Unit-3Network models and TechniquesOutline: Network, directed network, paths, cycles, connected network, tree and spanning tree connected shortest path model, minimum spanning tree problem, Max-flow problem and Min-cost problemProject Management:Phases of project management, guidelines for network construction, CPM (Critical Path Method) and PERT (Program Evaluation and Review Technique)Introduction:Application of Graph theory combined with optimization theory to practical problems is the domain of Network analysis. There is a multitude of OR situations that can be modeled and solved as networks. According to recent survey, as much as 70% of the real-world mathematical programming problems can be represented by network-related models.The following is the list illustrating the possible application areas of networks.1) Determination of shortest route between two cities in a network of roads2) Determination of the minimum-cost flow from oil fields to refineries through pipelines3) Determination of an effective time-schedule for completion of a projectMain Features of Network modelsThe following features are mainly responsible for the widespread use of network models in practice. Visual appealNetwork models allow us to depict the problems graphically. These network diagrams make the problems easily to understand.FlexibilityNetwork models are inherently flexible in the sense that models having similar structure can be analyzed, in addition to the application areas mentioned earlier. SolvabilitySeveral computationally efficient algorithms are available for solving most network flow problems. These algorithms attain efficiency due to special structure inherent in the network models.We present six of these algorithms.

1) Minimum spanning tree (MST)2) Shortest-route algorithm3) Maximum-flow algorithm4) Minimum-cost algorithm5) Critical path method (CPM)6) Programme Evaluation and Review Technique (PERT)

A brief Review of Graph TheoryDefinitions and basic conceptsA graph (network) is a pair (N,E) where N is the set of points(nodes) and E the set of lines(edges, arcs, branches, links). To illustrate consider the following sets:N={1,2,3,4,5), E={1-2,1-3,1-4,2-3,4-3,3-5,4-5,}The network is described by the following figure.

Network

Tree Spanning Tree

Directed or Oriented arcsAssociated with each network is some type of flow (i.e., oil flow in the pipeline, traffic flow on the highways). An arc is said to be directed or oriented if it allows positive flow in one direction and zero flow in the opposite direction. A directed network has all its arcs directed. Paths, Cycles and TreesA path is a sequence of distinct links that join two nodes through other nodes regardless of the direction of flow in each arc. A directed path from node i to node j is a sequence of connecting links having flows towards node j. A path forms a cycle if it connects a node to itself through other nodes. A cycle is directed if it consists of directed path.A connected network is a network where every two distinct nodes are connected by at least one path. Note that connecting path need not be directed even if the network is connected. A tree is a connected network that may involve only a subset of all the nodes of the network, without having any cycle in between. A spanning tree is a tree that links all the nodes of the network.Minimum Spanning Tree AlgorithmThe minimal spanning tree algorithm deals with linking the nodes of a network, using the shortest length of connecting branches. Suppose we are given a connected, undirected, weighted graph G = (N, E). We assume that the weights are real numbers. Our task is to find the minimum spanning tree of G, i.e., the spanning tree T minimizing the function

The steps of the procedure are given as follows.Let N={1,2,3, …,n} be the set of nodes of the network. At each iteration, one node is added to the set of nodes that have been permanently connected.

Step 1.The algorithm starts with node 1, connect it to the nearest distinct node. Tick (*) all nodes that have been connectedStep 2. Identify the unconnected node that is closest to the connected node and connect these two nodes. Step 3. Repeat step 2 until all the nodes are connected.The resulting tree is guaranteed to be a minimal spanning tree (optimal solution).Sample problemExample 6.2-1 ( From H.A.Taha)

Solution: The algorithm starts at node 1, which gives S ={1}, E = { }Iteration 0: Connect node 1 to 2 so that Iteration 1: Choose a neighbor 3 nearest to 1 and 2 so that

Repeat till number of edges in E is 5The minimum spanning tree is shown with red color and one of the edges with green color can be chosen.The length of the tree is 1+3+4+3+5=16The resulting minimum cable in miles needed to provide the desired cable services is 16 miles.

Practice problem:Build the lowest-cost road system that will connect the cities in the network below.

