1 Project Management Chapter 17 Lecture 5

2 Project Management  How is it different?  Limited time frame  Narrow focus, specific objectives  Why is it used?  Special needs  Pressures for new or improves products or services  Definition of a project  Unique, one-time sequence of activities designed to accomplish a specific set of objectives in a limited time frame

3 Project Management  What are the Key Metrics  Time  Cost  Performance objectives  What are the Key Success Factors?  Top-down commitment  Having a capable project manager  Having time to plan  Careful tracking and control  Good communications

4 Project Management  What are the tools?  Work breakdown structure  Network diagram  Gantt charts

5 Project Manager Responsible for: WorkQuality Human ResourcesTime CommunicationsCosts

6  Deciding which projects to implement  Selecting a project manager  Selecting a project team  Planning and designing the project  Managing and controlling project resources  Deciding if and when a project should be terminated Key Decisions

7  Temptation to understate costs  Withhold information  Misleading status reports  Falsifying records  Compromising workers’ safety  Approving substandard work  Ethical Issues

8 PERT and CPM PERT: Program Evaluation and Review Technique CPM: Critical Path Method  Graphically displays project activities  Estimates how long the project will take  Indicates most critical activities  Show where delays will not affect project  PERT and CPM have been used to plan, schedule, and control a wide variety of projects:  R&D of new products and processes  Construction of buildings and highways  Maintenance of large and complex equipment  Design and installation of new systems

9 PERT/CPM  PERT/CPM  used to plan the scheduling of individual activities that make up a project.  Projects may have as many as several thousand activities.  Complicating factor in carrying out the activities  some activities depend on the completion of other activities before they can be started.

10 PERT/CPM  Project managers rely on PERT/CPM to help them answer questions such as:  What is the total time to complete the project?  What are the scheduled start and finish dates for each specific activity?  Which activities are critical?  must be completed exactly as scheduled to keep the project on schedule?  How long can non-critical activities be delayed  before they cause an increase in the project completion time?

11 Planning and Scheduling Locate new facilities Interview staff Hire and train staff Select and order furniture Remodel and install phones Furniture setup Move in/startup Activity

12 Project Network  Project network  constructed to model the precedence of the activities.  Nodes represent activities  Arcs represent precedence relationships of the activities  Critical path for the network  a path consisting of activities with zero slack

13 Project Network – An Example A B C E F Locate facilities Order furniture Furniture setup Interview Remodel Move in D Hire and train GS 8 weeks 6 weeks 3 weeks 4 weeks 9 weeks 11 weeks 1 week

14 Management Scientist Solution Critical Path

15  Three-time estimate approach  the time to complete an activity assumed to follow a Beta distribution  An activity’s mean completion time is: t = (a + 4m + b)/6  a = the optimistic completion time estimate  b = the pessimistic completion time estimate  m = the most likely completion time estimate  An activity’s completion time variance is  2 = (( b - a )/6) 2 Uncertain Activity Times

16 Uncertain Activity Times  In the three-time estimate approach, the critical path is determined as if the mean times for the activities were fixed times.  The overall project completion time is assumed to have a normal distribution  with mean equal to the sum of the means of activities along the critical path, and  variance equal to the sum of the variances of activities along the critical path.

17 Activity Immediate Predecessor Optimistic Time (a) Most Likely Time (m) Pessimistic Time (b) A--468 B CA333 DA456 EA FB,C345 G HE,F567 I 258 JD,H KG,I357Example

18 Management Scientist Solution

19  Network activities  ES: early start  EF: early finish  LS: late start  LF: late finish  Used to determine  Expected project duration  Slack time  Critical path Key Terminology

20 The Network Diagram (cont’d)  Path  Sequence of activities that leads from the starting node to the finishing node  AON path: S  Critical path  The longest path; determines expected project duration  Critical activities  Activities on the critical path  Slack  Allowable slippage for path; the difference the length of path and the length of critical path

21 Advantages of PERT  Forces managers to organize  Provides graphic display of activities  Identifies  Critical activities  Slack activities

22 Limitations of PERT  Important activities may be omitted  Precedence relationships may not be correct  Estimates may include a fudge factor  May focus solely on critical path

