Appendix B — CPM Scheduling Workbook

Forward pass — ES = max(EF of predecessors), EF = ES + Dur.

Project duration = 19 work days.

Backward pass — LF = min(LS of successors), LS = LF − Dur. Set LF(E) = 19.

LS(A) = 0. The pass checks out.

Float

Critical path: A → C → D → E = 4 + 5 + 6 + 4 = 19 WD ✓

What this problem was teaching. The merge at D is the whole exercise. D waits for C at day 9, not for B at day 7, because a merging activity waits for its slowest predecessor. If you took an average, or the first predecessor you wrote down, you would have produced a project duration of 17 and been wrong in the direction that always favors you. That is how optimistic baselines get built without anybody lying.

Second point, quietly: erosion control is on the critical path and the fence is not. New project engineers assume the fence goes first because it is the visible thing. The stormwater controls have to be in before you disturb the ground, and on a site disturbing an acre or more that is a regulatory requirement, not a preference.

Forward pass

Project duration = 34 work days.

Backward pass, LF(G) = 34

Float

Critical path: A → B → D → E → G = 3 + 6 + 10 + 8 + 7 = 34 WD ✓

                               5   10   15   20   25   30
                               |    |    |    |    |    |
A  Mobilize                ###
B  Excavate footings          ######
C  Relocate 8" water main     ========..........
D  Form/place footings              ##########
E  Foundation walls / stem                    ########
F  Under-slab MEP                             ======..
G  Slab on grade                                      #######

   ###  critical (TF = 0)     ===  has float     ...  total float

Is it safe to start the water main late? On the arithmetic, yes — 10 days of total float, 8 of them free. On the job, no, and this is the judgment the calculation cannot make for you. A live 8-inch main relocation involves the water utility's schedule, a shutdown window, pressure testing, disinfection, and bacteriological sampling with a laboratory turnaround measured in days. None of that is inside your control, and none of it is in the 8-day duration. A chain whose float depends on a third party you cannot accelerate is a chain you manage as critical regardless of what the float column says.

What this problem was teaching. Two things. First, the double merge — D takes the max of two predecessors on the backward pass just as it takes the max on the forward pass, and the answer at A comes from the minimum of B and C's late starts. Second, that a float number is a fact about the network, not a permission slip.

Forward pass

Project duration = 30 work days.

Backward pass, LF(H) = 30

Float

Critical path: A → B → C → F → H = 5 + 6 + 8 + 5 + 6 = 30 WD ✓

The activity to watch is E — sheet-pile fabrication and delivery — and it carries one day of float. One day. It feeds both pile-driving operations, so a slip in E does not delay one chain, it delays both, and the second day of slip lands directly on the project. And unlike the abutment excavation, nobody on Del's crew can accelerate a mill. You cannot add a Saturday to a rolling schedule.

This is the Northgate steel lesson in a different pair of boots: "it's not on the critical path" is one of the most expensive sentences in construction when the chain ends at a fabricator.

What this problem was teaching. Read D and E together. Both have small total float and zero free float — meaning if either one slips at all, it pushes a successor's early start and eats the float of the activity behind it. G is the only activity holding genuine free float (2 days), and it holds it only because it is last in line before the merge. That distinction is Problem 4.

Forward and backward pass

Project duration = 50 work days.

Float

Critical path: A → D → E → F → G = 10 + 14 + 8 + 10 + 8 = 50 WD ✓

                               5   10   15   20   25   30   35   40   45   50
                               |    |    |    |    |    |    |    |    |    |
A  Dry-in north elevation  ##########
B  Set curtain wall, north           ============.....
C  Perimeter fire-safing                         =====.....
D  Overhead MEP rough, L2            ##############
E  In-wall rough-in, L2                            ########
F  Hang board, L2                                          ##########
G  Tape, finish, prime                                               ########

   ###  critical (TF = 0)     ===  has float     ...  total float

The answer. C's five days are free. B's five days are not.

• C has free float of 5. Perimeter fire-safing can run five days long and nothing moves — F is waiting on the in-wall rough-in anyway. Sofia's request costs the project nothing at all.
• B has free float of zero and interfering float of 5. If the curtain-wall subcontractor takes five extra days, the project still finishes on day 50 — but C's early start moves from 22 to 27, and C's free float goes to zero. The curtain-wall sub did not delay the project. It spent Sofia's room, and Sofia was not in the meeting.

Look at the dots on the Gantt chart. B's dots and C's dots are drawn separately and they are the same five days shown twice. If you have ever looked at a bar chart and added up the float column, this is the error you were making. Float belongs to the path, not to the activity.

Note also that C's independent float is zero. The five days C appears to own evaporate the moment B is late. C does not own them. C is merely standing where they are stored.

What this problem was teaching. Total float protects the project. Free float protects your successor. An activity with total float and no free float is spending somebody else's room, and the person losing it is never in the room when the extension is granted. Publish free float, not just total float, and grant extensions from the free float column.

Forward and backward pass

Project duration = 64 work days.

