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Lecture 5 Project Management Chapter 17 1 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 2 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 3 Project Management What are the tools? Work breakdown structure Network diagram Gantt charts 4 Project Manager Responsible for: Work Human Resources Communications Quality Time Costs 5 Key Decisions 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 6 Ethical Issues Temptation to understate costs Withhold information Misleading status reports Falsifying records Compromising workers’ safety Approving substandard work http://www.pmi.org/ 7 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 8 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. 9 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? 10 Planning and Scheduling Activity 0 2 4 6 8 10 12 14 16 18 Locate new facilities Interview staff Hire and train staff Select and order furniture Remodel and install phones Furniture setup Move in/startup 11 20 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 12 Project Network – An Example 8 weeks Locate facilities A 6 weeks Order furniture B F 11 weeks Remodel E S 4 weeks Interview C 3 weeks Furniture setup G Move in 1 week 9 weeks Hire and train D 13 Management Scientist Solution Critical Path Path Length Slack (weeks) A-B-F-G A-E-G C-D-G 18 20 14 2 0 6 14 Uncertain Activity Times 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 15 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. 16 Example Immediate Activity Predecessor Optimistic Time (a) Most Likely Time (m) Pessimistic Time (b) A -- 4 6 8 B -- 1 4.5 5 C A 3 3 3 D A 4 5 6 E A 0.5 1 1.5 F B,C 3 4 5 G B,C 1 1.5 5 H E,F 5 6 7 I E,F 2 5 8 J D,H 2.5 2.75 4.5 K G,I 3 5 7 17 Management Scientist Solution 18 Key Terminology Network activities ES: early start EF: early finish LS: late start LF: late finish Used to determine Expected project duration Slack time Critical path 19 The Network Diagram (cont’d) Path Critical path The longest path; determines expected project duration Critical activities Sequence of activities that leads from the starting node to the finishing node AON path: S-1-2-6-7 Activities on the critical path Slack Allowable slippage for path; the difference the length of path and the length of critical path 20 Advantages of PERT Forces managers to organize Provides graphic display of activities Identifies Critical activities Slack activities 4 2 1 5 6 3 21 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 22 Linear Programming George Dantzig – 1914 -2005 Concerned with optimal allocation of limited resources such as Materials Budgets Labor Machine time among competitive activities under a set of constraints George Dantzig – 1914 -2005 23 Product Mix Example (from session 1) Type 1 Type 2 Profit per unit $60 $50 Assembly time per unit 4 hrs 10 hrs Inspection time per unit 2 hrs Storage space per unit 3 cubic ft Resource Amount available Assembly time 100 hours Inspection time 22 hours Storage space 39 cubic feet 1 hr 3 cubic ft 24 Linear Programming Example Variables Maximize 60X1 + 50X2 Subject to Objective function 4X1 + 10X2 <= 100 2X1 + 1X2 <= 22 Constraints 3X1 + 3X2 <= 39 Non-negativity Constraints X1, X2 >= 0 What is a Linear Program? • A LP is an optimization model that has • continuous variables • a single linear objective function, and • (almost always) several constraints (linear equalities or inequalities) 25 Linear Programming Model Decision variables Objective Function unknowns, which is what model seeks to determine for example, amounts of either inputs or outputs goal, determines value of best (optimum) solution among all feasible (satisfy constraints) values of the variables either maximization or minimization Constraints restrictions, which limit variables of the model limitations that restrict the available alternatives Parameters: numerical values (for example, RHS of constraints) Feasible solution: is one particular set of values of the decision variables that satisfies the constraints Feasible solution space: the set of all feasible solutions Optimal solution: is a feasible solution that maximizes or minimizes the objective function There could be multiple optimal solutions 26 Another Example of LP: Diet Problem Energy requirement : 2000 kcal Protein requirement : 55 g Calcium requirement : 800 mg Food Energy (kcal) Protein(g) Calcium(mg) Oatmeal 110 4 2 Price per serving($) 3 Chicken Eggs Milk Pie 205 160 