Lecture Note
Safety Capacity
-
University:
California State University, Northridge -
Course:
ECON 500 | Survey of Economics Academic year:
2024
-
Views:
252
Pages:
57
Author:
frtlp20005yi
Safety Capacity
Make to stock vs. Make to Order
1
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 Safety Capacity
Examples
2
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) Safety Capacity
The DesiTalk Call Center
3
The Call Center Process
Incoming Calls
(Customer Arrivals)
Calls
on Hold
(Service Inventory)
Blocked Calls
Abandoned Calls
(Due to busy signal) (Due to long waits)
Sales Reps
Processing
Calls
(Service Process)
Calls In Process
(Due to long waits)
Answered Calls
(Customer Departures) Safety Capacity
Service Process Attributes
4
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 Safety Capacity
A GAP Store
5
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 Safety Capacity
Operational Performance Measures
6
Flow time T =
Inventory I =
Ti
Ii
+
+
Tp
Ip
Ti: waiting time in the inflow buffer
Ii: number of customers waiting in the inflow buffer
Waiting time in the servers (processors)
? Safety Capacity
Service Process Attributes
7
ρ = inflow rate / processing rate
ρ = throughout / process capacity
ρ = 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,
ρ = R/ Rp = 6/12 = 0.5
Safety Capacity = Rp – R = 12-6 = 6 Safety Capacity
Operational Performance Measures
8
Given a single server. And a utilization of ρ = 0.4
How many flow units are in the server ?
Given 2 servers. And a utilization of ρ = 0.4
How many flow units are in the servers ? Safety Capacity
Operational Performance Measures
9
Throughput = R
Flow time T
Inventory I
I = R× T
R = I/T
ρ = R/ Rp
ρ = Ip / c
=
= Ti
= Ii
+
+
Tp
Ip
Ii = R× Ti
Ip = R × Tp
Ii/Ti
= Ip/Tp Safety Capacity
Operational Performance Measures
10
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
ρ = R/ Rp = (Ip/Tp)/(c/Tp) = Ip/c
ρ = R/ Rp = Ip/c Safety Capacity
Financial Performance Measures
11
Sales
– Throughput Rate
– Abandonment Rate
– Blocking Rate
Cost
– Capacity utilization
– Number in queue / in system
Customer service
– Waiting Time in queue /in system Safety Capacity
Arrival Rate at an Airport Security Check Point
12
Arrival Time
Departure
Time
Time in
Process
1
0
5
5
2
4
10
6
3
8
15
7
4
12
20
8
5
16
25
9
6
20
30
10
7
24
35
11
8
28
40
12
9
32
45
13
10
36
50
14
10
9
8
Customer Number
Customer
Number
7
6
5
4
3
2
1
0
10
20
30
40
50
Time
What is the queue size?
What is the capacity utilization? Safety Capacity
Flow Times with Arrival Every 6 Secs
13
Arrival Time
Departure
Time
Time in
Process
1
0
5
5
2
6
11
5
3
12
17
5
4
18
23
5
5
24
29
5
6
30
35
5
7
36
41
5
8
42
47
5
9
48
53
5
10
54
59
5
10
9
8
Customer Number
Customer
Number
7
6
5
4
3
2
1
0
10
20
30
40
50
60
Time
What is the queue size?
What is the capacity utilization? Safety Capacity
Effect of Variability
14
Arrival Time
Processing Time
Time in
Process
1-A
0
7
7
2-B
10
1
1
3-C
20
7
7
4-D
22
2
7
5-E
32
8
8
10
9
8
7
Customer
Customer
Number
5
4
3
2
1
0
10
20
30
40
50
60
70
Time
6-F
33
7
14
7-G
36
4
15
8-H
43
8
16
9-I
52
5
12
4
10-J
54
1
11
3
Queue Fluctuation
Number
What is the queue size?
What is the capacity utilization?
6
2
1
0
1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64
Time Safety Capacity
Effect of Synchronization
15
Customer
Number
Arrival Time
Processing Time
Time in
Process
1-E
0
8
8
2-H
10
8
8
8
3-D
20
2
2
7
4-A
22
7
7
6
5-B
32
1
1
5
6-J
33
1
1
4
7-C
36
7
7
8-F
43
7
7
9-G
52
4
4
10-I
54
5
7
What is the queue size?
What is the capacity utilization?
