base	Less effective on clays

Concreting equipment follows the path of the material. A batching plant measures the ingredients; weigh batching is more accurate than volume batching. Mixers (drum or pan) blend them; a transit mixer carries ready-mixed concrete and keeps it agitated on the road. A concrete pump pushes the mix through a pipe to height or distance, which suits tall buildings and congested sites; a bucket and crane is the alternative. Vibrators consolidate the placed concrete: immersion (needle) vibrators for deep members such as columns, beams and walls, surface or screed vibrators for slabs, and form vibrators on formwork. Too little vibration leaves honeycombing; too much makes the mix segregate.

Hoisting equipment lifts materials vertically. A tower crane is fixed to a base or climbs with the building and serves a large circle; its rated load falls as the working radius grows, because the overturning moment (load × radius) is limited. A mobile crane (truck-mounted or crawler) is flexible and quick to set up, but needs firm level ground and outriggers. A material hoist or construction lift moves goods or workers vertically only. A derrick or gin pole suits light low-rise lifting. The choice depends on the heaviest load, the height, the radius, how often and how long the lift is needed, site access and whether hire is cheaper than purchase.

ℹ️ Match the machine to the task

Choose by function and constraint: depth and reach decide among backhoe, shovel, dragline and clamshell; soil type decides the roller; height and radius decide the crane. A mismatch, such as a sheepsfoot roller on sand or a mobile crane on soft ground, wastes money or causes accidents.

3. Equipment output: cycle time, swell and fleet balance

The hourly output of a digging machine is bucket capacity × fill factor × cycles per hour × job efficiency, where cycles per hour = 3600 ÷ cycle time in seconds. A 1.0 m³ bucket with fill factor 0.9, a 36-second cycle and efficiency 0.8 gives 1.0 × 0.9 × (3600 ÷ 36) × 0.8 = 72 m³ per hour of loose soil. The efficiency factor allows for delays such as waiting, repositioning and operator rest.

Soil has three volumes. Bank volume is in place, loose volume is after excavation (it swells) and compacted volume is after rolling (it is usually less than bank). Swell = (loose − bank) ÷ bank, and the load factor = bank ÷ loose = 1 ÷ (1 + swell). The 72 m³ per hour of loose soil at 25% swell is 72 ÷ 1.25 = 57.6 m³ per hour in bank measure. A 100 m³ bank excavation at 25% swell makes 125 m³ loose, so ten trucks of 10 m³ loose capacity carry only 80 m³ of bank in one round. Always state which volume a figure refers to.

Fleet balance. An excavator works well only if a truck is always waiting. The number of trucks = truck cycle time ÷ excavator loading time, rounded up. If loading takes 4 minutes and a truck’s cycle (loading 4, haul 6, dump 2, return 6) is 18 minutes, then 18 ÷ 4 = 4.5, so 5 trucks are needed. With 4 trucks the excavator idles; with 7 the trucks queue. The output of a chain of machines is set by its slowest link.

⚠️ Bank, loose and compacted are different volumes

Pay for excavation in bank measure, size trucks in loose measure, and fill embankments in compacted measure. Using loose capacity against a bank quantity without dividing by (1 + swell) makes the truck count too small.

4. Bar charts, milestone charts and network diagrams

The bar (Gantt) chart lists activities down the page and time across it, each activity a bar whose length is its duration. It is easy to draw, read and explain, which is why it remains the communication tool of every site. Its weakness is that it does not show which activity depends on which, which activities are critical, or how much float each has, and it becomes unreliable once the first delay forces a redraw. A milestone chart marks key events (foundation complete, roof cast, services tested, handover) as points in time; it shows what must be achieved by when, not the durations between, and suits reports to the client. A line of balance diagram suits repetitive work such as many identical houses, showing the rate of progress each trade must keep up to stay on schedule.

