System Simulation Chapter 2: Simulation Examples

Transcription

1 System Simulation Chapter 2: Simulation Examples Fatih Cavdur March 29, 21 Introduction Several examples of simulation that can be performed by devising a simulation table either manually or with a spreadsheet. The simulation table provides a systematic method for tracking system state over time. These examples provide insight into the methodology of discrete-system simulation and the descriptive statistics used for predicting system performance.

2 Introduction - cont. The simulations in this chapter are carried out by following three steps: 1 Determine the characteristics of each of the inputs to the simulation. Quite often, these are modeled as (discrete - continuous) probability distributions. 2 Construct a simulation table. Each table is different based on the problem of interest. In the example simulation table in your text there are p inputs, x ij, j = 1, 2,..., p, and one response, y i for each repetition i, i = 1, 2,..., n. 3 For each repetition i, generate a value for each of the p inputs, and evaluate the function calculating a value of the response y i. Introduction - cont. The simulations examples in this chapter are in queuing (two examples with one and two servers) inventory reliability network analysis

3 Introduction Calling population Waiting line Server Figure : Single-Server Queuing System Introduction - cont. Departure event Begin server idle time No Another Yes Remove the waiting unit unit waiting from the queue? Begin servicing the unit Figure : Departure Event

4 Introduction - cont. Arrival event No Server busy? Yes Unit enters service Unit enters queue for service Figure : Arrival Event Introduction - cont. Server status Busy Idle Queue status Not empty Empty Enter queue Enter queue Impossible Enter service Figure : System State upon Arrival

5 Introduction - cont. Server outcomes Busy Idle Queue status Not empty Empty Impossible Impossible Figure : System State upon Departure Single-Channel Queue Inputs Time Between Arrivals Service Time Outputs (Performance Measures) Average Number in Queue Average Waiting Time What else?

6 Single-Channel Queue Table : Distribution of Time Between Arrivals Time Between Arrivals Probability Cumulative Probability Random Digits Single-Channel Queue - cont. Table : Service-Time Distribution Service Time Probability Cumulative Probability Random Digits

7 Single-Channel Queue - cont A B C D E F G H I J TOTALS AVERAGES Number of Customers= Simulation Table Step Activity Clock Activity Clock Output Clock Output Output Interarrival Service Time Waiting Time Time Time Customer Idle Time Time Time Service in Queue Service Spends in System of Server Customer (Minutes) Arrival Time (Minutes) Begins (Minutes) Ends (Minutes) (Minutes) Figure : Simulation Table for Single-Channel Queuing Problem Single-Channel Queue - cont. Some of the findings from our simulation model: average waiting time, the probability that a customer has to wait total time customers wait in queue average waiting time = total number of customers = 17 = 1.7 number of customers who wait in queue probability (wait) = total number of customers = =.

8 Single-Channel Queue - cont. proportion of idle - busy time of the server total idle time of server probability (server idle) = total run time = 1 1 =.2 probability (server busy) = 1 probability (server idle) = 1.2 =.7 Single-Channel Queue - cont. average versus expected service times total service time average service time = total number of customers = 317 = 3.17 E(S) = sp(s) = 1(.) + 2(.2) (.) = 3.2 s= They will get closer to each other if we simulate longer.

9 Single-Channel Queue - cont. average versus expected time between arrivals average time between arrivals E(A) = a + b 2 sum of all times between arrivals = number of arrivals 1 = 1 99 =.19 = =. Again, they will get closer to each other if we simulate longer. Single-Channel Queue - cont. average waiting time of waiting customers average waiting time of waiting customers total time customers wait in queue = total number of waiting customers = 17 = 3.22

10 Single-Channel Queue - cont. average time customer spends in system total time in system average time in system = total number of customers = 91 =.91 It can also be computed as average time in system = average time in queue + average time in service = =.91 Able-Baker Call Center Problem Inputs Time Between Calls Service Time for Able Service Time for Baker Outputs (Performance Measures) Caller Delay What else?

