Project Planning and Scheduling

Work Breakdown Structure (WBS) is a hierarchical, deliverable-oriented decomposition of the total project scope into smaller manageable parts (work packages) down to the level at which cost, time and responsibility can be assigned.

Project
 +-- Major deliverable 1
 |     +-- Work package 1.1
 |     +-- Work package 1.2
 +-- Major deliverable 2
       +-- Work package 2.1

Importance (uses) in project planning

• Defines the full scope, so nothing is missed (100% rule).
• Makes a complex project manageable by dividing it.
• Basis for schedule: activities are derived from work packages for bar chart, CPM and PERT.
• Basis for cost estimate and budget at each level.
• Assigns responsibility for each work package (responsibility matrix).
• Helps resource planning and procurement.
• Helps control: progress and cost are tracked package by package.
• Improves communication and common understanding among team and client.
• Helps risk identification at the work-package level.
• Gives a coding system for reporting and accounting.

Network diagram (activity on arrow)

Events are numbered 1 to 7. Activity C has zero duration, so event 3 is reached when both A (via C) and B finish. Event times (EST = LST) are: 1: 0, 2: 3, 3: 3, 4: 10, 6: 18, 7: 19; event 5 has EST 8 and LST 12.

         +------D4-------+
         |               v
(1)-A3->(2)-C0->(3)-E7->(4)-G8->(6)-I1->(7)
 |              ^ |              ^
 +-----B2-------+ +-F5->(5)-H6---+

Event 4 follows D and E, event 6 follows G and H, and event 7 is the end.

Forward pass (EST, EFT)

$EST$ = largest $EFT$ of predecessors; $EFT = EST + D$.

Backward pass (LFT, LST)

$LFT$ = smallest $LST$ of successors; $LST = LFT - D$. Project duration $T = 19$ weeks.

Floats

$TF = LST - EST$; $FF = \min(EST\ \text{of successors}) - EFT$; $Int.F = TF - FF$; $Ind.F = \max\{0, \min(EST\ \text{of successors}) - \max(LFT\ \text{of predecessors}) - D\}$.

Sample check, activity F: $EST = \max(EFT_B, EFT_C) = \max(2,3) = 3$; $EFT = 8$; $LFT = LST_H = 12$; $LST = 7$; $TF = 7-3 = 4$; $FF = EST_H - EFT_F = 8-8 = 0$; $Int.F = 4$.

Answer: Project duration = 19 weeks. Critical path: A - C - E - G - I ($3+0+7+8+1 = 19$).

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (weeks). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(5)--> (2)
(1) --B(4)--> (3)
(2) --C(0)--> (3)
(2) --D(6)--> (4)
(3) --E(7)--> (4)
(3) --F(8)--> (5)
(4) --G(6)--> (6)
(5) --H(3)--> (6)
(6) --I(2)--> (7)

The network has 7 events and 0 dummy activities. Start event is (1) and the end event is (7).

Activity C has zero duration, so it is treated as a normal activity of 0 weeks (it only passes the logic on).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 5 = 5
• B: EST = 0; EFT = 0 + 4 = 4
• C: EST = EFT of A = 5; EFT = 5 + 0 = 5
• D: EST = EFT of A = 5; EFT = 5 + 6 = 11
• E: EST = max(B:4, C:5) = 5; EFT = 5 + 7 = 12
• F: EST = max(B:4, C:5) = 5; EFT = 5 + 8 = 13
• G: EST = max(D:11, E:12) = 12; EFT = 12 + 6 = 18
• H: EST = EFT of F = 13; EFT = 13 + 3 = 16
• I: EST = max(G:18, H:16) = 18; EFT = 18 + 2 = 20

Project duration = largest EFT = 20 weeks.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• I: LFT = project duration = 20; LST = 20 - 2 = 18
• H: LFT = LST of I = 18; LST = 18 - 3 = 15
• G: LFT = LST of I = 18; LST = 18 - 6 = 12
• F: LFT = LST of H = 15; LST = 15 - 8 = 7
• E: LFT = LST of G = 12; LST = 12 - 7 = 5
• D: LFT = LST of G = 12; LST = 12 - 6 = 6
• C: LFT = min(E:5, F:7) = 5; LST = 5 - 0 = 5
• B: LFT = min(E:5, F:7) = 5; LST = 5 - 4 = 1
• A: LFT = min(C:5, D:6) = 5; LST = 5 - 5 = 0

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): A, C, E, G, I
• Critical path: A - C - E - G - I (= 5+0+7+6+2 = 20 weeks)

Answer: project duration = 20 weeks; critical path = A - C - E - G - I

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (days). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(9)--> (2)
(1) --D(9)--> (3)
(1) --B(8)--> (4)
(1) --C(7)--> (5)
(2) --F(4)--> (4)
(2) --E(5)--> (6)
(3) --G(3)--> (5)
(3) --H(6)--> (8)
(4) - - dummy - -> (6)
(4) --I(7)--> (7)
(5) --J(8)--> (7)
(5) - - dummy - -> (8)
(6) --K(4)--> (9)
(7) --L(5)--> (9)
(8) --M(3)--> (9)

The network has 9 events and 2 dummy activities. Start event is (1) and the end event is (9).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 9 = 9
• B: EST = 0; EFT = 0 + 8 = 8
• C: EST = 0; EFT = 0 + 7 = 7
• D: EST = 0; EFT = 0 + 9 = 9
• E: EST = EFT of A = 9; EFT = 9 + 5 = 14
• F: EST = EFT of A = 9; EFT = 9 + 4 = 13
• G: EST = EFT of D = 9; EFT = 9 + 3 = 12
• H: EST = EFT of D = 9; EFT = 9 + 6 = 15
• J: EST = max(C:7, G:12) = 12; EFT = 12 + 8 = 20
• I: EST = max(B:8, F:13) = 13; EFT = 13 + 7 = 20
• K: EST = max(B:8, E:14, F:13) = 14; EFT = 14 + 4 = 18
• M: EST = max(C:7, G:12, H:15) = 15; EFT = 15 + 3 = 18
• L: EST = max(I:20, J:20) = 20; EFT = 20 + 5 = 25

Project duration = largest EFT = 25 days.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• L: LFT = project duration = 25; LST = 25 - 5 = 20
• M: LFT = project duration = 25; LST = 25 - 3 = 22
• K: LFT = project duration = 25; LST = 25 - 4 = 21
• I: LFT = LST of L = 20; LST = 20 - 7 = 13
• J: LFT = LST of L = 20; LST = 20 - 8 = 12
• H: LFT = LST of M = 22; LST = 22 - 6 = 16
• G: LFT = min(J:12, M:22) = 12; LST = 12 - 3 = 9
• F: LFT = min(I:13, K:21) = 13; LST = 13 - 4 = 9
• E: LFT = LST of K = 21; LST = 21 - 5 = 16
• D: LFT = min(G:9, H:16) = 9; LST = 9 - 9 = 0
• C: LFT = min(J:12, M:22) = 12; LST = 12 - 7 = 5
• B: LFT = min(I:13, K:21) = 13; LST = 13 - 8 = 5
• A: LFT = min(E:16, F:9) = 9; LST = 9 - 9 = 0

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): A, D, F, G, J, I, L
• Critical path: A - F - I - L (= 9+4+7+5 = 25 days)
• Critical path: D - G - J - L (= 9+3+8+5 = 25 days)
• There are two critical paths of the same length, so delay on either one delays the project.