A network diagram shows logic. In an activity-on-arrow (AOA, arrow diagram) network the arrows are activities and the circles are events (instants), numbered so that the tail number is smaller than the head number. In an activity-on-node (AON, precedence) network the nodes are activities and the arrows only show dependency; it needs no dummies and can carry the relations finish-to-start, start-to-start, finish-to-finish and start-to-finish, each with a lag. The rules: one start and one end, no loops, no dangling activities, and each activity identified by a unique pair of events.

The planning tools compared
Tool	Shows well	Does not show	Best for
Bar chart	Durations and overlaps on a calendar	Dependencies, criticality, float	Communicating a programme
Milestone chart	Key dates to be met	Durations and logic between events	Reports to the client and management
Network (AOA or AON)	Logic, critical path, float	Calendar view unless converted to a bar chart	Planning and controlling complex work

🧠 When an AOA network needs a dummy

A dummy (a dashed arrow of zero duration and no resources) is needed in two cases. First, when two parallel activities would start and end at the same pair of events, so they would share one number pair; a dummy gives one of them a separate end event. Second, when activity C depends on A and B but D depends on B only; a dummy from the end of B to the start of C carries the dependency of C on B, while D, which starts at the end of B, is not held up by A.

5. CPM: forward pass, backward pass, critical path and the four floats

The critical path method (CPM) gives each activity one fixed duration. In the forward pass the earliest start of an activity is the largest earliest finish among its predecessors, and ES + duration = EF; the largest EF is the project duration. In the backward pass, starting from the project duration, the latest finish is the smallest latest start among the successors, and LF − duration = LS. Take six activities: A (3 days, no predecessor), B (4, after A), C (2, after A), D (5, after B), E (6, after C) and F (2, after D and E). The forward pass gives a 14-day project, and the critical path is A–B–D–F, the longest path, with every activity on it at zero float. In arrow form it is events 1 to 6 with A = 1→2, B = 2→3, C = 2→4, D = 3→5, E = 4→5 and F = 5→6, and no dummy is needed.

Times and floats of the six-activity network (days)
Activity	Duration	ES	EF	LS	LF	Total float	Free float
A	3	0	3	0	3	0	0
B	4	3	7	3	7	0	0
C	2	3	5	4	6	1	0
D	5	7	12	7	12	0	0
E	6	5	11	6	12	1	1
F	2	12	14	12	14	0	0

Total float = LS − ES = LF − EF (in an arrow network, latest time of the head event − earliest time of the tail event − duration). It is the delay an activity can suffer without delaying the project, and it is shared with the other activities on its chain. Here C and E each show 1 day, but it is the same single day of the chain C–E.
Free float = earliest start of the earliest successor − EF (E of head event − E of tail event − duration). It is the delay that leaves every successor’s earliest start unchanged. C has total float 1 but free float 0, since E starts the moment C ends; E has free float 1.
Independent float = earliest time of the head event − latest time of the tail event − duration, taken as zero if negative. It is the delay available even if every predecessor finishes as late as possible and every successor starts as early as possible, so it is the most conservative float. Here it is 0 for both C and E.
Interfering float = total float − free float = the slack of the head event. C has 1 − 0 = 1 day of interfering float: using it does not delay the project but does eat into the float of E. The ordering is always independent ≤ free ≤ total.

⚠️ Float belongs to the chain, not to one activity

If C uses its 1 day of total float, E loses its float and becomes critical. Only free float can be consumed without touching any other activity. A path with all total floats equal to zero is critical, and more than one critical path is possible; a path with a total float of one or two days is near-critical and needs watching.

6. Crashing and the time–cost trade-off

Direct cost (labour, materials, equipment) of an activity rises when it is shortened below its normal time, by overtime, extra gangs or faster methods; indirect cost (site overheads, supervision, financing, penalties) falls as the project gets shorter. Their sum, the total cost, is U-shaped, so there is an optimum duration of least total cost. Each activity has a normal time and cost and a crash time (the shortest possible) and cost. The cost slope = (crash cost − normal cost) ÷ (normal time − crash time) is the price of one day saved. Only critical activities are crashed, because shortening a non-critical one saves no time; choose the one with the least cost slope that can still be crashed; and when a second path becomes critical, shorten both paths together, through a common activity or one activity on each.