Solution: The number of vertices is 8 and the minimum spanning tree would have 7 edges.We start from any vertex, say A Iteration 0: Connect a to the nearest neighbor C so that V = {A,C}, E = { AC}Iteration 1: Now B is the nearest neighbor to V = {A,C} and edge CB does not form a cycle with AC. Here the edge AB can also be chosen. Iteration 2: V = {A, C, B}, E = {AC, CB}Now E and F are both nearest to V = {A, C, B}. Let us choose F and the corresponding edge CF.Then V = {A, C, B, F} E = {AC, CB, CF}Iterate until the number of edges in E is 7. The total length of the road is 2+3+5+4+2+2+3=21Shortest-route (path) problemThis is a labeling algorithm. It finds the shortest route from the starting node to any other node in the net work. At each iteration, a new node is assigned a label. This label gives the shortest distance from the starting node to this node. Let s and t be two vertices of a connected weighted graph G. Shortest path problem is to find a path from s to t whose total weight is minimum. We now discuss an algorithm due to Dijkstra which efficiently solves this problem. The idea is to grow a tree, starting from the vertex s, by adding, at each iteration, a frontier edge, whose non-tree endpoint is as close as possible to s. The algorithm involves assigning labels to vertices.Dijkstra's algorithm, conceived by Dutch computer scientist Edsger Dijkstra in 1956 and published in 1959, is a graph search algorithm that solves the single-source shortest path problem for a graph with nonnegative edge path costs, producing a shortest path tree.For a given source vertex (node) in the graph, the algorithm finds the path with lowest cost (i.e. the shortest path) between that vertex and every other vertex. It can also be used for finding costs of shortest paths from a single vertex to a single destination vertex by stopping the algorithm once the shortest path to the destination vertex has been determined. For example, if the vertices of the graph represent cities and edge path costs represent driving distances between pairs of cities connected by a direct road, Dijkstra's algorithm can be used to find the shortest route between one city and all other cities.

Let the node at which we are starting be called the initial node. For each vertex x, let dist[x] denote the distance from vertex s to x and for each edge e, let w(e) be its edge weight.

After each iteration, the vertices in the tree with labeled vertices are those to which the shortest paths have been found. Priority of the frontier edge:Let e be a frontier edge and let its P value be given byP(e) = dist[x]+w(e) where x is the labeled endpoint of e and w(e) its weight.Then, the edge with the smallest P value is given the highest priority. The P value of this highest priority edge e gives the distance from the vertex s to the unlabeled endpoint of e.Dijkstra’s Shortest Path AlgorithmInitialize the algorithm at vertex s and the set of frontier edges as empty.Write label 0 on vertex s. Let e be the frontier edge that has smallest P value. Let x be the labeled endpoint of edge e and y the unlabeled endpoint of e.Add e to the tree. Define dist[y] = P(e)Write label dist[y] on vertex y.Repeat until all the vertices are covered. Sample problemExample 6.3-4 (From H.A.Taha)

Solution: Let us initialize at the vertex 1.Iteration 0: Initialize the tree as vertex 1 and write label 0 on vertex 1. Since dist[3]=1, label vertex 3 with 1. Iteration 1: P(1,2) = 5, P(3,2) = 1+2 = 3,P(1,4) = 5+7 = 12, P(3,4) = 1+6 = 7, P(1,5)= 5+1=6, P(3,4) =1 +7=8,P(1,6) = 5+6 = 11, P(1,7) = 5+6+2 = 8, P(3,5)= 1+7 = 8, P(3,6) = 1+7+4 = 13, P(3,7) = 17Since, P(3,2) = 1+2 = 3 minimum label vertex with 3.Iteration 2: P(2,4)= 5+7 = 12, P(3,4)=1+6 = 7 (minimum) , Label 4 with 7Iteration 3: P(1,5) = 1+7 = 8, P(2,5) = 5+1 =6, P(3,5) = 1+2+1 =4(minimum),P(4,5) = 1+6+3 = 10Label 5 with 4Iteration 4: P(1,6)=5+6 = 11, P(2,6)=5+7+4 = 16, P(3,6)=1+7+5= 12, P(4,6) = 1+6+4 = 11(minimum)Label 6 with 11Iteration 5: P(1,7)=5+6+2= 8(minimum), P(2,7)= 5+1+9 = 15,P(3,7) = 1+7+19 = 17, P(4,7) = 5+7+6 = 18, P(5,7)=5+1+9=15, P(6,7) = 5+6+2 = 13 Label 7 with 8.Shortest routes and distances of vertices 2, 3, 4, 5, 6, 7 from 1 are as under.Route Distance 3

2

1 7

4 9

13 Practice problem:Apply Dijkstra algorithm to find the shortest path from s to each other vertex in the graph given below.