Answer: project duration = 25 days; critical path = A - F - I - L ; D - G - J - L

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (days). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(2)--> (2)
(1) --B(4)--> (3)
(2) --C(3)--> (4)
(2) - - dummy - -> (6)
(3) --E(5)--> (5)
(3) - - dummy - -> (6)
(4) --F(1)--> (10)
(5) --I(7)--> (9)
(6) --D(5)--> (7)
(7) --G(4)--> (8)
(7) --H(5)--> (9)
(7) - - dummy - -> (10)
(8) --M(7)--> (12)
(9) --K(3)--> (11)
(10) --J(2)--> (12)
(11) --L(6)--> (12)

The network has 12 events and 3 dummy activities. Start event is (1) and the end event is (12).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 2 = 2
• B: EST = 0; EFT = 0 + 4 = 4
• C: EST = EFT of A = 2; EFT = 2 + 3 = 5
• D: EST = max(A:2, B:4) = 4; EFT = 4 + 5 = 9
• E: EST = EFT of B = 4; EFT = 4 + 5 = 9
• F: EST = EFT of C = 5; EFT = 5 + 1 = 6
• G: EST = EFT of D = 9; EFT = 9 + 4 = 13
• H: EST = EFT of D = 9; EFT = 9 + 5 = 14
• I: EST = EFT of E = 9; EFT = 9 + 7 = 16
• J: EST = max(D:9, F:6) = 9; EFT = 9 + 2 = 11
• M: EST = EFT of G = 13; EFT = 13 + 7 = 20
• K: EST = max(H:14, I:16) = 16; EFT = 16 + 3 = 19
• L: EST = EFT of K = 19; EFT = 19 + 6 = 25

Project duration = largest EFT = 25 days.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• L: LFT = project duration = 25; LST = 25 - 6 = 19
• K: LFT = LST of L = 19; LST = 19 - 3 = 16
• M: LFT = project duration = 25; LST = 25 - 7 = 18
• J: LFT = project duration = 25; LST = 25 - 2 = 23
• I: LFT = LST of K = 16; LST = 16 - 7 = 9
• H: LFT = LST of K = 16; LST = 16 - 5 = 11
• G: LFT = LST of M = 18; LST = 18 - 4 = 14
• F: LFT = LST of J = 23; LST = 23 - 1 = 22
• E: LFT = LST of I = 9; LST = 9 - 5 = 4
• D: LFT = min(G:14, H:11, J:23) = 11; LST = 11 - 5 = 6
• C: LFT = LST of F = 22; LST = 22 - 3 = 19
• B: LFT = min(D:6, E:4) = 4; LST = 4 - 4 = 0
• A: LFT = min(C:19, D:6) = 6; LST = 6 - 2 = 4

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): B, E, I, K, L
• Critical path: B - E - I - K - L (= 4+5+7+3+6 = 25 days)

Answer: project duration = 25 days; critical path = B - E - I - K - L

The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (weeks). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(1)--> (2)
(1) --B(6)--> (3)
(2) - - dummy - -> (4)
(2) --E(4)--> (8)
(3) - - dummy - -> (4)
(3) --D(2)--> (7)
(4) --C(2)--> (5)
(5) --F(3)--> (6)
(6) - - dummy - -> (7)
(6) - - dummy - -> (8)
(7) --H(2)--> (9)
(8) --G(4)--> (9)
(9) --I(5)--> (10)
(10) --J(3)--> (11)

The network has 11 events and 4 dummy activities. Start event is (1) and the end event is (11).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 1 = 1
• B: EST = 0; EFT = 0 + 6 = 6
• E: EST = EFT of A = 1; EFT = 1 + 4 = 5
• C: EST = max(A:1, B:6) = 6; EFT = 6 + 2 = 8
• D: EST = EFT of B = 6; EFT = 6 + 2 = 8
• F: EST = EFT of C = 8; EFT = 8 + 3 = 11
• G: EST = max(E:5, F:11) = 11; EFT = 11 + 4 = 15
• H: EST = max(D:8, F:11) = 11; EFT = 11 + 2 = 13
• I: EST = max(G:15, H:13) = 15; EFT = 15 + 5 = 20
• J: EST = EFT of I = 20; EFT = 20 + 3 = 23

Project duration = largest EFT = 23 weeks.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• J: LFT = project duration = 23; LST = 23 - 3 = 20
• I: LFT = LST of J = 20; LST = 20 - 5 = 15
• H: LFT = LST of I = 15; LST = 15 - 2 = 13
• G: LFT = LST of I = 15; LST = 15 - 4 = 11
• F: LFT = min(G:11, H:13) = 11; LST = 11 - 3 = 8
• D: LFT = LST of H = 13; LST = 13 - 2 = 11
• C: LFT = LST of F = 8; LST = 8 - 2 = 6
• E: LFT = LST of G = 11; LST = 11 - 4 = 7
• B: LFT = min(C:6, D:11) = 6; LST = 6 - 6 = 0
• A: LFT = min(C:6, E:7) = 6; LST = 6 - 1 = 5

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): B, C, F, G, I, J
• Critical path: B - C - F - G - I - J (= 6+2+3+4+5+3 = 23 weeks)

Answer: project duration = 23 weeks; critical path = B - C - F - G - I - J

The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (months). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(1)--> (2)
(1) --B(3)--> (3)
(2) --C(2)--> (4)
(2) --D(2)--> (5)
(3) --E(5)--> (6)
(4) --G(3)--> (6)
(4) --F(1)--> (8)
(5) --H(3)--> (6)
(6) --K(4)--> (7)
(6) - - dummy - -> (8)
(6) --J(1)--> (9)
(7) --M(2)--> (10)
(8) --I(5)--> (9)
(9) --L(1)--> (10)

The network has 10 events and 1 dummy activity. Start event is (1) and the end event is (10).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 1 = 1
• B: EST = 0; EFT = 0 + 3 = 3
• C: EST = EFT of A = 1; EFT = 1 + 2 = 3
• D: EST = EFT of A = 1; EFT = 1 + 2 = 3
• E: EST = EFT of B = 3; EFT = 3 + 5 = 8
• F: EST = EFT of C = 3; EFT = 3 + 1 = 4
• G: EST = EFT of C = 3; EFT = 3 + 3 = 6
• H: EST = EFT of D = 3; EFT = 3 + 3 = 6
• I: EST = max(E:8, F:4, G:6, H:6) = 8; EFT = 8 + 5 = 13
• J: EST = max(E:8, G:6, H:6) = 8; EFT = 8 + 1 = 9
• K: EST = max(E:8, G:6, H:6) = 8; EFT = 8 + 4 = 12
• M: EST = EFT of K = 12; EFT = 12 + 2 = 14
• L: EST = max(I:13, J:9) = 13; EFT = 13 + 1 = 14

Project duration = largest EFT = 14 months.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• L: LFT = project duration = 14; LST = 14 - 1 = 13
• M: LFT = project duration = 14; LST = 14 - 2 = 12
• K: LFT = LST of M = 12; LST = 12 - 4 = 8
• J: LFT = LST of L = 13; LST = 13 - 1 = 12
• I: LFT = LST of L = 13; LST = 13 - 5 = 8
• H: LFT = min(I:8, J:12, K:8) = 8; LST = 8 - 3 = 5
• G: LFT = min(I:8, J:12, K:8) = 8; LST = 8 - 3 = 5
• F: LFT = LST of I = 8; LST = 8 - 1 = 7
• E: LFT = min(I:8, J:12, K:8) = 8; LST = 8 - 5 = 3
• D: LFT = LST of H = 5; LST = 5 - 2 = 3
• C: LFT = min(F:7, G:5) = 5; LST = 5 - 2 = 3
• B: LFT = LST of E = 3; LST = 3 - 3 = 0
• A: LFT = min(C:3, D:3) = 3; LST = 3 - 1 = 2

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): B, E, I, K, M, L
• Critical path: B - E - I - L (= 3+5+5+1 = 14 months)
• Critical path: B - E - K - M (= 3+5+4+2 = 14 months)
• There are two critical paths of the same length, so delay on either one delays the project.

