1. Make to Stock (MTS) vs. Make to Order (MTO)
Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety inventory protects against stockouts due to variability of arrival time and processing time. Inventory also permits economies of scale.
Make-to-order (MTO) operations. Each order is specific, cannot be stored in advance. Ex. banks, restaurants, retail checkout counters, airline reservation, hospitals , repair shops, call centres. Production systems also try to follow Dell Computer model. We needs to maintain sufficient capacity to deal with uncertainty in both arrival and processing time. Safety Capacity vs. Safety Inventory.

2. A Call Centre The Call Centre Process Sales Reps Processing Calls
(Service Process)
Incoming Calls
(Customer Arrivals)
on Hold
(Service Inventory)
Answered Calls
(Customer Departures)
Blocked Calls
(Due to busy signal)
Abandoned Calls
(Due to long waits)
Calls In Process
The Call Centre Process

3. Characteristics of Waiting Lines (Queuing Systems)
The time of the arrival of an order is not known ahead of time. It is a random variable with estimated mean and standard deviation.
The time of the next telephone call is not known.
The service time is not known (precisely) ahead of time. It is a random variable with estimated mean and standard deviation.
The time a customers spends on the web page of amazon.com is not precisely known.
The time a customer spends speaking with the teller in the bank is unknown

4. Article: The Psychology of Waiting Lines
Unoccupied time feels longer than occupied time.
Pre-process waits feels longer than in-process waits.
Anxiety makes waits seem longer.
Uncertain waits are longer than known, finite waits.
Unexplained waits are longer than explained waits.
Unfair waits are longer than equitable waits.
The more valuable the service, the longer I will wait.
Solo waiting feels longer than group waiting.

5. Characteristics of Queuing Systems
Variability in arrival time and service time leads to
Idleness of resources
Waiting time of customers (orders) to be processed
We are interested in evaluating two measures:
Average waiting time of flow units. Average waiting time in the waiting line and in the system (Waiting line + Processor).
Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed).
Let us first look at the Servers or Processors

6. AVERAGE Processing Time Tp AVERAGE Processing Rate Rp
Tp : Processing time. Tp units of time. Ex. on average it takes 5 minutes to serve a customer. Rp : processing rate. Rp flow units are handled per unit of time. If Tp is 5 minutes. Compute Rp. Rp= 1/5 per minute, or 60/5 = 12 per hour.

7. More than One Server; c Servers
Tp : processing time. Rp : processing rate. What is the relationship between Rp and Tp? If we have one resource  Rp = 1/Tp What is the relationship between Rp and Tp when we have more than one resource; We have c recourses Rp = c/Tp Each customer always spends Tp unites of time with the server

8. Average Processing Rate of c Servers
Tp = 5 minutes. Processing time is 5 minute. Each customer on average is with the server for 5 minutes. c = 3, we have three servers. Processing rate of each server is 1/5 customers per minute, or 12 customer per hour. Rp is the processing rate of all three servers. Rp = c/Tp Rp = 3/5 customers/minute, or 36 customers/hour.

9. Inter-arrival Time (Ta) and Arrival Rate (Ra)
Ta : customer inter-arrival time. On average each 10 minutes one customer arrives. Ra: customer arrival (inflow) rate. What is the relationship between Ta and Ra Ta = every ten minutes one customer arrives How many customers in a minute? 1/10; Ra= 1/Ta= 1/10 Ra = 1/10 customers per min; 6 customers per hour Ra= 1/Ta

10. Throughput = Min (Ri,Rp)
Ra MUST ALWAYS <= Rp. We will show later that even Ra=Rp is not possible. Incoming rate must be less than processing rate. Throughput = Flow Rate R = Min (Ra, Rp) . Stable Process = Ra< Rp  R = Ra Safety Capacity Rs = Rp – Ra

11. Buffer (waiting line) and Processors (Servers)
Flow time T = Ti Tp
Inventory I = Ii Ip
Ti: waiting time in the inflow buffer
Ii: number of customers in the inflow buffer
What is the waiting time in the servers (processors)? Throughput?

