1. Make to stock vs. Make to Order
Made-to-stock (MTS) operations
Product is manufactured and stocked in advance of demand
Inventory permits economies of scale and protects against stockouts due to variability of inflows and outflows
Make-to-order (MTO) process
Each order is specific, cannot be stored in advance
Process manger needs to maintain sufficient capacity
Variability in both arrival and processing time
Role of capacity rather than inventory
Safety inventory vs. Safety Capacity
Example: Service operations

2. Examples Banks (tellers, ATMs, drive-ins)
Fast food restaurants (counters, drive-ins)
Retail (checkout counters)
Airline (reservation, check-in, takeoff, landing, baggage claim)
Hospitals (ER, OR, HMO)
Service facilities (repair, job shop, ships/trucks load/unload)
Some production systems- to some extend (Dell computer)
Call centers (telemarketing, help desks, 911 emergency)

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

4. Service Process Attributes
Ri : customer arrival (inflow) rate
inter-arrival time = 1/Ri
Tp : processing time
Rp : processing rate
If we have one resource  Rp = 1/Tp
In general when we have c c recourses, Rp = c/Tp

5. A GAP Store Ri = 6 customers per hour
inter-arrival time = 1/Ri = 1/6 hour or 10 minutes
Tp = processing time = 5 minutes = 5/60 =1/12 hour
Rp : processing rate
If we have one resource 
Rp = 1/Tp = 1/(1/12) = 12 customers per hour
If we have c c recourses
Rp = 2/Tp = 12 customers per hour

6. 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
Waiting time in the servers (processors) ?

7. Service Process Attributes
= inflow rate / processing rate
= throughout / process capacity
 = R/ Rp < 1
Safety Capacity = Rp – R
In the Gap example , R = 6 per hour, processing time for a single server is 6 min  Rp= 12 per hour,
 = R/ Rp = 6/12 = 0.5
Safety Capacity = Rp – R = 12-6 = 6

8. Operational Performance Measures
Given a single server. And a utilization of r = 0.4
How many flow units are in the server ?
Given 2 servers. And a utilization of r = 0.4
How many flow units are in the servers ?

9. Operational Performance Measures
Throughput = R
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
 = R/ Rp
 = Ip / c

10. Operational Performance Measures
I = R T Ii = R Ti Ip = R  Tp
R = I/T = Ii/Ti = Ip/Tp
Tp  if 1 server  Rp = 1/Tp
In general, if c servers  Rp = c/Tp
R = Ip/Tp
 = R/ Rp = (Ip/Tp)/(c/Tp) = Ip/c
 = R/ Rp = Ip/c

11. Financial Performance Measures
Sales
Throughput Rate
Abandonment Rate
Blocking Rate
Cost
Capacity utilization
Number in queue / in system
Customer service
Waiting Time in queue /in system

12. Arrival Rate at an Airport Security Check Point
Customer Number
Arrival Time
Departure Time
Time in Process 1 5 2 4 10 6 3 8 15 7 12 20 16 25 9 30 24 35 11 28 40 32 45 13 36 50 14
What is the queue size?
What is the capacity utilization?

13. Flow Times with Arrival Every 6 Secs
Customer Number
Arrival Time
Departure Time
Time in Process 1 5 2 6 11 3 12 17 4 18 23 24 29 30 35 7 36 41 8 42 47 9 48 53 10 54 59
What is the queue size?
What is the capacity utilization?

14. Effect of Variability What is the queue size?
Customer Number
Arrival Time
Processing Time
Time in Process
1-A 7
2-B 10 1
3-C 20
4-D 22 2
5-E 32 8
6-F 33 14
7-G 36 4 15
8-H 43 16
9-I 52 5 12
10-J 54 11
What is the queue size?
What is the capacity utilization?

15. Effect of Synchronization
Customer Number
Arrival Time
Processing Time
Time in Process
1-E 8
2-H 10
3-D 20 2
4-A 22 7
5-B 32 1
6-J 33
7-C 36
8-F 43
9-G 52 4
10-I 54 5
What is the queue size?
What is the capacity utilization?