Answer: project duration = 14 months; critical path = B - E - I - L ; B - E - K - M

A WBS divides the project into levels of increasing detail, each item carrying a code.

Four-level WBS of a rural road with a bridge project

Level 1  1.0 Rural road project
Level 2    1.1 Preliminary works
Level 3      1.1.1 Survey and design
Level 4        1.1.1.1 Topographic survey
Level 4        1.1.1.2 Geotechnical investigation
Level 3      1.1.2 Land and permits
Level 2    1.2 Road works
Level 3      1.2.1 Earthwork
Level 4        1.2.1.1 Clearing and grubbing
Level 4        1.2.1.2 Cutting and filling
Level 3      1.2.2 Pavement
Level 4        1.2.2.1 Sub-base and base
Level 4        1.2.2.2 Bituminous surfacing
Level 2    1.3 Bridge works
Level 3      1.3.1 Substructure
Level 4        1.3.1.1 Excavation and foundation
Level 4        1.3.1.2 Abutments and piers
Level 3      1.3.2 Superstructure
Level 4        1.3.2.1 Girders
Level 4        1.3.2.2 Deck slab
Level 2    1.4 Closure
Level 3      1.4.1 Testing and handover

Levels and coding

The code shows the parent of each item (1.2.1.1 belongs to 1.2.1, which belongs to 1.2). Cost, duration and responsible person are attached at level 3 or 4.

Resource allocation is the assigning of available resources (manpower, machines, materials, money) to the activities of a schedule so that the work can be done.

When activities are scheduled at their earliest times, the demand for a resource usually has high peaks and deep valleys, which is costly (hiring and firing, idle machines).

Resource smoothing (manpower levelling) rearranges the non-critical activities within their float so that the resource demand becomes as uniform as possible, without changing the project duration.

Process

1. Draw the network and bar chart at earliest start.
2. Prepare the resource histogram (resource vs time).
3. Find the peaks and valleys.
4. Shift non-critical activities within total float from peak periods to low periods.
5. Redraw the histogram; repeat until the profile is as flat as possible.
6. Critical activities are not moved.

Before        After
 |   ##          |  ## ##
 | ####          | ## ## ## 
 | ######        | ## ## ##
 +--------       +--------
 peak/valley     levelled

Benefits: lower peak demand, less idle time, better cost control, and easier procurement.

The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (weeks). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(3)--> (2)
(1) --B(2)--> (3)
(2) --C(4)--> (4)
(3) --E(3)--> (5)
(3) --D(3)--> (6)
(4) - - dummy - -> (6)
(4) --F(3)--> (7)
(5) --H(5)--> (7)
(6) --G(2)--> (7)
(7) --I(3)--> (8)

The network has 8 events and 1 dummy activity. Start event is (1) and the end event is (8).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 3 = 3
• B: EST = 0; EFT = 0 + 2 = 2
• D: EST = EFT of B = 2; EFT = 2 + 3 = 5
• E: EST = EFT of B = 2; EFT = 2 + 3 = 5
• C: EST = EFT of A = 3; EFT = 3 + 4 = 7
• H: EST = EFT of E = 5; EFT = 5 + 5 = 10
• F: EST = EFT of C = 7; EFT = 7 + 3 = 10
• G: EST = max(C:7, D:5) = 7; EFT = 7 + 2 = 9
• I: EST = max(F:10, G:9, H:10) = 10; EFT = 10 + 3 = 13

Project duration = largest EFT = 13 weeks.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• I: LFT = project duration = 13; LST = 13 - 3 = 10
• G: LFT = LST of I = 10; LST = 10 - 2 = 8
• F: LFT = LST of I = 10; LST = 10 - 3 = 7
• H: LFT = LST of I = 10; LST = 10 - 5 = 5
• C: LFT = min(F:7, G:8) = 7; LST = 7 - 4 = 3
• E: LFT = LST of H = 5; LST = 5 - 3 = 2
• D: LFT = LST of G = 8; LST = 8 - 3 = 5
• B: LFT = min(D:5, E:2) = 2; LST = 2 - 2 = 0
• A: LFT = LST of C = 3; LST = 3 - 3 = 0

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): A, B, E, C, H, F, I
• Critical path: A - C - F - I (= 3+4+3+3 = 13 weeks)
• Critical path: B - E - H - I (= 2+3+5+3 = 13 weeks)
• There are two critical paths of the same length, so delay on either one delays the project.

Answer: project duration = 13 weeks; critical path = A - C - F - I ; B - E - H - I

The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (weeks). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --B(3)--> (2)
(1) --A(1)--> (4)
(2) --D(5)--> (3)
(2) - - dummy - -> (4)
(2) --E(3)--> (5)
(3) - - dummy - -> (5)
(3) - - dummy - -> (6)
(3) --G(3)--> (7)
(4) --C(2)--> (6)
(5) --H(4)--> (7)
(6) --F(1)--> (7)
(7) --I(5)--> (8)

The network has 8 events and 3 dummy activities. Start event is (1) and the end event is (8).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 1 = 1
• B: EST = 0; EFT = 0 + 3 = 3
• C: EST = max(A:1, B:3) = 3; EFT = 3 + 2 = 5
• D: EST = EFT of B = 3; EFT = 3 + 5 = 8
• E: EST = EFT of B = 3; EFT = 3 + 3 = 6
• F: EST = max(C:5, D:8) = 8; EFT = 8 + 1 = 9
• G: EST = EFT of D = 8; EFT = 8 + 3 = 11
• H: EST = max(D:8, E:6) = 8; EFT = 8 + 4 = 12
• I: EST = max(F:9, G:11, H:12) = 12; EFT = 12 + 5 = 17

Project duration = largest EFT = 17 weeks.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• I: LFT = project duration = 17; LST = 17 - 5 = 12
• H: LFT = LST of I = 12; LST = 12 - 4 = 8
• G: LFT = LST of I = 12; LST = 12 - 3 = 9
• F: LFT = LST of I = 12; LST = 12 - 1 = 11
• E: LFT = LST of H = 8; LST = 8 - 3 = 5
• D: LFT = min(F:11, G:9, H:8) = 8; LST = 8 - 5 = 3
• C: LFT = LST of F = 11; LST = 11 - 2 = 9
• B: LFT = min(C:9, D:3, E:5) = 3; LST = 3 - 3 = 0
• A: LFT = LST of C = 9; LST = 9 - 1 = 8

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): B, D, H, I
• Critical path: B - D - H - I (= 3+5+4+5 = 17 weeks)

Answer: project duration = 17 weeks; critical path = B - D - H - I

Project planning is the process of defining the objectives and scope, deciding what work is to be done, how, by whom, when and with what resources, and preparing the schedule, budget and control methods to achieve the objectives. It answers what, why, how, who, when and how much.