Float

Critical path: A → B → C → E → F = 5 + 15 + 30 + 8 + 6 = 64 WD ✓

Independent float for D, worked:

   Independent float = max(0, smallest ES of successors
                              − largest LF of predecessors
                              − duration)
                     = max(0, ES(F) 58 − LF(B) 20 − 10)
                     = 28 work days

Why D's float is different in kind. In Problem 4, activity C had five days of total float and zero independent float — its slack could be consumed upstream by B before C ever got a vote. D's 28 days cannot be taken from it by anybody: its predecessor B is critical, so B cannot arrive late, and its successor F is driven by a different chain (the mass-fill work through C and E), so D cannot hurt anyone downstream. Twenty-eight days that are genuinely D's own.

What this problem was teaching. Independent float is the answer to a specific and common question in a delay argument: "how much room did that subcontractor have that nobody could have taken away?" It is usually zero in a tightly chained construction network, which is exactly why it is worth computing when it is not — it is the strongest form of the "you had room and you used it" argument, and it is the only float number an opposing scheduler cannot chip away at by re-arguing upstream logic.

Practical note: D's 28 days do not mean the duct bank should be scheduled late. Utility companies run on their own calendar, and the transformer set requires an inspection, a utility crew, and usually a scheduled outage window. Twenty-eight days of independent float is exactly the buffer you want in front of an activity you do not control.

Forward pass

Project duration = 58 work days.

Backward pass, LF(I) = 58

Float

Critical path: A → D → E → F → I = 4 + 6 + 15 + 25 + 8 = 58 WD ✓

What do you tell Colton? "Those two days are not yours. They belong to the CMU wall, the deck, and the roof — all three of you are on one chain with two days between you, and whoever gets there first spends them." The chain C → G → H has two days of total float, and all of it shows up as free float on H because H is last in line. If the CMU walls run two days long, C's slip pushes G, which pushes H, and H's free float is gone before Colton ever picks up a hammer.

What this problem was teaching. Three things worth carrying out.

First: the critical path runs entirely through paper. Submit → review → fabricate → deliver. Forty-six of the fifty-eight days involve nobody swinging a hammer, and Willow Street's field superintendent cannot accelerate any of it. This is why procurement is its own branch of the work breakdown structure and why a schedule that starts at "mobilize" cannot see the delay coming.

Second: B has ten days of total float and only eight of free float. Steel can slip eight days freely; the ninth and tenth days push G and eat the CMU chain's room.

Third: two days is not a buffer. The entire structural chain of a $6.8M building is protected by two work days — about three calendar days, worth 3 × $2,800 = $8,400 on Willow Street's total daily exposure of $1,600 extended general conditions plus $1,200 liquidated damages. Report that number in the owner-architect-contractor meeting before it is gone, not after.

Forward and backward pass

Project duration = 51 work days.

Float

Every path, ranked

                               5   10   15   20   25   30   35   40   45   50
                               |    |    |    |    |    |    |    |    |    |
A  HVAC duct mains, L3     ############
B  Sprinkler mains, L3                 ===========..
C  Plumbing / med gas, L3              =========...............
D  Conduit / cable tray                           ===========..
E  Frame partitions, L3                ##############
F  In-wall rough-in, L3                              ##########
G  Inspect + close walls                                       ######
H  Ceiling grid and tile                                             #########

   ###  critical (TF = 0)     ===  has float     ...  total float

Which goes on the two-week look-ahead? Both the critical path and the sprinkler/conduit chain. Kestrel's house rule — a company rule, not an industry standard — is that any path with total float of 10 work days or less is managed as critical: it goes on the two-week look-ahead, it goes in the OAC report, and it gets a named owner.

The reason is not caution for its own sake. The critical path is being watched by everybody. The chain with two days of float is being watched by nobody, and it takes two days to become critical. On Northgate that rule applied to 68 activities at zero float and another 106 at ten days or less, and Wei Chen reported both numbers every month.

What this problem was teaching. A float column is a ranking, not a pass/fail test. Sort by total float and manage from the top down until the number gets large enough that you stop caring — and decide in advance, in writing, what that number is. Note also that the plumbing and medical-gas chain at 15 days of float is genuinely comfortable and is the trade with the longest inspection lead time on a healthcare project. Float is not the only reason to watch something.

Forward and backward pass

Project duration = 54 work days.

Float

All four paths

Can Colton let the siding drift? On this house, yes — eleven days of float, all of it free and all of it independent, so nobody upstream can take it and nobody downstream feels it.

Across eleven houses, absolutely not, and this is the reason Chapter 14 says CPM is the wrong tool for Harbor Ridge. Colton's constraint is not the logic inside one house. It is the crew flowing through the lots: one framing crew moving at one house every 4.5 days, one drywall crew at one every 4.0 days, one siding crew that has to be somewhere every day of the year. A CPM network of 34 identical houses is 1,394 activities of noise. What Colton needs is a line-of-balance view — production-rate lines plotted against lot number — because the problem he actually has is that the faster crew catches the slower one and runs out of work around lot 22.

What this problem was teaching. Two things. First, that a network with a single obvious "main line" usually has three or four real paths, and you should list them all with their lengths before you form an opinion. The near-critical path here is one day away — dry-in versus windows, which the framing foreman treats as interchangeable and the calculation does not.