16. Conclusion
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

17. Causes of Delays and Queues
High, unsynchronized variability in
- Interarrival times
- Processing times
High capacity utilization ρ= 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

18. Drivers of Process Performance
Two key drivers of process performance, (1) Interarrival time and processing time variability, and (2) Capacity utilization
Variability in the interarrival and processing times 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.
Ci = coefficient of variation for interarrival times
Cp = coefficient of variation for processing times

19. The Queue Length Formula
Utilization effect
Variability effect x
 Ri / Rp, where Rp = c / Tp
Ci and Cp are the Coefficients of Variation
(Standard Deviation/Mean) of the inter-arrival and processing times (assumed independent)

20. Factors affecting Queue Length
This part factor captures the capacity utilization effect, which shows that queue length increases rapidly as the capacity utilization p increases to 1.
The second factor captures the variability effect, which shows that the queue length increases as the variability in interarrival and processing times increases.
Whenever there is variability in arrival or in processing queues will build up and customers will have to wait, even if the processing capacity is not fully utilized.

21. Throughput- Delay Curve
Variability
Increases
Average
Flow
Time T
Utilization (ρ) r
100% Tp

22. Example 8.4
A sample of 10 observations on Interarrival times in seconds
10,10,2,10,1,3,7,9, 2, 6
=AVERAGE ()  Avg. interarrival time = 6
Ri = 1/6 arrivals / sec.
=STDEV()  Std. Deviation = 3.94
Ci = 3.94/6 = 0.66
C2i = (0.66)2 =

23. Example 8.4 A sample of 10 observations on processing times in seconds
7,1,7 2,8,7,4,8,5, 1
Tp= 5 seconds
Rp = 1/5 processes/sec.
Std. Deviation = 2.83
Cp = 2.83/5 = 0.57
C2p = (0.57)2 =

24. Example 8.4 Ri =1/6 < RP =1/5  R = Ri
 = 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 - Ri = 1/5 - 1/6 = 1/30 passenger per second, or 2 per minutes

25. Example 8.4 Other performance measures:
Ti=Ii/R = (1.56)(6) = 9.4 seconds
Since TP= 5  T = Ti + TP = 14.4 seconds
Total number of passengers in the process is:
I = RT = (1/6) (14.4) = 2.4
C=2  Rp = 2/5  ρ = (1/6)/(2/5) = 0.42  Ii = 0.08 c ρ Rs Ii Ti T I 1
0.83
0.03
1.56
9.38
14.38
2.4 2
0.42
0.23
0.08
0.45
5.45
0.91

26. Exponential Model
In the exponential model, the interarrival and processing times are assumed to be independently and exponentially distributed with means 1/Ri and Tp.
Independence of interarrival and processing times means that the two types of variability are completely unsynchronized.
Complete randomness in interarrival and processing times.
Exponentially distribution is Memoryless: regardless of how long it takes for a person to be processed we would expect that person to spend the mean time in the process before being released.

27. The Exponential Model Poisson Arrivals
Infinite pool of potential arrivals, who arrive completely randomly, and independently of one another, at an average rate Ri  constant over time
Exponential Processing Time
Completely random, unpredictable, i.e., during processing, the time remaining does not depend on the time elapsed, and has mean Tp
Computations
Ci = Cp = 1
K = ∞ , use Ii Formula
K < ∞ , use Performance.xls

28. Example c ρ Rs Ii Formula Ti= Ri / Ii T= Ti+ 5/60 I= Ii + c ρ
Interarrival time = 6 secs  Ri = 10/min
Tp = 5 secs  Rp = 12/min for 1 server and 24 /min for 2 servers
Rs = = 2 c ρ Rs
Ii Formula
Ti= Ri / Ii
T= Ti+ 5/60
I= Ii + c ρ 1
0.83 2
4.16
0.42
0.5 5 14
0.18
0.02
0.1

29. t ≤ t in Exponential Distribution
Mean inter-arrival time = 1/Ri
Probability that the time between two arrivals t is less than or equal to a specific vaule of t
P(t≤ t) = 1 - e-Rit, where e = , the base of the natural logarithm
Example 8.5:
If the processing time is exponentially distributed with a mean of 5 seconds, the probability that it will take no more than 3 seconds is 1- e-3/5 =
If the time between consecutive passenger arrival is exponentially distributed with a mean of 6 seconds ( Ri =1/6 passenger per second)
The probability that the time between two consecutive arrivals will exceed 10 seconds is e-10/6 =

30. Performance Improvement Levers
Decrease variability in customer inter-arrival and processing times.
Decrease capacity utilization.
Synchronize available capacity with demand.

