Complete a critical path analysis

Which scan rule should be used first in a critical path analysis?

Here arrows represent one-way shuttle services. Can you travel from C to A using only the displayed services?

Vary D duration through 4,5,6 days. Edge labels give activity name and duration in days; D(d) uses the selected control value. Record completion time and D float. Trace every critical path at each setting, retaining the zero-duration dummies where required by the arrow representation

Why does increasing D from 4 to 5 days leave project completion at 18?

From directed routes to precedence constraints

The reused Year 10 direction check establishes that arrows have a specified meaning. Here an arrow is an activity, not a travel route or flow channel. Reaching an event requires completion of every incoming activity. Project completion is determined by the longest required path, not the shortest route

1. 1Use a maximum in a forward merge An activity can start only after every predecessor has finished \(\displaystyle \operatorname{EST}(C)=\max(0+4,0+6)=6\)
2. 2Use a minimum in a backward branch An activity must finish early enough for every following activity \(\displaystyle \operatorname{LST}(A)=\min(5,8)-2=3\)
3. 3Calculate total float Subtract the earliest start from the latest start \(\displaystyle F_A=\operatorname{LST}(A)-\operatorname{EST}(A)=3-0=3\)
4. 4Trace the critical path A critical path must be a continuous source-to-finish path with maximum duration. Its activities have zero total float, but two arbitrary zero-float activities need not be consecutive or form a complete path

Construct a complete precedence model

An invented installation project starts at time 0. Activities and durations in days are A4,B3,C5,D4,E6,F2,G3. A and B have no predecessors; C requires A; D requires A and B; E requires C and D; F requires D; G requires E and F. Durations are deterministic elapsed days on a continuous calendar. Initially assume sufficient independent resources, no additional setup or waiting, and non-preemptive activities

1. 1Translate the description Construct the complete immediate-predecessor table and a labelled activity network. Use either activity-on-node or activity-on-arrow, explain which objects represent activities/events, and preserve the exact dependencies without adding unrequired ones. For an arrow network, explain why zero-duration dummy arrows are needed
2. 2Audit the model Check both directions: every stated dependency appears in your graph, and each graph dependency is justified by the statement. Explain why merging A and B before C would add a false restriction. Calculate earliest starts of C and D using their different predecessor sets
3. 3Produce an independent record Choose a representation and checking method before opening the model response. Save the complete predecessor table, labelled activity network, forward/backward scans, start/finish/float table, complete critical paths, changed-case schedules, actual tool formulas or commands and qualified report. These independent constructions and external-tool operations are self/teacher-reviewed, not automatically marked by the short checks

Model response: Construct a complete precedence model

Compare the complete model, calculations and reasoning, not just a selected path or final time

1. 1Complete table and arrow construction The immediate predecessor rows are A: none; B: none; C:A; D:A,B; E:C,D; F:D; G:E,F. One activity-on-arrow model is A:1→2(4), B:1→3(3), d1:2→3(0), C:2→5(5), D:3→4(4), d2:4→5(0), F:4→6(2), E:5→6(6), G:6→7(3). d1 and d2 are dashed dummy arrows. The event numbers are labels, not a demand that arrows be drawn to scale
2. 2Why the dummies matter C starts after A alone at time 4. D starts after both A and B, at max(4,3)=4; d1 carries the A dependency into its start event without making C wait for B. E starts after C and D; d2 carries D into that merge while F requires D alone. Merging A and B before C would falsely require C to wait for B. The error is hidden by the original durations but exposed if B takes 6 days: C may still start at 4, whereas the wrong graph delays it to 6. Dummies take no time and are not optional real work

Given information

\(\displaystyle d_A=4,\quad \operatorname{EST}(A)=0,\quad \operatorname{pred}(C)=\{A\}\)

Find the earliest start of C, in days from time 0.

Given information

\(\displaystyle EF_A=4,\quad EF_B=3,\quad \operatorname{pred}(D)=\{A,B\}\)

Find the earliest start of D, in days from time 0.

In this arrow network, C requires A, while D requires A and B. A ends at event 2, D starts at event 3, and d1 is a zero-duration arrow from event 2 to event 3. What is the role of d1?