Second, that the right schedule type is the one that shows you the problem you actually have. The CPM here is arithmetically perfect and managerially useless to Colton, which is a good thing to learn from a problem rather than from a job.

(a) Forward and backward pass

Project duration = 55 work days. Every activity has zero total float.

Two critical paths:

   A → B → D → F → G  =  10 + 16 + 8 + 12 + 9 = 55 WD
   A → C → E → F → G  =  10 + 16 + 8 + 12 + 9 = 55 WD

This is not an error. It happens naturally on a schedule that has been compressed to fit a contract date, and it happens constantly on jobs where the finish date was fixed before the durations were.

(b) What does $9,600 buy? Nothing. Re-run the forward pass with B = 12:

Project duration: still 55 work days. F still waits for E at day 34. The mason's four days bought the east-and-south chain four days of float and bought the project nothing at all. $9,600 for a nicer-looking bar chart.

(c) To gain four days you have to buy both paths. Crash B and C from 16 to 12:

   A → B → D → F → G  =  10 + 12 + 8 + 12 + 9 = 51 WD
   A → C → E → F → G  =  10 + 12 + 8 + 12 + 9 = 51 WD

Project duration = 51 work days. Four days gained, for 2 × $9,600 = $19,200. Both paths are still critical, which is what you should expect — compressing symmetric paths symmetrically keeps them symmetric.

Is it worth it? Four work days is 4 × 1.4 = 5.6 calendar days. At Willow Street's total daily exposure of $2,800/CD ($1,600 extended general conditions + $1,200 liquidated damages), that is 5.6 × $2,800 = $15,680 of exposure avoided against $19,200 spent. On the arithmetic alone, no. You would need another reason — a weather window, an owner's occupancy date, a crew you need released for the next job — and you should say out loud what that reason is.

Alternatively, look at F. Crashing the roof trusses is on both paths, so every day you buy there is a day the project actually gets. Always look for the shared activity before you buy two chains.

What this problem was teaching. When somebody hands you an acceleration proposal, the first question is not "how much?" It is "how many critical paths are there, and does this proposal shorten all of them?" A proposal that shortens one of two parallel critical paths is a proposal to spend money and receive float.

(a) Forward and backward pass

Project duration = 49 work days. Every activity is critical.

(b) The lag arithmetic

   Sum of activity durations:  6 + 1 + 8 + 1 + 2 + 7 + 2 + 4 = 31 WD
   Sum of lags:                5 + 3 + 10                    = 18 WD
   Project duration:                                           49 WD

Eighteen of forty-nine work days — 37 percent of the pier sequence — is lag. Every one of those days is on the critical path. Every one of them has no responsible party, no resource, no cost code, and no way to report progress.

(c) Convert every lag to a real activity. Same 49 work days, same critical path — and now eleven activities instead of eight:

Nothing about the arithmetic changed. Everything about the management changed. Now CU3 has a name, an owner (Ingrid Sørensen, who schedules the testing laboratory), a status you can report, and a place to hang the field-cured cylinder results. It also shows up in the two-week look-ahead, which a lag never does — and it becomes visible to a cold-weather protection decision, because CU3 in January is not CU3 in July.

One more consequence worth noting: cure activities belong on a 7-day calendar. Concrete does not know it is Saturday. Ten work days of cure on a five-day calendar is 14 calendar days of real time, and if you model it as a work-day activity you have quietly added four calendar days of slop to your own schedule.

What this problem was teaching. Kestrel's house rule: a lag is a duration with no owner. Any lag longer than five days must either be converted into a real activity or carry a written justification in the activity notes. And never use a negative lag — an FS−10 is a start-to-start relationship written badly, and when the predecessor's duration changes the successor's start moves in ways that are hard to predict and harder to explain to an arbitrator.

(a) Forward pass

Project duration = 44 work days.

Backward pass, LF(G) = 44

Total float is zero on every activity. The entire ladder is critical.

(b) What the overlap saved

   Sum of durations (pure finish-to-start): 18 + 15 + 4 + 16 + 14 + 12 = 79 WD
   Actual duration with the ladder:                                      44 WD
   Saved:                                                                35 WD

Thirty-five work days — about 49 calendar days, worth 49 × $10,650 = $521,850 of Northgate exposure. That is what overlapping trades in a zone is worth, and it is also why coordination meetings on a floor like this are not optional: five trades are working inside each other's areas simultaneously, and every one of those SS lags is a promise about how far ahead one crew stays.

(c) The audit question — framing runs 24 days instead of 18

With the A → B FF + 2 link intact:

Project duration = 49 work days. Six extra days of framing cost the project five days — five, not six, because in the base case the FF link had one day of slack in it. The schedule correctly reports a delay.

With the FF + 2 link deleted — leaving only A → B SS + 6:

Project duration = 44 work days. The schedule reports zero impact.

Read that again. Framing ran six work days long, and the network says the project did not move — because a start-to-start relationship constrains the predecessor's start and says absolutely nothing about its finish. With SS+6 alone, the framing crew could run six weeks over and the schedule would keep reporting day 44, cheerfully, while drywall is scheduled to be hung on walls that do not exist.