Solution:Iteration 0: Label 0 on vertex s. and label 8 on z.Iteration 1: P(sy)=16, P(sw)=13, P(zy)=8+7=15, P(zx)=17+8=25,P(zw)=8+11=19, P(zv)=8+10=18 The edge sw has the highest priority and label 13 on w.Iteration 2: P(sy)=16, P(zy)=15, P(zx)=25, P(zv)=18, P(wx)=13+14=27The edge zy has the highest priority, Label 15 on y.Iteration 3: P(zv)=18, P(zx)=25, P(wx)=27, P(yx)=20The edge zv has the highest priority. Label 18 on v.Iteration 4: P(yx)=20, P(zx)=25, P(wx)=27yx is the edge of highest priority. Label 20 on x.All the verices are labeled. The following tree shows the shortest paths from s to any other vertex.

16

Maximum-flow and Minimum cost problemsThe maximum flow problem is again structured on a network; but here the arc capacities, or upper bounds, are the only relevant parameters. The problem is to find the maximum flow possible from some given source node to a given sink node. Maximum flow problem and minimum cost problem are part of a broad topic Transport Networks.Transport NetworksIn graph theory, a flow network (also known as a transportation network) is a directed graph where each edge has a capacity and each edge receives a flow. The amount of flow on an edge cannot exceed the capacity of the edge. Often in Operations Research, a directed graph is called a network, the vertices are called nodes and the edges are called arcs. A flow must satisfy the restriction that the amount of flow into a node equals the amount of flow out of it, except when it is a source, which has more outgoing flow, or sink, which has more incoming flow. A network can be used to model traffic in a road system, fluids in pipes, currents in an electrical circuit, or anything similar in which something travels through a network of nodes.An important application of weighted directed graphs is to model transport networks like water supply network, traffic network of highways, communication networks of cables, power supply network of wires etc. The purpose of modeling these networks is to maximize the flow of water, electricity, traffic, information etc. We desire that the flow through the network should be maximum. Such a flow is called maximum flow. Definition: A transport network is a simple weighted directed graph with the following properties.

1) There is a designated vertex called the source, that has no incoming edge2) There is a designated vertex called the sink, that has no outgoing edge 3) There is a nonnegative weight called the capacity of the edge

Since the graph is simple each edge is unidirectional. Definition: A flow in a network is a nonnegative number assigned to each edge not exceeding its capacity such that the total flow leaving the source is equal to the total flow entering the sink.For example consider the oil pipeline network as shown below.

The capacities of the flow are cOA= 5, cAB= 4, cBS= 5, cOC= 3, cCD= 4, cDS= 5, cCB= 2, cDB= 2The flows into the edges are fOA= 3, fAB= 3, fBS= 5, fOC= 3, fCD= 2, fDS= 1, fCB= 1, fDB= 1We note that, flow into the vertex B = 3+1+1 = 5, and flow out of the vertex B = 5 Also, the total flow leaving the source is 3+3= 5+1 = 6, the total flow entering the sink. Therefore, Conservation of flow is satisfied.

The value of the flow is defined as the amount of flow passing from the source to the sink. The maximum flow problem is to maximize the value of flow, which is to route as much flow as possible from O to S.Should we need to model a network with more than one source, a super source is introduced to the graph. This consists of a vertex connected to each of the sources with edges of infinite capacity, so as to act as a global source. A similar construct for sinks is called a super sink.In a minimum cost flow problem, each edge has a given cost, and the cost of sending the flow across the edge is cij. The objective is to send a given amount of flow from the source to the sink, at the lowest possible price.ExampleConsider the water pipeline network shown below for two towns A and B in which water is supplied from four bore wells w1, w2, w3, w4 and a, b, c, d represent intermediate pumping stations. Determine the value of the flow.

Solution:To obtain the equivalent the required network we tie together the given sources into a super source O and the towns A and B into a super sink S as shown below.

The value of the flow is f = 3+2+3+2=10.Definition: A maximal flow in a network is a flow with maximum value.In general, there will be several flows having the same maximum value. For example, consider the network shown below.

Consider the following two flows for the network:The value of the flow in fig (a) is 10 and three of the five edges in the network attain their maximum capacity. But for the same network the value of the flow in fig,(b) is 12 and four out of the five edges are carrying maximum capacity. Therefore, the flow in the second case is better than the first one.