Crash data for the six-activity network (cost units; time in days)
Activity	Normal time	Normal cost	Crash time	Crash cost	Cost slope per day
A	3	3000	2	3500	500
B	4	4000	3	4300	300
C	2	2000	1	2400	400
D	5	5000	3	5800	400
E	6	6000	4	6900	450
F	2	2500	2	2500	cannot be crashed

Crashing step by step, with indirect cost 450 per day
Duration (days)	Step taken	Direct cost	Indirect cost	Total cost
14	Normal; critical path A–B–D–F	22500	6300	28800
13	Crash B by 1 day (slope 300); A–C–E–F is now also critical	22800	5850	28650
12	Crash A by 1 day (500), which shortens both paths; cheaper than D + C at 800	23300	5400	28700
11	Crash D and C by 1 day each (400 + 400)	24100	4950	29050
10	Crash D by a second day and E by 1 day (400 + 450); the limit	24950	4500	29450

Total cost falls from 28,800 to 28,650 at 13 days, then rises, so the optimum is 13 days. The rule behind it: keep crashing while the cost slope is less than the indirect saving per day. B (300) is below 450 and pays; A (500) is above 450 and does not. At 10 days the critical path A–B–D–F has every activity at its crash time, so no further reduction is possible whatever the budget.

7. PERT: three-point estimates, variance and the probability of completion

PERT (programme evaluation and review technique) treats activity times as uncertain. Each activity gets three estimates: optimistic a (everything goes well), most likely m and pessimistic b (serious but not catastrophic trouble). The time is assumed to follow a beta distribution, giving the expected time te = (a + 4m + b) ÷ 6, standard deviation σ = (b − a) ÷ 6 and variance σ² = ((b − a) ÷ 6)². The critical path is found with the te values; the expected project duration Te is the sum of te along it, and, treating activities as independent, the project variance is the sum of the variances of the critical activities (the standard deviations are not added).

A three-activity critical path (days)
Activity	a	m	b	te = (a + 4m + b) ÷ 6	Variance ((b − a) ÷ 6)²
X	2	5	8	5	1
Y	4	7	16	8	4
Z	3	6	15	7	4

Here Te = 5 + 8 + 7 = 20 days, the project variance is 1 + 4 + 4 = 9, so σ = 3 days. For a target date Ts the standard normal deviate is Z = (Ts − Te) ÷ σ, and the probability of finishing by Ts is Φ(Z) from the normal table. For Ts = 23 days, Z = 1 and the probability is Φ(1) = 0.8413. For Ts = 17, Z = −1 and the probability is 0.1587. For Ts = 20 it is exactly 0.50. To be 95% sure, Z = 1.645, so Ts = 20 + 1.645 × 3 ≈ 24.9 days. The normal curve is justified because the sum of many independent activity times tends to be normal.

PERT and CPM compared
Feature	PERT	CPM
Time estimate	Three estimates, probabilistic	One estimate, deterministic
Orientation	Event-oriented	Activity-oriented
Suited to	Uncertain, non-repetitive work such as research and development	Familiar construction work with well-known durations
Typical output	Probability of meeting a date	Time–cost trade-off and crashing

⚠️ PERT probabilities are optimistic

The calculation uses the variance of the critical path alone, but a near-critical path with a large variance can also finish late. The true chance that the whole project meets the date is therefore lower than Φ(Z) of the critical path. Also note that te is not the most likely time m: when b − m exceeds m − a the distribution is skewed to the right and te is larger than m.