What this problem was teaching. Every SS relationship needs a companion FF relationship wherever the physical work requires the predecessor to finish. Frame → rough-in is the textbook case: the electrician can start six days behind the framers, but the electrician cannot finish until the framers finish. Model only the start and you have built a schedule that cannot see its own delay.

When you audit somebody's schedule, filter for activities that have SS successors and no FF successors, and ask about every one. It is a five-minute check and it finds real defects. See Chapter 14 §14.5.1 for why negative lags fail for a closely related reason.

Forward pass with E = 22

Project duration = 65 work days — exactly seven days longer.

Float, before and after

Critical path: A → D → E → F → I = 4 + 6 + 22 + 25 + 8 = 65 WD ✓ — unchanged.

What this problem was teaching. Three results worth having in your hands.

One: seven days added to a critical activity cost the project exactly seven days. No absorption, no cushion, no negotiation. That one-for-one relationship is the definition of the critical path and it is the basis of the verification test in §A.9.4 — add 100 days to a critical activity and the finish must move 100 days.

Two: the critical path did not move, because there was no competing path close enough to take over. The nearest rival was 56 days (A → C → G → H → I) against 58, and pushing the critical path to 65 pushed it further away, not closer.

Three — and this is the one people miss — everybody else got more float. The structural chain went from 2 days to 9. That is not good news. It is a redistribution, and it is exactly the condition in which a project team relaxes about the CMU walls, spends the nine days on something else, and discovers in month six that the architect caught up on the review and the structure is critical again. Float that appears because somebody else is late is borrowed float. Report it as such.

Convert it to money before you leave: seven work days is 7 × 1.4 = 9.8 calendar days. At Willow Street's $2,800/CD that is $27,440 of exposure from one understaffed review desk — and the entitlement question ("who owns a review that ran seven days past the contractual period?") is Chapter 33's territory. Give notice on time either way.

(a) and (b) Forward pass with C = 24

Project duration = 60 work days — two days longer, not four.

Backward pass, LF(I) = 60, and the float table:

The critical path moved. It was A → D → E → F → I (the procurement chain, 58 days). It is now A → C → G → H → I = 4 + 24 + 14 + 10 + 8 = 60 WD ✓

                               5   10   15   20   25   30   35   40   45   50   55   60
                               |    |    |    |    |    |    |    |    |    |    |    |
A  Award and mobilize      ####
B  Steel + joists, L1          ============............
C  CMU walls, L1               ########################
D  Submit gym floor shops      ======..
E  Architect review                  ===============..
F  Fabricate and deliver                            =========================..
G  2nd-floor framing/deck                              ##############
H  Roof framing + dry-in                                             ##########
I  Install gym floor                                                           ########

   ###  critical (TF = 0)     ===  has float     ...  total float

(d) How many days C could absorb. C had two days of total float. Run it at 22 days:

Project duration = 58 work days — unchanged. And now every activity except B has zero total float: two simultaneous critical paths, exactly at the knee.

So: days one and two are free. Day three costs a day. Day four costs a day. The masonry subcontractor's four-day slip cost the project two days, and the fourth day cost exactly as much as the third. There is no further absorption available anywhere.

2 WD × 1.4 = 2.8 CD × $2,800/CD = $7,840 of exposure.

What this problem was teaching. This is the perturbation that matters, and there are three lessons in it.

One: a "non-critical" activity is only non-critical up to its float. Two days of float means two days of protection, not a category.

Two: the critical path is a property of the current model, not of the job. Yesterday it ran through a submittal, an architect's desk, and a factory. Today it runs through masons, carpenters, and a roof. Nobody chose that. The arithmetic did.

Three, and this is the management point: when the critical path moves off procurement and onto field work, your options change completely. Ray cannot accelerate an architect's review or a factory slot. Margo Deacon can add a mason crew, add a Saturday, and re-sequence the deck by half. A critical path you can influence is worth more than one you cannot — which is why the news in this problem is genuinely better than the news in Problem 12, even though the project got longer.

(a) Backward pass with LF(I) = 55 instead of 58. Early dates do not change — the forward pass knows nothing about imposed finishes.

Check A both ways: TF = LS − ES = −3 − 0 = −3 and LF − EF = 1 − 4 = −3 ✓

(b) The critical path under an imposed date is the path with the most negative float, not the path with zero. Here that is A → D → E → F → I at −3 work days. The structural chain (C → G → H) sits at −1 and is also late — just less late.

That distinction matters practically. A schedule with an imposed date has no zero-float path at all, and a filter set to "total float = 0" will return an empty list. Filter on the minimum, and report every path that is negative.

(c) What free float does: nothing. Compare the free float column above with Problem 6's — B is 8, H is 2, everything else is 0, exactly as before. Free float is computed entirely from early dates, and the forward pass is unaffected by an imposed finish. This is a useful diagnostic: if somebody shows you a schedule where the constraint changed the free float, the file has something else going on in it.