Importance (advantages)

• Gives clear direction and common understanding of objectives.
• Reduces uncertainty by anticipating problems and risks.
• Ensures effective use of money, manpower, machines and materials.
• Helps to finish on time and within the budget by setting a schedule and a cost baseline.
• Provides a standard for control: actual progress is compared with the plan.
• Improves coordination between departments, contractors and consultants.
• Supports quality, safety and communication planning.
• Helps in decision making and in obtaining finance and approval.
• Gives confidence and motivation to the team and stakeholders.
• Reduces waste, conflicts and rework.

Resource leveling is the scheduling technique used when the resources available are limited. Activities are delayed, according to priority, so that the demand for a resource never exceeds its limit. The project duration may increase.

Process

1. Prepare the network and the earliest-start schedule.
2. Prepare the resource histogram for each scarce resource.
3. Compare demand with the availability limit; find the periods where demand exceeds it.
4. Delay (shift) non-critical activities using their float, in order of priority (least float first, shorter duration first).
5. If the limit is still exceeded, delay critical activities or split activities; this extends the project.
6. Recompute the schedule and histogram; repeat until demand is within the limit.
7. Check the new project duration and cost.

Priority rules

Least total float first; earliest late start; smallest duration; most resources.

Result: a feasible resource-constrained schedule. Compare with resource smoothing, which keeps the project duration fixed and only evens out the demand.

Linear programming (LP) is a mathematical technique to find the best (maximum or minimum) value of a linear objective function, subject to linear constraints on limited resources. In project work it is used for optimum allocation of resources, time-cost trade-off (crashing) and product mix.

Components

• Decision variables: quantities to be decided ($x$, $y$).
• Objective function: to maximise profit or minimise cost.
• Constraints: limits on resources such as labour, machine hours, material.
• Non-negativity: $x, y \ge 0$.

Methods

Graphical method (two variables), simplex method (many variables), and software (Excel Solver).

Example

A contractor makes two products, $x$ and $y$, with profit $40 and $30 per unit. Maximise $Z = 40x + 30y$ subject to $2x + y \le 100$, $x + y \le 80$, $x \le 40$, $x, y \ge 0$.

Corner points of the feasible region and values of $Z$:

Answer: maximum $Z = 2600$ at $x = 20$, $y = 60$.

Assumptions and limits

Linearity, certainty of data, divisibility, and a single objective. Real project data are often uncertain.

The components of CPM are: the network, event times (TE, TL), activity times (EST, EFT, LST, LFT), floats (TF, FF, Int.F, Ind.F), the critical path and the project duration. All are found below with the activity-on-arrow (AOA) method.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (time units). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(3)--> (2)
(1) --B(6)--> (3)
(1) --C(2)--> (4)
(2) --D(2)--> (6)
(3) --E(1)--> (5)
(3) - - dummy - -> (6)
(3) - - dummy - -> (7)
(4) --F(3)--> (7)
(5) --H(3)--> (8)
(6) --G(7)--> (8)
(7) --I(4)--> (8)

The network has 8 events and 2 dummy activities. Start event is (1) and the end event is (8).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 3 = 3
• B: EST = 0; EFT = 0 + 6 = 6
• C: EST = 0; EFT = 0 + 2 = 2
• F: EST = EFT of C = 2; EFT = 2 + 3 = 5
• D: EST = EFT of A = 3; EFT = 3 + 2 = 5
• E: EST = EFT of B = 6; EFT = 6 + 1 = 7
• G: EST = max(B:6, D:5) = 6; EFT = 6 + 7 = 13
• I: EST = max(B:6, F:5) = 6; EFT = 6 + 4 = 10
• H: EST = EFT of E = 7; EFT = 7 + 3 = 10

Project duration = largest EFT = 13 time units.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• H: LFT = project duration = 13; LST = 13 - 3 = 10
• I: LFT = project duration = 13; LST = 13 - 4 = 9
• G: LFT = project duration = 13; LST = 13 - 7 = 6
• E: LFT = LST of H = 10; LST = 10 - 1 = 9
• D: LFT = LST of G = 6; LST = 6 - 2 = 4
• F: LFT = LST of I = 9; LST = 9 - 3 = 6
• C: LFT = LST of F = 6; LST = 6 - 2 = 4
• B: LFT = min(E:9, G:6, I:9) = 6; LST = 6 - 6 = 0
• A: LFT = LST of D = 4; LST = 4 - 3 = 1

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): B, G
• Critical path: B - G (= 6+7 = 13 time units)

Answer: project duration = 13 time units; critical path = B - G

A bar chart (Gantt chart) is a graphical schedule in which activities are listed down the left side and time is shown across the top; each activity is a horizontal bar whose start, length and end show its start date, duration and finish date.

Activity   W1 W2 W3 W4 W5 W6 W7 W8
Survey     ####
Design        ######
Procure             ####
Build                  ##########
Test                            ####

Progress is shown by shading the completed part of each bar and a vertical "today" line.

Advantages

• Simple to prepare, read and understand, even for non-technical people.
• Shows the schedule and progress at a glance.
• Good for communication, reporting and small projects.
• Easily used for resource and cash-flow planning below the chart.
• Low cost; needs little training.

Limitations

• Does not show the dependency (logical relation) between activities clearly.
• Does not show the critical path or float, so priorities are unclear.
• Hard to update when changes occur; poor for large, complex projects.
• Gives no probability or uncertainty of time.
• Not suitable for trade-off analysis of time and cost.
• Result depends on the planner's judgement.

A resource schedule shows how many units of each resource (manpower, equipment, materials, money) are needed in each time period of the project, based on the activity schedule. It is shown by a resource histogram or table.

Microsoft Project (MS Project) is a widely used project planning and management software for building schedules, assigning resources and tracking progress.

Main features

• Create the WBS and task list with durations and links (finish-to-start, start-to-start and so on) and lag.
• Automatic CPM calculation: critical path, float, early and late dates.
• Gantt chart, network diagram and calendar views.
• Resource sheet: people, equipment, material, cost rates; assignment and over-allocation warnings; resource levelling.
• Budget and cost tracking; cash flow reports.
• Baseline saving, then tracking of actual progress and variance (earned value).
• Reports, filters, dashboards and export to Excel or PDF.

Advantages

Fast updating, handles large networks, reduces calculation errors, gives what-if analysis and clear reports.

Limitations

Needs training; results depend on correct input; may be too heavy for very small projects. Alternatives: Primavera P6, Excel.

Network diagram (activity on node)

Each box is: activity (duration). Arrows show precedence.

        +--> C(2) ---------> F(5) ------+
 A(1) --+                               |
        +--> D(4) ---+---> G(1) --------+--> K(2) --> L(4)
 B(3) --+            |                  |           ^
        |            +---> H(2) --+--> I(6) --------+
        |                         |                 |
        +--> E(3) ----------------+--> J(3) --------+

Links: A, B start; C after A; D after A and B; E after B; F after C and D; G and H after D; I and J after E and H; K after F, G and I; L after K and J.

Forward and backward passes

$EST$ = largest $EFT$ of predecessors; $LFT$ = smallest $LST$ of successors. Project duration $T = 21$ weeks. $TF = LST - EST$; $FF = \min(EST_{succ}) - EFT$; $Int.F = TF - FF$; $Ind.F = \max\{0, \min(EST_{succ}) - \max(LFT_{pred}) - D\}$.