Key takeaways

Planning decides what, in what order and with what; scheduling dates it; control compares and corrects. Programmes run from master to stage to short-term, built on a WBS and on the logic set by technology and resources.
Match machine to task: backhoe digs below, front shovel above, dragline and clamshell for reach or depth, sheepsfoot for clay, vibratory for granular soil; a tower crane’s capacity falls as radius grows.
Output = capacity × fill × (3600 ÷ cycle s) × efficiency; bank = loose ÷ (1 + swell); trucks = truck cycle ÷ loading time, rounded up.
CPM: forward pass ES and EF, backward pass LS and LF; total float = LS − ES, free float = earliest successor ES − EF, interfering = total − free, and independent ≤ free ≤ total; the critical path is the longest path with zero total float.
Crashing: cost slope = (crash cost − normal cost) ÷ (normal time − crash time); crash critical activities by least slope, shorten parallel critical paths together, and stop when the slope exceeds the indirect saving per day.
PERT: te = (a + 4m + b) ÷ 6, σ² = ((b − a) ÷ 6)², path variance is the sum of variances, Z = (Ts − Te) ÷ σ and the probability is Φ(Z), which is optimistic because near-critical paths are ignored.

Practice questions (20)

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.

Which machine, with its bucket carried on a rigid arm, digs below its own standing level and is the usual choice for trenches and foundation pits?
1. Front shovel, whose bucket is pushed away from the machine to cut into a bank
2. Backhoe, whose bucket is drawn back toward the machine below ground level
3. Bulldozer, whose blade pushes loose soil ahead of the tractor
4. Scraper, which cuts, hauls and spreads the soil in a single pass
Show answer
Answer: B — Backhoe, whose bucket is drawn back toward the machine below ground level
The backhoe digs below its standing level by drawing the bucket back toward the machine, which makes it the standard trench and foundation machine. The front shovel is the tempting answer because it is also an excavator, but it digs above its standing level. A bulldozer and a scraper are earthmoving, not trenching, machines.
A road job needs a clean sand-and-gravel sub-base compacted and, elsewhere, a layer of cohesive clay. Which pairing of roller and soil is most appropriate?
1. A sheepsfoot roller on both the sand-and-gravel and the clay
2. A vibratory roller on both the sand-and-gravel and the clay
3. A sheepsfoot roller on the sand-and-gravel and a vibratory roller on the clay
4. A vibratory roller on the sand-and-gravel and a sheepsfoot roller on the clay
Show answer
Answer: D — A vibratory roller on the sand-and-gravel and a sheepsfoot roller on the clay
Vibration rearranges loose grains into a dense state, which is why vibratory rollers suit clean sand and gravel; the kneading action of the projecting feet of a sheepsfoot roller suits cohesive clay, where vibration does little. The reverse pairing is the tempting slip, and using one roller for both soils wastes effort on one of them.
A hydraulic excavator has a bucket of 1.5 m³, a fill factor of 0.8, a cycle time of 30 s and a job efficiency of 0.75. The soil swells by 20% on excavation. Its hourly output in bank cubic metres is ____.
Numerical answer — type the value.
Show answer
Answer: 90
Cycles per hour = 3600 ÷ 30 = 120. Loose output = 1.5 × 0.8 × 120 × 0.75 = 108 m³ per hour. Bank output = loose ÷ (1 + swell) = 108 ÷ 1.20 = 90 m³ per hour. Quoting 108 is the usual slip: it is the loose volume, not the bank volume. Multiplying by 1.20 instead of dividing gives 129.6, which is wrong in the opposite direction.
An excavator loads one truck in 3 minutes. A truck’s full cycle (loading, haul, dumping and return) is 21 minutes. The number of trucks needed to keep the excavator working without waiting is ____.
Numerical answer — type the value.
Show answer
Answer: 7
Trucks needed = truck cycle time ÷ excavator loading time = 21 ÷ 3 = 7. With 6 trucks the excavator would stand idle for part of every cycle, and more than 7 would only make trucks queue. Dividing by the cycle without including the loading time inside it, for example 21 ÷ (3 + 21), confuses the cycle with the loading time.
A 100 m³ (bank) excavation is in soil that swells 25% on digging. Each truck carries 10 m³ of loose soil. The minimum number of truck loads needed to carry it all away is
1. 8, found by dividing 100 m³ by 1.25 and then by the 10 m³ capacity of a truck
2. 10, found by dividing the 100 m³ directly by the 10 m³ truck capacity and ignoring swell
3. 12, found by rounding down the loose volume of 125 m³ divided by the truck capacity