(d) What you do. Negative float is not a plan. It is the arithmetic telling you that you are already late against something. Three moves, in order:

1. Convert it to money before the meeting. Three work days is 3 × 1.4 = 4.2 calendar days. At Willow Street's $2,800/CD — $1,600 extended general conditions plus $1,200 liquidated damages — that is $11,760. That is the number the conversation is actually about.
2. Look at the −1 path too. Fixing only the −3 path leaves you at −1, which is still late. Recovery has to address every negative path or it is theater. This is Problem 9 in a different costume.
3. Then decide among three priced options — resequence (free, but what does it cost in risk?), accelerate a specific activity (name it, price the crew), or request contract time (name the entitlement and give the notice). Chapter 14's Willow Street checkpoint asks for exactly those three.

What this problem was teaching. Negative float ranks urgency, and the ranking is the useful part. A schedule carrying −45 days across a hundred activities is not a schedule; it is an alarm. A schedule carrying −3 on one path and −1 on another is a work plan for the next two weeks.

⚠️ One caution. A schedule can carry two different negative floats against two different dates at the same time, and they mean different things. On Northgate after the steel delay, substantial completion forecast at −23 CD against the contract date and −10 CD against Meridian's interim clinic lease expiry. The −23 was a money problem worth 23 × $10,650 = $244,950. The −10 was an operational catastrophe for a health system with patients scheduled into a clinic that had no building. Both numbers were correct. Only one of them decided anything.

(a) No constraint

Duration 34 WD. Critical path A → B → C → D → F. E carries 28 days of float, which is honest — the MRI could arrive any time in a four-week window and nothing would move.

(b) With Start No Earlier Than day 40 on E

Duration 46 WD, and the entire concrete-and-shielding chain now shows twelve days of total float that does not exist.

The tell: LS(A) = 12, not 0. In an unconstrained network the first activity's late start must come back to zero. When it does not, either you made an arithmetic error or there is an imposed date somewhere in the file. That single check, applied to every schedule you are handed, finds constraints faster than any report.

Why the float is phantom: it exists only because somebody typed a date. If the MRI actually shows up on day 34, the twelve days evaporate instantly — and by then the crew, reading the float column, has taken them.

(c) What to do: convert the constraint into logic. "Cannot start before the equipment arrives" is not a date. It is a predecessor. Model it:

Same 46-day answer today. Completely different behavior tomorrow, because now:

• The delivery chain has a duration, an owner, and a status you can report.
• If the vendor slips, the delay propagates and the schedule tells you what it costs.
• The 12 days of float on the concrete chain are visibly sponsored by the delivery chain, so when the delivery date moves, the float moves with it and everyone can see it happen.

(d) The killer — a Mandatory Start. Suppose the schedule carries a Mandatory Start on D at day 17 (its originally calculated early start), because someone wanted the shielding date to "hold." Now Ray's demolition and re-forming (A) runs 16 work days instead of 6.

Without the mandatory constraint:

Duration 44 WD — the ten-day delay shows up as ten days. Correct.

With the Mandatory Start on D at day 17:

Duration 34 WD. The schedule reports that a ten-day delay had zero impact.

That is not a rounding artifact. It is the mechanism: CPM works because delay propagates. A slip in one activity moves the next, which moves the next, until it reaches the finish or runs out of float. A mandatory constraint is a wall. Delay hits it and stops. Everything downstream keeps showing its original dates, and the finish does not move.

I have watched a project team celebrate that report. Six weeks later they were thirty days behind and could not say when it happened.

What this problem was teaching — the five-minute constraint audit. Run this on every schedule you are handed:

1. Count the constraints. Kestrel's house standard is fewer than 1 percent of activities constrained and zero mandatory constraints. Numbers vary by owner and specification; what does not vary is that the count should be small and every one should be explainable.
2. List them with the reason. If nobody can say why one exists, delete it and re-run.
3. Convert what you can into logic. A permit is an activity, not a date.
4. Check whether the constraint is creating float. Run the schedule with it removed and compare total float upstream. If float collapses, that float was never real.
5. Look at the project finish milestone. This is where the worst one hides. If the finish milestone is constrained to a date later than the contract date, every path in the schedule is carrying phantom float equal to the difference.

(a) Base network

Project duration = 106 work days.

(b) Cost slopes

Cost slope = (crash cost − normal cost) ÷ (normal duration − crash duration)

One extension shown in full:

   A:  ($166,000 − $148,000) ÷ (12 − 9) = $18,000 ÷ 3 = $6,000 per day

(c) The compression, step by step

Step 1 — crash A by 3 days ($6,000/day, the cheapest critical activity).

   Cost: 3 × $6,000 = $18,000
   Path 1: 9 + 16 + 20 + 18 + 10 + 30 = 103    ← still critical
   Path 2: 9 + 16 + 20 + 24 + 30      =  99    ← still 4 days of float
   Duration: 103 WD.  Gained 3.

A is fully crashed. Next cheapest on the critical path is B at $11,000.

Step 2 — crash B by 4 days.

   Cost: 4 × $11,000 = $44,000   (cumulative $62,000)
   Path 1: 9 + 12 + 20 + 18 + 10 + 30 = 99     ← still critical
   Path 2: 9 + 12 + 20 + 24 + 30      = 95     ← still 4 days of float
   Duration: 99 WD.  Gained 7.

