This course contains the basics of Quantitative Techniques For Management

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Pragnya Meter Exam

##### List of Topics

- QUANTITATIVE TECHNIQUES – INTRODUCTION
- Measures of Central Tendency
- Mathematical Model
- Linear Programming: Graphical Method
- Linear Programming: Simplex Method
- Transportation Model
- ASSIGNMENT MODEL
- NETWORK MODEL
- WAITING MODEL (QUEUING THEORY)
- PROBABILITY
- THEORETICAL PROBABILITY DISTRIBUTIONS
- PROBABILITY DISTRIBUTION OF A RANDOM VARIABLE
- INVENTORY MODEL
- Game Theory
- Simulation

**NETWORK MODEL**

Network models are possibly still the most important of the special structures in linear programming. Any project involves planning, scheduling and controlling a number of interrelated activities with use of limited resources, namely, men, machines, materials, money and time. The projects may be extremely large and complex such as construction of a power plant, a highway, a shopping complex, ships and aircraft, introduction of new products and research and development projects.

It is required that managers must have a dynamic planning and scheduling system to produce the best possible results and also to react immediately to the changing conditions and make necessary changes in the plan and schedule. A convenient analytical and visual technique of PERT and CPM prove extremely valuable in assisting the managers in managing the projects.

Basically, CPM (Critical Path Method) and PERT (Program Evaluation Review Technique) are project management techniques, which have been created out of the need of Western industrial and military establishments to plan, schedule and control complex projects. Both the techniques use similar terminology and have the same purpose.

*PERT: *

This was developed by the US Navy for the planning and control of the Polaris missile program and the emphasis was on completing the program in the shortest possible time. In addition PERT had the ability to cope with uncertain activity completion times**. **

In PERT activities are shown as a network of precedence relationships using activity-on-arrow network construction

- Multiple time estimates
- Probabilistic activity times

PERT is used in project management for non-repetitive jobs (research and development work), where the time and cost estimates tend to be quite uncertain. This technique uses probabilistic time estimates.

*CPM:*

CPM (Critical Path Method) was developed by Du Pont and the emphasis was on the trade-off between the cost of the project and its overall completion time (e.g. for certain activities it may be possible to decrease their completion times by spending more money - how does this affect the overall completion time of the project.

In CPM activities are shown as a network of precedence relationships using activity-on-node network construction

- Single estimate of activity time
- Deterministic activity times

CPM is used in production management for the jobs which are repetitive in nature where the activity time estimates can be predicted with considerable certainty due to the existence of past experience.

Project management generally consists of three phases.

** Planning: **Planning involves setting the objectives of the project. Identifying various activities to be performed and determining the requirement of resources such as men, materials, machines, etc. The cost and time for all the activities are estimated, and a network diagram is developed showing sequential interrelationships (predecessor and successor) between various activities during the planning stage.

** Scheduling: **Based on the time estimates, the start and finish times for each activity are worked out by applying forward and backward pass techniques, critical path is identified, along with the slack and float for the non-critical paths.

** Controlling: **Controlling refers to analyzing and evaluating the actual progress against the plan. Reallocation of resources, crashing and review of projects with periodical reports are carried out.

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?

PERT / CPM networks contain two major components

- Activities, and
- Events

*Activity**: *An activity represents an action and consumption of resources (time, money, energy) required to complete a portion of a project. Activity is represented by an arrow.

*An Activity*

*Event**: *An event (or node) will always occur at the beginning and end of an activity. The event has no resources and is represented by a circle. The ith event and jth event are the tail event and head event respectively.

*An Event*

*Merge and Burst Events*

One or more activities can start and end simultaneously at an event.

**Preceding and Succeeding Activities **

Activities performed before given events are known as ** preceding activities**, and activities performed after a given event are known as

*Preceding and Succeeding Activities*

Activities A and B precede activities C and D respectively.

*Dummy Activity*

An imaginary activity which does not consume any resource and time is called a ** dummy activity. **Dummy activities are simply used to represent a connection between events in order to maintain a logic in the network. It is represented by a dotted line in a network, see Figure.

*Dummy Activity*

a. Two activities starting from a tail event must not have a same end event. To ensure this, it is absolutely necessary to introduce a dummy activity, as shown in Figure.

*Correct and Incorrect Activities*

b. **Looping error **should not be formed in a network, as it represents performance

of activities repeatedly in a cyclic manner, as shown below in Figure.

*Looping Error*

c. In a network, there should be only one start event and one ending event as shown

below, in Figure.

