Operations Management WaitingLine Models Module D. Outline. Characteristics of a WaitingLine System. Arrival characteristics. WaitingLine characteristics. Service facility characteristics. Waiting Line (Queuing) Models. M/M/1: One server. M/M/2: Two servers. M/M/S: S servers.
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You’ve Been There Before!The average person spends 5 years waiting in line!!
‘The other line always moves faster.’
Situation Arrivals Servers Service Process
Bank Customers Teller Deposit etc.
Doctor’s Patient Doctor Treatmentoffice
Traffic Cars Traffic Controlledintersection Signal passage
Assembly line Parts Workers Assembly
Example: Number of customers that arrive each halfhour.
Discrete distribution with mean =
Example: Mean arrival rate = 5/hour .
Probability:
Time between arrivals has a negative exponential distribution.
Poisson DistributionServed units
Arrivals
Queue
Service facility
Ship unloading system
Ships at sea
Empty ships
Waiting ship line
Dock
Single Channel, Single PhaseService system
Served units
Arrivals
Queue
Service facility
Service facility
McDonald’s drivethrough
Cars in area
Cars& food
Waiting cars
Pay
Pickup
Served units
Service facility
Queue
Arrivals
Service facility
Example: Bank customers wait in single line for one of several tellers.
MultiChannel, Single PhaseServed units
Service facility
Service facility
Queue
Arrivals
Service facility
Service facility
Example: At a laundromat, customers use one of several washers, then one of several dryers.
MultiChannel, MultiPhaseMean servicerate =
6 customers/hr.
Mean service time = 1/
1/6 hour = 10 minutes.
Nonrandom:May be constant.
Example: Automated car wash.
Service TimesProbability:
Example: Time between arrivals.
Mean servicerate =
6 customers/hr.
Mean service time = 1/
1/6 hour = 10 minutes
=1
=2
=3
=4
Probability t>x
Negative Exponential Distributiona/b/S
Number of servers or channels.
Service time distribution.
Arrival time distribution.
=
S
1
=
=
W
L
L
L
L
W
+
+
q
s
s
s
q
q
=
=
W
W
s
q
General Queuing EquationsGiven one of Ws ,Wq ,Ls,or Lq you can use these equations to find all the others.
Average # of customers in system: L
=
s

1
=
Average time in system: W
s

2
Average # of customers in queue: L
=
q
(  )
Average time in queue: W
=
q
(  )
=
M/M/1 Model EquationsSystem utilization

1
M/M/1 Probability EquationsProbability of 0 units in system, i.e., system idle:
=

=
P
1
0
Probability of more than k units in system:
( )
k+1
l
=
P
n>k
This is also probability of k+1 or more units in system.
Average arrival rate is 10 per hour. Average service time is 5 minutes.
= 10/hr and = 12/hr
(1/ = 5 minutes = 1/12 hour)
Q1: What is the average time between departures?
5 minutes? 6 minutes?
Q2: What is the average wait in the system?
1
=
W
=
0.5 hour or 30 minutes
s
12/hr10/hr
1
1
1
=
W
W
W
+
s
q
2
12
q
Also note:
so
O.41667 hours

=
=
W

=
s
M/M/1 Example 1 = 10/hr and = 12/hr
Q3: What is the average wait in line?
10
O.41667 hours = 25 minutes
=
=
W
q
12 (1210)
L
q
s
M/M/1 Example 1 = 10/hr and = 12/hr
Q4: What is the average number of customers in line and in the system?
102
4.1667 customers
=
=
L
q
12 (1210)
10
5 customers
=
=
L
s
1210
Also note:
=
W
=
10 0.41667 = 4.1667
q
=
W
=
10 0.5 = 5
s

