Time Management

Master and milestone schedules, CPM network logic and calculations, float, schedule updates and revisions, contractor schedule review, recovery schedules, delays and time impact analysis.

5 minutes 5 Questions

Time Management is one of the core knowledge domains in the Certified Construction Manager (CCM) body of knowledge, as defined by the Construction Management Association of America (CMAA). It encompasses the processes, tools, and techniques a construction manager uses to plan, monitor, and control …

Concepts covered
Time Impact Analysis and FragnetsCPM Network Logic and RelationshipsContractor Baseline Schedule ReviewProgress Measurement and Percent CompleteCritical Path DeterminationMaster and Milestone SchedulesDesign Schedule and Long-Lead ProcurementSchedule Updates Versus Schedule RevisionsCost- and Resource-Loaded SchedulesTotal Float and Free FloatForward and Backward Pass CalculationsRecovery Schedules and AccelerationExcusable, Compensable and Concurrent DelaysScheduling System Selection and Level of DetailFloat Ownership
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CMAA CCM - Time Management Example Questions

Test your knowledge of Time Management

Question 1

Under a contract clause that assigns total float exclusively to the contractor, what is the usual effect of an owner-caused delay to a noncritical activity?

Correct Answer: The contractor can seek relief for float consumed by the owner's delay

When the contract assigns total float exclusively to the contractor, float becomes a contractor-owned resource. An owner-caused delay that consumes float on a noncritical activity takes something that belongs to the contractor, so the contractor can seek compensation or time relief for the float lost.
The claim that the delay uses float first, leaving no relief, describes the common "float belongs to the project" or first-come rule, not exclusive contractor ownership. The idea that the owner can use float because it controls the contract date contradicts the clause entirely. Pro rata sharing describes a shared-float clause, not exclusive assignment.
Key concept: contractual float ownership and its effect on delay entitlement.

Question 2

A university laboratory building contract makes total float a shared project resource, available to whichever party needs it first. An earlier owner-approved change used 12 of the 15 days of float on the fume hood exhaust path. The latest update shows that path at -8 days. The cause is the general contractor's late ductwork fabrication. The contractor argues that the owner spent the float and requests an 8-day extension. What should the construction manager recommend?

Correct Answer: Deny the extension and require a contractor recovery schedule

Under a shared-float clause, the owner's change properly used 12 days of float, leaving 3 days. The contractor's late ductwork fabrication consumed those 3 days and caused the 8-day overrun, so this is contractor-caused delay with no entitlement to time, and the contractor must submit a recovery schedule.
Granting 8 days wrongly treats the owner's legitimate, first-come use of shared float as a delay. Granting 12 days restores float that was never reserved for the contractor and was correctly spent. Splitting the delay assumes concurrency, but the only critical delay is the contractor's.
Key concept: float ownership under a shared-float, first-come clause and excusable versus non-excusable delay.

Question 3

During pre-design for a fire station, the construction manager checks a draft network. All links are finish-to-start, and the first activity starts at day 0. A (5 days) has no predecessor. B (8 days) and C (6 days) each follow A. E (4 days) follows both B and C, and its finish is the project finish. The owner then adds an interim milestone: B must finish by day 12 so the apparatus bay can be released early. The project finish stays where the network places it. What are the total floats of A and C?

Correct Answer: A: −1 day; C: 2 days

The forward pass gives A at 0 to 5, B at 5 to 13, C at 5 to 11 and E at 13 to 17. On the backward pass, B's late finish is capped at day 12, so B's late start is 4. That makes A's late finish 4 and its total float negative one day. C is still driven by E's late start of 13, so C's total float is 13 minus 11, or 2 days.
A: 0 days; C: 2 days ignores the imposed milestone. C: 1 day wrongly carries B's negative float onto C, which has its own path. A: −4 days miscalculates the backward pass through B.
Key concept: imposed constraints creating negative float on the backward pass.

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