Sample check, activity C: $ES = EF_A = 1$; $EF = 3$; $LF = LS_F = 10$; $LS = 8$; $TF = 8-1 = 7$; $FF = ES_F - EF_C = 7-3 = 4$; $Int.F = 3$; $Ind.F = 7 - 3 - 2 = 2$ (using $LF_A = 3$).

(i) Critical path

Activities with $TF = 0$ form the critical path B - D - H - I - K - L: $3+4+2+6+2+4 = 21$ weeks.

Answer: Critical activities: B, D, H, I, K, L. Project duration = 21 weeks.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (time units). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(2)--> (2)
(1) --B(4)--> (3)
(2) --C(5)--> (6)
(3) --D(5)--> (4)
(3) --E(3)--> (5)
(3) - - dummy - -> (6)
(4) --G(2)--> (7)
(5) --H(5)--> (7)
(6) --F(4)--> (7)
(7) --I(3)--> (8)

The network has 8 events and 1 dummy activity. Start event is (1) and the end event is (8).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 2 = 2
• B: EST = 0; EFT = 0 + 4 = 4
• C: EST = EFT of A = 2; EFT = 2 + 5 = 7
• D: EST = EFT of B = 4; EFT = 4 + 5 = 9
• E: EST = EFT of B = 4; EFT = 4 + 3 = 7
• F: EST = max(B:4, C:7) = 7; EFT = 7 + 4 = 11
• H: EST = EFT of E = 7; EFT = 7 + 5 = 12
• G: EST = EFT of D = 9; EFT = 9 + 2 = 11
• I: EST = max(F:11, G:11, H:12) = 12; EFT = 12 + 3 = 15

Project duration = largest EFT = 15 time units.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• I: LFT = project duration = 15; LST = 15 - 3 = 12
• G: LFT = LST of I = 12; LST = 12 - 2 = 10
• H: LFT = LST of I = 12; LST = 12 - 5 = 7
• F: LFT = LST of I = 12; LST = 12 - 4 = 8
• E: LFT = LST of H = 7; LST = 7 - 3 = 4
• D: LFT = LST of G = 10; LST = 10 - 5 = 5
• C: LFT = LST of F = 8; LST = 8 - 5 = 3
• B: LFT = min(D:5, E:4, F:8) = 4; LST = 4 - 4 = 0
• A: LFT = LST of C = 3; LST = 3 - 2 = 1

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): B, E, H, I
• Critical path: B - E - H - I (= 4+3+5+3 = 15 time units)

Answer: project duration = 15 time units; critical path = B - E - H - I

IF is taken as independent float.

Notation: EST/EFT = earliest start/finish, LST/LFT = latest start/finish, TF = total float, FF = free float, Int.F = interfering float, Ind.F = independent float.

Network diagram (activity-on-arrow)

Each activity is an arrow with its duration in brackets (days). Events are circled numbers; dashed arrows are dummy activities (zero duration) that show dependencies only.

(1) --A(5)--> (2)
(2) --B(4)--> (3)
(2) --C(2)--> (4)
(2) --D(3)--> (5)
(3) --E(2)--> (6)
(4) - - dummy - -> (6)
(4) --F(1)--> (7)
(5) --G(3)--> (7)
(6) --H(1)--> (8)
(7) --I(2)--> (8)

The network has 8 events and 1 dummy activity. Start event is (1) and the end event is (8).

Forward pass (EST and EFT)

Rule: EST = largest EFT of the preceding activities (0 for start activities); EFT = EST + D.

• A: EST = 0; EFT = 0 + 5 = 5
• B: EST = EFT of A = 5; EFT = 5 + 4 = 9
• C: EST = EFT of A = 5; EFT = 5 + 2 = 7
• D: EST = EFT of A = 5; EFT = 5 + 3 = 8
• F: EST = EFT of C = 7; EFT = 7 + 1 = 8
• G: EST = EFT of D = 8; EFT = 8 + 3 = 11
• E: EST = EFT of B = 9; EFT = 9 + 2 = 11
• H: EST = max(C:7, E:11) = 11; EFT = 11 + 1 = 12
• I: EST = max(F:8, G:11) = 11; EFT = 11 + 2 = 13

Project duration = largest EFT = 13 days.

Backward pass (LFT and LST)

Rule: LFT = smallest LST of the succeeding activities (project duration for end activities); LST = LFT - D.

• I: LFT = project duration = 13; LST = 13 - 2 = 11
• H: LFT = project duration = 13; LST = 13 - 1 = 12
• E: LFT = LST of H = 12; LST = 12 - 2 = 10
• G: LFT = LST of I = 11; LST = 11 - 3 = 8
• F: LFT = LST of I = 11; LST = 11 - 1 = 10
• D: LFT = LST of G = 8; LST = 8 - 3 = 5
• C: LFT = min(F:10, H:12) = 10; LST = 10 - 2 = 8
• B: LFT = LST of E = 10; LST = 10 - 4 = 6
• A: LFT = min(B:6, C:8, D:5) = 5; LST = 5 - 5 = 0

Floats

• Total float $TF = LST - EST = LFT - EFT$
• Free float $FF = \min(\text{EST of successors}) - EFT$
• Interfering float $IntF = TF - FF$
• Independent float $IndF = \max\left(0,\ \min(\text{EST of successors}) - \max(\text{LFT of predecessors}) - D\right)$

Result table

Event times (TE from the forward pass, TL from the backward pass):

Critical path and project duration

• Critical activities (TF = 0): A, D, G, I
• Critical path: A - D - G - I (= 5+3+3+2 = 13 days)

Answer: project duration = 13 days; critical path = A - D - G - I

What is a plan?

A plan is a pre-decided course of action that states what is to be done, how, by whom, when and with which resources so that the project objectives are achieved within time, cost and quality limits. For a building project, the plan contains the scope, WBS, schedule, resource and cost estimates, and responsibilities.

Importance of project planning

• Gives direction. Everyone knows the objectives and the route to reach them.
• Reduces uncertainty and risk. Problems are foreseen and handled in advance.
• Uses resources economically. Manpower, machines and materials are arranged when needed, so idle time and waste fall.
• Basis for control. The plan is the baseline against which actual progress, cost and quality are compared.
• Improves coordination. Departments, contractors and suppliers work to one programme.
• Helps decisions and finance. Cash flow, loan drawdown and procurement are planned from it.
• Saves time and cost. Critical activities are identified and given attention.
• Motivation and accountability. Clear targets and responsibilities.

Systematic process of project planning

1. Define objectives and scope - fix what the project must deliver, with its time, cost and quality targets.
2. Collect data and study constraints - site, market, law, funds, resources, weather.
3. Prepare the Work Breakdown Structure (WBS) - divide the project into manageable work packages.
4. List activities and fix their logical sequence - decide which activity precedes or follows which.
5. Estimate duration, resources and cost of each activity (quantity x productivity, rates).
6. Prepare the schedule - bar chart, milestone chart, CPM or PERT network; find the critical path.
7. Assign responsibility - responsibility matrix, organisation structure, contract packages.
8. Prepare resource, material, cash-flow and budget plans from the schedule.
9. Plan quality, safety, risk and communication.
10. Review, approve and issue the plan as the baseline; revise it when the actual progress differs (feedback).

