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Module 6 · Networks and Critical Paths

Forward & Backward Scan

Earliest start times · Latest start times · Scanning a network

MS12-7 MS-N3 Lesson 8 of 12

Think First

In a building project, Activity A (foundations, 4 days) must finish before Activities B (framing, 3 days) and C (plumbing, 5 days) can start. B and C both feed into D (roofing, 2 days).

Before reading on, what is the earliest day roofing (D) can start? And: what is the latest day framing (B) can start without delaying the final completion date?

Learning Intentions

Define Earliest Start Time (EST) and Latest Start Time (LST) for each activity
Apply the forward scan to find EST for every node, working left to right
Apply the backward scan to find LST for every node, working right to left
Record EST and LST in split node boxes on a network diagram
Interpret which activities are on the critical path using EST and LST values

Key Terms

EST (Earliest Start Time): The earliest a node can be reached, the maximum of all incoming path totals at that node.
LST (Latest Start Time): The latest a node can be reached without delaying the project, found by working backwards from the final node.
Forward scan: Working left to right through the network, calculating EST at each node by taking the maximum path total from the start.
Backward scan: Working right to left, calculating LST at each node by taking the minimum of (successor LST − activity duration).
Node box: A split box on each node showing EST (left) and LST (right). On the critical path, EST = LST.
Float: The slack in a non-critical activity: LST − EST. (Studied in detail in Lesson 9.)

Earliest Start Time (EST), Forward Scan

The Earliest Start Time (EST) at any node is the earliest moment (in days, weeks, etc.) that the work after that node can begin. You find it by running a forward scanmoving left to right through the network.

Forward scan rules

Set EST = 0 at the start node.
For each subsequent node, add each incoming activity's duration to the EST of the node it came from.
If two or more paths arrive at the same node, take the maximumyou must wait for all predecessors to finish.
The EST of the final node is the minimum project duration (same as the critical path length found in Lesson 7).

Why maximum? All predecessors must be complete before the next activity can start. If path 1 arrives at day 5 and path 2 at day 8, you cannot start until day 8.

Worked example, forward scan

Network: Start → A(3) → B(4) → End; Start → C(5) → End (C also feeds into End).

Network: Two parallel paths from S to End.

Path 1: S → A (dur 3) → B (dur 4) → End
Path 2: S → C (dur 5) → End

Step 1, EST(S) = 0

Step 2, EST(A) = EST(S) + dur(SA) = 0 + 3 = 3

Step 3, EST(C) = EST(S) + dur(SC) = 0 + 5 = 5

Step 4, EST(B) = EST(A) + dur(AB) = 3 + 4 = 7

Step 5, EST(End): Two paths arrive: via B → 7; via C → 5. Take max: EST(End) = 7

Minimum project duration = 7 days.

EST (Earliest Start Time) at each node is found by the forward scan: set EST = 0 at start, then at each node take the maximum of (predecessor EST + activity duration). The final node's EST is the minimum project duration.

Pause, copy the forward scan rule: EST at each node = maximum of (predecessor EST + activity duration), and explain why maximum is used, you can only start when ALL predecessors have finished into your book.

Latest Start Time (LST), Backward Scan

The forward scan sets EST = 0 at the start node and works left to right, taking the maximum of (predecessor EST + activity duration) at each merge node. This gives the earliest each event can occur. To find the latest each activity can start without pushing the project over time, the backward scan reverses the process: start at the final node with LST = EST, then subtract activity durations right to left, taking the minimum at each burst node.

The Latest Start Time (LST) at any node is the latest the work after that node can begin without pushing out the project completion date. You find it by running a backward scanmoving right to left through the network.

Backward scan rules

Set LST = EST at the final node (so the project is not delayed).
For each node, subtract each outgoing activity's duration from the LST of the node it leads to.
If two or more paths leave the same node, take the minimumyou must start in time for the most urgent successor.

Why minimum? If one successor needs you by day 5 and another by day 8, you must start by day 5 to avoid delay.

Network: EST values found above: S=0, A=3, C=5, B=7, End=7.

Step 1, LST(End) = EST(End) = 7

Step 2, LST(B) = LST(End) − dur(B→End) = 7 − 4 = 3

Step 3, LST(C) = LST(End) − dur(C→End) = 7 − 5 = 2
Wait, but we calculated EST(C) = 5. Check: LST = 2, EST = 5? That cannot be right because LST < EST is impossible in a valid network. Let me re-examine: C has duration 5 and goes directly to End (LST 7). LST(C) = 7 − 5 = 2.

Note: In this network C is NOT on the critical path (it has float = 2 − 0 = 2 days at the start node side; the actual float equals LST(node after C) − EST(node before C) − dur(C) = 7 − 0 − 5 = 2). This is fine, float means slack.

Step 4, LST(A) = LST(B) − dur(A→B) = 3 − 4 = −1? No, check again. B has duration 4 coming from A. LST(B) is the LST of node B. LST(A) = LST(B) − dur(AB) = 3 − 3 = 0.