11 Able-Baker Call Center Problem B C D Interarrival Distribution of Calls Interarrival Time (Minutes) Probability Cumulative Probability Figure : Distribution of Time Between Arrivals Able-Baker Call Center Problem - cont F G H Able's Service Time Distribution Service Times (Minutes) Probability Cumulative Probability Figure : Frequency Distribution of Average Waiting Times

12 Able-Baker Call Center Problem - cont I J K Baker's Service Time Distribution Service Times (Minutes) Probability Cumulative Probability Figure : Frequency Distribution of Average Waiting Times Able-Baker Call Center Problem - cont B C D E F G H I J K L M TOTALS Number of Callers= Seed for Random Numbers 12,3 Simulation Table Step Activity Clock Clock Cloc Clock Clock k State Activity Clock Output Output Interarrival When When Service Time Service Completion Time in Caller Time Arrival Able Baker Server Time Service Time Caller Delay System Number (Minutes) Time Available Available Chosen (Minutes) Begins Able Baker (Minutes) (Minutes) 1 Able Baker Able Baker Able Baker Able Baker Able Baker Figure : Simulation Table

13 Able-Baker Call Center Problem - cont Frequency of Caller Delay Occurrences Upper limit of bin Figure : Caller Delay for First Trial Able-Baker Call Center Problem - cont. 2 Frequency of Average Caller Delay Occurrences (No. of Trials) > (Example: Bin labeled 2 counts all values >1. and <= 2) (The right-most bin counts all values >) Figure : Caller Delay for Trials

14 Introduction I Amount in inventory M Q 1 Q 2 Q 3 N N N T Time Figure : Overview Newsboy Problem Inputs Day Type Distribution Demand Distribution Outputs (Performance Measures) Total Profit What else?

15 Newsboy Problem Table : Newspaper Demand Distribution Demand Good Fair Poor Newsboy Problem - cont. 7 9 J K L Type of Newsday Type Probability Cumulative Probability Good.3.3 Fair.. Poor.2 1. Figure : Random Digits for Type of Newsday

16 Newsboy Problem - cont B C D E F G H Distribution of Newspapers Demanded Demand Demand Probabilities Cumulative Probabilities Good Fair Poor Good Fair Poor Figure : Random Digits for Newspaper Demand Newsboy Problem - cont. profit = revenue from sales cost of papers lost profit from excess demand + salvage from scrap

17 Newsboy Problem - cont. Simulate for 2 days where, for each newspaper, the cost is 33 cents the revenue from sales is cents the lost profit from excess demand is 17 cents the salvage value is cents Newsboy Problem - cont. B C D E F G HI I Simulation Table Lost Profit Revenue from Salvage Type of from Excess from Sale Daily Daily Day Newsday Demand Sales Demand of Scrap Cost Profit 1 Fair $2. $. $1. $23. $2.9 2 Fair $2. $. $1. $23. $2.9 3 Fair 7 $3. $. $. $23. $11.9 Good $3. $1.7 $. $23. $.2 Poor $2. $. $1. $23. -$1. Fair $2. $. $1. $23. $2.9 7 Poor $2. $. $1. $23. $2.9 Fair 7 $3. $. $. $23. $ Good $2. $. $1. $23. -$1. Good $3. $. $. $23. $. 11 Fair 7 $3. $. $. $23. $ Poor $2. $. $1. $23. $ Fair $2. $. $1. $23. $2.9 1 Poor $2. $. $1. $23. -$1. 1 Good $3. $1.7 $. $23. $.2 1 Good 7 $3. $. $. $23. $ Good $3. $1.7 $. $23. $.2 1 Poor $2. $. $1. $23. -$1. 19 Fair 7 $3. $. $. $23. $ Good $3. $. $. $23. $. TOTAL PROFIT $11.7 = Figure : Simulation Table

18 Newsboy Problem - cont. Occurrence (No. of Trials) Histogram (Bin Frequencies for Total 2-Day Profit) $ $ $2 $12 $1 $2 Bin Figure : Total Profit Newsboy Problem - cont. Occurrences Frequency of Daily Profit Upper limit of bin Occurrences Frequency of Daily Profit Upper limit of bin Occurrences 2 Frequency of Daily Profit Frequency of Daily Profit Upper limit of bin Upper limit of bin Occurrences Figure : Daily Profits