10
9
3
2
1
0
10
20
30
40
50
60
70 Safety Capacity
Conclusion
16
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 Safety Capacity
Causes of Delays and Queues
17
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 smallscale c and/or slow speed 1 / Tp Safety Capacity
Drivers of Process Performance
18
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 Safety Capacity
The Queue Length Formula
19
Ii
ρ
=
Utilization
effect
2(c +1)
2
i
C +C
2
p
2
1− ρ
x
Variability
effect
ρ = 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) Safety Capacity
Factors affecting Queue Length
20
Ii
ρ
=
2(c+1)
1− ρ
2
i
C +C
2
2
p
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. Safety Capacity
Throughput- Delay Curve
21
Average
Flow
Time T
Variability
Increases
Tp
Utilization (ρ)
100%
ρ Safety Capacity
Example 8.4
22
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 = 0.4312 Safety Capacity
Example 8.4
23
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 = 0.3204 Safety Capacity
Example 8.4
24
Ri =1/6 < RP =1/5 R = Ri
ρ = 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)] ×[(0.662+0.572)/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 Safety Capacity
Example 8.4
25
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 Safety Capacity
Exponential Model
26
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. Safety Capacity
The Exponential Model
27
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 Safety Capacity
Example
28
Interarrival time = 6 secs Ri = 10/min
Tp = 5 secs Rp = 12/min for 1 server and 24 /min for 2 servers
Rs = 12-10 = 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
2
0.42
14
0.18
0.02
0.1
1 Safety Capacity
t ≤ t in Exponential Distribution
29
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 = 2.718282, 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 = 0.451188
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 = 0.1888 Safety Capacity
Performance Improvement Levers
30
– Decrease variability in customer inter-arrival
and processing times.
– Decrease capacity utilization.
– Synchronize available capacity with
demand. Safety Capacity
Variability Reduction Levers
31
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 Safety Capacity
Capacity Utilization Levers
32
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) Safety Capacity
Synchronizing Capacity with Demand
33
Capacity Adjustment Strategies
– Personnel shifts, cross training, flexible resources
– Workforce planning & season variability
– Synchronizing of inputs and outputs Safety Capacity
Effect of Pooling
34
Ri/2
Queue 1
Ri
Ri/2
Queue 2
Server 1
Server 2
Server 1
Ri
Queue
Server 2 Safety Capacity
Effect of Pooling
35
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 Safety Capacity
Effect of Buffer Capacity
36
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
ρ
0.749
0.768
0.782 Safety Capacity
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
– Increases with competition
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
37 Safety Capacity
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
20
154.2
182.6
386.4
6
35
20
120.6
214.8
390.4
38 Safety Capacity
Optimal Processing Capacity
39
c
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
3
0.0009
0.021
$94.2
4
2
0.0004
0.003
$110.8 Safety Capacity
Performance Variability
40
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) Safety Capacity
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
41 Safety Capacity
Effect of Blocking and Abandonment
Net Rate:
Ri(1- Pb)(1- Pa)
Throughput Rate:
R=min[Ri(1- Pb)(1- Pa),Rp]
42 Safety Capacity
Example 8.8 - DesiCom Call Center
43
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 Safety Capacity
Example 8.8 - DesiCom Call Center
Throughput Rate
R=min[Ri(1- Pb),Rp]= min[20*(1-0.0771),24]
R=18.46 calls/hour
Server utilization:
R/ Rp=18.46/24=0.769
44 Safety Capacity
Example 8.8 - DesiCom Call Center
45
Number of lines
5
6
7
8
9
10
Number of servers c
1
1
1
1
1
1
Buffer Capacity K
4
5
6
7
8
9
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 Safety Capacity
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 $386.6/hour
46 Example 8.9 - Effect of Buffer Capacity on
Total Cost
Safety Capacity
47
Number of lines n
5
6
7
8
9
Number of CSR’s c
1
1
1
1
1
Buffer capacity K=n-c
4
5
6
7
8
Cost of servers ($/hr)=20c
20
20
20
20
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 Safety Capacity
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
3
0.09%
0.018
0.02
30+60+(0.018x100)+(0.02x120)=94.2
4
2
0.04%
0.008
0.00
30+80+(0.008x100)+(0.00x120)110.8
48 Safety Capacity
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.
49 Safety Capacity
Example 8.11 - 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
50 Safety Capacity
Example 8.11 - Probability distribution of the actual
time customer spends in process (obtained by
simulation)
14000
10000
8000
6000
4000
2000
Total Time in Process
74
68
62
56
50
44
38
32
26
20
14
8
0
2
Frequency
12000
51 Safety Capacity
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
52 Tduedate
Service Promise:
, Service Level & Safety Time
Safety Capacity
53
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. Safety Capacity
Due Date Quotation
Due Date Quotation is the practice of promising a time frame within
which the product will be delivered.
54
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) Safety Capacity
Example 8.12 - 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
55 Safety Capacity
Relating Utilization and Safety Time:
Safety Time Vs. Capacity Utilization
Capacity utilization ρ
Waiting time Ti
Total flow time T= Ti + Tp
Promised time Tduedate
Safety time Tsafety = Tduedate – T
60 %
1.5Tp
2.5Tp
7.7Tp
5Tp
70%
2.33Tp
3.33Tp
10Tp
6.67Tp
56
80%
4Tp
5Tp
15Tp
10Tp
90%
9Tp
10Tp
30Tp
20Tp
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 Safety Capacity
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
57
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