Post on 19-Dec-2015
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Shortest Path ProblemShortest 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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Example: Shortest RouteExample: Shortest Route Find the Shortest Route From Node 1 to All
Other Nodes in the Network:
66664444
44
77
33
55 11
88
66
22
55
33
66
22
33331111
2222 5555
7777
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Example Solution SummaryExample 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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ApplicationsApplications
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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Optimal Equipment Replacement Optimal Equipment Replacement PolicyPolicy
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.
Year Purchase Cost (`000)
1 170
2 190
3 210
4 250
5 300
Age Maintenance cost (`000)
Salvage value (`000)
1 50 20
2 97 15
3 182 10
4 380 0
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Project ManagementProject 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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Project Scheduling: PERT/CPMProject Scheduling: PERT/CPM
Project Scheduling with Known Activity Times Project Scheduling with Uncertain Activity
Times
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PERT/CPMPERT/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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PERT/CPMPERT/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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PERT/CPMPERT/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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Project NetworkProject 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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Planning and SchedulingPlanning 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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Project Network – An ExampleProject Network – An Example
A
B
C
E
F
Locatefacilities
Orderfurniture
Furnituresetup
Interview
RemodelMove in
D
Hire andtrain
GS
8 weeks
6 weeks
3 weeks
4 weeks9 weeks
11 weeks
1 week
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Management Scientist SolutionManagement Scientist Solution
Path Length (weeks)
Slack
A-B-F-G A-E-G C-D-G
18 20 14
2 0 6
Critical PathCritical Path
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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 An activity’s completion time variancecompletion time variance is is 22 = (( = ((bb--aa)/6))/6)22
Uncertain Activity TimesUncertain Activity Times
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Uncertain Activity TimesUncertain 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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ActivityImmediate
PredecessorOptimisticTime (a)
Most LikelyTime (m)
PessimisticTime (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
ExampleExample
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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 TerminologyKey Terminology
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Immediate Immediate CompletionCompletion
ActivityActivity DescriptionDescription PredecessorsPredecessors Time (wks)Time (wks) A Overhaul machine I ---A Overhaul machine I --- 7 7 B Adjust machine I B Adjust machine I A A 3 3 C C Overhaul machine II ---Overhaul machine II --- 6 6
D Adjust machine IID Adjust machine II C C 3 3 E Test systemE Test system B,D B,D 2 2
Example: Two Machine Example: Two Machine Maintenance ProjectMaintenance Project
Start
A 0
7
7 0
7C 0
6
6 1
7
B 7
10
3 7
10D 6
9
3 7
10
E 10
12
2 10
12
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Normal Costs and Crash Costs
Activity
Normal Time
Normal Cost ($)
Crash Time
Crash Cost ($)
Maximum Reduction
in TimeCrash Cost per
day ($)
A Overhaul Machine I 7 500 4 800 3(800-500)/3
= 100
B Adjust machine I 3 200 2 350 1 150
C Overhaul Machine II 6 500 4 900 2 200
D Adjust machine II 3 200 1 500 2 150
E Test System 2 300 1 550 1 250
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Linear Program for Minimum-Cost Linear Program for Minimum-Cost CrashingCrashing
Let: Let: XXii = earliest finish time for activity = earliest finish time for activity ii YYii = the amount of time activity = the amount of time activity ii is crashed 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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Forecasting - TopicsForecasting - 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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Time Series ForecastsTime 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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Forecast VariationsForecast Variations
Trend
Irregularvariation
Seasonal variations
908988
Cycles
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Smoothing/Averaging MethodsSmoothing/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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Sales of gasoline for the past 12 weeks at your Sales of gasoline for the past 12 weeks at your local Chevron (in ‘000 gallons). If the dealer local Chevron (in ‘000 gallons). If the dealer uses a 3-period moving average to forecast uses a 3-period moving average to forecast sales, what is the forecast for Week 13?sales, what is the forecast for Week 13?
Example of Moving Average
Past Sales
WeekWeek SalesSales WeekWeek SalesSales 1 17 7 201 17 7 20 2 21 8 182 21 8 18 3 19 9 223 19 9 22 4 23 10 204 23 10 20 5 18 11 155 18 11 15
6 166 16 12 12 22 22
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Management Scientist SolutionsManagement 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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MA(5) versus MA(3)MA(5) versus MA(3)
Week Actual MA(3) MA(5)1 172 213 194 23 195 18 216 16 20 19.67 20 19 19.48 18 18 19.29 22 18 19
10 20 20 18.811 15 20 19.212 22 19 19
MA Forecast Graph
0
5
10
15
20
25
1 2 3 4 5 6 7 8 9 10 11 12
Week
Actu
al/M
A Fo
reca
st s
ale
valu
es
Actual
MA(3)
MA(5)
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Exponential SmoothingExponential 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.
Ft+1 = Ft + (At - Ft), Formula 16.3
10
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Linear Trend EquationLinear Trend Equation
Ft = Forecast for period t t = Specified number of time periods a = Value of Ft at t = 0 b = Slope of the line
Ft = a + bt
0 1 2 3 4 5 t
Ft
a
Suitable for time series data that exhibit a long term linear trend
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Linear Trend ExampleLinear 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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Actual vs ForecastActual vs Forecast
Linear Trend Example
0
5
10
15
20
25
30
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1 2 3 4 5 6 7 8 9 10
Week
Act
ual
/Fo
reca
sted
sal
es
Actual
Forecast
F(t) = 20.4 + 1.1t
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Measure of Forecast AccuracyMeasure of Forecast Accuracy MSE = Mean Squared Error
Week # Actual (A) Forecast(F) Error =E =A-F E(squared)1 21.6 21.5 0.1 0.012 22.9 22.6 0.3 0.093 25.5 23.7 1.8 3.244 21.9 24.8 -2.9 8.415 23.9 25.9 -2 46 27.5 27 0.5 0.257 31.5 28.1 3.4 11.568 29.7 29.2 0.5 0.259 28.6 30.3 -1.7 2.89
10 31.4 31.4 0 0
Sum of E(squared) 30.7
MSE= 3.07
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Forecasting with Trends and Seasonal Forecasting with Trends and Seasonal Components – An ExampleComponents – 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 seasonsduring 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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Interpretation of Seasonal IndicesInterpretation 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