CS代考 Variability and Its Impact on Process Performance: Waiting Time Problems

Variability and Its Impact on Process Performance: Waiting Time Problems
An Example of a Simple Queuing System
Call center
 At peak, 80% of calls dialed received a busy signal.

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 Customers getting through had to wait on average 10 minutes
 Extra telephone expense per day for waiting was $25,000.
Financial consequences
Incoming calls
Blocked calls (busy signal)
Answered Calls
Calls on Hold
Abandoned calls (tired of waiting)
Lost throughput Holding cost Cost of capacity $$$ Revenue $$$
Lost goodwill
Lost throughput (abandoned)
Cost per customer
Sales reps processing calls

Managerial Responses to Variability: Priority Rules in Waiting Time Systems
Total wait time: 4+13+21=38 min
Service times: A: 9 minutes B: 10 minutes C: 4 minutes D: 8 minutes
Total wait time: 9+19+23=51min
• Flow units are sequenced in the waiting area (triage step) • Shortest Processing Time Rule
– Minimizes average waiting time • First-Come-First-Serve
– easy to implement
– perceived fairness
• Sequence based on importance
– emergency cases
– identifying profitable flow units
4 min. D 12 min.
A Somewhat Odd Service Process
Arrival Time
Service Time
7:00 7:10 7:20 7:30 7:40 7:50 8:00

A More Realistic Service Process
Patient 1 Patient 3 Patient 5 Patient 7 Patient 9 Patient 11
Patient 2 Patient 4 Patient 6 Patient 8 Patient 10 Patient 12
Time 7:00 7:10 7:20 7:30 7:40 7:50 8:00
Arrival Time
Service Time
A More Realistic Service Process
Patient 1 Patient 3 Patient 5 Patient 7 Patient 9 Patient 11
Patient 2 Patient 4 Patient 6 Patient 8 Patient 10 Patient 12
Time 7:00 7:10 7:20 7:30 7:40 7:50 8:00
Arrival Time
Service Time
Service times
Number of cases