Construct and verify the full critical-path analysis

Use the original seven activities, durations and predecessor table. All otherwise-ready activities may overlap under the initial unlimited-resource assumption

1. 1Forward and backward records Calculate earliest and latest event times on the arrow network. Produce EST,EFT,LST,LFT and total float for every real activity. Explain maxima at forward merges and minima at backward branches, and show the calculations at the D and E starts
2. 2Certificate and limits Trace every continuous source-to-finish path, including any dummy arrows, and compare durations. State the project completion time and complete critical-path set. Explain why summing all durations or joining arbitrary zero-float activities is not a valid path certificate

Model response: Construct and verify the full critical-path analysis

Compare the complete model, calculations and reasoning, not just a selected path or final time

1. 1Event scans For events 1,2,3,4,5,6,7, earliest times are 0,4,4,8,9,15,18. Latest times are 0,4,5,9,9,15,18. At event 3, max(B finish 3,A finish 4)=4; at event 5, max(C finish 9,D finish 8)=9. Backward, event 4 is min(9−0,15−2)=9, and event 2 is min(5−0,9−5)=4. An event waits for all incoming requirements; a latest event time must satisfy every outgoing requirement
2. 2Complete real-activity table In order EST,EFT,LST,LFT,float, the rows are A:(0,4,0,4,0), B:(0,3,2,5,2), C:(4,9,4,9,0), D:(4,8,5,9,1), E:(9,15,9,15,0), F:(8,10,13,15,5), G:(15,18,15,18,0). For an arrow i→j of duration d, EST=earliest(i), LFT=latest(j), LST=LFT−d and float=LST−EST. Distinguish an activity’s latest start from its latest finish
3. 3All path certificates The five complete arrow paths are A–C–E–G:18; A–d1–D–d2–E–G:17; A–d1–D–F–G:13; B–D–d2–E–G:16; B–D–F–G:12. The unique critical path is A–C–E–G and minimum completion time is 18 days under the stated resources. Adding every real duration gives 27, but parallel work need not be consecutive. A and E both have zero float yet A–E alone is neither a direct arrow path nor a complete project path

Given information

\(\displaystyle \text{Complete path lengths: }18,17,13,16,12\)

Give the original project completion time, in days

Given information

\(\displaystyle L(3)=5,\quad d_B=3\)

Find the latest start of B, in days from time 0.

Given information

\(\displaystyle EST_D=4,\quad LST_D=5\)

Find the total float of D, in days

Given information

\(\displaystyle EST_F=8,\quad LST_F=13\)

Find the total float of F, in days

Which is a valid critical-path certificate?

Investigate delays, ties and a resource limit

Consider changes separately, each starting from the original model. First vary D duration through 4,5,6. Next delay D’s start by 2 days from its original earliest start, or delay B’s start by 2 then 3 days. Finally, change the resource model: A and B require the same single crew and cannot overlap, while other activities still have independent resources

1. 1Recompute the changed schedules For each duration/delay case, calculate the new earliest times and every critical path. State how much original float is used or exceeded. Do not calculate a negative project increase when a delay is within float
2. 2Add the actual resource constraint Under the single-crew condition, construct complete feasible schedules for A then B and for B then A, allowing other work to start as soon as its own conditions permit. Compare both orders and explain why the original 18-day schedule is no longer feasible

Model response: Investigate delays, ties and a resource limit

Compare the complete model, calculations and reasoning, not just a selected path or final time

1. 1Duration change and critical paths D durations 4,5,6 produce project times 18,18,19 and D floats 1,0,0. At D=5, A–C–E–G and A–d1–D–d2–E–G both take 18. At D=6 only the latter is critical at 19. In real-activity shorthand these are A–C–E–G and A–D–E–G; the full arrow trace still includes the required dummies. The explorer’s bounded formula is T=max(18,13+d) for d=4,5,6.
2. 2Start delays D originally starts at 4; delaying it to 6 uses its one day of float and exceeds it by one, so E starts at 10 and completion becomes 19. B starts at 2 in the two-day-delay case, finishes at 5, and D starts at 5; E still starts at 9, so completion remains 18. Delaying B to 3 makes it finish at 6, D finish at 10, E finish at 16 and G finish at 19. For one activity with otherwise unchanged conditions, project increase=max(0,delay−total float). A delay less than float gives zero increase, not a negative increase or a false subtraction equality
3. 3Complete resource-feasible schedules With A then B on the one crew: A0–4,B4–7,C4–9,D7–11,E11–17,F11–13,G17–20. Completion is 20. With B then A: B0–3,A3–7,C7–12,D7–11,E12–18,F11–13,G18–21. Completion is 21. These are the only two non-preemptive orders for the crew’s two tasks; starting their first task at 0 and avoiding crew idle time cannot worsen any precedence finish. Thus 20 is best under this particular added resource model. The original schedule overlaps A and B from 0 to 3 and violates the new condition

Given information

\(\displaystyle \text{Path lengths: }18,18,14,17,13\)

How many critical paths are there when only D duration becomes 5?

Given information

\(\displaystyle T=18,\quad F_D=1,\quad \Delta_D=2\)

If only D starts 2 days later than its original EST, give project completion

Given information

\(\displaystyle T=18,\quad F_B=2,\quad \Delta_B=2\)

If only B starts 2 days later than its original EST, give project completion

Given information

\(\displaystyle EF_E=17,\quad EF_F=13,\quad d_G=3\)

With A then B on the shared crew, give project completion

Given information

\(\displaystyle EF_E=18,\quad EF_F=13,\quad d_G=3\)

With B then A on the shared crew, give project completion

Why does the original 18-day result not apply to the shared-crew case?

