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Appendix E

Arrow Notation Scheduling Calculations

E.1. CPM CALCULATIONS (ARROWNOTATION)

In making calculations with arrow notation, the arrow and its two associated nodes have

attributes that are formally defined as symbols for mathematical purposes. This formal

notation associated with the arrow is shown in Figure E.1.

The left-hand node on the arrow represents the event time at which the activity begins.

It is referred to as the i node. The right-hand node represents the end time of the activity. It

is referred to as the j node. Associated with each node is an earliest time, which is shown as

TE

i for the i node and TE

j for the j node. Similarly, each node can have a latest event time,

which is shown in the figure as TL

i for the i node and TL

j for the j node. This establishes

four events, two associated with starting and two with ending nodes, which are of interest

in calculating the critical path of the network. The duration of the activity (as shown in the

figure) is given as tij. Because the starting and ending nodes in arrow notation are referred

to as i and j, arrow notation is sometimes referred to as i-j notation.

A schematic diagram representing the application of the forward-pass algorithm using

arrow notation is shown in Figure E.2. The objective of the forward-pass algorithm is to

calculate the earliest point in time at which a given event can occur. That is, the algorithm

calculates the earliest event time of a given node. The earliest event time for a given node

is controlled by the earliest event times of each of the set of events that precede it. The

algorithm is given as follows:

All i

TE

j = max[TE

i + tij]

iM

where M is the set of all i events that immediately precede j.

The earliest event time for a given node j is controlled by the earliest event times of

each of the i nodes that precede it. Each i node plus the duration of the associated activity,

tij which links it to the j node, must be investigated. The maximum of the preceding i node

early event times plus the durations of the appropriate activity ij controls the earliest time

at which a given event j, can occur.

To demonstrate this, consider Figure E.2. Node 30 is preceded by nodes 22, 25, and

26. The durations of the activities emanating from each of these nodes are as follows:

Act 22,30 t22,30 = 7 days

Act 25,30 t25,30 = 2 days

Act 26,30 t26,30 = 6 days

The earliest event times for each of the preceding nodes are as follows:

TE

22 = 10

TE

25 = 13

TE

26 = 15

347

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348 Appendix E

Figure E.1 Arrow notation symbols.

The algorithm for the early event time for node 30 is then

TE

30 = max(TE

22 + t22,30, TE

25 + t25,30, TE

26 + t26,30)

or

TE

30 = max(10 + 7, 13 + 2, 15 + 6) = 21

E.2. CALCULATING THE EARLY EVENT TIMES (ARROWNOTATION)

In order to understand howthe forward-pass algorithm is applied, consider the arrownotation

network model of the small gas station in Figure E.3. In order to record the calculated values

of the early event time (TE

i), a partitioned oval is located above each node. The calculated

early event time is recorded in the left side of the oval. During the backward pass, the late

event times for each node will be recorded in the right side.

The forward-pass algorithm is applied repetitively starting with the source node (node

A) and moving from left to right in a ÒbootstrappingÓ fashion. The starting node A is given

an early event time of zero (0). Moving to node B, the set of preceding events consists of

only one event. Therefore, TE

B is max (TE

A + tAB) = max (0 + 10) = 10. Calculations for

all of the nodes are shown in Table E.1. The values for each node are shown in Figure E.3.

The earliest time at which each activity can begin is given by the TE

i value for the i

node associated with the activity of interest. In addition to this information, it is now clear

that the minimum duration of the project is 96 days since the earliest time at which node S

can be realized has been calculated as 96 time units.

E.3. BACKWARD-PASS ALGORITHM (ARROWNOTATION)

A schematic diagram representing the application of the backward-pass algorithm is shown

in Figure E.4. The backward-pass algorithm calculates the latest time at which each event

can occur. The latest event time for a node i is controlled by the latest event times of the

set of events that follow it. The late event time of each j node minus the duration of the

associated activity, ij, must be investigated. The minimum of the following j node late event

times minus the duration of activity ij controls the latest time at which the i event can occur.

To demonstrate this, consider Figure E.4.

Figure E.2 Schematic of forward-pass calculation.

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Appendix E 349

Figure E.3 Expanded Project Model with Early Event Times.

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350 Appendix E

Table E.1 Calculation of Early Event Times

Node Formula Numerical Value T E

i

A N/A N/A 0

B TE

B = max (TE

A + tAB) max (0 + 10) 10

C TE

C = max (TE

B + tBC, TE

A + tAC ) max (10 + 8, 0 + 15) 18

D TE

D = max (TE

C + tCD) max (18 + 0) 18

E TEE

= max (TE

C + tCE) max (18 + 0) 18

F TE

F = max (TE

B + tBF, TE

C + tCF, TE

D + tDF,

TEE

+ tEF)

max (10 + 12, 18 + 2, 18

+ 5, 18 + 6)

24

G TEG

= max (TE

F + tFG) max (24 + 8) 32

H TE

H = max (TEG

+ tGH) max (32 + 10) 42

I TEI

= max (TE

H + tHI ) max (42 + 0) 42

J TEJ

= max (TE

H + tHJ) max (42 + 15) 57

K TEK

= max (TE

H + tHK, TEI

+ tIK ) max (42 + 14, 42 + 4) 56

L TE

L = max (TEJ

+ tJL, TEK

+ tKL) max (57 + 10, 56 + 0) 67

M TEM

= max (TE

L + tLM) max (67 + 0) 67

N TEN

= max (TEO

+ tON) max (77 + 0) 77

O TEO

= max (TE

L + tLO, TEM

+ tMO) max (67 + 6, 67 + 10) 77

P TE

P = max (TEO

+ tOP) max (77 + 0) 77

Q TE

Q = max (TEK

+ tKQ, TEN

+ tNQ, TEO

+ tOQ,

TE

P + tPQ, TE

L + tLO)

max (56 + 12, 77 + 3,

77 + 6, 77 + 8, 67 + 25)

92

R TE

R = max (TE

Q + tQR) max (92 + 1) 93

S TE

S = max (TE

R + tS ) max (93 + 3) 96

Node i, labeled 18, is followed by nodes 21, 23, and 25. The durations of the associated

ij activities are:

Act 18,21 t18,21 = 12

Act 18,23 t18,23 = 3

Act 18,25 t18,25 = 10

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