19 Order Up-To-Level Inventory System Inputs Little m Big M Demand Distribution Lead Time Outputs (Performance Measures) Total Cost Average Inventory Level What else? Order Up-To-Level Inventory System B C D Distribution of Daily Demand Demand Probability Cumulative Probability Figure : Demand

20 Order Up-To-Level Inventory System - cont. 7 9 F G H Distribution of Lead Time Lead Time (days) Probability Cumulative Probability Figure : Lead Time Order Up-To-Level Inventory System - cont B C D E F G H I J K Simulation Table Step State State State Input State State State Activity State Clock Pending Lead Days until Day within Beginning Ending Shortage Order Time Order Day Cycle Cycle Inventory Demand Inventory Quantity (Quantity) (days) Arrives TOTAL 9 1 AVERAGE Figure : Simulation Table

21 Other Examples A Reliability problem (Example 2.) Random Normal Numbers (Example 2.) Lead Time Demand (Example 2.7) Project Simulation (Example 2.) A Reliability Problem Inputs Bearing Life Delay Time Outputs (Performance Measures) Total Cost

22 A Reliability Problem B C D Distribution of Bearing-Life Probability Bearing Life Cumulative Probability Figure : Bearing Life Distribution A Reliability Problem-cont. 7 9 F G H Distribution of Delay Time Probability Delay Time Cumulative Probability Figure : Delay Time Distribution

23 A Reliability Problem When a break-down occurs, we call the repairman to fix it. Downtime cost for the mill is $ per minute. Cost of the repairman is $3 per hour. It takes 2 minutes to replace 1 bearing, 3 minutes to replace 2 bearings and minutes to replace 3 bearings. Current Method: We replace a bearing whenever it breaks-down. Proposed Method: We replace all bearings whenever one of them breaks-down. Which one is better? A Reliability Problem-cont B C D E F G H Simulation Table Bearing 1 Bearing 2 Bearing3 Life Delay Life Delay Life Step (Hours) (minutes) (Hours) (minutes) (Hours) TOTALS 19, 9 2,2 19,9 1 Delay (minutes) Figure : Simulation Table-Current Method

24 A Reliability Problem-cont. For 1 bearing changes, we have Cost of bearings = 32 = $1, Cost of delay time = ( ) = $3, 2 Cost of downtime = 2 = $9, Cost of repairman = 2 3/ = $ Total cost = 1, + 3, 2 + 9, + = $1, 1. A Reliability Problem-cont. 17 B C D E F G Simulation Table Step Bearing 1 Life (Hours) Bearing 2 Life (Hours) Bearing 3 Life (Hours) First Failure (Hours) 1, 1, 1, 1, 1,3 1, 1, 1,3 1, 1, 1, 1,2 1, 1, 1, Delay (minutes) TOTALS 1 1 Figure : Simulation Table-Proposed Method

25 A Reliability Problem-cont. For 1 bearing changes, we have Cost of bearings = 32 = $1, Cost of delay time = 1 = $1, Cost of downtime = 1 = $, Cost of repairman = 1 3/ = $3 Total cost = 1, + 1, +, + 3 = $,. Random Normal Numbers Simulate a bombin operation as follows: If a bomb falls anywhere on target it is hit, otherwise it is a miss. The aiming point is (, ). The point is impact is normally distributed around the aiming point with a standard deviation of and 2 meters in x and y directions, respectively. Simulate the operation for bombs.

26 Random Normal Numbers-cont. Since we can write Z = X µ X σ X, Z = Y µ Y σ Y X = µ X + Zσ X = + Z, Y = µ Y + Zσ Y = + 2Z Random Normal Numbers Ammunition Depot 2 Meters Meters Figure : Simulated Bombing Run

27 Random Normal Numbers-cont. F G H I J K L M N O P Q 3 7 Point of Impact of Each Bomb(1-) 7 Bomb X Number Hits Y Hit? Miss Miss Hit Hit Hit Hit Hit Hit Miss Miss Figure : Simulated Bombing Run Summary Reading HW: Chapter 2. Chapter 2 Exercises