4. 13, because the loose volume is 125 m³ and 12.5 loads must be rounded up
Show answer
Answer: D — 13, because the loose volume is 125 m³ and 12.5 loads must be rounded up
Loose volume = bank × (1 + swell) = 100 × 1.25 = 125 m³. Loads = 125 ÷ 10 = 12.5, and since a partial load still needs a trip the count is rounded up to 13. The figure of 8 comes from dividing by the swell instead of multiplying, 10 ignores swell altogether, and 12 would leave 5 m³ behind.
As the working radius of a tower crane is increased, its safe lifting capacity
1. rises, because a longer jib spreads the weight over a larger area of the slab
2. falls, because the overturning moment, load multiplied by radius, is limited
3. stays constant, because the counterweight is moved to compensate for the radius exactly
4. rises until the jib is horizontal and then falls as the hook descends toward the ground
Show answer
Answer: B — falls, because the overturning moment, load multiplied by radius, is limited
A crane is limited by stability and structure: the overturning moment (load × radius) cannot exceed a fixed value, so the permitted load is largest at the smallest radius and falls as the hook moves out, which is why every tower crane carries a load–radius chart. A longer jib does not spread the weight; it lengthens the lever arm.
Which of the following statements about construction equipment are correct?
1. A backhoe digs below its standing level, so it suits trenches and foundation pits
2. A dragline suits soft or wet ground where a long reach is needed
3. A grader is the machine chosen for bulk excavation of very large volumes of earth
4. Needle (immersion) vibrators are used to consolidate concrete in deep members such as columns and beams
Show answer
Answer: A — A backhoe digs below its standing level, so it suits trenches and foundation pits; B — A dragline suits soft or wet ground where a long reach is needed; D — Needle (immersion) vibrators are used to consolidate concrete in deep members such as columns and beams
The backhoe, the dragline and the needle vibrator are correctly matched to trenching, soft or wet ground with long reach, and deep concrete members. A grader carries a long blade for fine grading and shaping road bases, not for bulk excavation, which is the tempting overstatement because it is a major earthmoving machine.
What is the main limitation of a bar chart as a planning tool?
1. It does not show the interdependence of activities, so critical activities and float cannot be read
2. It cannot show how long each activity takes, because the bars carry no length scale at all
3. It cannot be drawn against a calendar, so the dates have to be written out beside every single bar
4. It needs three separate time estimates for every activity before any bar can be drawn
Show answer
Answer: A — It does not show the interdependence of activities, so critical activities and float cannot be read
In a bar chart the length of a bar is the duration and the horizontal axis is the calendar, so durations and dates are shown clearly. What is missing is the logic: which activity must precede which, which activities are critical and how much float exists. Three time estimates belong to PERT, not to the bar chart.
In an activity-on-arrow network, a dummy activity is used
1. to represent an activity whose duration is very uncertain and so needs a time margin added
2. to mark which path is critical so that it stands out clearly in the network diagram
3. to show a logical dependency or keep event numbers unique, using no time or resources
4. to add a safety allowance of time at the end of the project to absorb delays
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Answer: C — to show a logical dependency or keep event numbers unique, using no time or resources
A dummy has zero duration and consumes no resources; it exists only to carry a dependency or to give parallel activities different end events so that each has a unique pair of numbers. It is not a margin for uncertainty, which is a different idea, and it does not mark the critical path.
A project has these activities: A takes 2 days and has no predecessor; B takes 5 days, after A; C takes 3 days, after A; D takes 4 days, after B; E takes 6 days, after B and C; F takes 3 days, after D and E. The total float of activity C, in days, is ____.
Numerical answer — type the value.
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Answer: 2
Forward pass: A 0–2, B 2–7, C 2–5, D 7–11, E starts at the larger of 7 and 5, so 7–13, F starts at the larger of 11 and 13, so 13–16. Backward pass: F 13–16, E has latest finish 13 and latest start 7, so C has latest finish 7 and latest start 4. Total float of C = LS − ES = 4 − 2 = 2 days. The critical path is A–B–E–F of 16 days; taking the start of E minus the start of C, 7 − 2 = 5, forgets to subtract C’s own duration of 3 days; that is the usual slip.