Fig. (a) Fig. (b)

Example:Find the maximum possible increase in the following network.

Solution:We cannot increase the flow in the path OABS because the capacity and the flow of the edge AB are equal. But if we increase flow in the path OCDS by 1, we get the network shown in Fig.(a). Further, if we increase the flow in the path OCBS by 1, we get the network shown in Fig. (b).No further increase in flow is possible since capacities of all edges leading to the sink have been exhausted. The value of the flow is 7.

(5,5)

Fig (a) Fig. (b)

Example: Find the maximal flow in the network shown below.

Solution: Let the initial flow be 0 in each edge. Thus we have the network as shown in Fig. (a). Increase the flow in the path OAS to get the network as shown in Fig. (b). Now increase the flow by 2 in the path OBS to get the network as shown in Fig. (c). Finally, increase the flow by 2 in the path OABS to get the network as shown in Fig. (d).

Any further increase in the flow is not possible. Thus the maximal flow in the network has been reached. The value of the flow = 2+4 = 6.

Minimum cuts

A cut of a network N divides it into two non-empty parts. Formally speaking, a cut is a partition (U,V) of the node set V into two non-empty subsets U and V such that O

The following figure shows a cut of the transport network.

The dashed line divides the vertex set {O,A,B,C,D,S}into sets U= {O,A}, V={C,B,D,S}

In fact, a cut does cut network into two pieces, one containing the source and the other sink, such that nothing can flow from source to sink if edges of a cut are removed.

Note that if the edges in the cut-set of N are removed, | f | = 0.

A cut edge is an edge that is crossed by the cut, that is, whose tail belongs to U and whose head belongs to V, or vice-versa.

OA, AB, OC are the cut edges.

The value of a cut is the sum of all weights (capacities) of all edges that cross the cut from U to V, that is, whose tail belongs to U and whose head belongs to V.

C (U, V) =5+2+3=10

A minimum cut is a cut of minimum value.

A well-known theorem from graph theory states that the maximum flow is always equal to the minimum capacity of over all the cuts.

Max flow Min cut theorem: The maximum value of flow is equal to the minimum value of the cut.

The minimum cut problem is minimizing the capacity that is; to determine U and V such that the capacity of the U-V cut is minimal. This is the dual of maximum flow problem.

Example: In the network shown below, find the maximum flow and verify by appealing to Max flow Min cut theorem.

Solution: The initial labeling yields the network (a). Increase the flow by 1 in the path OAS to get network (b). Increase the flow by 1 in the path OCS to get network (c). Further increase the flow by 1 in the path OABS to get network (d). Now increase the flow by 1 in the path OCBS to get network (e). Lastly increase the flow by 2 in the path OBS to get network (f). In fact, the capacities in the edges leading to the sink are exhausted so that no further increase is possible.

(a) (b) (c)

(d) (e) (f)

We note that the total value f=2+2+2 = 1+4+1= 6

We now verify this using Max flow Min cut theorem. We consider the following cuts.

(i) K1 = {OC,OA, OB} with C(K1) = 2+3+2 = 7(ii) K2 = {OC, OB, AB, AS} with C(K2) = 2+3+1+1 = 7(iii) K3 = {OC, BC, BS, AS} with C(K3) = 2+1+4+1 = 8(iv) K4 = {CS, BS, AS} with C(K4) = 1+4+1 = 6We note that the minimum cut value is 6. Hence the maximum flow is 6.

The minimum-cost flow problem

The minimum-cost flow problem is finding the cheapest possible way of sending a certain amount of flow through a flow network.

Travelling Salesman Problem (TSP)

TSP can be modeled as an undirected weighted graph, such that cities are the graph's vertices, paths are the graph's edges, and a path's distance is the edge's length. It is a minimization problem starting and finishing at a specified vertex after having visited each other vertex exactly once. Often, the model is a complete graph (i.e. each pair of vertices is connected by an edge). If no path exists between two cities, adding an arbitrarily long edge will complete the graph without affecting the optimal tour.An equivalent formulation in terms of graph theory is: Given a complete weighted graph (where the vertices would represent the cities, the edges would represent the roads, and the weights would be the cost or distance of that road), find a Hamiltonian cycle with the least weight.

CPM and PERT Techniques

Basically, CPM (Critical Path Method) and PERT (Programme Evaluation Review Technique) are project management techniques, which have been created out of the need of industrial and military establishments to plan, schedule and control complex projects.