31. 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

32. Capacity Utilization Levers
If the capacity utilization can be decreased, there will also be a decrease in delays and queues.
Since ρ=Ri/RP, to decrease capacity utilization there are two options:
Manage Arrivals: Decrease inflow rate Ri
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)

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

34. Effect of Pooling Ri/2 Server 1 Queue 1 Ri Ri/2 Server 2 Queue 2

35. Effect of Pooling Under Design A,
We have Ri = 10/2 = 5 per minute, and TP= 5 seconds, c =1 and K =50, we arrive at a total flow time of 8.58 seconds
Under Design B,
We have Ri =10 per minute, TP= 5 seconds, c=2 and K=50, we arrive at a total flow time of 6.02 seconds
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 only partially dependent on each preceding customer’s processing time
Combining queues reduces variability and leads to reduce waiting times

36. Effect of Buffer Capacity
Process Data
Ri = 20/hour, Tp = 2.5 mins, c = 1, K = # Lines – c
Performance Measures K 4 5 6 Ii
1.23
1.52
1.79 Ti
4.10
4.94
5.72 Pb
0.1004
0.0771
0.0603 R
17.99
18.46
18.79 r
0.749
0.768
0.782

37. Economics of Capacity Decisions
Cost of Lost Business Cb
$ / customer
Increases with competition
Cost of Buffer Capacity Ck
$/unit/unit time
Cost of Waiting Cw
$ /customer/unit time
Cost of Processing Cs
$ /server/unit time
Increases with 1/ Tp
Tradeoff: Choose c, Tp, K
Minimize Total Cost/unit time
= Cb Ri Pb + Ck K + Cw I (or Ii) + c Cs

38. Optimal Buffer Capacity
Cost Data
Cost of telephone line = $5/hour, Cost of server = $20/hour, Margin lost = $100/call, Waiting cost = $2/customer/minute
Effect of Buffer Capacity on Total Cost K
$5(K + c)
$20 c
$100 Ri Pb
$120 Ii
TC ($/hr) 4 25 20
200.8
147.6
393.4 5 30
154.2
182.6
386.4 6 35
120.6
214.8
390.4

39. Optimal Processing Capacity
K = 6 – c Pb Ii
TC ($/hr) = $20c + $5(K+c) + $100Ri Pb+ $120 Ii 1 5
0.0771
1.542
$386.6 2 4
0.0043
0.158
$97.8 3
0.0009
0.021
$94.2
0.0004
0.003
$110.8

40. Performance Variability
Effect of Variability
Average versus Actual Flow time
Time Guarantee
Promise
Service Level
P(Actual Time  Time Guarantee)
Safety Time
Time Guarantee – Average Time
Probability Distribution of Actual Flow Time
P(Actual Time  t) = 1 – EXP(- t / T)

41. Effect of Blocking and Abandonment
Blocking: the buffer is full = new arrivals are turned away
Abandonment: the customers may leave the process before being served
Proportion blocked Pb
Proportion abandoning Pa

42. Net Rate: Ri(1- Pb)(1- Pa) Throughput Rate: R=min[Ri(1- Pb)(1- Pa),Rp]
Effect of Blocking and Abandonment
Net Rate: Ri(1- Pb)(1- Pa) Throughput Rate: R=min[Ri(1- Pb)(1- Pa),Rp]

43. Example 8.8 - DesiCom Call Center
Arrival Rate Ri= 20 per hour=0.33 per min
Processing time Tp =2.5 minutes (24/hr)
Number of servers c=1
Buffer capacity K=5
Probability of blocking Pb=0.0771
Average number of calls on hold Ii=1.52
Average waiting time in queue Ti=4.94 min
Average total time in the system T=7.44 min
Average total number of customers in the system I=2.29

44. Example 8.8 - DesiCom Call Center
Throughput Rate
R=min[Ri(1- Pb),Rp]= min[20*( ),24]
R=18.46 calls/hour
Server utilization:
R/ Rp=18.46/24=0.769

45. Example 8.8 - DesiCom Call Center
Number of lines 5 6 7 8 9 10
Number of servers c 1
Buffer Capacity K 4
Average number of calls in queue
1.23
1.52
1.79
2.04
2.27
2.47
Average wait in queue Ti (min)
4.10
4.94
5.72
6.43
7.08
7.67
Blocking Probability Pb (%)
10.04
7.71
6.03
4.78
3.83
3.09
Throughput R (units/hour)
17.99
18.46
18.79
19.04
19.23
19.38
Resource utilization
.749
.769
.782
.793
.801
.807

46. Capacity Investment Decisions
The Economics of Buffer Capacity
Cost of servers wages =$20/hour
Cost of leasing a telephone line=$5 per line per hour
Cost of lost contribution margin =$100 per blocked call
Cost of waiting by callers on hold =$2 per minute per customer
Total Operating Cost is $386.6/hour

