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The critical path method calculates an earliest schedule, a latest schedule for a chosen finish time, and each activity's task float. A critical path is a start-to-finish graph path whose total duration is greatest. There can be several such graph paths, and every one constrains completion. In the six-activity project scheduling example, the critical graph path is A-C-D-F, of duration $3+4+2+1=10$.

Add a zero-duration source $s$ before every initial activity and a zero-duration sink $z$ after every terminal activity. Give each edge $i\to j$ weight ${\tau }_i$, including zero weight on edges leaving $s$. This produces a weighted graph in which a graph path's weight counts every activity on it exactly once. The possible source-to-sink graph paths in our example, omitting the dummy endpoints, are A-B-D-F with weight 8, A-C-D-F with weight 10, and E-F with weight 4.

Every feasible schedule must accommodate each of these sequential chains. Consequently,This bound applies to all source-to-sink graph paths, not just the one we happen to inspect. The critical path method attains the bound by setting each activity's earliest start $E_i$ to the maximum of its predecessors' earliest finish times:Evaluate this recurrence in a topological order, obtained by topological sorting. Each $E_j$ satisfies every incoming optimization constraint, so these times form a feasible solution. The same recurrence calculates the greatest weight of a graph path from $s$ to $j$; hence the feasible finish $E_z$ matches the lower bound. This is dynamic programming on a Directed acyclic graph, rather than an enumeration of all graph paths. See Paul Johnson, Scheduling and longest paths, longest-path algorithm. The linear programming formulation and its certificate are developed in project scheduling duality and unit-flow certificate for project duration.

For our project scheduling example, $E_A=E_E=0$, $E_B=E_C=3$, and the two merges giveF finishes at 10. All six activities are completed, even though E is absent from the critical graph path. Recording which predecessor attains each maximum lets us trace a critical graph path backward from the sink; retaining tied choices exposes multiple critical graph paths.

Now fix the project start at zero and the finish at $T_{\ast }=10$. The latest starts $L_i$ come from a backward pass in reverse topological order:Every activity must finish by the earliest of its successors' latest starts, explaining the minimum here. This gives $L_F=9$, $L_D=7$, $L_E=6$, $L_B=5$, $L_C=3$, and $L_A=\min (5-3,3-3)=0$. These latest times themselves form a feasible solution finishing at 10. The backward calculation depends on the chosen finish time; a later deadline would give different latest times and task floats. The forward and backward passes are described in Olivier de Weck, MIT ESD.36 Lecture 2, slides 22–29.

Changing a duration can change which graph paths are critical. If B's duration becomes 4, A-B-D-F and A-C-D-F both take 10: there are two critical graph paths. Shortening C alone then cannot reduce the finish time, because the chain through B still takes 10. If B's duration instead becomes 5, the earliest starts of D and F become 8 and 10, and the finish becomes 11. A-B-D-F is now the unique critical graph path; C has task float 1 relative to that new optimal finish. These calculations show why the critical path method is useful for understanding where a duration change matters, even in a project containing many parallel activities.

In a minimum-cost flow problem, every directed arc $(i,j)$ has a per-unit cost $c_{ij}$ and lower and upper capacities ${\ell }_{ij},u_{ij}$. Each vertex has a prescribed net supply $b_i$, with ${\sum }_i{b}_i=0$. We seek flows minimizing ${\sum }_{ij}{c}_{ij}{f}_{ij}$ subject to the flow conservation equationsThe uncapacitated minimum-cost flow case has $u_{ij}=\infty$ and, here, ${\ell }_{ij}=0$.

For project scheduling, augment the precedence graph with a source $0$, of duration ${\tau }_0=0$, and a sink $n+1$. Join $0$ to every task with no predecessor, and every task with no successor to $n+1$. Extend the precedence matrix to these dummy arcs. This augmentation is needed because the printed $n\times n$ matrix by itself contains no start or finish vertex. Assuming the precedence graph is a Directed acyclic graph, every task lies on a source-to-sink path. Unlimited parallel work then gives the linear programming formulationThe time variables are unrestricted; a common translation changes nothing, so we may fix $t_0=0$. These inequalities say that each successor starts after its predecessor finishes. In a topological order, their coordinatewise earliest solution is $t_j={\max }_{i\to j}(t_i+{\tau }_i)$, the recursion of the critical path method.

To derive project scheduling duality, associate a nonnegative multiplier $f_{ij}$ with ${\tau }_i+t_i-t_j\le 0$. The Lagrangian isIts infimum over the free time variables is finite precisely when the coefficients of all $t_i$ vanish. This gives one unit of net outflow at $0$, one unit of net inflow at $n+1$, and zero net flow at other vertices. Hence the dual isEquivalently, minimize $\sum c_{ij}{f}_{ij}$ with $c_{ij}=-{\tau }_i$. The project duration is the negative of the optimal minimum-cost flow value. The sign is essential: the dual of the minimization problem maximizes total duration, selecting a longest precedence path.