23  George Dantzig –  Concerned with optimal allocation of limited resources such as  Materials  Budgets  Labor  Machine time  among competitive activities  under a set of constraints Linear Programming George Dantzig –

24 Product Mix Example (from session 1) Type 1Type 2 Profit per unit $60$50 Assembly time per unit 4 hrs10 hrs Inspection time per unit 2 hrs1 hr Storage space per unit 3 cubic ft ResourceAmount available Assembly time100 hours Inspection time22 hours Storage space39 cubic feet

25 Maximize 60X X 2 Subject to 4X X 2 <= 100 2X 1 + 1X 2 <= 22 3X 1 + 3X 2 <= 39 X 1, X 2 >= 0 Linear Programming Example Variables Objective function Constraints What is a Linear Program? A LP is an optimization model that has continuous variables a single linear objective function, and (almost always) several constraints (linear equalities or inequalities) Non-negativity Constraints

26  Decision variables  unknowns, which is what model seeks to determine  for example, amounts of either inputs or outputs  Objective Function  goal, determines value of best (optimum) solution among all feasible (satisfy constraints) values of the variables  either maximization or minimization  Constraints  restrictions, which limit variables of the model  limitations that restrict the available alternatives  Parameters: numerical values (for example, RHS of constraints)  Feasible solution: is one particular set of values of the decision variables that satisfies the constraints  Feasible solution space: the set of all feasible solutions  Optimal solution: is a feasible solution that maximizes or minimizes the objective function  There could be multiple optimal solutions Linear Programming Model

27 Another Example of LP: Diet Problem  Energy requirement : 2000 kcal  Protein requirement : 55 g  Calcium requirement : 800 mg FoodEnergy (kcal)Protein(g)Calcium(mg)Price per serving($) Oatmeal Chicken Eggs Milk Pie Pork

28 Example of LP : Diet Problem  oatmeal: at most 4 servings/day  chicken: at most 3 servings/day  eggs: at most 2 servings/day  milk: at most 8 servings/day  pie:at most 2 servings/day  pork: at most 2 servings/day Design an optimal diet plan which minimizes the cost per day

29 Step 1: define decision variables  x 1 = # of oatmeal servings  x 2 = # of chicken servings  x 3 = # of eggs servings  x 4 = # of milk servings  x 5 = # of pie servings  x 6 = # of pork servings Step 2: formulate objective function In this case, minimize total cost minimize z = 3x x x 3 + 9x x x 6

30 Step 3: Constraints  Meet energy requirement 110x x x x x x 6  2000  Meet protein requirement 4x x x 3 + 8x 4 + 4x x 6  55  Meet calcium requirement 2x x x x x x 6  800  Restriction on number of servings 0  x 1  4, 0  x 2  3, 0  x 3  2, 0  x 4  8, 0  x 5  2, 0  x 6  2

31 So, how does a LP look like? minimize 3x x x 3 + 9x x x 6 subject to 110x x x x x x 6  x x x 3 + 8x 4 + 4x x 6  55 2x x x x x x 6   x 1  4 0  x 2  3 0  x 3  2 0  x 4  8 0  x 5  2 0  x 6  2

32 Optimal Solution – Diet Problem Using LINDO 6.1  Cost of diet = $96.50 per day Food# of servings Oatmeal4 Chicken0 Eggs0 Milk6.5 Pie0 Pork2

33 Optimal Solution – Diet Problem Using Management Scientist  Cost of diet = $96.50 per day Food# of servings Oatmeal4 Chicken0 Eggs0 Milk6.5 Pie0 Pork2

34 Guidelines for Model Formulation  Understand the problem thoroughly.  Describe the objective.  Describe each constraint.  Define the decision variables.  Write the objective in terms of the decision variables.  Write the constraints in terms of the decision variables  Do not forget non-negativity constraints

35 Product Mix Problem Floataway Tours has $420,000 that can be used to purchase new rental boats for hire during the summer. The boats can be purchased from two different manufacturers. Floataway Tours would like to purchase at least 50 boats. They would also like to purchase the same number from Sleekboat as from Racer to maintain goodwill. At the same time, Floataway Tours wishes to have a total seating capacity of at least 200. Formulate this problem as a linear program