*Only One Start and End Event*

d. The direction of arrows should flow from left to right avoiding mixing of direction

as shown in Figure.

*Wrong Direction of Arrows*

- No single activity can be represented more than once in a network. The length of an arrow has no significance.
- The event numbered 1 is the start event and an event with highest number is the end event. Before an activity can be undertaken, all activities preceding it must be completed. That is, the activities must follow a logical sequence (or – interrelationship) between activities.
- In assigning numbers to events, there should not be any duplication of event numbers in a network.
- Dummy activities must be used only if it is necessary to reduce the complexity of a network.
- A network should have only one start event and one end event.

Some conventions of network diagram are shown in Figures below:

** Some Conventions followed in making Network Diagrams**

*Step1***: **Number the start or initial event as 1.

*Step2***: **From event 1, strike off all outgoing activities. This would have made one or more events as initial events (event which do not have incoming activities).

Number that event as 2.

*Step3***: **Repeat step 2 for event 2, event 3 and till the end event. The end event must have the highest number.

** Example : **Draw a network for a house construction project. The sequence of activities with their predecessors is given in the following table, below.

*Sequence of Activities for House Construction Project*

*Solution:*

*Network diagram representing house construction project.*

The network diagram in the above figure shows the procedure relationship between the activities. Activity A (preparation of house plan), has a start event 1 as well as an ending event 2. Activity B (Construction of house) begins at event 2 and ends at event 3.

The activity B cannot start until activity A has been completed. Activities C and D cannot begin until activity B has been completed, but they can be performed simultaneously. Similarly, activities E and F can start only after completion of activities C and D respectively. Both activities E and F finish at the end of event 6.

** Example : **Consider the project given in the following table and construct a network diagram.

*Sequence of Activities for Building Construction Project*

** Solution: **The activities C and D have a common predecessor A. The network representation shown in Figure

*Network representing the Error*

**Correct representation of Network using Dummy Activity**

** Example : **Construct a network for a project whose activities and their predecessor relationship are given in Table.

*Activity Sequence for a Project*

** Solution: **The network diagram for the given problem is shown in Figure with activities A, B and C starting simultaneously.

* Network Diagram*

** Example : **Draw a network diagram for a project given in Table.

*Project Activity Sequence*

** Solution: **An activity network diagram describing the project is shown in figure below:

*Network Diagram*

The critical path for any network is the longest path through the entire network. Since all activities must be completed to complete the entire project, the length of the critical path is also the shortest time allowable for completion of the project. Thus if the project is to be completed in that shortest time, all activities on the critical path must be started as soon as possible.

These activities are called **critical activities**. If the project has to be completed ahead of the schedule, then the time required for at least one of the critical activity must be reduced. Further, any delay in completing the critical activities will increase the project duration.

The activity, which does not lie on the critical path, is called **non-critical activity. **These non-critical activities may have some slack time. The slack is the amount of time by which the start of an activity may be delayed without affecting the overall completion time of the project. But a critical activity has no slack. To reduce the overall project time, it would require more resources (at extra cost) to reduce the time taken by the critical activities to complete.

*Scheduling of Activities: Earliest Time and Latest Time*

Before the critical path in a network is determined, it is necessary to find the earliest and latest time of each event to know the earliest expected time (TE) at which the activities originating from the event can be started and to know the latest allowable time (TL) at which activities terminating at the event can be completed.

*Forward Pass Computations (to calculate Earliest, Time TE)*

*Procedure*

*Step 1***: **Begin from the start event and move towards the end event.

*Step 2***: **Put TE = 0 for the start event.

** Step 3: **Go to the next event (i.e node 2) if there is an incoming activity for event 2, add calculate TE of previous event (i.e event 1) and activity time.

** Note: **If there are more than one incoming activities, calculate TE for all incoming activities and take the maximum value. This value is the TE for event 2.

*Step 4***: **Repeat the same procedure from step 3 till the end event.

*Backward Pass Computations (to calculate Latest Time T **Network Model L)*

*Procedure*

*Step 1***: **Begin from end event and move towards the start event. Assume that the direction of arrows is reversed.

*Step 2***: **Latest Time TL for the last event is the earliest time. TE of the last event.

*Step 3***: **Go to the next event, if there is an incoming activity, subtract the value of TL of previous event from the activity duration time. The arrived value is TL for that event. If there are more than one incoming activities, take the minimum TE value.

*Step 4***: **Repeat the same procedure from step 2 till the start event.