The network simplex algorithm is a general method for minimum-cost flow. After shifting lower bounds to zero, first find a feasible flow, using artificial arcs and a phase-one objective if needed. A network simplex tree basis fixes each non-tree arc at a bound and determines the tree flows by flow conservation. Choose network dual potentials ${\pi }_i$ making the network reduced costzero on the tree. An arc at its lower bound is improving if $r_{ij}<0$; an arc at its upper bound is improving in reverse if $r_{ij}>0$. Adding that residual arc creates one fundamental cycle. Increase flow around this cycle until an arc reaches a bound, then exchange the entering and leaving tree arcs, or flip the entering arc to its other bound. The flow conservation equations remain satisfied. Use an anti-cycling rule for degenerate pivots. If a negative-cost residual cycle has no finite limiting capacity, the objective is unbounded below; if phase one cannot remove artificial flow, the problem is infeasible. Otherwise, termination with no negative residual network reduced costs certifies optimality, since potentials cancel around cycles and every residual network circulation has nonnegative cost.

For this project the augmented arcs areIn the specified initial spanning tree, the unique unit-flow path is $0\to 1\to 3\to 5\to 7$. The flow on its arcs is one and that on all other tree arcs is zero. Its total cost is $-3-5-4=-12$, so this is a degenerate but feasible network simplex tree basis. Normalizing ${\pi }_0=0$, the tree equations giveThere are two nonbasic arcs. Their network reduced costs areBring $(0,2)$ into the tree. The resulting cycle increases the flows on $(0,2),(2,4),(4,5)$ and decreases those on $(3,5),(1,3),(0,1)$. The three decreasing arcs initially carry one unit, so the maximal step is one. Resolve the leaving-arc tie by removing $(0,1)$. The resulting unit flow follows $0\to 2\to 4\to 5\to 7$ and has cost $-4-6-4=-14$.

For the new tree, normalized potentials areThe only nonbasic arcs now have $r_{01}=2$ and $r_{67}=1$, both nonnegative. All tree network reduced costs vanish, proving optimality after this single pivot. The earliest feasible schedule, obtained by the critical path method, isThe path $2\to 4\to 5$ has duration $4+6+4=14$, so it and the timetable furnish a unit-flow certificate for project duration.

Project scheduling asks how quickly a collection of dependent activities can finish. Adding their durations usually overestimates the answer, because independent activities can run together; taking only the longest single activity usually underestimates it, because some activities must wait. The critical path method identifies the chain of dependencies that determines the finish time. Its simplest model assumes known, fixed durations, unlimited parallel processing, and precedence requirements saying that each predecessor must finish before its successor starts. These are the assumptions of the basic model in Hendrickson, Project Management for Construction, §10.2.

Represent each activity by a vertex of a finite Directed acyclic graph; an edge $i\to j$ means that activity $j$ waits for activity $i$. Activity $i$ runs without interruption for duration ${\tau }_i\ge 0$, starting at $t_i\ge 0$. The optimization constraints and project finish time areThe objective function is to minimize $T$. All durations and start times use the same time unit, and the project starts at time zero. The Directed acyclic graph assumption allows a topological order in which every predecessor is processed first.

Consider this small, abstract project:

Activities A and E may start together. After A finishes, B and C may run together, but D must wait for both. F must similarly wait for both D and E: a merge means all incoming requirements must hold. This activity-as-vertex convention keeps durations on the activities and uses arrows only to express precedence.

Unlimited parallel processing is essential to the equality between the longest graph path duration and the optimal finish time. For example, two independent chains, each containing two activities of duration 3, have longest graph path duration 6. Two processors can finish both chains in 6, but one processor needs 12 because all four activities must run sequentially. Thus with limited resources the longest graph path still gives a lower bound, but it need not be achievable. Such resource requirements belong to the broader subject of scheduling.

// Target: mathematical-optimization.bigb

Task float, also called total float or task slack, measures an activity's available start-time delay relative to its earliest start, while allowing a feasible schedule that meets the fixed project finish. For earliest start $E_i$, latest start $L_i$, and fixed duration ${\tau }_i$, it isIn the six-activity project scheduling example with start zero and finish 10, B has task float 2 and E has task float 6. B can move from the interval $[3,5]$ to $[5,7]$, and E from $[0,3]$ to $[6,9]$, without changing D's or F's earliest starts. A, C, D, and F have zero task float. With the finish fixed to the optimal duration and the basic critical path method assumptions, zero task float identifies exactly the activities lying on at least one critical graph path. See Hendrickson, §10.4, Activity Float and Schedules.

Task float is not generally an independent allowance that can be spent separately on every activity. For a small example, let two duration-1 activities X then Y form one chain, running in parallel with a separate duration-4 activity. The optimal finish is 4. X has earliest/latest starts 0/2 and Y has earliest/latest starts 1/3, so each has task float 2. Delaying X to 2 forces Y to start at 3, using Y's allowance too. Choosing X at its latest start 2 and Y at its earliest start 1 violates precedence, although both chosen times lie inside their individual ranges. The optimization constraints still couple the decisions. Here B and E can both use their full task floats together because they lie on separate branches; that freedom must be checked from the Directed acyclic graph.

// Target: mathematical-optimization.bigb