12. Utilization is Always Less than 1
U = Utilization U =inflow rate / processing rate U = throughout / process capacity U = R/ Rp < 1 Safety Capacity = Rp – R For example , R = 6 per hour, processing time for a single server is 6 min  Rp= 12 per hour, U = R/ Rp = 6/12 = 0.5 Safety Capacity = Rp – R = 12-6 = 6

13. Given the Utilization, How Many Flow Units are in the Processor(s)
Given a single server. And a utilization of U= 0.5 How many flow units are in the server ?
U = 0.5 means
50% of time there is 1 flow unit in the server
50% of time there is 0 flow unit in the server
0.5   0 = 0.5
Average Inventory in the server is equal to utilization
Ip= 1U = U

14. Given the Utilization, How Many Flow Units are in the Processor(s)
Given 2 servers. And a utilization of U = 0.5
How many flow units are in the servers ?
U = 0.5 means
50% of time there is 1 flow unit in each server
50% of time there is 0 flow unit in each server
0.5   0 = 0.5 flow unit in each server
Average Inventory in the server is equal to utilization timesthe number of servers Ip= 2U = cU

15. What We Have Learned Without Looking for any Formula
Processing time: Tp, Ex. Tp = 5 minutes Number of servers: c, Ex. c=3 Tp is also waiting time in the server, no mater one server or c servers. Tp in this example is always 5 min. Processing rate Rp= c/Tp. Ex. Rp =3/5 per min; 36/hr Utilization: U. Ex. U = 0.8 in our example Number of the flow units in all servers, Ip = cU In our example, Ip = 3  0.8 = 2.4 Can we compute R?  TR = I Tp  R = cU  R = cU/Tp 5  R = 2.4  R = 0.48 flow units per minute or 28.8 / hr We learned it without looking at any formula

16. Characteristics of Queuing Systems
Variability in arrival time and service time leads to
Idleness of resources
Waiting time of customers (orders) to be processed
We are interested in evaluating two measures:
Average waiting time of flow units. Average waiting time in the waiting line and in the system (Waiting line + Processor).
Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed).

17. Utilization and Variability
Two key drivers of process performance are Utilization and Variability.
The higher the utilization the longer the waiting line.
High capacity utilization U= R/Rp or low safety capacity Rs =R – Rp, due to
High inflow rate R
Low processing rate Rp = c / Tp, which may be due to small-scale c and/or slow speed 1 /Tp
The higher the variability, the longer the waiting line.

18. Drivers of Process Performance
Variability in the interarrival time and processing time can be measured using standard deviation. Higher standard deviation means greater variability.
Coefficient of Variation: the ratio of the standard deviation of interarrival time (or processing time) to the mean.
Ca = coefficient of variation for interarrival times
Cp = coefficient of variation for processing times

19. Now Let’s Look at the Rest of the System; The Little’s Law Applies Everywhere
Flow time T Ti Tp
Inventory I = Ii Ip
I = R T
Ii = R Ti
Ip = R  Tp
R = I/T = Ii/Ti = Ip/Tp
We have already learned
Rp = c/Tp
U= R/Rp = Ip/c

20. Operational Performance Measures
Flow time T = Ti Tp
Inventory I = Ii Ip
Ti: waiting time in the inflow buffer = ?
Ii: number of customers waiting in the inflow buffer =?
A set of excel sheet are provided at the end of this lecture.
We relay on the approximation formula, given shortly, and our understanding of the waiting line logics.
We also provide an exact table for a special case.

21. The Queue Length APPROXIMATION Formula
Utilization effect U-part
Variability effect
V-part
U= R /Rp,where Rp = c / Tp
Ca and Cp are the Coefficients of Variation
Standard Deviation/Mean of the inter-arrival or processing times (assumed independent)

22. Factors affecting Queue Length
This part captures the capacity utilization effect. It shows that queue length increases rapidly as U approaches 1.
This part captures the variability effect. The queue length increases as the variability in interarrival and processing times increases.
Even if the processing capacity is not fully utilized, whenever there is variability in arrival or in processing times, queues will build up and customers will have to wait.