An activity has a total float of 6 days and a free float of 2 days. It is delayed by 4 days. Which statement is correct?
1. The project finishes 4 days late, because the delay is larger than the free float of the activity
2. The project is not delayed, but a successor’s earliest start is pushed back by 2 days, using up its float
3. Nothing else is affected at all, because the delay is smaller than the total float of 6 days
4. The project is delayed by 2 days, which is the part of the delay beyond the free float of 2 days
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Answer: B — The project is not delayed, but a successor’s earliest start is pushed back by 2 days, using up its float
A delay up to the total float (6 days) cannot delay the project, so the project date is safe. But only the first 2 days (the free float) leave all successors untouched; the remaining 2 days come out of the interfering float (total minus free, 6 − 2 = 4 days), pushing some successor’s earliest start back and consuming its float. The claim that nothing else is affected ignores this, and a project delay of 2 days would need a delay beyond the total float.
On the critical path, activity P has a cost slope of 800 per day with two days of crashing left, and activity Q has slope 500 but is already at its crash time. Activity R has slope 600 with one day left. Non-critical activity S has slope 200. To shorten the project by one day at least cost, crash
1. activity S, because it has the lowest cost slope of all the activities in the network
2. activity Q, because it has the lowest slope among the critical activities listed
3. activity P, because it has the most crashing time left on the critical path
4. activity R, because it is the cheapest critical activity that can still be crashed
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Answer: D — activity R, because it is the cheapest critical activity that can still be crashed
Only critical activities shorten the project, and only those that still have crash time left can be crashed. Of P (800) and R (600), R is cheaper. S is cheap but non-critical, so crashing it saves no project time, and Q is already at its crash time, so its low slope is irrelevant.
A critical activity normally takes 10 days at ₹ 50,000 and can be crashed to 7 days at ₹ 65,000. The project’s indirect cost is ₹ 6,000 per day. The activity is crashed by 2 days and no other path becomes critical. The net saving in total project cost, in ₹, is ____.
Numerical answer — type the value.
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Answer: 2000
Cost slope = (65,000 − 50,000) ÷ (10 − 7) = 5,000 per day. Crashing 2 days costs 2 × 5,000 = 10,000 extra direct cost. Two days shorter saves 2 × 6,000 = 12,000 of indirect cost. Net saving = 12,000 − 10,000 = 2,000. Quoting 12,000 forgets the added direct cost, and taking the slope as 15,000 ÷ 10 uses the wrong denominator.
Which of the following statements about floats in a CPM network are correct?
1. Every activity on the critical path has zero total float when the project is to finish at its earliest completion
2. The free float of an activity can never exceed its total float
3. Delaying an activity by exactly its free float always delays the project completion date
4. Independent float is never less than free float
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Answer: A — Every activity on the critical path has zero total float when the project is to finish at its earliest completion; B — The free float of an activity can never exceed its total float
With the project finishing at its earliest date, critical activities have no total float, and free float is the part of total float that can be used without touching any successor, so it cannot exceed it. Using exactly the free float disturbs no successor, let alone the project. The ordering is independent ≤ free ≤ total, so independent float is never greater than free float.
Which of the following statements about crashing and the time–cost trade-off are correct?
1. As a project is crashed, its direct cost rises
2. As a project is shortened, its indirect cost falls
3. The optimum duration is the one of least total cost, which may lie between the normal and the crash duration
4. Crashing a non-critical activity with a low cost slope is the cheapest way to shorten the project