Seven days. One more to go. Next cheapest on the critical path is D at $12,000.

Step 3 — crash D by 1 day.

   Cost: 1 × $12,000 = $12,000   (cumulative $74,000)
   Path 1: 9 + 12 + 20 + 17 + 10 + 30 = 98     ← still critical
   Path 2: 9 + 12 + 20 + 24 + 30      = 95     ← float now 3
   Duration: 98 WD.  Gained 8.  ✓

Result: eight work days for $74,000, an average of $9,250 per day — against a "sticker" range of $6,000 to $25,000 per day. Note that F, at $22,000 a day, was never a candidate: it is not on the critical path, so crashing it buys nothing at any price.

Is it worth $74,000? Eight work days is 8 × 1.4 = 11.2 calendar days. At Willow Street's $2,800/CD that is $31,360 of exposure avoided. On the schedule arithmetic alone, no — you would be spending $74,000 to avoid $31,360. The city has to want those eleven days for a reason worth $42,640, and if the grant-funded programming start is genuinely worth that, this becomes a change order conversation about who pays for acceleration, not a Kestrel decision.

What this problem was teaching. Four rules.

1. Only crash activities on the critical path. Everything else is a donation.
2. Crash the cheapest one first, and take all of its available days before moving on — unless the path shifts first, which is Problem 17.
3. Re-run the network after every step. Not at the end. After every step.
4. Watch the float on the parallel path shrink. It went 4 → 4 → 3. When it hits zero you have a second critical path and the arithmetic changes completely.

Steps 1 and 2 are unchanged from Problem 16 — A by 3 ($18,000) and B by 4 ($44,000), reaching 99 work days for $62,000 with path 2 still holding 4 days of float.

Step 3 — crash D all four days ($12,000/day).

   Cost: 4 × $12,000 = $48,000   (cumulative $110,000)
   Path 1: 9 + 12 + 20 + 14 + 10 + 30 = 95
   Path 2: 9 + 12 + 20 + 24 + 30      = 95
   Duration: 95 WD.  Gained 11.

★ Path shift #1. There are now two critical paths. Every activity in the network has zero total float. From here, any day you buy on path 1 alone is a day you do not get, because path 2 will govern.

Step 4 — crash C by 4 days ($17,000/day). C is on both paths, so it is the only single activity that still shortens the project.

   Cost: 4 × $17,000 = $68,000   (cumulative $178,000)
   Path 1: 9 + 12 + 16 + 14 + 10 + 30 = 91
   Path 2: 9 + 12 + 16 + 24 + 30      = 91
   Duration: 91 WD.  Gained 15.

Step 5 — crash G by 5 days ($25,000/day). Also on both paths.

   Cost: 5 × $25,000 = $125,000  (cumulative $303,000)
   Path 1: 9 + 12 + 16 + 14 + 10 + 25 = 86
   Path 2: 9 + 12 + 16 + 24 + 25      = 86
   Duration: 86 WD.  Gained 20.

Every shared activity is now fully crashed: A at 9, B at 12, C at 16, D at 14, G at 25. To go further you must buy both paths simultaneously.

Step 6 — crash E by 2 days and F by 2 days, together.

   Combined slope: $14,000 (E) + $22,000 (F) = $36,000 per day of project time
   Cost: 2 × $36,000 = $72,000   (cumulative $375,000)
   Path 1: 9 + 12 + 16 + 14 + 8 + 25 = 84
   Path 2: 9 + 12 + 16 + 22 + 25     = 84
   Duration: 84 WD.  Gained 22.

★ The crash point: 84 work days. E is fully crashed at 8 days, so path 1 cannot be shortened further at any price. F still has two crashable days left — and buying them is pure waste: take F to 20 and path 2 drops to 82 while path 1 stays at 84, so the project stays at 84 and you have spent $44,000 to give F two days of float.

The complete compression curve

Look at the marginal cost column and nothing else for a moment: $6,000 → $11,000 → $12,000 → $17,000 → $25,000 → $36,000. The last two days cost six times what the first three did. That convexity is not a quirk of these numbers; it is the shape of every crash curve ever drawn, and it is the reason "how much would it cost to finish a month early?" has no single answer.

Where the money stops making sense. At Willow Street's $2,800/CD, a work day of schedule is worth 1.4 × $2,800 = $3,920. Every step on this curve costs more than that — even step 1 at $6,000. Pure liquidated-damages-and-general-conditions arithmetic never justifies acceleration on this job. Acceleration gets justified by something outside the schedule: an owner's operational date, a weather window, a crew you need for the next job, or an owner willing to pay for it.

That is exactly the Northgate steel decision. Absorbing 23 days cost $244,950; accelerating cost $168,000 and recovered 17 days, so acceleration saved $13,050 — noise on a $47.5M job. Nadia Haddad approved it in four minutes anyway, and the number she was looking at was not $13,050. It was Meridian's interim clinic lease expiring October 1, Year 2.