A Brief Introduction and Historical background of CPM/PERT

CPM/PERT or Network Analysis as the technique is sometimes called, developed along two parallel streams, one industrial and the other military. The two techniques were almost developed simultaneously (1956-58) by two different groups of engineers working on the Polar’s Missile Programmes of the U.S Navy. Both methods have been applied to wide variety of problems in the industry and commerce. The methods have also found applications in Government organizations.

In CPM firstly the network diagram is drawn and the required time to do each operation is written to the left side of each operation circle. Then these times are combined to develop a schedule which minimizes or maximizes the measure of performance for each operation. The schedule thus obtained is used to determine the required critical path. The CPM can be effectively used in production planning, scheduling traffic, communication networks etc.

Planning, Scheduling & Control

Planning, Scheduling (or organizing) and Control are considered to be basic Managerial functions, and CPM/PERT has been rightfully accorded due importance in the literature on Operations Research and Quantitative Analysis.

Far more than the technical benefits, it was found that PERT/CPM provided a focus around which managers could brain-storm and put their ideas together. It proved to be a great communication medium by which thinkers and planners at one level could communicate their ideas, their doubts and fears to another level. Most important, it became a useful tool for evaluating the performance of individuals and teams.

There are many variations of CPM/PERT which have been useful in planning costs, scheduling manpower and machine time. CPM/PERT can answer the following important questions:

How long will the entire project take to be completed? What are the risks involved?

Which are the critical activities or tasks in the project which could delay the entire project if they were not completed on time?

Is the project on schedule, behind schedule or ahead of schedule?

If the project has to be finished earlier than planned, what is the best way to do this at the least cost?

The Framework for PERT and CPM

Essentially, there are six steps which are common to both the techniques. The procedure is listed below:

I. Define the Project and all of it’s significant activities or tasks. The Project (made up of several tasks) should have only a single start activity and a single finish activity.

II. Develop the relationships among the activities. Decide which activities must precede and which must follow others.

III. Draw the "Network" connecting all the activities. Each Activity should have unique event numbers. Dummy arrows are used where required to avoid giving the same numbering to two activities.

IV. Assign time and/or cost estimates to each activityV. Compute the longest time path through the network. This is called the critical

path.VI. Use the Network to help plan, schedule, monitor and control the project.

The Key Concept used by CPM/PERT is that a small set of activities, which make up the longest path through the activity network control the entire project. If these "critical" activities could be identified and assigned to responsible persons, management resources could be optimally used by concentrating on the few activities which determine the fate of the entire project.

Non-critical activities can be re-planned, rescheduled and resources for them can be reallocated flexibly, without affecting the whole project.

Five useful questions to ask when preparing an activity network are:

Is this a Start Activity? Is this a Finish Activity? What Activity Precedes this? What Activity Follows this? What Activity is Concurrent with this?

Some activities are serially linked. The second activity can begin only after the first activity is completed. In certain cases, the activities are concurrent, because they are independent of each other and can start simultaneously. This is especially the case in organizations which have supervisory resources so that work can be delegated to various departments which will be responsible for the activities and their completion as planned. When work is delegated like this, the need for constant feedback and co-ordination becomes an important senior management pre-occupation.

Drawing the CPM/PERT Network

At Event 3, we have to evaluate two predecessor activities - Activity 1-3 and Activity 2-3, both of which are predecessor activities. Activity 1-3 gives us an Earliest Start of 3 weeks at Event 3. However, Activity 2-3 also has to be completed before Event 3 can begin. Along this route, the Earliest Start would be 4+0=4. The rule is to take the longer (bigger) of the two Earliest Starts. So the Earliest Start at event 3 is 4.

Similarly, at Event 4, we find we have to evaluate two predecessor activities - Activity 2-4 and Activity 3-4. Along Activity 2-4, the Earliest Start at Event 4 would be 10 wks, but along Activity 3-4, the Earliest Start at Event 4 would be 11 wks. Since 11 wks is larger than 10 wks, we select it as the Earliest Start at Event 4.We have now found the longest path through the network. It will take 11 weeks along activities 1-2, 2-3 and 3-4. This is the Critical Path.

The Backward Pass - Latest Finish Time Rule

To make the Backward Pass, we begin at the sink or the final event and work backwards Each activity (or sub-project) in a PERT/CPM Network is represented by an arrow symbol. Each activity is preceded and succeeded by an event, represented as a circle and

numbered.

to the first event.