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Answer: A — As a project is crashed, its direct cost rises; B — As a project is shortened, its indirect cost falls; C — The optimum duration is the one of least total cost, which may lie between the normal and the crash duration
Shortening raises direct cost and lowers indirect cost, so the total cost curve has a minimum, the optimum duration, which usually lies between the normal and crash durations. Crashing a non-critical activity does not shorten the project at all, so spending on it is pure waste however low its slope.
An activity has an optimistic time of 4 days, a most likely time of 9 days and a pessimistic time of 20 days. Its PERT expected time, in days, is ____.
Numerical answer — type the value.
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Answer: 10
te = (a + 4m + b) ÷ 6 = (4 + 4 × 9 + 20) ÷ 6 = 60 ÷ 6 = 10 days. It exceeds the most likely time of 9 days because the pessimistic tail (20 − 9 = 11) is longer than the optimistic one (9 − 4 = 5). The simple mean (4 + 9 + 20) ÷ 3 = 11 is the tempting slip, since PERT weights the most likely value four times.
The critical path of a project has an expected duration of 50 days and a variance of 25 days². Given Φ(1.20) = 0.8849, the probability, in per cent correct to one decimal place, that the project is completed within 56 days is ____.
Numerical answer — type the value.
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Answer: 88.5
σ = √25 = 5 days. Z = (Ts − Te) ÷ σ = (56 − 50) ÷ 5 = 1.20. The probability is Φ(1.20) = 0.8849 = 88.5%. Dividing by the variance, 6 ÷ 25 = 0.24, instead of the standard deviation is the common slip, and quoting Z itself, 1.2, forgets to read the probability from the table.
Which statement correctly contrasts PERT with CPM?
1. PERT uses a single deterministic duration and is chiefly used for crashing a project, while CPM uses three time estimates
2. PERT is activity-oriented and suited to repetitive work, while CPM is event-oriented and suited to research projects
3. PERT uses three estimates to give the chance of meeting a date; CPM uses one duration and allows time–cost trade-off
4. PERT and CPM differ only in drawing style, since both use one duration and give identical outputs for any project
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Answer: C — PERT uses three estimates to give the chance of meeting a date; CPM uses one duration and allows time–cost trade-off
PERT is probabilistic: three estimates per activity, so it answers how likely a date is, and it suits uncertain work. CPM is deterministic: one duration, so it supports crashing and the time–cost trade-off, and suits familiar construction work. The claim that PERT uses a single duration simply swaps the two methods, which is the tempting slip; they also differ in orientation, PERT being event-oriented.
The probability of finishing a project by a given date, calculated in PERT from the critical path alone, tends to be
1. higher than the true probability, because other near-critical paths can also finish late
2. lower than the true probability, because the beta distribution always understates the likely time
3. exactly equal to the true probability whenever the three time estimates are unbiased
4. unrelated to the true probability, since PERT ignores the variance of the activities
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Answer: A — higher than the true probability, because other near-critical paths can also finish late
The project is late if any path is late. PERT looks only at the critical path, so it counts the chance of that one path finishing in time and ignores the chance that a near-critical path with large variance overruns, which makes the stated probability optimistic. Unbiased estimates do not remove this, and PERT very much uses variance.
Which of the following statements about PERT time estimates are correct?
1. The expected time is the simple average (a + m + b) ÷ 3 of the three estimates
2. The variance of an activity time is ((b − a) ÷ 6)²
3. The variance of a path is the sum of the variances of its activities, if they are independent
4. The standard deviation of a path is the sum of the standard deviations of its activities
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Answer: B — The variance of an activity time is ((b − a) ÷ 6)²; C — The variance of a path is the sum of the variances of its activities, if they are independent
Activity variance is ((b − a) ÷ 6)², and for independent activities variances add along the path. Standard deviations do not add; the path σ is the square root of the summed variance, so for variances 1, 4 and 4 it is 3, not 1 + 2 + 2 = 5. The expected time weights the most likely value four times, (a + 4m + b) ÷ 6, not equally.