As discussed earlier, the non – critical activities have some slack or float. The ** float **of an activity is the amount of time available by which it is possible to delay its completion time without extending the overall project completion time.

For an activity i = j, let

t_{ij} = duration of activity

TE = earliest expected time

T_{L} = latest allowable time

ES_{ij} = earliest start time of the activity

EF_{ij} = earliest finish time of the activity

LS_{ij} = latest start time of the activity

LF_{ij} = latest finish time of the activity

** Total Float TF_{ij}: **The total float of an activity is the difference between the latest start

time and the earliest start time of that activity.

TF_{ij} = LS_{ij}– ES_{ij} ....................(1)

or

TF_{ij} = (T_{L}– TE) – t_{ij} ....................(2)** Free Float FF_{ij}: **The time by which the completion of an activity can be delayed from its earliest finish time without affecting the earliest start time of the succeeding activity is called free float.

FF_{ij} = (E_{j}– E_{i}) – t_{ij} ....................(3)

FF_{ij} = Total float – Head event slack

** Independent Float IF_{ij}: **The amount of time by which the start of an activity can be delayed without affecting the earliest start time of any immediately following activities, assuming that the preceding activity has finished at its latest finish time.

IF

IF

Where tail event slack = L

The negative value of independent float is considered to be zero.

** Critical Path: **After determining the earliest and the latest scheduled times for various activities, the minimum time required to complete the project is calculated. In a network, among various paths, the longest path which determines the total time duration of the project is called the

An activity is said to be critical only if both the conditions are satisfied.

1. T_{L}– T_{E} = 0

2. T_{Lj}– t_{ij}– T_{Ej} = 0

** Example : **A project schedule has the following characteristics as shown in the table

**Project Schedule**

i. Construct PERT network.

ii. Compute T_{E} and T_{L} for each activity.

iii. Find the critical path.

*Solution:*

(i) From the data given in the problem, the activity network is constructed as shown in the following figure.

**Activity Network Diagram**

(ii) To determine the critical path, compute the earliest, time T Network Model E and latest time T_{L} for each of the activity of the project. The calculations of TE and T_{L} are as follows:

To calculate T_{E}for all activities,

T_{E1} = 0

T_{E2} = T_{E1} + t_{1}, 2 = 0 + 4 = 4

T_{E3} = T_{E1} + t_{1}, 3 = 0 + 1 =1

T_{E4} = max (T_{E2} + t_{2}, 4 and T_{E3} + t_{3}, 4)

= max (4 + 1 and 1 + 1) = max (5, 2)

= 5 days

T_{E5} = T_{E3} + t 3, 6 = 1 + 6 = 7

T_{E6} = T_{E5} + t 5, 6 = 7 + 4 = 11

T_{E7} = T_{E5} + t_{5}, 7 = 7 + 8 = 15

T_{E8} = max (T_{E6} + t 6, 8 and T_{E7} + t_{7}, 8)

= max (11 + 1 and 15 + 2) = max (12, 17)

= 17 days

T_{E9} = T_{E4} + t_{4}, 9 = 5 + 5 = 10

T_{E10 }= max (T_{E9} + t_{9}, 10 and T_{E8} + t_{8}, 10)

= max (10 + 7 and 17 + 5) = max (17, 22)

= 22 days

To calculate T_{L} for all activities

T_{L10} = T_{E10} = 22

T_{L9 }= T_{E10}– t_{9,10} = 22 – 7 = 15

T_{L8} = T_{E10}– t_{ 8, 10} = 22 – 5 = 17

T_{L7} = T_{E8}– t _{7, 8} = 17 – 2 = 15

T_{L6} = T_{E8}– t _{6, 8} = 17 – 1 = 16

T_{L5 }= min (T_{E6}– t_{5, 6} and T_{E7}– t_{5, 7})

= min (16 – 4 and 15 –8) = min (12, 7)

= 7 days

T_{L4} = T_{L9 }– t_{ 4, 9} = 15 – 5 =10

T_{L3} = min (T_{L4 }– t_{3, 4 }and T_{L55}– t _{3, 5})

= min (10 – 1 and 7 – 6) = min (9, 1)

= 1 day

T_{L2} = T_{L4}– t_{2}, 4 = 10 – 1 = 9

T_{L1} = Min (T_{L2}– t_{1, 2} and T_{L3}– t_{1, 3})

= Min (9 – 4 and 1 – 1) = 0

*Various Activities and their Floats*

(iii) From the table, we observe that the activities 1 – 3, 3 – 5, 5 – 7,7 – 8 and 8 – 10 are critical activities as their floats are zero.