Objectives -> WBS -> Activities & logic
     -> Estimates -> Schedule -> Resources
     -> Budget -> Baseline plan -> Control
            ^                        |
            +------ feedback -------+

Planning

Planning is the process of deciding in advance what to do, how to do it, when to do it, who will do it and with what resources, so as to achieve the project objectives in the best possible way. It bridges the gap between where the project is now and where it wants to be.

Features of good project planning

1. Objective-oriented - every part of the plan serves the stated project objectives.
2. Forward looking - it deals with the future and forecasts conditions such as price, weather and funds.
3. Systematic and logical - it follows a clear sequence from scope to WBS, schedule and budget.
4. Realistic and practical - durations, costs and resources are based on facts, past data and productivity norms.
5. Flexible - it can be revised when conditions change.
6. Clear and simple - it is easy to understand by all the project team.
7. Comprehensive and integrated - it covers time, cost, quality, resources and risk together.
8. Participative - managers, engineers, contractors and users give inputs.
9. Measurable - it has milestones, targets and indicators so that progress can be checked.
10. Continuous - it is reviewed and updated throughout the project life.

Work Breakdown Structure (WBS) as a planning tool

A WBS is a hierarchical, deliverable-oriented breakdown of the total project work into smaller and smaller parts (levels) down to work packages that can be estimated, scheduled, assigned and controlled. Level 1 is the project, level 2 the major parts, and the lowest level the work packages. Each element gets a code number.

House project
|-- 1 Site work
|     |-- 1.1 Clearing and layout
|     `-- 1.2 Excavation
|-- 2 Substructure
|     |-- 2.1 PCC and footing
|     `-- 2.2 Plinth beam and DPC
|-- 3 Superstructure
|     |-- 3.1 Columns and walls
|     `-- 3.2 Roof slab
`-- 4 Finishing
      |-- 4.1 Plastering
      |-- 4.2 Flooring, doors, windows
      `-- 4.3 Painting

How it works as a planning tool:

• The whole scope is covered, so no work is forgotten (100 % rule).
• Each work package is given duration, cost, resources and a responsible person.
• It is the starting point for the activity list, network diagram, bar chart, cost budget and Responsibility matrix.
• Costs and progress of lower levels can be rolled up to higher levels for reporting.

Importance of project planning

• Gives clear direction and fixes the objectives, scope, time and cost targets.
• Foresees problems and reduces risk and uncertainty.
• Gives economical use of manpower, machines, materials and money.
• Improves coordination between designers, contractors, suppliers and client.
• Provides the baseline for monitoring and control.
• Helps in arranging finance and cash flow in time.
• Reduces delay, cost overrun and rework.

Need of planning in a dynamic environment

A project works in an environment that keeps changing: prices of material and labour, exchange rates, government policy, weather, technology, availability of funds and the needs of the client. Without planning the project would be run by guesswork. Planning is necessary because it:

• forecasts the likely changes and prepares alternatives in advance;
• sets objectives and a baseline so that deviations can be measured;
• allocates scarce resources properly and avoids idle time and shortage;
• coordinates many parties (client, consultant, contractor, suppliers);
• allows quick re-planning when a change occurs, by updating the schedule and resources;
• reduces risk, delay and cost overrun.

Linked bar chart as a planning tool

A linked bar chart is a bar chart in which the bars are joined by arrows (links) that show the dependence between activities: the finish of a predecessor is linked to the start of its successor. It therefore combines the simple view of a bar chart with the logic of a network.

Example: a small building project (time in weeks).

                 0   4   8   12  16  20  24  (wk)
                +------------------------
A Site clearing |#>                      
B Excavation    | ##>                    
C Foundation    |   ###>                 
D Plinth beam   |      ##>               
E Walls+columns |        #####>          
F Roof slab     |             ###>       
G Plumb/electric|             ###>       
H Plastering    |                ###>    
I Floor,doors   |                   ###> 
J Painting      |                      ##
                +------------------------

Reading: arrow > means the successor starts when the predecessor finishes. The project takes 24 weeks. Roof (F) and plumbing (G) run in parallel after walls (E); plastering (H) waits for both. The links show that delay in E, F or G will push H, I and J, i.e. these are critical.

Justification: the chart is easy to prepare and read, shows start, finish and duration, and the links show which activities control the others, so it is suitable for planning and for explaining the schedule to site staff.

Limitations

• Becomes crowded and unclear when the number of activities is large.
• Does not clearly show the critical path or the float of each activity.
• Does not show uncertainty of durations (no probability).
• Difficult to revise: a change needs redrawing.
• Shows only simple finish-to-start dependence; resource and cost are not shown.

Planning and scheduling

• Planning decides what work is to be done, how (method and sequence), by whom and with what resources, to achieve the project objectives. It answers what, how, who.
• Scheduling puts the planned activities on a time scale. It fixes the start and finish date of each activity, the project duration and the float, answering when. Planning comes first; scheduling follows from it.

Linked bar chart: construction of a house (10 activities)

Linked bar chart (time in weeks; > shows the link from the finish of a predecessor to the start of the successor):

                 0   4   8   12  16  20  24  (wk)
                +------------------------
A Site clearing |#>                      
B Excavation    | ##>                    
C Foundation    |   ###>                 
D Plinth beam   |      ##>               
E Walls+columns |        #####>          
F Roof slab     |             ###>       
G Plumb/electric|             ###>       
H Plastering    |                ###>    
I Floor,doors   |                   ###> 
J Painting      |                      ##
                +------------------------

Start times: A 0, B 1, C 3, D 6, E 8, F 13, G 13, H 16, I 19, J 22. Project duration = 22 + 2 = 24 weeks. Activities F and G are done in parallel; all others are in a series, so the chain A-B-C-D-E-F(or G)-H-I-J controls the completion.

Planning tools

• Work Breakdown Structure (WBS) and Organisation Breakdown Structure (OBS)
• Bar chart (Gantt chart) and linked bar chart
• Milestone chart
• Network techniques: CPM and PERT (also precedence diagram method)
• Line of balance (for repetitive work) and S-curve
• Resource histogram, resource levelling and smoothing
• Responsibility matrix, checklists

Milestone chart as an improved bar chart

A milestone is a key event (a point in time, zero duration) such as completion of foundation or handing over. A bar chart shows many activities and their durations but not which events are the important control points. A milestone chart marks only these key events on a time scale, so the manager and the client see major targets at a glance and compare the planned date with the forecast date. It also removes the clutter of detailed bars, and a slip in a milestone is easily noticed. Hence it is an improvement over the bar chart.

Example: house project (24 weeks), milestones taken from the bar chart.

                       0   4   8   12  16  20  24 (wk)
                      +------------------------
M1 Site ready         |*
M2 Foundation done    |     *
M3 Structure done     |            *
M4 Roof+services done |               *
M5 Handover           |                       *
                      +------------------------

If M3 is forecast at week 15 instead of 13, the chart immediately warns that all later milestones are in danger.

Work Breakdown Structure

WBS divides the whole project into smaller manageable parts in levels (project, major parts, work packages) so that each part can be planned, estimated, assigned and controlled.

