1 safety capacity make to stock vs. make to order made-to-stock (mts) operations product is...
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1
Safety Capacity
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
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Safety Capacity
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)
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Safety Capacity
Sales RepsProcessing
Calls
(Service Process)
Incoming Calls(Customer Arrivals) Calls
on Hold(Service Inventory)
Answered Calls(Customer Departures)
Blocked Calls(Due to busy signal)
Abandoned Calls(Due to long waits)
The DesiTalk Call Center
Calls In Process(Due to long waits)
The Call Center Process
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Safety Capacity
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
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Safety Capacity
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
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Safety Capacity
Operational Performance Measures
Waiting time in the servers (processors)
?
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
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Safety Capacity
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
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Safety Capacity
Operational Performance Measures
Given a single server. And a utilization of = 0.4
How many flow units are in the server ?
Given 2 servers. And a utilization of = 0.4
How many flow units are in the servers ?
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Safety Capacity
Operational Performance Measures
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
Throughput = R
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Safety Capacity
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
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Safety Capacity
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
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Safety Capacity
Arrival Rate at an Airport Security Check Point
Customer Number 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
0 10 20 30 40 50
Time
1
2
3
4
5
6
7
8
9
10
Cust
omer
Num
ber
What is the queue size?What is the capacity utilization?
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Safety Capacity
Customer Number 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
0 10 20 30 40 50 60
Time
1
2
3
4
5
6
7
8
9
10
Cust
omer
Num
ber
Flow Times with Arrival Every 6 Secs
What is the queue size?What is the capacity utilization?
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Safety Capacity
Customer Number 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
6-F 33 7 14
7-G 36 4 15
8-H 43 8 16
9-I 52 5 12
10-J 54 1 11
0 10 20 30 40 50 60 70
Time
1
2
3
4
5
6
7
8
9
10
Cu
sto
mer
Queue Fluctuation
0
1
2
3
4
1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64
Time
Nu
mb
er
Effect of Variability
What is the queue size?What is the capacity utilization?
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Safety Capacity
Customer Number Arrival Time Processing Time
Time in Process
1-E 0 8 8
2-H 10 8 8
3-D 20 2 2
4-A 22 7 7
5-B 32 1 1
6-J 33 1 1
7-C 36 7 7
8-F 43 7 7
9-G 52 4 4
10-I 54 5 7 0 10 20 30 40 50 60 70
1
2
3
4
5
6
7
8
9
10
Effect of Synchronization
What is the queue size?What is the capacity utilization?
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Safety Capacity
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
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Safety Capacity
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
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Safety Capacity
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 timesCp = coefficient of variation for processing times
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Safety Capacity
The Queue Length Formula
ρ1
ρ1)2(c
iI
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)
2
22pi CC
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Safety Capacity
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.ρ1
ρ1)2(c
iI
2
22pi CC
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.
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Safety Capacity
VariabilityIncreases
AverageFlowTime T
Utilization (ρ) 100%
Tp
Throughput- Delay Curve
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Safety Capacity
Example 8.4
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
A sample of 10 observations on Interarrival times in seconds
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Safety Capacity
Example 8.4
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
A sample of 10 observations on processing times in seconds
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Safety Capacity
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)] ×[(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
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Safety Capacity
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
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Safety Capacity
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.
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Safety Capacity
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
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Safety Capacity
Example
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
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Safety Capacity
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 = 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
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Safety Capacity
Performance Improvement Levers
– Decrease variability in customer inter-arrival and processing times.
– Decrease capacity utilization.
– Synchronize available capacity with demand.
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Safety Capacity
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
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Safety Capacity
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
servicesManaging Capacity
– Increase scale of the process (the number of servers)– Increase speed of the process (lower processing time)
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Safety Capacity
Synchronizing Capacity with Demand
Capacity Adjustment Strategies
– Personnel shifts, cross training, flexible resources
– Workforce planning & season variability
– Synchronizing of inputs and outputs
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Safety Capacity
Server 1Queue 1
Server 2Queue 2
Server 1
Queue
Server 2
Effect of Pooling
Ri
Ri
Ri/2
Ri/2
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Safety Capacity
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 secondsUnder 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
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Safety Capacity
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
0.749 0.768 0.782
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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
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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/minuteEffect 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
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Safety Capacity
Optimal Processing Capacity
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
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Safety Capacity
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)
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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
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Safety Capacity
Net Rate: Ri(1- Pb)(1- Pa)
Throughput Rate:R=min[Ri(1- Pb)(1- Pa),Rp]
Effect of Blocking and Abandonment
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Safety Capacity
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=1Buffer 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 minAverage total time in the system T=7.44 minAverage total number of customers in the system I=2.29
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Safety Capacity
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
Example 8.8 - DesiCom Call Center
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Safety Capacity
Example 8.8 - DesiCom Call Center
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
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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 $386.6/hour
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Safety Capacity
Example 8.9 - Effect of Buffer Capacity on Total Cost
Number of lines n 5 6 7 8 9
Number of CSR’s c 1 1 1 1 1
Buffer capacity K=n-c 4 5 6 7 8
Cost of servers ($/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 = $100RiPb200.8 154.2 120.6 95.6 76.6
Average number of calls in queue Ii1.23 1.52 1.79 2.04 2.27
Hourly cost of waiting=120Ii147.6 182.4 214.8 244.8 272.4
Total cost of service, blocking and waiting ($/hr)
393.4 386.6 390.4 400.4 414
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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
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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.
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Safety Capacity
Example 8.11 - WalCo Drugs
One pharmacist, DaveAverage of 20 customers per hourDave takes Average of 2.5 min to fill prescriptionProcess rate 24 per hourAssume 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
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Safety Capacity
Example 8.11 - Probability distribution of the actual time customer spends in process
(obtained by simulation)
0
2000
4000
6000
8000
10000
12000
14000
Total Time in Process
Fre
qu
ency
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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
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Safety Capacity
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.
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Safety Capacity
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)
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Safety Capacity
Example 8.12 - WalCo Drug
WalCo has set SL = 0.95Assuming total time for customers is exponential
Tduedate = -T ln(1 – SL)
Tduedate = -T ln(0.05) = 3TFlow time for 95 percentile of exponential distribution is three times
the average T
Tduedate = 3 * 15 = 4595% of customers will get served within 45 min
Tduedate = T + Tsafety
Tsafety = 45 – 15 = 30 min30 min is the extra margin that WalCo should allow as protection
against variability
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Safety Capacity
Relating Utilization and Safety Time: Safety Time Vs. Capacity Utilization
Capacity utilization ρ 60 % 70% 80% 90%
Waiting time Ti 1.5Tp 2.33Tp 4Tp 9Tp
Total flow time T= Ti + Tp 2.5Tp 3.33Tp 5Tp 10Tp
Promised time Tduedate 7.7Tp 10Tp 15Tp 30Tp
Safety time Tsafety = Tduedate – T 5Tp 6.67Tp 10Tp 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
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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
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Safety Capacity
Thank you
Questions?