1 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Made-to-stock (MTS) operations ...

1Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2

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

Make to stock vs. Make to Order


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)

Examples


The DesiTalk Call Center

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)

Calls In Process(Due to long waits)

The Call Center Process


Tp : processing time. Tp units of time. Example, 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.

Service Process Attributes


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 TpRp = c/Tp

Each customer always stay for Tp unites of time with the server

In general when we have c recourses


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.




Ta : customer inter-arrival time.

On average each 10 minutes one customer arrives. Ri : customer arrival (inflow) rate.

What is the relationship between Ta and RiTa = every ten minutes one customer arrivesHow many customers in a minute? 1/10; Ri = 1/Ta= 1/10Ri = 1/10 customers per min; 6 customers per hourRi = 1/Ta



Ri MUST ALWAYS <= RpWe will show later that even Ri=Rp is not possibleIncoming rate must be less than processing rateThroughput = Flow Rate R = Min (Ri, Rp) Stable Process = Ri < Rp,, so that R = Ri

Safety Capacity Rs = Rp – Ri


Waiting time in the servers (processors)?Throughput?

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


= Utilization= inflow rate / processing rate = throughout / process capacity = R/ Rp < 1Safety Capacity = Rp – R

For example , R = 6 per hour, processing time for a single server is 6 min Rp= 12 per hour,

= R/ Rp = 6/12 = 0.5Safety Capacity = Rp – R = 12-6 = 6



Given a single server. And a utilization of r = 0.5How many flow units are in the server ?


Given 2 servers. And a utilization of r = 0.5How many flow units are in the servers ?



Flow time T Ti + Tp


R

I = R T

R = I/T = Ii/Ti = Ip/Tp

Ii = R Ti Ip = R Tp

Tp if 1 server Rp = 1/Tp In general, if c servers Rp = c/Tp



= R/ Rp = (Ip/Tp)/(c/Tp) = Ip/c

And it is obvious that = Ip/cBecause on average there are Ip people in the

processors and the capacity of the servers is to serve c costomer at a time.

= R/Rp = Ip/c

From the Little’s Law

Rp = c/Tp

R = Ip/Tp

From definition of Rp


Sales Throughput Rate

Cost Capacity utilization Number in queue / in system

Customer service Waiting Time in queue /in system

Financial Performance Measures


What is the queue size?What is the capacity utilization?

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



Flow Times with Arrival Every 6 Secs

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




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




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 70 10 20 30 40 50 60 70

1

2

3

4

5

6

7

8

9

10


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

Conclusion


Two key drivers of process performance are (i) Capacity utilization, and (ii) interarrival time and processing time variability

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 High, unsynchronized variability in

Interarrival times Processing times

Causes of Delays and Queues


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

Drivers of Process Performance



Flow time T = Ti + Tp


Ti: waiting time in the inflow buffer = ?

Ii: number of customers waiting in the inflow buffer =?


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)

The Queue Length Formula

ρ1

ρ1)2(c

Utilization effect

Variability effect

2

22pi CC

iI


This part factor captures the capacity utilization effect, which shows that queue length increases rapidly as the capacity utilization p increases to 1

Factors affecting Queue Length

ρ1

ρ1)2(c

iI

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.

2

22pi CC


Throughput- Delay Curve

VariabilityIncreases

AverageFlowTime T

Utilization (ρ) r100%

Tp


Coefficient of Variations

Interarrival Time or Processing Time distribution General Poisson Exponential Constant

Mean Interarrival Time or Processiong Time m m m m

Standard Deviation of interarrival or Processing Time s m m 0

Coeffi cient of Varriation of Interarrival or Processing Time s/m 1 1 0


A sample of 10 observations on Interarrival times in seconds

10,10,2,10,1,3,7,9, 2, 6=AVERAGE () Avg. interarrival time = 6Ri = 1/6 arrivals / sec.

=STDEV() Std. Deviation = 3.94Ci = 3.94/6 = 0.66

C2i = (0.66)2 = 0.4312

Example 8.4


A sample of 10 observations on processing times in seconds

7,1,7 2,8,7,4,8,5, 1Tp= 5 secondsRp = 1/5 processes/sec.Std. Deviation = 2.83Cp = 2.83/5 = 0.57C2

p = (0.57)2 = 0.3204

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:

Example 8.4

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

Ii = [(0.832)/(1-0.83)] ×[(0.662+0.572)/2] = 1.56

ρ1

ρ1)2(c

2

22pi CC

iI 0.831

0.831)2(1

2

57.066.0 22


Other performance measures: Ti=Ii/R = (1.56)(6) = 9.4 seconds

Compute T? T = Ti+TpSince TP= 5 T = Ti + TP = 14.4 seconds

Total number of passengers in the process is: I = RT = (1/6) (14.4) = 2.4Alternatively, 1.56 in the buffer. How many in the process?I = 1.56 + 0.83 = 2. 39

Example 8.4


Compute R, Rp and : Ta= 6 min, Tp = 5 minR = Ri= 1/6 per minuteProcessing rate for one processor 1/5 for 2

processorsRp = 2/5 = R/Rp = (1/6)/(2/5) = 5/12 = 0.417

Now suppose we have two servers.

On average 0.076 passengers waiting in line.safety capacity is Rs= RP - Ri = 2/5 - 1/6 = 7/30

passenger per second, or 14 passengers per minutes

ρ1

ρ1)2(c

2

22pi CC

iI 0.4171

0.4171)2(2

2

57.066.0 22


Other performance measures: Ti=Ii/R = (0.076)(6) = 0.46 seconds

Compute T? T = Ti+TpSince TP= 5 T = Ti + TP = 0.46+5 = 5.46 seconds

Total number of passengers in the process is: I= 0.08 in the buffer and 0.417 in the process. I = 0.076 + 2(0.417) = 0.91

Example 8.4

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


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?

Assignment- Problem 1: M/M/1 Performance Evaluation


Assignment- Problem 2: M/M/1 Performance Evaluation

What if the arrival rate is 11 per hour?


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 ?

Assignment- Problem 3: M/G/c


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)?

Assignment- Problem 4: M/M/c Example


Suppose the service time is a constant What is the answer of the previous question?

Assignment- Problem 5: M/D/c Example

1 Ardavan Asef-Vaziri Sep-09Operations Management: Waiting Lines2 Made-to-stock (MTS) operations ...

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