⚠️ And one cost that is on none of these tables. Kestrel's acceleration produced trade stacking, a rework event on deck-edge detailing, and a measurable spike in near-misses in weeks 34 through 36 — including the scaffold event Bea Salgado investigated, whose third finding was "a crew running behind after the steel acceleration, with an unwritten 'make it up' pressure." When you price acceleration, price it honestly: premium time, a second crew, and the elevated risk of putting more people in less space under more pressure. Sometimes the answer is still yes. It should never be an easy yes.

(a) CPM table

Project duration = 37 work days. Critical path: A → B → E → F → G = 10 + 8 + 4 + 5 + 10 = 37 WD ✓ The first-floor chain C → D carries 8 days of total float, all of it appearing as free float on D.

(b) Early-start histogram

   Carpenters      0    2    4    6    8   10   12
                   |----|----|----|----|----|----|
   Days  1-10       ############                    6   (A)
   Days 11-14       ####################            10  (B 6 + C 4)   *** OVER ***
   Days 15-18       ####################            10  (B 6 + D 4)   *** OVER ***
   Day     19       ################                8   (D 4 + E 4)
   Days 20-27       ########                        4   (E, then F)
   Days 28-37       ##########                      5   (G)

   PEAK = 10 carpenters, days 11 through 18

(c) Level to a crew of 8. Shift C and D later, inside their float: C to ES 18 (work days 19–22) and D to ES 22 (work days 23–27). Both land exactly at their late starts, which is the most you can shift them.

   Days  1-18       ############                    6   (A, then B)
   Days 19-22       ################                8   (C 4 + E 4)
   Days 23-27       ################                8   (D 4 + F 4)
   Days 28-37       ##########                      5   (G)

   PEAK = 8 carpenters.  PROJECT DURATION = 37 WD, unchanged.

Cost: zero dollars and zero days — but all eight days of float on the C → D chain are now spent. That is the honest price. Leveling looked free and it was not: it converted flexibility into a commitment. If the first-floor blocking now slips one day, the project slips one day, and the schedule has no room left anywhere.

(d) Level to a crew of 7. Now the arithmetic turns.

Any pair of these activities that could run concurrently exceeds 7:

C and D cannot run concurrently with anything. They have to run alone, and they cannot start before A finishes or finish after G starts. The best available sequence — either A → C → D → B → E → F → G or A → B → E → F → C → D → G — gives the same answer:

   Days  1-10   A   6 carpenters
   Days 11-18   B   6
   Days 19-22   E   4
   Days 23-27   F   4
   Days 28-31   C   4
   Days 32-36   D   4
   Days 37-46   G   5

   PEAK = 6 carpenters.  PROJECT DURATION = 46 WD.

Cost of leveling to 7: nine additional work days.

   9 WD × 1.4 = 12.6 calendar days
   12.6 CD × $2,800/CD = $35,280

What this problem was teaching. The rule that matters:

Leveling is free only as long as you are spending float. Once the float on an activity is gone, the only way to level further is to extend the project.

Every scheduling tool will happily run a leveling routine and hand you back a longer duration without making a noise about it. When somebody runs resource leveling, always ask what happened to the finish date.

And note the second-order lesson: the difference between a crew of 8 and a crew of 7 is one carpenter, and it is worth $35,280 on a $6.8M job. That is a staffing decision worth having a real conversation about with Margo Deacon before the leveling routine makes it for you.

Daily burn rate for each activity = cost ÷ duration:

   A: $148,000 ÷ 12 = $12,333.33/day    E: $186,000 ÷ 10 = $18,600.00/day
   B: $305,000 ÷ 16 = $19,062.50/day    F: $735,000 ÷ 24 = $30,625.00/day
   C: $520,000 ÷ 20 = $26,000.00/day    G: $940,000 ÷ 30 = $31,333.33/day
   D: $410,000 ÷ 18 = $22,777.78/day

(a) Early-start curve

One month worked in full, so the pattern is unmistakable:

   Month 3 (work days 41 through 60):
     C, days 41-48:   8 × $26,000.00 = $208,000
     D, days 49-60:  12 × $22,777.78 = $273,333
     F, days 49-60:  12 × $30,625.00 = $367,500
                                       --------
                                       $848,833

The curve, drawn

                      0%       20%       40%       60%       80%      100%
                      |---------|---------|---------|---------|---------|
   M1  $0.30M cum     ##### 9.3%
   M2  $0.77M cum     ############ 23.6%
   M3  $1.61M cum     ######################### 49.7%
   M4  $2.43M cum     ##################################### 74.9%
   M5  $3.06M cum     ############################################### 94.2%
   M6  $3.24M cum     ################################################## 100.0%

That is the S — slow at the start when one or two trades are mobilizing, steep in the middle when structure, enclosure, and interiors overlap, flattening at the end as the trade count collapses to punch and closeout. Any project whose cash-flow curve is a straight line has been cost-loaded by dividing the contract sum by the number of months, which tells you nobody loaded anything.

(b) Late-start curve. F is the only activity with float (4 days), so it is the only bar that moves: LS 52 instead of ES 48, running work days 53–76.