*Critical Path of the Project*

The critical path is 1-3-5-7-8-10 (shown in double line in the above figure) with the project duration of 22 days.

Go to **MAIN MENU **and select **PROJECT PLANNING **and **CPM – CRITICAL PATH METHOD **. Enter the values of the network problem as shown in the above figure.

*Solving Network Problem on Computer Using TORA (Input Screen)*

Now select **SOLVE MENU **and **GO TO OUTPUT SCREEN**. There are two options for output, select CPM calculations. For step-by-step calculation of earliest time and latest time using forward pass and backward pass procedure click **NEXT STEP **button. To get all the values instantly, then press **ALL STEPS **button. The screen gives all the required values to analyze the problem. You may note that at the bottom of the table, the critical activities are highlighted in red colour. The output screen is shown in the below figure, below:

*Solving Network Problem on Computer Using TORA (Output Screen)*

** Example: **The following table gives the activities in construction project and time duration.

*Project Schedule with Time Duration*

a. Draw the activity network of the project.

b. Find the total float and free float for each activity.

*Solution:*

a. From the activity relationship given, the activity network is shown in figure below:

* Activity Network Diagram*

b. The total and free floats for each activity are calculated as shown in table

*Calculation of Total and Free Floats*

** Example: **Draw the network for the following project given in the table.

*Project Schedules*

Number the events by Fulkerson’s rule and find the critical path. Also find the time for completing the project.

** Solution: **The network is drawn as shown in the following figure using the data provided. Number the events using Fulkerson’s rule and find the Earliest and Latest time and total float is computed for each activity to find out the critical path as given the following table.

*TL, TL and TFij Calculated*

**Activity Network Diagram**

The critical path is a – c – f – h– j and the minimum time for the completion of the project is 42 weeks.

In the critical path method, the time estimates are assumed to be known with cert_{a}inty. In cert_{a}in projects like research and development, new product introductions, it is difficult to estimate the time of various activities. Hence PERT is used in such projects with a probabilistic method using three time estimates for an activity, rather than a single estimate, as shown in Figure.

*PERT Using Probabilistic Method with 3 Time Estimates*

*Optimistic time tO:*

It is the shortest time t_{a}ken to complete the activity. It means that if everything goes well then there is more chance of completing the activity within this time.

*Most likely time t _{m}:*

It is the normal time t_{a}ken to complete an activity, if the activity were frequently repeated under the same conditions.

*Pessimistic time t _{p}:*

It is the longest time that an activity would t_{a}ke to complete. It is the worst time estimate that an activity would t_{a}ke if unexpected problems are faced.

Taking all these time estimates into consideration, the expected time of an activity is arrived at.

The average or mean (t_{a}) value of the activity duration is given by,

T_{a}= t_{0}+4t_{m}+t_{p}/6 .....................(5)

The variance of the activity time is calculated using the formula,

T_{a}= t_{0}+4t_{m}+t_{p}/6 ...................(6)

*Probability for Project Duration *

The probability of completing the project within the scheduled time (Ts) or contracted time may be obt_{a}ined by using the st_{a}ndard normal deviate where Te is the expected

time of project completion.

...............(7)

Probability of completing the project within the scheduled time is,

P (T≤ Ts) = P ( Z≤ Z0 ) (from normal t_{a}bles) .................(8)

** Example : **An R & D project has a list of t

are given in the t

Time expected for each activity is calculated using the formula (5):

T_{a}= t_{0}+4t_{m}+t_{p}/6

= 4+4(6)+8/6 = 36/6 = 6 days for activity A

Similarly, the expected time is calculated for all the activities. The variance of activity time is calculated using the formula (6).

Similarly, variances of all the activities are calculated. Construct a network diagram and calculate the time earliest, TE and time Latest TL for all the activities.

*Network Diagram*

* Time Estimates for R & D Project*

- Draw the project network.
- Find the critical path.
- Find the probability that the project is completed in 19 days. If the probability is less that 20%, find the probability of completing it in 24 days.

*Solution:*

Calculate the time average t_{a} and variances of each activity as shown in the following t_{a}ble.

*Te & s2 Calculated*

From the network diagram Figure, the critical path is identified as 1-4, 4-6, 6-7, with project duration of 22 days. The probability of completing the project within 19 days is given by,

Thus, the probability of completing the R & D project in 19 days is 9.01%. Since the probability of completing the project in 19 days is less than 20%, we find the probability of completing it in 24 days.