House project
|-- 1 Site work
|     |-- 1.1 Clearing and layout
|     `-- 1.2 Excavation
|-- 2 Substructure
|     |-- 2.1 PCC and footing
|     `-- 2.2 Plinth beam and DPC
|-- 3 Superstructure
|     |-- 3.1 Columns and walls
|     `-- 3.2 Roof slab
`-- 4 Finishing
      |-- 4.1 Plastering
      |-- 4.2 Flooring, doors, windows
      `-- 4.3 Painting

Need of project planning

• It sets objectives, scope, time and cost targets and gives direction to all parties.
• It foresees problems and reduces uncertainty and risk.
• It arranges resources (labour, machines, material, money) in time and avoids idle time and waste.
• It coordinates client, consultant, contractor and suppliers.
• It gives the baseline for monitoring and control.
• It reduces delay and cost overrun, and helps to arrange finance.

Planning tools

• Work Breakdown Structure (WBS) and Organisation Breakdown Structure (OBS)
• Bar chart (Gantt chart) and linked bar chart
• Milestone chart
• Network techniques: CPM and PERT (also precedence diagram method)
• Line of balance (for repetitive work) and S-curve
• Resource histogram, resource levelling and smoothing
• Responsibility matrix, checklists

Limitations of the conventional bar chart

• Does not show the logical dependence between activities, so the effect of a delay on other activities is not seen.
• Critical activities and float are not shown, so the manager cannot say where attention is needed.
• Not suitable for large and complex projects; the chart becomes crowded.
• Difficult to update; any change in logic needs redrawing.
• Durations are single-point values; uncertainty is not considered.
• Does not show resource or cost relations.

Example: bar chart of a house project (weeks).

                 0   4   8   12  16  20  24  (wk)
                +------------------------
A Site clearing |#                       
B Excavation    | ##                     
C Foundation    |   ###                  
D Plinth beam   |      ##                
E Walls+columns |        #####           
F Roof slab     |             ###        
G Plumb/electric|             ###        
H Plastering    |                ###     
I Floor,doors   |                   ###  
J Painting      |                      ##
                +------------------------

In this chart, the roof slab (F) and plumbing (G) start together, but the chart does not say that plastering (H) cannot start until both are finished. If F is delayed by 2 weeks, one cannot tell from the chart how H, I and J are affected, or whether the 24-week completion is in danger. It also does not show that G has free time while F does not, unlike a network.

A schedule converts the plan into a time-table: it gives the start and finish time of every activity and the project completion date. It is important because:

• Fixes the completion date. The project duration and the critical activities are known.
• Co-ordinates work. Each party knows when its work starts and what must be finished before it.
• Resource planning. Labour, equipment and material are arranged in time from the schedule, avoiding idle time and shortage.
• Cash flow and finance. Payments, loan withdrawals and bills depend on when the work is done.
• Basis for control. Actual progress is compared with the schedule; delay is detected early and corrective action taken.
• Float is known. Non-critical activities can be shifted to level resources.
• Contract and claims. The approved schedule is used to assess delay, extension of time and liquidated damages.
• Communication. Gives clients and management a clear picture of progress.

The method is chosen according to the size of the project, nature of the activities and the certainty of the time estimates.

Bar chart

Use it when:

• the project is small, simple or has few activities (a house, a small culvert);
• a quick, easy-to-read schedule is needed for site staff or for a presentation to top management;
• the work is at the preliminary stage and the logic is not yet detailed;
• progress is to be shown to people without technical training.

CPM (Critical Path Method)

Use it when:

• the project is large and complex, with many inter-related activities, but the durations can be estimated with fair certainty from experience (roads, buildings, bridges, canals);
• the critical path and floats are needed;
• time-cost trade-off (crashing) and resource scheduling are required;
• the project is repetitive or similar projects were done before.

PERT (Programme Evaluation and Review Technique)

Use it when:

• the work is new, research-type or has no past data (first-of-its-kind hydropower, tunnelling, software, defence projects);
• activity durations are uncertain and three estimates (optimistic, most likely, pessimistic) are needed;
• the probability of finishing the project by a given date is to be found;
• the main interest is time, not cost.

Gantt chart

A Gantt chart is a bar chart in which each activity is drawn as a horizontal bar on a time scale; the bar is filled in as the work progresses, so that planned and actual progress can be compared. Example: house project of 10 activities (progress shown at the end of week 10; # = planned, = = work done, | = status date).

                 0   4   8   12  16  20  24 (wk)
                +------------------------
A Site clearing |=         |             
B Excavation    | ==       |             
C Foundation    |   ===    |             
D Plinth beam   |      ==  |             
E Walls+columns |        ==###           
F Roof slab     |          |  ###        
G Plumb/electric|          |  ###        
H Plastering    |          |     ###     
I Floor,doors   |          |        ###  
J Painting      |          |           ##
                +------------------------
                           ^ status (wk 10)

Reading at week 10: A, B, C and D are complete (on plan); E (walls) has done 2 of its 5 weeks (planned: it started at week 8, so it is on schedule); the rest have not started. Total duration is 24 weeks.

Advantages

• Very simple to prepare, read and understand.
• Shows start, finish and duration of each activity and the total project time on one page.
• Planned and actual progress are compared directly, so delay is seen quickly.
• Good tool for communication with the client, management and site staff.
• Helps in planning of manpower, machine and material on a time scale.
• Cheap; no training or computer is needed.

Example project: part of a house project (weeks).

Gantt chart

A Gantt chart shows every activity as a bar against a calendar scale and records progress inside the bar. It is used to compare planned and actual progress. Here = shows work done up to week 6 and # the work still to do.

                 0   4   8   12  16 (wk)
                +----------------
A Site clearing |=     |         
B Excavation    | ==   |         
C Foundation    |   ===|         
D Plinth beam   |      ##        
E Walls+columns |      | #####   
F Roof slab     |      |      ###
                +----------------
                       ^ status (wk 6)

At week 6, A, B and C are done, D is just finished and E, F have not yet started.

Linked (link) bar chart

A bar chart in which arrows or links join the finish of a predecessor to the start of its successor. It shows the dependence among the activities. In the figure > shows that the next activity starts when the previous one finishes.

                 0   4   8   12  16  (wk)
                +----------------
A Site clearing |#>              
B Excavation    | ##>            
C Foundation    |   ###>         
D Plinth beam   |      ##>       
E Walls+columns |        #####>  
F Roof slab     |             ###
                +----------------

Milestone chart

It shows only the key events (milestones) of the project on the time scale, such as site ready, foundation complete and structure complete. It is used by top management to see the major targets and to compare planned and forecast dates.

                       0   4   8   12  16 (wk)
                      +----------------
M1 Site ready         |*
M2 Foundation done    |     *
M3 Structure done     |            *
                      +----------------

Assumption: a small irrigation scheme (weir, main canal and field channels) of 16 months.

Activities and durations

Bar chart (months)

                       0   4   8   12  16  (wk)
                      +----------------
A Survey & design     |##              
B Land acquisition    | ###            
C Headworks (weir)    |   ######       
D Main canal earthwork|    #####       
E Canal lining        |        ####    
F Canal structures    |      ######    
G Field channels      |           #### 
H Testing, handover   |               #
                      +----------------

Milestone chart

Milestones are key events that mark completion of a major stage; * shows the planned date.

                       0   4   8   12  16 (wk)
                      +----------------
M1 Design approved    | *
M2 Land ready         |   *
M3 Headworks done     |        *
M4 Main canal done    |           *
M5 Distribution done  |              *
M6 Handover           |               *
                      +----------------

The project is finished in 16 months. Land acquisition must end by month 4 and design approval by month 2; these early milestones are the main control points.

Bar chart

A bar chart shows each activity as a horizontal bar on a time scale; the length of the bar is the duration. Example: house project (weeks).