23. Utilization – Variability - Delay Curve
Average
Time in System T
Utilization (U) U
100% Tp
Variability
Increases

24. Lessons Learned
If inter-arrival and processing times are constant, queues will build up if and only if the arrival rate is greater than the processing rate.
If there is (unsynchronized) variability in inter-arrival and/or processing times, queues will build up even if the average arrival rate is less than the average processing rate.
If variability in interarrival and processing times can be synchronized (correlated), queues and waiting times will be reduced.

25. Coefficient of Variations for Alternative Distributions

26. Exact Ii for Poisson Arrival Rate, Exponential Service Time

27. Terminology and Classification of Waiting Lines
Terminology: The characteristics of a queuing system is captured by five parameters; Arrival pattern, Service pattern, Number of server, Restriction on queue capacity, The queue discipline.
M/M/1; Poisson arrival rate, Exponential service times, one server, No capacity limit.
M/G/12/23; Poisson arrival rate, General service times, 12 servers, Queue capacity is 23

28. Example; Coefficient of Variation of Interarrival Time
A sample of 10 observations on Interarrival times in minutes  10,10,2,10,1,3,7,9, 2, 6 minutes.
=AVERAGE ()  Avg. interarrival time = 6 min. Ra = 1/6 arrivals / min. =STDEV()  Std. Deviation = 3.94 Ca = 3.94/6 = 0.66 C2a = (0.66)2 =

29. Example: Coefficient of Variation of Processing Time
A sample of 10 observations on Processing times in minutes  7,1,7 2,8,7,4,8,5, 1 minutes.
Tp= 5 minutes Rp = 1/5 processes/min. Std. Deviation = 2.83 Cp = 2.83/5 = 0.57 C2p = (0.57)2 =

30. Example: Utilization and Safety Capacity
Ra=1/6 < RP =1/5  R = Ra U= R/ RP = (1/6)/(1/5) = 0.83 With c = 1, the average number of passengers in queue is as follows:
Ii = [(0.832)/(1-0.83)] ×[( )/2] = 1.56
On average 1.56 passengers waiting in line, even though safety capacity is Rs= RP - Ra= 1/5 - 1/6 = 1/30 passenger per minute, or 2 per hour.

31. Example: Other Performance Measures
Waiting time in the line? Ti=Ii/R = (1.56)(6) = 9.4 minutes. Waiting time in the system? T = Ti+Tp Since Tp= 5  T = Ti+ Tp= 14.4 minutes Total number of passengers in the process is: I = RT = (1/6) (14.4) = 2.4 Alternatively, 1.56 in the buffer. How many in the process? I = = 2. 39

32. Example: Now suppose we have two servers.
Compute R, Rp and U: Ta= 6 min, Tp = 5 min, c=2 R = Ra= 1/6 per minute Processing rate for one processor 1/5 for 2 processors Rp = 2/5 U = R/Rp = (1/6)/(2/5) = 5/12 = 0.417
On average passengers waiting in line.
safety capacity is Rs= RP - Ra= 2/5 - 1/6 = 7/30 passenger per minute, or 14 passengers per hour

33. Other Performance Measures for Two Servers
Other performance measures: Ti=Ii/R = (0.076)(6) = 0.46 minutes Compute T? T = Ti+Tp Since TP = 5  T = Ti + Tp= = 5.46 minutes Total number of passengers in the process is: I= 0.08 in the buffer and in the process. I = (0.417) = 0.91 c ρ Rs Ii Ti T I 1
0.83
0.03
1.56
9.38
14.38
2.4 2
0.417
0.23
0.077
0.46
5.46
0.91

34. Processing Time, Waiting Time; Long Waiting Line
Ra =R= 10/hour Tp = 5 minutes Interarrival time Poisson Service time exponential
Ra =
10/hour
Increasing Capacity
Polling

35. More Servers + Specialization
Ra/2 =
10/hour
Ra/2 =
10/hour

36. Polling: Lower Waiting Time, Longer Processing Time (Perhaps)
Ra =
10/hour

37. Effect of Pooling
Server 1
Queue 1
Ra/2 Ra
Server 2
Queue 2
Ra/2
Ra =R= 10/hour Tp = 5 minutes Interarrival time Poisson Service time exponential
Server 1
Queue
Server 2 Ra