Example is solved using computer with TORA. Go to **MAIN MENU**, **SELECT PROJECT PLANNING **and Click **PERT **– Program Evaluation Review Technique.

Enter the values as shown in figure below.

*Solving PERT Problem Using Computer with TORA (Input Screen)*

Now, go to solve menu and click. In the output screen, select Activity mean / Variance option in select output option. The following screen appears as shown in the following figure.

*TORA (Output Screen), Select PERT Calculations.*

Selecting the PERT calculations option. The following screen appears. This shows the average duration and standard deviation for the activities.

*TORA (Output Screen) Showing Average Durations and Standard Deviation for Activities*

The two important components of any activity are the cost and time. Cost is directly proportional to time and vice versa. For example, in constructing a shopping complex, the expected time of completion can be calculated using be time estimates of various activities. But if the construction has to the finished earlier, it requires additional cost to complete the project. We need to arrive at a time / cost trade-off between total cost of project and total time required to complete it.

*Normal time**: *Normal time is the time required to complete the activity at normal conditions and cost.

*Crash time**: *Crash time is the shortest possible activity time; crashing more than the normal time will increase the direct cost.

**Cost Slope**

Cost slope is the increase in cost per unit of time saved by crashing. A linear cost curve is shown in Figure.

* Linear Cost Curve*

Cost slope=Crash cost C_{c}– Normal cost N_{c}/Normal time Ntt

** Example : **An activity takes 4 days to complete at a normal cost of Rs. 500.00. If it is possible to complete the activity in 2 days with an additional cost of Rs. 700.00, what is the incremental cost of the activity?

*Solution:*

Incremental Cost or Cost Slope = C_{c}– N_{c} / Ntt

= 700-500/4-2 = Rs. 100.00

It means if one day is reduced we have to spend Rs. 100/- extra per day.

*Project Crashing*

*Procedure for crashing*

*Step1***: **Draw the network diagram and mark the Normal time and Crash time.

** Step2**: Calculate TE and TL for all the activities.

*Step3***: **Find the critical path and other paths.

** Step 4: **Find the slope for all activities and rank them in ascending order.

** Step 5: **Establish a tabular column with required field.

** Step 6: **Select the lowest ranked activity; check whether it is a critical activity. If so, crash the activity, else go to the next highest ranked activity.

** Note**: The critical path must remain critical while crashing.

** Step 7: **Calculate the total cost of project for each crashing.

** Step 8: **Repeat Step 6 until all the activities in the critical path are fully crashed.

** Example: **The following Table 8.13 gives the activities of a construction project and other data.

*Construction Project Data*

If the indirect cost is Rs. 20 per day, crash the activities to find the minimum duration of the project and the project cost associated.

** Solution: **From the data provided in the table, draw the network diagram and find the critical path.

**Network Diagram**

From the diagram, we observe that the critical path is 1-2-5 with project duration of 14 days The cost slope for all activities and their rank is calculated as shown in table below

Cost slope=Crash cost C_{c}– Normal cost N_{c}/Normal time Ntt

Cost Slope for activity 1– 2 = 80 – 50/6 – 4 = 30/2 = 15.

*Cost Slope and Rank Calculated*

The available paths of the network are listed down in Table indicating the sequence of crashing.

* Sequence of Crashing*

** Network Diagram Indicating Sequence of Crashing**

The sequence of crashing and the total cost involved is given in the following table

Initial direct cost = sum of all normal costs given

= Rs. 490.00

* Sequence of Crashing & Total Cost*

It is not possible to crash more than 10 days, as all the activities in the critical path are fully crashed. Hence the minimum project duration is 10 days with the total cost of Rs. 970.00.

Network models have three main advantages over linear programming:

- They can be solved very quickly. Problems whose linear program would have 1000 rows and 30,000 columns can be solved in a matter of seconds. This allows network models to be used

in many applications (such as real-time decision making) for which linear programming would be inappropriate. - They have naturally integer solutions. By recognizing that a problem can be formulated as a network program, it is possible to solve special types of integer programs without resorting

to the ineffective and time consuming integer programming algorithms. - They are intuitive. Network models provide a language for talking about problems that is much more intuitive than the variables, objective, and constraints" language of linear and

integer programming.

Of course these advantages come with a drawback: network models cannot formulate the wide range of models that linear and integer programs can. However, they occur often enough that they form an important tool for real decision making.