47. Example 8.9 - Effect of Buffer Capacity on Total Cost
Number of lines n 5 6 7 8 9
Number of CSR’s c 1
Buffer capacity K=n-c 4
Cost of servers ($/hr)=20c 20
Cost of tel.lines ($/hr)=5n 25 30 35 40 45
Blocking Probability Pb (%)
10.04
7.71
6.03
4.78
3.83
Lost margin = $100RiPb
200.8
154.2
120.6
95.6
76.6
Average number of calls in queue Ii
1.23
1.52
1.79
2.04
2.27
Hourly cost of waiting=120Ii
147.6
182.4
214.8
244.8
272.4
Total cost of service, blocking and waiting ($/hr)
393.4
386.6
390.4
400.4
414

48. Example 8.10 - The Economics of Processing Capacity
The number of line is fixed: n=6
The buffer capacity K=6-c c K
Blocking Pb(%)
Lost Calls RiPb
(number/hr)
Queue length Ii
Total Cost ($/hour) 1 5
7.71%
1.542
1.52
30+20+(1.542x100)+(1.52x120)=386.6 2 4
0.43%
0.086
0.16
30+40+(0.086x100)+(0.16x120)=97.8 3
0.09%
0.018
0.02
30+60+(0.018x100)+(0.02x120)=94.2
0.04%
0.008
0.00
30+80+(0.008x100)+(0.00x120)110.8

49. Variability in Process Performance
Why considering the average queue length and waiting time as performance measures may not be sufficient?
Average waiting time includes both customers with very long wait and customers with short or no wait.
We would like to look at the entire probability distribution of the waiting time across all customers.
Thus we need to focus on the upper tail of the probability distribution of the waiting time, not just its average value.

50. Average of 20 customers per hour
Example WalCo Drugs
One pharmacist, Dave
Average of 20 customers per hour
Dave takes Average of 2.5 min to fill prescription
Process rate 24 per hour
Assume exponentially distributed interarrival and processing time; we have single phase, single server exponential model
Average total process is;
T = 1/(Rp – Ri) = 1/(24 -20) = 0.25 or 15 min

51. Example Probability distribution of the actual time customer spends in process (obtained by simulation)

52. Example 8.11 - Probability Distribution Analysis
65% of customers will spend 15 min or less in process
95% of customers are served within 40 min
5% of customers are the ones who will bitterly complain. Imagine if they new that the average customer spends 15 min in the system.
35% may experience delays longer than Average T,15min

53. Service Promise: Tduedate , Service Level & Safety Time
SL; The probability of fulfilling the stated promise. The Firm will set the SL and calculate the Tduedate from the probability distribution of the total time in process (T).
Safety time is the time margin that we should allow over and above the expected time to deliver service in order to ensure that we will be able to meet the required date with high probability
Tduedate = T + Tsafety
Prob(Total time in process <= Tduedate) = SL
Larger SL results in grater probability of fulfilling the promise.

54. Due Date Quotation
Due Date Quotation is the practice of promising a time frame within which the product will be delivered.
We know that in single-phase single server service process; the Actual total time a customer spends in the process is exponentially distributed with mean T.
SL = Prob(Total time in process <= Tduedate) = 1 – EXP( - Tduedate /T)
Which is the fraction of customers who will no longer be delayed more than promised.
Tduedate = -T ln(1 – SL)

55. 95% of customers will get served within 45 min
Example WalCo Drug
WalCo has set SL = 0.95
Assuming total time for customers is exponential
Tduedate = -T ln(1 – SL)
Tduedate = -T ln(0.05) = 3T
Flow time for 95 percentile of exponential distribution is three times the average T
Tduedate = 3 * 15 = 45
95% of customers will get served within 45 min
Tduedate = T + Tsafety
Tsafety = 45 – 15 = 30 min
30 min is the extra margin that WalCo should allow as protection against variability

56. Higher the utilization; Longer the promised time and Safety time
Relating Utilization and Safety Time: Safety Time Vs. Capacity Utilization
Capacity utilization ρ % % % %
Waiting time Ti Tp Tp Tp Tp
Total flow time T= Ti + Tp Tp Tp Tp Tp
Promised time Tduedate Tp Tp Tp Tp
Safety time Tsafety = Tduedate – T Tp Tp Tp Tp
Higher the utilization; Longer the promised time and Safety time
Safety Capacity decreases when capacity utilization increases
Larger safety capacity, the smaller safety time and therefore we can promise a shorter wait

57. Managing Customer Perceptions and Expectations
Uncertainty about the length of wait (Blind waits) makes customers more impatient.
Solution is Behavioral Strategies
Making the waiting customers comfortable
Creating distractions
Offering entertainment

58. Thank you
Questions?