At Event 3 there is only one activity, Activity 3-4 in the backward pass, and we find that the value is 11-7 = 4 weeks. However at Event 2 we have to evaluate 2 activities, 2-3 and 2-4. We find that the backward pass through 2-4 gives us a value of 11-6 = 5 while 2-3 gives us 4-0 = 4. We take the smaller value of 4 on the backward pass.

Tabulation & Analysis of Activities

We are now ready to tabulate the various events and calculate the Earliest and Latest Start and Finish times. We are also now ready to compute the SLACK or TOTAL FLOAT, which is defined as the difference between the Latest Start and Earliest Start.

Event Duration(Weeks) Earliest Start

Earliest Finish

Latest Start

Latest Finish

Total Float

1-2 4 0 4 0 4 0

2-3 0 4 4 4 4 0

3-4 7 4 11 4 11 0

1-3 3 0 3 1 4 1

2-4 6 4 10 5 11 1

The Earliest Start is the value in the rectangle near the tail of each activity The Earliest Finish is = Earliest Start + Duration The Latest Finish is the value in the diamond at the head of each activity The Latest Start is = Latest Finish - Duration

There are two important types of Float or Slack. These are Total Float and Free Float.

TOTAL FLOAT is the spare time available when all preceding activities occur at the earliest possible times and all succeeding activities occur at the latest possible times.

Total Float = Latest Start - Earliest Start

Activities with zero Total float are on the Critical Path

FREE FLOAT is the spare time available when all preceding activities occur at the earliest possible times and all succeeding activities occur at the earliest possible times.

When an activity has zero Total float, Free float will also be zero.

There are various other types of float (Independent, Early Free, Early Interfering, Late Free, Late Interfering), and float can also be negative. We shall not go into these situations at present for the sake of simplicity and be concerned only with Total Float for the time being.

Having computed the various parameters of each activity, we are now ready to go into the scheduling phase, using a type of bar chart known as the Gantt Chart.

There are various other types of float (Independent, Early Free, Early Interfering, Late Free, Late Interfering), and float can also be negative. We shall not go into these situations at present for the sake of simplicity and be concerned only with Total Float for the time being. Having computed the various parameters of each activity, we are now ready to go into the scheduling phase, using a type of bar chart known as the Gantt Chart.

Scheduling of Activities Using a Gantt Chart

Once the activities are laid out along a Gantt Chart (Please see chart below), the concepts of Earliest Start & Finish, Latest Start & Finish and Float will become very obvious.

Activities 1-3 and 2-4 have total float of 1 week each, represented by the solid timeline which begins at the latest start and ends at the latest finish. The difference is the float, which gives us the flexibility to schedule the activity.

For example, we might send the staff on leave during that one week or give them some other work to do. Or we may choose to start the activity slightly later than planned, knowing that we have a week’s float in hand. We might even break the activity in the middle (if this is permitted) for a week and divert the staff for some other work, or declare a National or Festival holiday as required under the National and Festival Holidays Act.

These are some of the examples of the use of float to schedule an activity. Once all the activities that can be scheduled are scheduled to the convenience of the project, normally reflecting resource optimization measures, we can say that the project has been scheduled.

Exercise

A Social Project manager is faced with a project with the following activities:

Activity-id Activity - Description Duration

1-2 Social Work Team to live in Village 5 Weeks

1-3 Social Research Team to do survey 12 Weeks

3-4 Analyse results of survey 5 Weeks

2-4 Establish Mother & Child Health Program 14 Weeks

3-5 Establish Rural Credit Programme 15 Weeks

4-5 Carry out Immunisation of Under Fives 4 Weeks

Draw the arrow diagram, using the helpful numbering of the activities, which suggests the following logic: Unless the Social Work team lives in the village, the Mother and Child Health Programme cannot be started due

to ignorance and superstition of the villagers The Analysis of the survey can obviously be done only after the survey is complete. Until rural survey is done, the Rural Credit Programme cannot be started Unless Mother and Child Programme is established, the Immunisation of Under Fives cannot be started - Calculate the Earliest and Latest Event Times - Tabulate and Analyse the Activities - Schedule the Project Using a Gantt Chart

The PERT (Probabilistic) Approach

So far we have talked about projects, where there is high certainty about the outcomes of activities. In other words, the cause-effect logic is well known. This is particularly the case in Engineering projects.