36 Maximum Expected Daily Boat Builder Cost Seating Profit Speedhawk Sleekboat $ $ 70 Silverbird Sleekboat $ $ 80 Catman Racer $ $ 50 Classy Racer $ $110 Product Mix Problem

37  Define the decision variables x 1 = number of Speedhawks ordered x 2 = number of Silverbirds ordered x 3 = number of Catmans ordered x 4 = number of Classys ordered  Define the objective function Maximize total expected daily profit: Max: (Expected daily profit per unit) x (Number of units) Max: 70x x x x 4 Product Mix Problem

38  Define the constraints (1) Spend no more than $420,000: 6000x x x x 4 < 420,000 (2) Purchase at least 50 boats: x 1 + x 2 + x 3 + x 4 > 50 (3) Number of boats from Sleekboat equals number of boats from Racer: x 1 + x 2 = x 3 + x 4 or x 1 + x 2 - x 3 - x 4 = 0 (4) Capacity at least 200: 3x 1 + 5x 2 + 2x 3 + 6x 4 > 200 Nonnegativity of variables: x j > 0, for j = 1,2,3,4 Product Mix Problem

39 Max 70x x x x 4 s.t. 6000x x x x 4 < 420,000 x 1 + x 2 + x 3 + x 4 > 50 x 1 + x 2 - x 3 - x 4 = 0 3x 1 + 5x 2 + 2x 3 + 6x 4 > 200 x 1, x 2, x 3, x 4 > 0 Product Mix Problem - Complete Formulation  Daily profit = $5040 Boat# purchased Speedhawk28 Silverbird0 Catman0 Classy28

40 Marketing Application: Media Selection  Advertising budget for first month = $30000  At least 10 TV commercials must be used  At least customers must be reached  Spend no more than $18000 on TV adverts  Determine optimal media selection plan Advertising Media# of potential customers reached Cost ($) per advertisement Max times available per month Exposure Quality Units Day TV Evening TV Daily newspaper Sunday newspaper Radio

41 Media Selection Formulation  Step 1: Define decision variables  DTV = # of day time TV adverts  ETV = # of evening TV adverts  DN = # of daily newspaper adverts  SN = # of Sunday newspaper adverts  R = # of radio adverts  Step 2: Write the objective in terms of the decision variables  Maximize 65DTV+90ETV+40DN+60SN+20R  Step 3: Write the constraints in terms of the decision variables DTV<=15 ETV<=10 DN<=25 SN<=4 R DTV+3000ETV+400DN+1000SN+100R<=30000 DTV+ETV>= DTV+3000ETV<= DTV+2000ETV+1500DN+2500SN+300R>=50000 Budget Customers reached TV Constraints Availability of Media DTV, ETV, DN, SN, R >= 0 Exposure = 2370 units VariableValue DTV10 ETV0 DN25 SN2 R30

42 Applications of LP  Product mix planning  Distribution networks  Truck routing  Staff scheduling  Financial portfolios  Capacity planning  Media selection: marketing

43 Possible Outcomes of a LP  A LP is either  Infeasible – there exists no solution which satisfies all constraints and optimizes the objective function  or, Unbounded – increase/decrease objective function as much as you like without violating any constraint  or, Has an Optimal Solution  Optimal values of decision variables  Optimal objective function value

44 Infeasible LP – An Example  minimize 4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16 x33+5x34 Subject to  x11+x12+x13+x14=100  x21+x22+x23+x24=200  x31+x32+x33+x34=150  x11+x21+x31=80  x12+x22+x32=90  x13+x23+x33=120  x14+x24+x34=170  xij>=0, i=1,2,3; j=1,2,3,4 Total demand exceeds total supply

45 Unbounded LP – An Example maximize 2x 1 + x 2 subject to - x 1 + x 2  1 x 1 - 2x 2  2 x 1, x 2  0 x 2 can be increased indefinitely without violating any constraint => Objective function value can be increased indefinitely

46 Multiple Optima – An Example maximize x x 2 subject to 2x 1 + x 2  4 x 1 + 2x 2  3 x 1, x 2  0 x 1 = 2, x 2 = 0, objective function = 2 x 1 = 5/3, x 2 = 2/3, objective function = 2