Variability Leads to Waiting Time
Arrival Time
Service Time
Service time
Inventory (Patients at lab)
5 4 3 2 1 0
7:00 7:10 7:20 7:30
7:00 7:10 7:20 7:30
7:40 7:50 8:00
7:40 7:50 8:00 9-7
Modeling Variability in Flow
Processing
No loss, waiting only This requires u < 100% Outflow = Inflow Demand process is “random” What generates queues?  An arrival rate that predictably exceeds the service rate (i.e., capacity) of a process.  e.g., toll booth congestion on the NJ Turnpike during the Thanksgiving Day rush.  Variation in the arrival and service rates in a process where the average service rate is more than adequate to process the average arrival rate.  e.g., calls to a brokerage are unusually high during a particular hour relative to the same hour in other weeks (i.e., calls are high by random chance). • Random arrivals (randomness is the rule, not the exception) • Incoming quality • Product Mix Activity times: • Inherent variation • Lack of operating procedures • Quality (scrap / rework) Resources: • Breakdowns / Maintenance • Operator absence • Set-up times Processing Figure 6.4.: Variability and where it comes from • Variable routing • Dedicated machines Defining interarrival times and a stationary process  An interarrival time is the amount of time between two arrivals to a process.  An arrival process is stationary over a period of time if the number of arrivals in any subinterval depends only on the length of the interval and not on when the interval starts.  For example, if the process is stationary over the course of a day, then the expected number of arrivals within any three hour interval is about the same no matter which three hour window is chosen (or six hour window, or one hour window, etc.).  Processes tend to be nonstationary (or seasonal) over long time periods (e.g., over a day or several hours) but stationary over short periods of time (say one hour, or 15 minutes). Arrivals to An-ser over the day are nonstationary Number of customers Per 15 minutes 160 140 120 100 80 60 40 20 The number of arrivals in a 3-hour interval clearly depends on which 3-hour interval your choose ... ... and the peaks and troughs are predictable (they occur roughly at the same time each day. 0:15 2:00 3:45 5:30 7:15 9:00 10:45 12:30 14:15 16:00 17:45 19:30 21:15 23:00 How to describe (or model) interarrival times  We will use two parameters to describe interarrival times to a process:  The average interarrival time.  The standard deviation of the interarrival times.  What is the standard deviation?  Roughly speaking, the standard deviation is a measure of how variable the interarrival times are.  Two arrival processes can have the same average interarrival time (say 1 minute) but one can have more variation about that average, i.e., a higher standard deviation. Relative and absolute variability  The standard deviation is an absolute measure of variability.  Two processes can have the same standard deviation but one can seem much more variable than the other:  Below are random samples from two processes that have the same standard deviation. The left one seems more variable. 50 45 40 35 30 25 20 15 10 Average = 10 Stdev = 10 Observation 120 100 80 60 40 20 0 Average = 100 Stdev = 10 Observation Inter-arrival time (min) Inter-arrival time (min) Relative and absolute variability (continued)  The previous slide plotted the processes on two different axes.  Here, the two are plotted relative to their average and with the same axes.  Relative to their average, the one on the left is obviously more variable (sometimes more than 400% above the average or less than 25% of the average). Average = 10 Stdev = 10 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 Average = 100 Stdev = 10 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 Observation Observation Coefficient of variation, arrival  The coefficient of variation is a measure of the relative variability of a process – it is the standard deviation divided by the average.  The coefficient of variation of the arrival process: CV􏰀 = Standard deviation of interarrival time Average interarrival time CVa =10/10=1 CVa = 10 /100 = 0.1 Average = 10 Stdev = 10 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 Average = 100 Stdev = 10 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 Observation Observation Inter-arrival time /Average Inter-arrival time /Average Inter-arrival time /Average Inter-arrival time /Average Service time (process) variability  The coefficient of variation of the service time process: CV􏰁 = Standard deviation of activity times 800 600 400 200  Standard deviation of activity times = 150 seconds.  Average call time (i.e., activity time) = 120 seconds.  CVp =150/120=1.25 Average activity time Call durations (seconds) Process configuration: A multi-server queue Arriving customers  Assumptions: Departing customers  All servers are equally skilled, i.e., they all take p time to process each customer.  Each customer is served by only one server.  Customers wait until their service is completed.  Thereissufficientcapacityto serve all demand.  a = average interarrival time  Flowrate=1/a  p = average activity time  Capacity of each server = 1 / p  m = number of servers  Capacity=mx1/p  Utilization = Flow rate / Capacity =(1/a)/(mx1/p) = p / (a x m) Customers waiting for service Multiple servers A multi-server queue – some data and analysis  Suppose:  a = 35 seconds per customer  Flow rate = 1 / 35 customers per second  p = 120 seconds per customer  Capacity of each server = 1 / 120 customers per second  m = number of servers = 4  Capacity=mx1/p=4x1/120=1/30customerspersecond  Utilization = Flow rate / Capacity = p / (a x m) = 120 / (35 x 4) = 85.7% The implications of utilization A 85.7% utilization means:  At any given moment, there is a 85.7% chance a server is busy serving a customer and a 1 - 0.857 = 14.3% chance the server is idle.  At any given moment, on average 85.7% of the servers are busy serving customers and 1 - 0.857 = 14.3% are idle. If utilization < 100% then:  The system is stable which means that the size of the queue does not keep growing over time. (Capacity is sufficient to serve all demand.) If utilization >= 100% then:
 The system is unstable, which means that the size of the queue will continue to grow over time.

Stable and unstable queues
difference of customers in system
p = 120 160
140 120 100
80 60 40 20
p = 120 160
Utilization = 114%
5000 10000 15000 20000 Time (seconds)
140 120 100
80 60 40 20
Utilization = 86%
10000 15000 20000 Time (seconds)
25000 30000
25000 30000
Time spent in the two stages of the system
Tq = Time in queue p = Time in service T=Tq +p=Flowtime
 2(m1)1  CV2 CV2  Activitytime Utilization   a p 
Time in queue   m  1-Utilization  2  
The above Time in queue equation works only for a stable system, i.e., a
system with utilization less than 100%
p = Activity time
m = number of servers
Total number of customers in the system
Total number of customers in the system

Three factors that determine time in queue in a stable system
The capacity factor:
Average processing time of the system = (p/m). Demand does not influence this factor.
Time in queue   m  1- Utilization 
 2(m1)1  CV2 CV2 
 Activity time   Utilization 
p   
The variability factor:
This is how variability influences time in queue – the more variability (holding average demand and capacity constant) the more time in queue.
The utilization factor:
Average demand influences this factor because utilization is the ratio of demand to capacity.
Example: Evaluating Time in queue
 Suppose:
 a=35seconds,p=120seconds,m=4  Utilization=p/(axm)=85.7%
CVa =1,CVp =1
 2(m1)1 CV2CV2
120 0.857 2(41)1  12 12 
     150
 Activity time   Utilization  Time in queue   m  1-Utilization 
4 1-0.857  2   So Flow time = 150 + 120 = 270 seconds.
 In other words, the average customer will spend 270 seconds in the system (waiting for service plus time in service)

