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1 Lecture 4 MGMT 650 Network Models – Shortest Path Project Scheduling Forecasting
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2 Shortest Path Problem Belongs to class of problems typically known as network flow models What is the “best way” to traverse a network to get from one point to another as cheaply as possible? Network consists of nodes and arcs For example, consider a transportation network Nodes represent cities Arcs represent travel distances between cities Criterion to be minimized in the shortest path problem not limited to distance Other criteria include time and cost
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3 Example: Shortest Route Find the Shortest Route From Node 1 to All Other Nodes in the Network: 6644 4 7 3 5 1 8 6 2 5 3 6 2 3311 2255 77
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4 Management Scientist Input
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5 Example Solution Summary Node Minimum Distance Shortest Route 2 4 1-2 3 6 1-4-3 4 5 1-4 5 8 1-4-3-5 6 11 1-4-3-5-6 7 13 1-4-3-5-6-7
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6 Applications Stand alone applications Emergency vehicle routing Urban traffic planning Telecommunications Sub-problems in more complex settings Allocating inspection effort in a production line Scheduling operations Optimal equipment replacement policies Personnel planning problem
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7 Optimal Equipment Replacement Policy The Erie County Medical Center allocates a portion of its budget to purchase newer and more advanced x-ray machines at the beginning of each year. As machines age, they break down more frequently and maintenance costs tend to increase. Furthermore salvage values decrease. Determine the optimal replacement policy for ECMC that minimizes the total cost of buying, selling and operating the machine over a planning horizon of 5 years, such that at least one x-ray machine must be in service at all times. YearPurchase Cost (`000) 1170 2190 3210 4250 5300 AgeMaintenance cost (`000) Salvage value (`000) 15020 29715 318210 43800
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8 Project Scheduling Chapter 10 Lecture 4
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9 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
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10 Project Scheduling: PERT/CPM Project Scheduling with Known Activity Times Project Scheduling with Uncertain Activity Times
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11 PERT/CPM PERT Program Evaluation and Review Technique CPM Critical Path Method 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
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12 PERT/CPM PERT/CPM is used to plan the scheduling of individual activities that make up a project. Projects may have as many as several thousand activities. A complicating factor in carrying out the activities is that some activities depend on the completion of other activities before they can be started.
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13 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?
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14 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
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15 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 0 2 4 6 8 10 12 14 16 18 20
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16 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
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17 Management Scientist Solution Critical Path
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18 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
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19 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 along the critical path, and variance equal to the sum of the variances along the critical path.
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20 Activity Immediate Predecessor Optimistic Time (a) Most Likely Time (m) Pessimistic Time (b) A--468 B 14.55 CA333 DA456 EA0.511.5 FB,C345 G 11.55 HE,F567 I 258 JD,H2.52.754.5 KG,I357Example
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21 Management Scientist Solution
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22 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
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23 Immediate Completion Immediate Completion Activity Description Predecessors Time (wks) Activity Description Predecessors Time (wks) A Overhaul machine I --- 7 A Overhaul machine I --- 7 B Adjust machine I A 3 B Adjust machine I A 3 C Overhaul machine II --- 6 C Overhaul machine II --- 6 D Adjust machine II C 3 D Adjust machine II C 3 E Test system B,D 2 E Test system B,D 2 Example: Two Machine Maintenance Project Start A07707A07707 C06617C06617 B710 3710 D69 3 710 E1012 21012
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24 Normal Costs and Crash Costs Activity Normal Time Normal Cost ($) Crash Time Crash Cost ($) Maximum Reduction in Time Crash Cost per day ($) A Overhaul Machine I 750048003 (800-500)/3 = 100 B Adjust machine I 320023501150 C Overhaul Machine II 650049002200 D Adjust machine II 320015002150 E Test System 230015501250
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25 Linear Program for Minimum-Cost Crashing Let: X i = earliest finish time for activity i Y i = the amount of time activity i is crashed 10 variables, 12 constraints Crash activity A by 2 days Crash activity D by 1 day Crash cost = 200 + 150 = $350 Crash activity A by 1 day Crash activity E by 1 day Crash cost = 100 + 250 = $350
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26 Lecture 4 Forecasting Chapter 16
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27 Forecasting - Topics Quantitative Approaches to Forecasting The Components of a Time Series Measures of Forecast Accuracy Using Smoothing Methods in Forecasting Using Trend Projection in Forecasting
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28 Time Series Forecasts Trend - long-term movement in data Seasonality - short-term regular variations in data Cycle – wavelike variations of more than one year’s duration Irregular variations - caused by unusual circumstances
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29 Forecast Variations Trend Irregular variatio n Seasonal variations 90 89 88 Cycles
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30 Smoothing/Averaging Methods Used in cases in which the time series is fairly stable and has no significant trend, seasonal, or cyclical effects Purpose of averaging - to smooth out the irregular components of the time series. Four common smoothing/averaging methods are: Moving averages Weighted moving averages Exponential smoothing
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31 n Sales of gasoline for the past 12 weeks at your local Chevron (in ‘000 gallons). If the dealer uses a 3- period moving average to forecast sales, what is the forecast for Week 13? Example of Moving Average n Past Sales Week Sales Week Sales Week Sales Week Sales 1 17 7 20 1 17 7 20 2 21 8 18 2 21 8 18 3 19 9 22 3 19 9 22 4 23 10 20 4 23 10 20 5 18 11 15 5 18 11 15 6 16 12 22 6 16 12 22
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32 Management Scientist Solutions MA(3) for period 4 = (17+21+19)/3 = 19 Forecast error for period 3 = Actual – Forecast = 23 – 19 = 4
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33 MA(5) versus MA(3)
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34 Exponential Smoothing Premise - The most recent observations might have the highest predictive value. Therefore, we should give more weight to the more recent time periods when forecasting. F t+1 = F t + ( A t - F t ), Formula 16.3
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35 Linear Trend Equation F t = Forecast for period t t = Specified number of time periods a = Value of F t at t = 0 b = Slope of the line F t = a + bt 0 1 2 3 4 5 t FtFt a Suitable for time series data that exhibit a long term linear trend
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36 Linear Trend Example F11 = 20.4 + 1.1(11) = 32.5 Linear trend equation Sale increases every time period @ 1.1 units
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37 Actual vs Forecast Linear Trend Example 0 5 10 15 20 25 30 35 12345678910 Week Actual/Forecasted sales Actual Forecast F(t) = 20.4 + 1.1t
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38 Measure of Forecast Accuracy MSE = Mean Squared Error
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39 Forecasting with Trends and Seasonal Components – An Example Business at Terry's Tie Shop can be viewed as falling into three distinct seasons: (1) Christmas (November-December); (2) Father's Day (late May - mid-June); and (3) all other times. Average weekly sales ($) during each of the three seasons during the past four years are known and given below. Determine a forecast for the average weekly sales in year 5 for each of the three seasons. Year Season 1 2 3 4 1 1856 1995 2241 2280 2 2012 2168 2306 2408 3 985 1072 1105 1120
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40 Management Scientist Solutions
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41 Interpretation of Seasonal Indices Seasonal index for season 2 (Father’s Day) = 1.236 Means that the sale value of ties during season 2 is 23.6% higher than the average sale value over the year Seasonal index for season 3 (all other times) = 0.586 Means that the sale value of ties during season 3 is 41.4% lower than the average sale value over the year
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