Correction with proper labels: Let nodes be numbered 1(S), 2(A), 3(C merge), 4(End). Activity SA has dur 3 (1→2), AB has dur 4 (2→4), SC has dur 5 (1→4). Forward: EST₁=0, EST₂=3, EST₄=max(3+4, 0+5)=max(7,5)=7. Backward: LST₄=7, LST₂=7−4=3, LST₁=min(3−3, 7−5)=min(0,2)=0. Result: Node 1: EST=0,LST=0 ✓; Node 2: EST=3,LST=3 ✓ (critical); Node 4: EST=7,LST=7 ✓. Path S→C→End has float = 7−5−0 = 2 days.

LST (Latest Start Time) is found by the backward scan: set LST = EST at the final node, then work right to left subtracting activity durations. When paths diverge backwards, take the minimum of outgoing LST − duration.

Pause, copy the backward scan rule: LST at each node = minimum of (successor LST − activity duration), and explain why minimum is used, you must not push any successor activity past its latest start time into your book.

Which statement does NOT describe the backward scan?

Recording EST and LST in Node Boxes

The forward scan gives EST and the backward scan gives LST at every node. Recording both in split boxes (EST on the left, LST on the right) immediately shows which nodes are on the critical path: wherever EST = LST, there is zero float, that node must occur at exactly that time or the project is delayed. The difference LST − EST at any other node is the float available for those activities.

On network diagrams, each node is drawn as a split box:

Left compartment → EST (found during forward scan)
Right compartment → LST (found during backward scan)
When EST = LST, the node is on the critical path
When LST > EST, the activity leading into this node has float (LST − EST days of slack)

Full worked example, 5 activities

Precedence table:

Activity	Duration (days)	Depends on
A	2
B	4	A
C	3	A
D	5	B
E	2	C, D

Forward scan (EST):

Node 1 (Start): EST = 0
Node 2 (after A): EST = 0 + 2 = 2
Node 3 (after B, from node 2): EST = 2 + 4 = 6
Node 4 (after C, from node 2): EST = 2 + 3 = 5
Node 5 (after D and E): paths arriving: via D → 6+5=11; via C → 5+2 is not right, E depends on both C and D. So node 5 gets: max(EST₃+dur(D)=6+5=11, EST₄+dur(C but C feeds E...).
Re-mapping: S→A(2)→node2. node2→B(4)→node3. node2→C(3)→node4. node3→D(5)→node5. node4→E but E also needs D, so node5 = max(node3+5, node4+2) = max(11, 7) = 11
EST(End) = 11 days

Backward scan (LST):

Node 5 (End): LST = 11
Node 3 (before D): LST = 11 − 5 = 6
Node 4 (before E): LST = 11 − 2 = 9
Node 2 (before B and C): min(LST₃−4, LST₄−3) = min(6−4, 9−3) = min(2, 6) = 2
Node 1 (Start): LST = 2 − 2 = 0

Node boxes:

Node	EST	LST	On critical path?
1 (Start)	0	0	Yes (EST=LST)
2 (after A)	2	2	Yes (EST=LST)
3 (after B)	6	6	Yes (EST=LST)
4 (after C)	5	9	No (float = 4)
5 (End)	11	11	Yes (EST=LST)

Critical path: Start → A → B → D → End (total = 2+4+5 = 11 days). Activity C has 4 days float.

Each node box shows EST on the left and LST on the right. When EST = LST, the node is on the critical path (no float). Float = LST − EST at any node (activity can be delayed this many time units without delaying the project).

Pause, copy the split-box format (EST | LST), the critical-path condition EST = LST at a node, and the float formula: float = LST − EST (zero on critical path) into your book.

In a node box, when EST = LST, the node is on the _______. When LST > EST, the activity has _______.

Activities

Multiple Choice

Short Answer

Answers

Activity	Duration	Depends on
P	3
Q	6	P
R	4	P
S	2	Q, R

Show MC Answers

Q1 → D (11)Forward scan takes the maximum of incoming paths.

Q2 → CBackward scan starts at the final node, setting LST equal to EST of that node.

Q3 → C (3)Float = LST − EST = 7 − 4 = 3.

Q4 → B (2)Float = LST(end node) − EST(start node) − duration = 10 − 3 − 5 = 2.

Q5 → DEST = LST means zero float, so the node is on the critical path.

Show SAQ Model Answers

SAQ 1: Forward: Start=0, after A=5, after B=8, after C=9, End=max(8+2, 9+2)=max(10,11)=11. Backward: End=11, before D (from B side)=11−2=9, before D (from C side)=11−2=9. Before B: 9−3=6. Before C: 9−4=5. Before A: min(6−5, 5−5)=min(1,0)=0. Node table: Start(0,0)✓, after A(5,5)✓, after B(8,6)float2, after C(9,9)✓, End(11,11)✓. Critical path: Start→A→C→D→End = 5+4+2 = 11.

SAQ 2: Forward scan, at a merge node all predecessors must be complete before work can continue, so we wait for the last one: maximum ensures every required activity is done. Backward scan, at a fork node the activity must start early enough for the most urgent branch to meet its deadline: minimum ensures no deadline is missed.