1
M/M/1 Example 1 = 10/hr and = 12/hr
Q5: What is the fraction of time the system is empty (server is idle)?
10
=

=
=
=
16.67% of the time
P
1

1
0
12
Q6: What is the fraction of time there are more than 5 customers in the system?
( )
6
10
=
33.5% of the time
=
P
n>5
12
All are P
n>5
5
10
1  P
=
1  0.402 = 0.598 or 59.8%
= 1 
n>4
12
M/M/1 Example 1 = 10/hr and = 12/hr
Q7: How much time per day (8 hours) are there 6 or more customers in line?
=
0.335 so 33.5% of time there are 6 or more in line.
P
n>5
0.335 x 480 min./day = 160.8 min. = ~2 hr 40 min.
Q8: What fraction of time are there 3 or fewer customers in line?
Five copy machines break down at UM St. Louis per eight hour day on average. The average service time for repair is one hour and 15 minutes.
= 5/day ( = 0.625/hour)
1/ = 1.25 hours = 0.15625 days
= 1 every 1.25 hours = 6.4/day
Q1: What is the number of “customers” in the system?
5/day
= 3.57 broken copiers
=
L
S
6.4/day5/day
= 5/day (or = 0.625/hour)
= 6.4/day (or = 0.8/hour)
Q2: How long is the average wait in line?
5
W
= 0.558 days (or 4.46 hours)
=
q
6.4(6.4  5)
0.625
W
= 4.46 hours
=
q
0.8(0.8  0.625)
= 5/day (or = 0.625/hour)
= 6.4/day (or = 0.8/hour)
Q3: How much time per day are there 2 or more broken copiers waiting for the repair person?
2 or more “in line” = more than 2 in the system
( )
3
5
P
=
= 0.477 (47.7% of the time)
n>2
6.4
0.477x 480 min./day = 229 min. = 3 hr 49 min.
Probability of zero people or units in the system:
Average number of people or units in the system:
Average time a unit spends in the system:
Note: M = number of servers in these equations
Average number of people or units waiting for service:
Average time a person or unit spends in the queue:
W
=
s
42  2
L
L
2
s
q
W
=
q
(2 + )(2 )
=
=
W
W
q
s
2 
P
=
0
2 +
M/M/2 Model EquationsAverage time in system:
Average time in queue:
Average # of customers in queue:
Average # of customers in system:
Probability the system is empty:
Average arrival rate is 10 per hour. Average service time is 5 minutes for each of 2 servers.
= 10/hr, = 12/hr, and S=2
Q1: What is the average wait in the system?
412
W
=
0.1008 hours = 6.05 minutes
=
s
4(12)2 (10)2
1
=
W
W
W
+
q
s
q
M/M/2 Example = 10/hr, = 12/hr, and S=2
Q2: What is the average wait in line?
(10)2
W
=
0.0175 hrs = 1.05 minutes
=
q
12 (212 + 10)(212  10)
Also note:
so
0.1008  0.0833 =0.0175 hrs
=
W

=
s
L
q
s
M/M/2 Example = 10/hr, = 12/hr, and S=2
Q3: What is the average number of customers in line and in the system?
=
W
=
10/hr 0.0175 hr = 0.175 customers
q
=
W
=
10/hr 0.1008 hr = 1.008 customers
s
= 10/hr and = 12/hr
Q4: What is the fraction of time the system is empty (server is idle)?
212  10
P
=
41.2% of the time
=
0
212 + 10
1 server2 servers3 servers
Wq 25 min. 1.05 min. 0.1333 min. (8 sec.)
0.417 hr 0.0175 hr 0.00222 hr
WS 30 min. 6.05 min. 5.1333 min.
Lq 4.167 cust. 0.175 cust. 0.0222 cust.
LS 5 cust. 1.01 cust. 0.855 cust.
P0 16.7% 41.2% 43.2%
= 10/hr and = 12/hr
Suppose servers are paid $7/hr and work 8 hours/day and the marginal cost to serve each customer is $0.50.
M/M/1 Service cost per day
= $7/hr x 8 hr/day + $0.5/cust x 10 cust/hr x 8 hr/day
= $96/day
M/M/2 Service cost per day
= 2 x $7/hr x 8hr/day + $0.5/cust x 10 cust/hr x 8 hr/day
= $152/day
= 10/hr and = 12/hr
Suppose customer waiting cost is $10/hr.
M/M/1 Waiting cost per day
= $10/hr x 0.417 hr/cust x 10 cust/hr x 8 hr/day = $333.33/day
M/M/1 total cost = 96 + 333.33 = $429.33/day
M/M/2 Waiting cost per day
= $10/hr x 0.0175 hr/cust x 10 cust/hr x 8 hr/day =$14/day
M/M/2 total cost = 152 + 14 = $166/day
Suppose customer waiting cost is not known = C.
M/M/1 Waiting cost per day
= Cx 0.417 hr/cust x 10 cust/hr x 8 hr/day = 33.33C $/day
M/M/1total cost = 96 + 33.33C
M/M/2 Waiting cost per day
= Cx 0.0175 hr/cust x 10 cust/hr x 8 hr/day =1.4C $/day
M/M/2 total cost = 152 + 1.4C
M/M/2 is preferred when152 + 1.4C < 96 + 33.33C or
C > $1.754/hr
Q: How large must customer waiting cost be for M/M/3 to be preferred over M/M/2?
M/M/2 total cost = 152 + 1.4C
M/M/3 Waiting cost per day
= Cx 0.00222 hr/cust x 10 cust/hr x 8 hr/day = 0.1776C $/day
M/M/3 total cost = 208 + 0.1776C
M/M/3 is preferred over M/M/2 when
208 + 0.1776C < 152 + 1.4C
C > $45.81/hr
If average service time is 15 minutes, then μ is 4 customers/hour