Investigate bounded shortening and report a justified plan

Return to the original unlimited-resource model. Only whole-day reductions are allowed: A at most 1 day for $50/day; C at most 2 for $30/day; D at most 1 for $20/day; E at most 2 for $80/day; G at most 1 for $60/day. B and F cannot be shortened. Costs are additive invented planning values, with no indirect costs or change to precedence. Seek project completion no later than day 16 at minimum extra cost

1. 1Construct and compare Investigate shortening C by one day and then two, and shortening D alone by one day. Explain when a second critical path blocks a further saving. Enumerate every permitted combination or give an equivalent complete optimality argument, recording new durations, project time and extra cost. Find every minimum-cost feasible plan
2. 2Choose technology and report Choose and justify a spreadsheet or short program for the complete bounded enumeration. Record actual formulas/commands, inputs, all feasible outputs and how constraints are checked. Compare your results with independent path calculations. Communicate the objective, assumptions, selected plans, limits and missing real evidence; do not turn invented costs into a purchasing recommendation

Model response: Investigate bounded shortening and report a justified plan

Compare the complete model, calculations and reasoning, not just a selected path or final time

1. 1Why a greedy claim fails Shortening C by one day gives A–C–E–G=17 and A–D–E–G=17. A second day removed from C alone makes the first path 16 but leaves the second at 17, so the project still takes 17. Shortening D alone saves no project time because A–C–E–G remains 18. Recompute every relevant path, not only the originally unique critical path
2. 2Complete bounded enumeration There are 72 bounded reduction combinations. Let a, c, d, e and g be whole-day reductions; B and F stay fixed. The maximum complete path length is the project time. Enumerate the stated integer ranges, retain combinations with project time at most 16 days, then minimise extra cost \(\displaystyle \begin{aligned}A&=4-a,&C&=5-c,&D&=4-d,&E&=6-e,&G&=3-g,\\ B&=3,&F&=2\\ P_1&=18-a-c-e-g,&P_2&=17-a-d-e-g,\\ P_3&=13-a-d-g,&P_4&=16-d-e-g,&P_5&=12-d-g,\\ T&=\max(P_1,P_2,P_3,P_4,P_5),&\text{extra cost}&=50a+30c+20d+80e+60g\\ 0&\le a\le1,&0&\le c\le2,&0&\le d\le1,&0&\le e\le2,&0&\le g\le1\end{aligned}\)
3. 3Two optimal plans and verification The minimum extra cost is $80, achieved by exactly two plans: reduce A and C by one day each (50+30); or reduce C by two days and D by one (60+20). In the first plan the two long paths are 3+4+6+3=16 and 3+4+6+3=16. In the second they are 4+3+6+3=16 and 4+3+6+3=16. The other three path lengths are respectively 12,16,12 and 12,15,11, so neither plan leaves a hidden path above 16. In the first plan B–D–E–G is also critical at 16, giving three critical paths; the second plan has two. No plan costing less than $80 can shorten E or G. If A is shortened by one day, the A–C–E–G path requires at least one day of C reduction, bringing the cost to at least $80. If A is not shortened, that path requires two days of C reduction and the A–D–E–G path requires one day of D reduction, again costing at least $80. The two listed plans are therefore the only feasible plans at the minimum cost
4. 4Reproducible technology and conclusion A spreadsheet can store a,c,d,e,g in five columns, the five displayed path formulas in the next five, MAX of those five in the next, and 50a+30c+20d+80e+60g in the last. List all 72 bounded integer rows and filter project time≤16; retain both rows tied at the minimum cost. An equivalent program loops over the five integer ranges, evaluates those expressions and sorts feasible rows by cost. All stated inputs and results are exact integers, so no rounding is needed. Preserve actual formulas or commands and output. Under these assumptions either $80 plan meets day 16, but uncertain durations, calendars, availability, safety, quality, real cost quotes or resource coupling could change the conclusion. No equipment purchase follows from this hypothetical model alone

Given information

\(\displaystyle \text{Path lengths: }17,17,13,16,12\)

After shortening only C by 1 day, give project completion

Given information

\(\displaystyle \text{Path lengths: }16,17,13,16,12\)

After shortening only C by 2 days, give project completion

Given information

\(\displaystyle \text{Path lengths: }18,16,12,15,11\)

After shortening only D by 1 day, give project completion

Given information

\(\displaystyle \text{Feasible minimum plans: }(a,c,d,e,g)=(1,1,0,0,0)\text{ or }(0,2,1,0,0);\quad \text{extra cost}=50a+30c+20d+80e+60g\text{ dollars}\)

Give the minimum extra cost, in dollars, to finish by day 16.

Which report is supported by this investigation?