                 0   4   8   12  16  20  24  (wk)
                +------------------------
A Site clearing |#                       
B Excavation    | ##                     
C Foundation    |   ###                  
D Plinth beam   |      ##                
E Walls+columns |        #####           
F Roof slab     |             ###        
G Plumb/electric|             ###        
H Plastering    |                ###     
I Floor,doors   |                   ###  
J Painting      |                      ##
                +------------------------

Advantages

• Simple to prepare, read and understand, even for non-technical persons.
• Shows start, finish, duration and overlap of activities on a time scale.
• Cheap and quick; needs no special training.
• Good for presenting the summary to top management and site staff.
• Progress can be marked on the bars and compared with the plan.
• Useful for resource scheduling on a time scale.

Disadvantages

• Does not show the logical dependence between activities, so the effect of a delay on other activities is not seen.
• Critical activities and float are not shown, so the manager cannot say where attention is needed.
• Not suitable for large and complex projects; the chart becomes crowded.
• Difficult to update; any change in logic needs redrawing.
• Durations are single-point values; uncertainty is not considered.
• Does not show resource or cost relations.

Expected time and variance (PERT)

Given: optimistic time $t_o = 6$, most likely time $t_m = 8$, pessimistic time $t_p = 12$.

$$\begin{aligned} t_e &= \frac{t_o + 4t_m + t_p}{6} = \frac{6 + 4(8) + 12}{6} = \frac{50}{6} = 8.33 \\ \sigma &= \frac{t_p - t_o}{6} = \frac{12 - 6}{6} = 1 \\ \text{Variance} &= \sigma^2 = 1^2 = 1 \end{aligned}$$

Answer: $t_e = 8.33$ and variance $= 1$ (standard deviation 1).

Gantt chart

A Gantt chart is a bar chart in which each activity is a horizontal bar on a time scale; the bar is filled in as work progresses, so that planned and actual progress are seen together. It is used for scheduling and monitoring.

Example (days): A excavation (3 days, 4 men), C foundation (3 days, 2 men, after A) and B fencing and store (2 days, 2 men, independent). = shows work done up to the status at the end of day 3.

                  0 2 4 6 (day)
                 +------
A Excavation     |===|  
C Foundation     |   ###
B Fencing, store |== |  
                 +------
                     ^ status (day 3)

Resource allocation

Resource allocation is the assignment of the available manpower, machines and materials to the activities of the schedule. The daily requirement of every activity (early start schedule) is added to get a resource aggregation chart, drawn as a resource histogram. If the demand is uneven or higher than the supply, the schedule is adjusted. Two cases occur:

1. Resource levelling (time-limited) - project duration is fixed; activities are shifted inside their float to get a smooth demand.
2. Limited resource allocation (resource-limited) - the resource available is fixed; activities are delayed, even beyond float, so the duration may increase.

Resource smoothing

Smoothing is the adjustment of non-critical activities within their total float to remove peaks and valleys of the histogram without changing the project duration or the critical path.

In the example the critical chain is A-C (6 days); B has a float of 4 days. Early start histogram (B on days 1-2):

  6 |# #         
  5 |# #         
  4 |# # #       
  3 |# # #       
  2 |# # # # # # 
  1 |# # # # # # 
    +------------
     1 2 3 4 5 6  day
     6 6 4 2 2 2  men

Peak = 6 men. Starting B on day 4 (inside its float) gives:

  4 |# # # # #   
  3 |# # # # #   
  2 |# # # # # # 
  1 |# # # # # # 
    +------------
     1 2 3 4 5 6  day
     4 4 4 4 4 2  men

The peak falls from 6 to 4 men; total man-days (22) and the completion time (6 days) are unchanged. Benefits: steady employment, less hiring and firing, better use of equipment.

Project planning

Project planning is the process of defining the objectives and scope of a project and deciding in advance the work, methods, sequence, time, cost and resources needed to complete it. The result is a plan used as baseline for execution and control.

Work Breakdown Structure (WBS)

A WBS divides the whole project into smaller, manageable work packages in a hierarchy of levels.

House project
|-- 1 Site work
|     |-- 1.1 Clearing and layout
|     `-- 1.2 Excavation
|-- 2 Substructure
|     |-- 2.1 PCC and footing
|     `-- 2.2 Plinth beam and DPC
|-- 3 Superstructure
|     |-- 3.1 Columns and walls
|     `-- 3.2 Roof slab
`-- 4 Finishing
      |-- 4.1 Plastering
      |-- 4.2 Flooring, doors, windows
      `-- 4.3 Painting

Importance of WBS for project planning

1. Covers the complete scope. Every work item appears once, so nothing is forgotten and nothing is repeated.
2. Base for the activity list. Work packages are expanded into activities for the bar chart or network diagram.
3. Better estimates. Cost, duration and resource are easier to estimate for small packages than for the whole project.
4. Fixes responsibility. Each package is assigned to one person, contractor or section (responsibility matrix).
5. Basis for budget and cost control. Costs are collected package-wise and rolled up to higher levels.
6. Progress measurement. Percent complete of packages is combined into overall progress.
7. Improves communication. The team and client share one understanding of the project.
8. Helps in contract packaging and procurement.

A dummy activity is an imaginary activity of zero duration and zero resource, shown by a dashed arrow in an arrow (AOA) network. It only shows a logical dependence.

Applications in scheduling:

1. Showing correct logic. When D depends on A and B but C depends on A only, a dummy from the end of A to the start of D keeps the logic right.
2. Unique numbering. When two activities start and end at the same events (parallel activities), a dummy gives them different end events so that each has a unique i-j number.
3. Connecting events so that the network has one start and one end event.

Rule: C depends on A only; D depends on A and B

(1) --A--> (2)
(1) --B--> (3)
(2) --C--> (4)
(2) - - dummy - -> (3)
(3) --D--> (5)

Dummy (2)->(3): D starts only after A and B end,
while C starts after A alone.

Critical path

The critical path is the longest continuous chain of activities from the start to the end of the project network; it fixes the shortest time in which the project can be completed. Activities on it have zero total float.

Significance:

• It gives the minimum project duration.
• Any delay in a critical activity delays the whole project, so these activities need the closest monitoring and the first call on resources.
• To shorten the project, only critical activities should be crashed.
• Non-critical activities have float and can be shifted to level resources.

Dummy activity

A dummy activity is a zero-duration, zero-resource activity (dashed arrow) used in the arrow diagram.

Significance:

• It maintains the correct logical dependence between activities that cannot be shown by arrows alone.
• It gives a unique i-j number to parallel activities.
• It does not consume time or resource, so it does not change the duration, but it may lie on the critical path.

Definition

The critical path is the longest path (sequence of connected activities) from the start event to the end event of the network. Its length is the project duration, and it is the path with zero total float.

Characteristics

1. It is the longest path in the network; no other path takes more time.
2. All activities on it are critical: TF = 0, so EST = LST and EFT = LFT.
3. Events on it have zero slack: TE = TL.
4. It runs continuously from the start to the end event.
5. A network may have more than one critical path.
6. A delay in any critical activity delays the whole project by the same time.
7. It can change when activity durations change (a non-critical path may become critical when its float is used).

Importance

• Gives the minimum project completion time.
• Shows where management attention and best resources should go.
• Only critical activities need to be shortened for crashing the project.
• Float on other activities allows shifting and resource levelling.
• Base for monitoring, updating and claims for time extension.