160 420 32 13 8 4 12 54 285 22 24 13 9 24 Pork 260 14 80 13 27 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 28 Step 1: define decision variables x1 = # of oatmeal servings x2 = # of chicken servings x3 = # of eggs servings x4 = # of milk servings x5 = # of pie servings x6 = # of pork servings Step 2: formulate objective function • In this case, minimize total cost minimize z = 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6 29 Step 3: Constraints Meet energy requirement 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 Meet protein requirement 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55 Meet calcium requirement 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800 Restriction on number of servings 0x14, 0x23, 0x32, 0x48, 0x52, 0x62 30 So, how does a LP look like? minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6 subject to 110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55 2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800 0x14 0x23 0x32 0x48 0x52 0x62 31 Optimal Solution – Diet Problem Using LINDO 6.1 Food Oatmeal # of servings 4 Chicken Eggs Milk Pie 0 0 6.5 0 Pork 2 Cost of diet = $96.50 per day 32 Optimal Solution – Diet Problem Using Management Scientist Food Oatmeal # of servings 4 Chicken Eggs Milk Pie 0 0 6.5 0 Pork 2 Cost of diet = $96.50 per day 33 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 34 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 35 Product Mix Problem Boat Maximum Builder Cost Speedhawk Sleekboat Silverbird Sleekboat Catman Racer Classy Racer $6000 $7000 $5000 $9000 Expected Seating 3 5 2 6 Daily Profit $ 70 $ 80 $ 50 $110 36 Product Mix Problem Define the decision variables x1 = number of Speedhawks ordered x2 = number of Silverbirds ordered x3 = number of Catmans ordered x4 = number of Classys ordered Define the objective function Maximize total expected daily profit: Max: (Expected daily profit per unit) x (Number of units) Max: 70x1 + 80x2 + 50x3 + 110x4 37 Product Mix Problem Define the constraints (1) Spend no more than $420,000: 6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000 (2) Purchase at least 50 boats: x1 + x2 + x3 + x4 > 50 (3) Number of boats from Sleekboat equals number of boats from Racer: x1 + x2 = x3 + x4 or x1 + x2 - x3 - x4 = 0 (4) Capacity at least 200: 3x1 + 5x2 + 2x3 + 6x4 > 200 Nonnegativity of variables: xj > 0, for j = 1,2,3,4 38 Product Mix Problem - Complete Formulation Max 70x1 + 80x2 + 50x3 + 110x4 s.t. 6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000 x1 + x2 + x3 + x4 > 50 Boat # purchased x1 + x2 - x3 - x4 = 0 Speedhawk 28 3x1 + 5x2 + 2x3 + 6x4 > 200 Silverbird 0 x1, x2, x3, x4 > 0 Catman 0 Classy 28 Daily profit = $5040 39 Marketing Application: Media Selection Advertising Media # of potential customers reached Cost ($) per advertisement Max times available per month Exposure Quality Units Day TV 1000 1500 15 65 Evening TV 2000 3000 10 90 Daily newspaper 1500 400 25 40 Sunday newspaper 2500 1000 4 60 Radio 300 100 30 20 Advertising budget for first month = $30000 At least 10 TV commercials must be used At least 50000 customers must be reached Spend no more than $18000 on TV adverts Determine optimal media selection plan 40 Media Selection Formulation Step 1: Define decision variables Step 2: Write the objective in terms of the 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 Maximize 65DTV+90ETV+40DN+60SN+20R Step 3: Write the constraints in terms of the decision variables DTV ETV DN SN R + 25 <= 4 <= 30 <= 30000 0 DN 25 SN 2 R 30 Exposure = 2370 units Availability of Media Budget + ETV >= 10 1500DTV + 3000ETV <= 18000 TV Constraints 1000DTV + 2000ETV >= 50000 Customers reached + 100R <= ETV DTV 2500SN + 10 10 3000ETV + 1000SN <= DTV + 1500DN + 15 Value 1500DTV + 400DN <= Variable 300R DTV, ETV, DN, SN, R >= 0 41 Applications of LP Product mix planning Distribution networks Truck routing Staff scheduling Financial portfolios Capacity planning Media selection: marketing 42 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 43 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 44 Unbounded LP – An Example maximize 2x1 + x2 subject to -x1 + x2 1 x1 - 2x2 2 x1 , x 2 0 x2 can be increased indefinitely without violating any constraint => Objective function value can be increased indefinitely 45 Multiple Optima – An Example maximize x1 + 0.5 x2 subject to 2x1 + x2 4 x1 + 2x2 3 x1 , x2 0 • x1= 2, x2= 0, objective function = 2 • x1= 5/3, x2= 2/3, objective function = 2 46