(c) What the gap is for. The two curves form an envelope — schedulers call it the banana curve — and your actual cost should live inside it. Three uses:

1. It is your billing forecast. The City of Rivermont's finance office needs to know when the draws land, and so does Owen Baptiste, Kestrel's CFO, who has to fund payroll and subcontractor payments against a 30-day owner payment cycle with retention held.
2. It is your first early-warning system. Actual billings running consistently below the curve mean the job is behind. Actual billings running consistently above it may mean you are ahead — or that the schedule of values is front-loaded, which is a different conversation entirely.
3. Actual cost above the early-start curve is a genuine alarm. The early-start curve is the fastest the money can legitimately be spent. Spending faster than that means either the work is costing more than budgeted or the cost is being booked against the wrong codes.

Two observations worth carrying to a real job. First, the money curve and the headcount curve peak at different times. On Northgate the money peaks around month 10 and the craft headcount peaks around month 14, because steel, curtain wall, switchgear, and air handlers are enormous dollar values installed by small crews, while partitions, drywall, ceilings, and paint are enormous headcounts installing low unit value. Staff your general conditions off the headcount curve, not the money curve, or you will be short of supervision exactly when you have the most people on site.

Second, the point where the curve crosses 50 percent is a date, not a trivia item. On Northgate, retention drops from 10 percent to 5 percent at 50 percent complete, which put roughly $1.3 million of cash back in Kestrel's hands in month 10. Find that crossing on your own curve and circle it.

(a) Update 8 baseline — the remaining-work forecast before either event

Remaining duration = 42 work days, against 45 work days available. Project float = +3 WD.

The longest path runs P1 → P2 → P3 → P7 → P8. Note that nothing shows zero float, because the late dates are set by the contract date rather than by the calculated finish. That is normal on an update and it is what "project float" means.

(b) Owner event alone — insert the fragnet and recalculate

Remaining duration = 67 work days.

   Time impact = 67 − 42 = 25 work days of delay to the forecast completion
   Against the contract date at 45 WD:  45 − 67 = −22 WD

Read that carefully, because the two numbers are different and both matter. The event pushed the forecast 25 work days. But 3 of those days were absorbed by the project float that existed before the event, so the delay past the contract date is 22 work days. The critical path has moved from the second-floor MEP chain onto the flooring procurement chain — F1 → F2 → F3 → P4 → P5 → P7 → P8.

(c) Contractor event alone — P1 = 40, no fragnet

Remaining duration = 72 work days. Against the contract date: −27 WD.

(d) Both events together

Remaining duration = 72 work days. Against the contract date: −27 WD.

(e) The concurrent period

Now apportion, using the but-for test on each event separately:

• Total delay past the contract date: 27 WD.
• The owner event, on its own, would have caused 22 WD of that.
• The contractor event, on its own, would have caused 27 WD of that.
• The two overlap for 22 work days — a period during which either cause, standing alone, would have delayed completion. That is genuine concurrency.
• The remaining 5 work days are caused by the contractor's delay alone. The owner event does not reach that far.

In calendar days, on a five-day calendar:

   Total delay:               27 WD × 1.4 = 37.8  →  38 CD
   Concurrent period:         22 WD × 1.4 = 30.8  →  31 CD
   Contractor-only period:     5 WD × 1.4 =  7.0  →   7 CD

(f) Entitlement, and the money

The usual treatment where true concurrency is found — and treatments vary by jurisdiction, by contract form, and by tribunal — is that the contractor gets time but not money, and the owner gets no liquidated damages, for the concurrent period.

The money, on Willow Street's canonical rates — extended general conditions $1,600/CD, liquidated damages $1,200/CD:

The concurrency cost Kestrel $49,600 of recovery it would otherwise have had — and it cost it because of a subcontractor manpower failure that Kestrel could see coming and did not fix.

What this problem was teaching. Five things, and every one of them is a thing that decides real money.

One: a time impact analysis inserts the fragnet into the schedule update that was in effect when the event occurred, not into the baseline and not into the as-built. That is what makes it contemporaneous and it is why many contracts require this method by name. See Chapter 33.

Two: project float gets spent before entitlement starts. The event pushed the forecast 25 work days and produced 22 work days of delay past the contract date, because three days of float were consumed first. Under the common "float belongs to the project" position, nobody gets paid for that.

Three: concurrency is proved with the float column, not with a calendar. Two things happening in the same month are not concurrent. Two things each of which would independently have delayed completion are concurrent. If Kestrel's mechanical slip had run 24 days instead of 30, the MEP chain would have finished at day 48 against the flooring chain's 49 — it would have consumed all its float and driven nothing, and there would be no concurrency at all. The same subcontractor failure is worth $0 or an owner's entire compensable claim depending on a number that exists only in a schedule update somebody had to run at the time.

Four: entitlement, causation, and damages are three separate proofs, and you must win all three. This problem gave you causation. Entitlement lives in the contract's changes clause, its delay clause, and — the one that kills more claims than any tribunal ever has — its notice provision. Give the notice on time, in writing, in the form the contract requires, even when you do not yet know the number.

Five: run this analysis every month, not when the claim starts. The reason this problem is solvable at all is that Update 8 exists, with real logic, real remaining durations, and a float column somebody computed on a date nobody thought was interesting. The paper trail is the project's memory, and contemporaneous records are worth ten times reconstructed ones.