How many customers are in the system?
Iq = Inventory in queue
I = Iq + Ip = Inventory in the system
 UseLittle’sLaw,I=RxT  R = Flow Rate = (1/a)
 The flow rate through the system equals the demand rate because we are demand constrained (utilization is less than 100%)
 Iq = RXT=(1/a)xTq =Tq /a
 Ip =RXT=(1/a)xp=p/a
 Note, time in service does not depend on the number of servers because when a customer is in service they are processed by only one server no matter how many servers are in the system.
Ip = Inventory in service
Utilization and system performance
 2(m1)1  CV2 CV2 
 Activity time   Utilization  Timeinqueue m  1-Utilization  2 
 a p 
 Time in queue increases
200 180 160 140 120
dramatically as the utilization 100
Activity time = 1
approaches 100%
80 60 40 20
80% 82% 84% 86% 88% 90% 92% 94% 96% 98% 100%
Utiliz ation
Time in queue

Which system is more effective?
Pooled system: One queue, four
Partially pooled system:
Two queues, two servers with each queue.
Separate queue system:
Four queues, one server with each queue.
a = 35 m=4
For each of the 2 queues: a = 35 x 2 = 70 m=2
For each of the 4 queues:
a = 35 x 4 = 140 m=1
Across these three types of systems:
Variability is the same: CVa =1,CVp =1
Total demand is the same: 1/35 customers per second.
Activity time is the same: p = 120
Utilization is the same: p / a x m = 85.7%
The probability a server is busy is the same = 0.857
Pooling can reduce Time in queue considerably
Pooled system
Partially pooled system
Separate queue system
a=35,p=120,m=4,CVa =1,CVp =1 Tq = 150
Flow Time = 270
For each of the two queues: a=70,p=120,m=2,CVa =1,CVp =1 Tq = 336
Flow Time = 456
For each of the four queues: a=140,p=120,m=1,CVa =1,CVp =1 Tq = 720
Flow Time = 840

Pooling reduces the number of customers waiting but not the number in service
Pooled system
Partially pooled system
Separate queue system
Iq =4.3 Ip = 3.4
Iq = 4.8 Ip = 1.7
Iq =4.3 Ip = 3.4
Iq = 9.6 Ip = 3.4
( 2 times the inventory in each queue )
( 4 times the inventoryin each queue )
Inventories for each queue within the system
Inventories for the total system
Why does pooling work?
= A customer waiting
= A customer in service
Separate queue system
Pooled system
 Both systems currently have 6 customers.
 In the pooled system, all servers are busy and only 2 customers are waiting.
 In the separate queue system, 2 servers are idle and 4 customers are waiting.
 The separate queue system is inefficient because there can be customers waiting and idle servers at the same time.

Limitations to pooling
 Pooling may require workers to have a broader set of skills, which may require more training and higher wages:
 Imagine a call center that took orders for McDonalds and Wendys … now the order takers need to be experts in two menus.
 Suppose cardiac surgeons need to be skilled at kidney transplants as well.
 Pooling may disrupt the customer – server relationship:
 Patients like to see the same physician.
 Pooling may increase the time-in-queue for one customer class at the expense of another:
 Removing priority security screening for first-class passengers may decrease the average time-in-queue for all passengers but will likely increase it for first-class passengers.
Stationary Arrivals?
Exponentially distributed inter-arrival times?
Break arrival process up into smaller time intervals
• Compute a: average interarrival time • CVa=1
• All results of chapters 6 and 7 apply
• Compute a: average interarrival time
• CVa= St.dev. of interarrival times / a
• All results of chapter 6 apply
• Results of chapter 7 do not apply, require
simulation or more complicated models
Figure 6.10: How to analyze a demand / arrival process?

Managing Waiting Systems: Points to Remember
• Variability is the norm, not the exception
• Variability leads to waiting times although utilization < 100% • Use the Waiting Time Formula to - get a qualitative feeling of the system - analyze specific recommendations / scenarios • Managerial response to variability: - understand where it comes from and eliminate what you can - accommodate the rest by holding excess capacity • Difference between variability and seasonality - seasonality is addressed by staffing to (expected) demand  Even when a process is demand constrained (utilization is less than 100%), waiting time for service can be substantial due to variability in the arrival and/or service process.  Waiting times tend to increase dramatically as the utilization of a process approaches 100%.  Pooling multiple queues can reduce the time-in-queue with the same amount of labor (or use less labor to achieve the same time-in-queue). 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com