However, in Research & Development projects, or in Social Projects which are defined as "Process Projects", where learning is an important outcome, the cause-effect relationship is not so well established.

In such situations, the PERT approach is useful, because it can accommodate the variation in event completion times, based on an expert’s or an expert committee’s estimates.

For each activity, three time estimates are taken

The Most Optimistic The Most Likely The Most Pessimistic

The Duration of an activity is calculated using the following formula:

Where te is the Expected time, to is the Optimistic time, tm is the most probable activity time and tp is the Pessimistic time.

It is not necessary to go into the theory behind the formula. It is enough to know that the weights are based on an

approximation of the Beta distribution.

The Standard Deviation, which is a good measure of the variability of each activity is calculated by the rather simplified formula:

The Variance is the Square of the Standard Deviation.

PERT Calculations for the Social Project

In our Social Project, the Project Manager is now not so certain that each activity will be completed on the basis of the single estimate he gave. There are many assumptions involved in each estimate, and these assumptions are illustrated in the three-time estimate he would prefer to give to each activity.

In Activity 1-3, the time estimates are 3,12 and 21. Using our PERT formula, we get:

The Standard Deviation (s.d.) for this activity is also calculated using the PERT formula

We calculate the PERT event times and other details as below for each activity:

Event to tm tp te ES EF LS LF TF s.d. Var.

1-3 3 12 21 12 0 12 0 12 0 3 9

3-5 6 15 30 16 12 28 12 28 0 4 16

1-2 2 5 14 6 0 6 5 11 5 2 4

2-4 5 14 17 13 6 19 11 24 5 2 4

3-4 2 5 8 5 12 17 19 24 7 1 1

4-5 1 4 7 4 19 23 24 28 5 1 1

5. Estimating Risk

Having calculated the s.d. and the Variance, we are ready to do some risk analysis. Before that we should be aware of two of the most important assumptions made by PERT.

The Beta distribution is appropriate for calculation of activity durations.

Activities are independent, and the time required to complete one activity has no bearing on the completion times of it’s successor activities in the network. The validity of this assumption is questionable when we consider that in practice, many activities have dependencies.

Expected Length of a Project

PERT assumes that the expected length of a project (or a sequence of independent activities) is simply the sum of their separate expected lengths.

Thus the summation of all the te's along the critical path gives us the length of the project.

Similarly the variance of a sum of independent activity times is equal to the sum of their individual variances.

In our example, the sum of the variance of the activity times along the critical path, VT is found to be equal to (9+16) = 25.

The square root VT gives us the standard deviation of the project length. Thus, ST=Ö 25=5. The higher the standard deviation, the greater the uncertainty that the project will be completed on the due date.

Although the te's are randomly distributed, the average or expected project length Te approximately follows a Normal Distribution.

Since we have a lot of information about a Normal Distribution, we can make several statistically significant conclusions from these calculations.

A random variable drawn from a Normal Distribution has 0.68 probability of falling within one standard deviation of the distribution average. Therefore, there is a 68% chance that the actual project duration will be within one standard

deviation, ST of the estimated average length of the project, te.

In our case, the te = (12+16) = 28 weeks and the ST = 5 weeks. Assuming te to be normally distributed, we can state that there is a probability of 0.68 that the project will be completed within 28 ± 5 weeks, which is to say, between 23 and 33 weeks.

Since it is known that just over 95% (.954) of the area under a Normal Distribution falls within two standard deviations, we can state that the probability that the project will be completed within 28 ± 10 is very high at 0.95.

Probability of Project Completion by Due Date

Now, although the project is estimated to be completed within 28 weeks (t e=28) our Project Director would like to know what is the probability that the project might be completed within 25 weeks (i.e. Due Date or D=25).

For this calculation, we use the formula for calculating Z, the number of standard deviations that D is away from te.

By looking at the following extract from a standard normal table, we see that the probability associated with a Z of -0.6 is 0.274. This means that the chance of the project being completed within 25 weeks, instead of the expected 28 weeks is about 2 out of 7. Not very encouraging.

On the other hand, the probability that the project will be completed within 33 weeks is calculated as follows:

The probability associated with Z= +1 is 0.84134. This is a strong probability, and indicates that the odds are 16 to 3 that the project will be completed by the due date.

If the probability of an event is p, the odds for its occurrence are a to b, where: