PROJECT MANAGEMENT CHAPTER 1

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1 PROJECT MANAGEMENT CHAPTER 1 Project management is the process and activity of planning, organizing, motivating, and controlling resources, procedures and protocols to achieve specific goals in scientific or daily problems. A project is an endeavor ( او محولة (مسعى to accomplish a specific objective through a unique set of interrelated tasks and the effective utilization of resources. It has a clearly defined objective stated in terms of scope, schedule, and cost. The responsibility of the project manager is to make sure that the project objective is accomplished and that the work scope is completed in a quality manner, within budget, and on time, to the customer s satisfaction. A project has a well-defined objective - an expected result or product. The objective of a project is usually defined in terms of scope, schedule, and cost. For example, the objective of a project might be to introduce to the market- in 10 months and within a budget of $500,000-a new food preparation appliance that meets certain predefined performance specifications. Furthermore, it is expected that the work scope will be accomplished in a quality manner and to the customer s satisfaction. ATTRIBUTES OF A PROJECT A project is carried out through a series of interdependent tasks-that is, a number of nonrepetitive tasks that need to be accomplished in a certain sequence in order to achieve the project objective. A project utilizes various resources to carry out the tasks. Such resources can include different people, organizations, equipment, materials, and facilities. A project has a specific time frame, or finite life span. It has a start time and a date by which the objective must be accomplished. A project may be a unique or one-time endeavor. Some projects like designing and building a space station are unique because they have never before been attempted. 1

2 PROJECT MANAGEMENT CHAPTER 1 Other projects, such as developing a new product, building a house, or planning a wedding, are unique because of the customization they require. A project has a customer. The customer is the entity that provides the funds (capital) necessary to accomplish the project. It can be a person, an organization, or a partnership of two or more people or organizations. Finally, a project involves a degree of uncertainty. Before a project is started, a plan is prepared based on certain assumptions and estimates. It is important to document these assumptions, because they will influence the development of the project budget, schedule, and work scope. A project is based on a unique set of tasks and estimates of how long each task should take, various resources and assumptions about the availability and capability of those resources, and estimates of the costs associated with the resources. This combination of assumptions and estimates causes a degree of uncertainty that the project objective will be completely accomplished. The following are some examples of projects: Developing and introducing a new product Planning a wedding Designing and implementing a computer system Modernizing a factory Consolidating two manufacturing plants Hosting a conference Building a shopping mall Performing a series of surgeries Rebuilding a town after a natural disaster Hosting a dinner PROJECT LIFE CYCLE Figure 1 shows the four phases of the project lifecycle and the relative amount of effort and time devoted to each phase. The first phase of the project life cycle involves the identification of a need, problem, or opportunity and can result in the customer s requesting proposals from individuals, a project team, or organizations (contractors) to address the identified need or solve the problem. The need and requirements are usually written up by the 2

3 PROJECT MANAGEMENT CHAPTER 1 customer in a document called a request for proposal (RFP). Through the RFP, the customer asks individuals or contractors to submit proposals on how they might solve the problem, along with the associated cost and schedule. FIGURE 1 Project Life Cycle Efforts The second phase of the project life cycle is the development of a proposed solution to the need or problem. In this phase, the contractor effort is dominant. Contractors interested in responding to the RFP may spend several weeks developing approaches to solving the problem, estimating the types and amounts of resources that would be needed as well as the time it would take to design and implement the proposed solution. Each contractor documents this information in a written proposal. All of the contractors submit their proposals to the customer. The third phase of the project life cycle is the implementation of the proposed solution. This phase begins after the customer decides which of the proposed solutions will best fulfill the need and an agreement is reached between the customer and the individual or contractor who submitted the proposal. This phase, referred to as performing the project, involves doing the detailed planning for the project and then implementing that plan to accomplish the project objective. During the course of performing the project, different types of resources will be utilized. The final phase of the project life cycle is terminating the project. An important task during this phase is evaluating performance of the project in order to learn what could be improved if a similar project were to be carried out in the future. This phase should include obtaining feedback from the customer to determine the level of the customers satisfaction and whether the project meet the customer s expectations. Also, feedback should be obtained from the project team in the form of recommendations for improving performance of projects in the future. 3

9 , and are Linear Programming The method linear programming was used to get optimal solution. General Linear Programming Problem A general mathematical way of representing a Linear Programming Problem (L.P.P.) is as given below: Where all crjr brir arijr s are constants and xrjr decision variables. UExample1:U A company manufactures two products X and Y, which require, the following resources. The resources are the capacities machine M1, M2, and M3. The available capacities are 50, 25 and 15 hours respectively in the planning period. Product X requires 1 hour of machine M2 and 1 hour of machine M3. Product Y requires 2 hours of machine M1, 2 hours of machine M2 and 1 hour of machine M3. The profit contribution of products X and Y are Rs.5/ per unit and Rs.4/ per unit respectively. Formulate the problem Solution: The contents of the statement of the problem can be summarized as follows: Maximize Z = 5x + 4y OBJECTIVE FUNCTION. Mohammad Hameed 1 D.U.C.

10 For Machine M1 0x + 2y 50 For Machine M2 1x + 2y 25 CONSTRAINTS For machine M3 1x + 1y 15 x, y 0 UExample2:U A retail store stocks two types of shirts A and B. These are packed in attractive cardboard boxes. During a week the store can sell a maximum of 400 shirts of type A and a maximum of 300 shirts of type B. The storage capacity, however, is limited to a maximum of 600 of both types combined. Type A shirt fetches a profit of Rs. 2/ per unit and type B a profit of Rs. 5/ per unit. How many of each type the store should stock per week to maximize the total profit? Formulate a mathematical model of the problem. Solution: Maximize Z = 2a + 5b s.t. 1a + 0b 400 0a + 1b 300 1a + 1b 600 a, b 0 Graphical Method For Solution Linear Programming Model (LP) Only two variable problems are considered, because we can draw straight lines in two-dimensional plane (X- axis and Y-axis). UExample3:U A company manufactures two products, X and Y by using three machines A, B, and C. Machine A has 4 hours of capacity available during the coming week. Similarly, the available capacity of machines B and C during the coming week is 24 hours and 35 hours respectively. One unit of product X requires one hour of Machine A, 3 hours of machine B and 10 hours of machine C. Similarly one unit of product Y requires 1 hour, 8 hour and 7 hours of machine A, B and C respectively. When one unit of X is sold in the market, it yields a profit of Rs. 5/per product and that of Y is Rs. 7/per unit. Solve the problem by using graphical method to find the optimal product mix. Mohammad Hameed 2 D.U.C.

11 USolution:U The details given in the problem is given in the table below: Let the company manufactures x units of X and y units of Y, and then the L.P. model is: Maximise Z = 5x + 7y s.t. 1x + 1y 4 machine A 3x + 8y 24 machine B 10x + 7y 35 machine C x,y 0 As we cannot draw graph for inequalities, let us consider them as equations. Maximise Z = 5x + 7y s.t. 1x + 1y = 4 3x + 8y = 24 10x + 7y = 35 x, y are 0 machine A machine B machine C Let us take machine A. and find the boundary conditions. If x = 0, machine A can manufacture 4/1 = 4 units of y. Similarly, if y = 0, machine A can manufacture 4/1 = 4 units of x. Machine B When x = 0, y = 24/8 = 3 and when y = 0 x = 24/3 = 8 Machine C When x = 0, y = 35/10 = 3.5 and when y = 0, x = 35 / 7 = 5. Mohammad Hameed 3 D.U.C.

12 Graph for machine A Graph for machine B Graph for machine C Graph for machine A, B and C combined Method 1. Here we find the co-ordinates of corners of the closed polygon ROUVW and substitute the values in the objective function. In maximization problem, we select the coordinates giving maximum value. And in minimization problem, we select the coordinates, which gives minimum value. In the problem the co-ordinates of the corners are: R = (0, 3.5), O = (0,0), U = (3.5,0), V = (2.5, 1.5) and W = (1.6,2.4). Substituting these values in objective function: Z( 0,3.5) = = Rs , at point R Z (0,0) = = Rs , Z(3.5,0) = = Rs Z (2.5, 1.5) = = Rs Z (1.6, 2.4) = = Rs at point O at point U at point V at point W Mohammad Hameed 4 D.U.C.

13 Method 2. Isoprofit Line Method: Isoprofit line, a line on the graph drawn as per the objective function, assuming certain profit. On this line any point showing the values of x and y will yield same profit. UFor exampleu in the given problem, the objective function is Maximise Z = 5x + 7y. we draw line ZZ (preferably dotted line) for 5x + 7y = 35. Then draw parallel line to this line ZZ at origin. The line at origin indicates zero rupees profit. Hence slowly move this line away from origin. While moving it touches corners of the polygon showing certain higher profit. Finally, it touches the farthermost corner covering all the area of the closed polygon. This point where the line passes (farthermost point) is the UOPTIMAL SOLUTIONU of the problem. In the figure the line ZZ passing through point W covers the entire area of the polygon, hence it is the point that yields highest profit. Now point W has co-ordinates (1.6, 2.4). Now Optimal profit Z = = Rs UExample4:U A company produces both interior and exterior paints from two raw materials, Ml and M.1. The following table provides the basic data of the problem: A market survey indicates that the daily demand for interior paint cannot exceed that for exterior paint by more than 1 ton. Also, the maximum daily demand for interior paint is 2 tons. The A company wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit. Mohammad Hameed 5 D.U.C.

14 USolutionU: Mohammad Hameed 6 D.U.C.

18 = = 0, = 0

= 0 = = 0 = 0 = + = 0 + 4 = 4 = 4 = = 25 7 = 18 = 18 = + = 0 + 1 = 1 = { + } = {( + ) ( + )} {(4 + 1), (1 + 1)} = + = 1 + 6 = 7 = + = 7 + 4 = 11 = + = 7 + 8 = 15 = { + } = {( + ) ( + )} {(11 + 1),

19 = 0 = = 0 = 0 = + = = 4 = 4 = = 25 7 = 18 = 18 = + = = 1 = { + } = {( + ) ( + )} {(4 + 1), (1 + 1)} = + = = 7 = + = = 11 = + = = 15 = { + } = {( + ) ( + )} {(11 + 1), (15 + 2)} = { + } = {( + ) ( + )} {(5 + 5), (17 + 1)} = { + } = {( + ) ( + )} {(8 + 7), (17 + 8)} = 5 = 7 = = {( ) ( )} {(18 1), (25 8)} = = 17 2 = 15 = = 17 1 = 16 = = {( ) ( )} {(15 8), (16 4)} = = 18 5 = 13 = = 7 6 = 11 = = 13 1 = 12 = = {( ) ( )} {(12 4), (1 1)} = 15 = 16 = 13 = 11 = 12 = = 17 = 12 )

20 = + ( ) ( ) =

21 = = 0, = 0 = 0 = = 46 = 46 = + = = 20 = 20 = = = 36 = 36 = + = = 30 = = 36 6 = 30 = 30 = { + } = {( + ) ( + )} = 36 {( ), (30 + 6)} = = {( ) ( )} {(30 10), (36 12)} = 20 = + = = 46 = 46 = = {( ) ( )} {(20 20), (30 25)} = = + ( ) = ( )

22 = 0 = = 15 = 0 = + = = 2 = + = = 2 = + = = 1 = + = = 6 = + = = 7 = 2 = 1 = 7 = { + } = {( + ) ( + )} {(2 + 8), (1 + 3)} = { + } = {( + ) ( + )} {(6 + 1), (7 + 4)} = { + } = {( + ) ( + )} {(11 + 3), (10 + 5)} = = {( ) {(15 3) = (12) = {( ) (15 5) } = {( ) (12 1) } = {( ) (12 4) } = = {( ) ( )} {(10 8), (8 5)} = {( ) (10 3) } = {( ) (11 4) } = 12 = = {( ) ( ) ( )} {(7 2), (2 2)(7 1)} = 10 = 11 = 8 = 2 = 7 = 7 == 0

23 = 7 = 0, = 0 = + ( ) = ( )

24 = =. = =.

26 ) ) ) ) ( ( ) ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = = 0 = = 32.5 = 32.5 = { + } = {( + ) = + = = 3.5 = { + } = = 2,1 {( + ) ( + ) = max{ ) ( )} = 1 = { + } = = 3,4 {( + ) ( + ) = max{ ) ( )} = + = = 25.8 = { + } = = 5,3 {( + ) ( + ) = max{ ) ( )} = 24.5 = + = = 34.9 = + = = 32.5 = 5.3 = 3.5 = = 34.9 = 34.9 = {( ) (32.5 8) } = {( ) ( ) } = = {( ) ( )} = 25.8 {( ), (24.5 5)} = {( ) (19.5 7) } = = {( ) = 10 ( )} {( ), ( )} = {( ) (11 4) } = = {( ) ( )} {( ), ( )} = 24.5 = 11 = 12.5 = 7 = 0

27 ) = 0, = 0 = 5.3

28 ( ) ( (. ) ( ( ) = ( ( ) = ( ( ) = ( ( ) = 3 ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = ( ( ) = 2 ( ( ) = = 0, = 0

29 = 0 = = = 32.5 = { + } = {( + ) = + = = 4 = + = = 5 = { + } = = 3,4 {( + ) ( + ) = max{5 + 3) (4 + 4)} = (8,8) = + = = 10 = { + } = = 5,6 {( + ) ( + ) = max{8 + 5) (10 + 7)} = 17 = + = = 19 = + = 17 + = = { + } = = 8,9 {( + ) ( + ) = max{19 + 2) ( + 5)} = = 4 = 5 = 10 = 17 = = 2 = 8 = = 5 = = = = = {( ) ( )} 2, ( } = {( ) (17 7) } = {( ) (17 5) } = {( ) (10 5) } = { } = {12 4) = 8 = = {( ) ( )} {(8 2), (5 3)} = { } = {2 2) = 0 = 0 = = = 17 = 10 = 12 = 5 = 8 = 2

30 = 0 = 3

LP Definition and Introduction to Graphical Solution Active Learning Module 2

LP Definition and Introduction to Graphical Solution Active Learning Module 2 J. René Villalobos and Gary L. Hogg Arizona State University Paul M. Griffin Georgia Institute of Technology Background Material