38. Effect of Pooling : 2M/M/1
Server 1
Queue 1
Ra/2
Ra/2 = R= 5/hour
Tp = 5 minutes
c = 1  Rp = 12 / hour
U= 5/12 = 0.417 Ra
Server 2
Queue 2
Ra/2

39. Comparison of 2M/M/1 with M/M/2
Server 1
Queue 1
Ra/2 Ra
Server 2
Queue 2
Ra/2
Server 1
Queue
Server 2 Ra

40. Effect of Pooling: M/M/2
Ra=R= 10/hour Tp = 5 minutes c = 2 Rp = 24 /hour U= 10/24 U= AS BEFORE for each processor
Server 1
Queue
Server 2 Ra

41. Effect of Pooling Under Design A, Under Design B,
We have Ra = 10/2 = 5 per hour, and Tp = 5 minutes, c =1, we arrive at a total flow time of 8.6 minutes
Under Design B,
We have Ra =10 per hour, Tp= 5 minutes, c=2, we arrive at a total flow time of 6.2 minutes
So why is Design B better than A?
Design A the waiting time of customer is dependent on the processing time of those ahead in the queue
Design B, the waiting time of customer is partially dependent on each preceding customer’s processing time
Combining queues reduces variability and leads to reduce waiting times

42. Performance Measures Financial Performance Measures
Sales: Throughput Rate
Cost: Capacity utilization, Number in queue / in system
Customer service: Waiting Time in queue /in system
Performance Improvement Levers
Decrease variability in customer inter-arrival and processing times.
Decrease capacity utilization.
Synchronize available capacity with demand.

43. 1. Variability Reduction Levers
Customers arrival are hard to control
Scheduling, reservations, appointments, etc….
Variability in processing time
Increased training and standardization processes
Lower employee turnover rate  more experienced work force
Limit product variety, increase commonality of parts

44. 2. Capacity Utilization Levers
If the capacity utilization can be decreased, there will also be a decrease in delays and queues.
Since U=R/Rp, to decrease capacity utilization there are two options
Manage Arrivals: Decrease inflow rate Ra
Manage Capacity: Increase processing rate Rp
Managing Arrivals
Better scheduling, price differentials, alternative services
Managing Capacity
Increase scale of the process (the number of servers)
Increase speed of the process (lower processing time)

45. 3. Synchronizing Capacity with Demand
Capacity Adjustment Strategies
Personnel shifts, cross training, flexible resources
Workforce planning & season variability
Synchronizing of inputs and outputs, Better scheduling

46. The M/M/c Model EXACT Formulas

47. The M/G/1 Model EXACT Formulas

48. The M/G/1 Model EXACT Formulas

49. Assignment 1: M/M/1 Performance Evaluation
Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has Exponential distribution.
On average how many customers are in the waiting line?
How long a customer stays in the line?
How long a customer stays in the processor (with the server)?
On average how many customers are with the server?
On average how many customers are in the system?
On average how long a customer stay in the system?

50. Assignment 2: M/M/1 Performance Evaluation
What if the arrival rate is 11 per hour?

51. Assignment 3: M/G/c
A local GAP store on average has 10 customers per hour for the checkout line. The inter-arrival time follows the exponential distribution. The store has two cashiers. The service time for checkout follows a normal distribution with mean equal to 5 minutes and a standard deviation of 1 minute.
On average how many customers are in the waiting line?
How long a customer stays in the line?
How long a customer stays in the processors (with the servers)?
On average how many customers are with the servers?
On average how many customers are in the system ?
On average how long a customer stay in the system ?

52. Assignment 4: M/M/c Example
A call center has 11 operators. The arrival rate of calls is 200 calls per hour. Each of the operators can serve 20 customers per hour. Assume interarrival time and processing time follow Poisson and Exponential, respectively. What is the average waiting time (time before a customer’s call is answered)?

53. Assignment 5: M/D/c Example
Suppose the service time is a constant
What is the answer of the previous question?