Identifying Models Using Kendall Notation

Identifying Models Using Kendall Notation
• D. G. Kendall
• arrival distribution/service time distribution/number of service channels open
• M = Poisson distribution for number of occurrences (or exponential times)
D = constant (deterministic rate)
G = General distribution with mean and variance known
• M/M/1 – single channel model with Poisson arrivals and exponential service times
M/M/2 – two-channel system with Poisson arrivals and exponential service times
M/M/m – m distinct channels in the queuing system with Poisson arrivals and exponential service
times
M/D/3 – three-channel system with Poisson arrivals and constant service time
M/G/4 –four-channel system with Poisson arrivals and service times that are normally distributed
Single Channel Queuing Model with Poisson Arrivals & Exponential Service Times (M/M/1)
Assumptions
• Arrivals are served on a FIFO basis.

• There is no balking or reneging.
• Arrivals are independent of preceding arrivals, but the average number of arrivals (arrival rate) does
not change over time.
• Arrivals are described by a Poisson distribution and come from an infinite or very large population.
• Service times also vary from one customer to the next and are independent of one another, but their
average rate is known.
• Service times occur according to the negative exponential probability distribution.
• The average service rate is greater than the average arrival rate
Operating Characteristics
Let λ = mean number of arrivals per time period

µ = mean number of items or people served per time period
These seven queuing equations for the single-channel, single-phase model describe the important
operating characteristics of the service system.
1. Average number of customers or units in the system (number on line plus number being served):
l l
L= = Lq +
µ -l µ
2. Average time a customer spends in the system (waiting + being served):
1 1
W= = Wq +
µ -l µ
3. Average number of customers in the queue:
l2
Lq =
µ (µ - l )
4. Average time a customer spends waiting in the queue:

l
Wq =
µ (µ - l )
5. Utilization factor for the system, that is, the probability that the service facility is being used:
l
r=
µ
6. Percent idle time, that is, the probability that no one is in the system:
Po = 1 - r
7. Probability that the number of customers in the system is greater than k:
k +1
ælö
Pn > k = çç ÷÷
èµø
µ > l otherwise the waiting line will continue to grow without limit because the service facility
does not have sufficient capacity to handle the arriving units.
Example:
Reid Blank, the mechanic at Arnold’s Muffler Shop, is able to install new mufflers at an average rate of
3 per hour or about one every 20 minutes according to a negative exponential distribution. Customers
seeking this service arrive at the shop on the average of 2 per hour, following a Poisson distribution.
The customers are served on a FIFO basis, and come from a very large (almost infinite) population of
possible buyers.
What are the operating characteristics of Arnold’s Muffler Shop’s queuing system?
l = 2 cars arriving per hour

µ = 3 cars serviced per hour
l
L= = 2 cars in the system on the average
µ -l
1
W= = 1 hour that an average car spends in the system
µ -l
l2
Lq = = 4 / 3 cars waiting on line on the average
µ (µ - l )
l 2
Wq = = hour = average waiting time per car
µ (µ - l ) 3
l 2
r = = = percent of time mechanic is busy, or the probability that the server is busy
µ 3
1
Po = 1 - r = = probability that there are 0 cars in the system
3
k +1
ælö
Pn > k = çç ÷÷ , k = 0, 1, 2, 3, 4, 5, 6, 7
èµø
Introducing costs into the model
Let m = number of channels

C s = labor cost of each channel
C w = cost of waiting
Total service cost:

total service cost = (number of channels)(cost per channel)
= mC s
Waiting cost:
If the waiting time is based on time in the system
total waiting cost = (total time spent waiting by all arrivals)(cost of waiting)
= (number of arrivals)(average wait per arrival) C w
= (lW )C w
If the waiting time cost is based on time in the queue, this becomes
total waiting cost = (lWq )C w
Total cost:
If the waiting time is based on time in the system
total cost = total service cost + total waiting cost

= mC s + lWC w
If the waiting time is based on time in the queue
total cost = mC s + lWq C w
Example (continued)
Larry Arnold, the owner of Arnold’s Muffler estimates that the cost of customer waiting time, in terms
of customer dissatisfaction and lost goodwill, is $10 per hour of time spent waiting in line. What is the
total customer waiting time cost per day?
total daily waiting cost = (8 hours per day)(2)(2/3)($10) = $106.67
The only other major cost that Arnold can identify in the queuing situation is the salary of Blank, who
earns $7 per hour, what is the total daily cost?
Total daily service cost = (8 hours per day) mC s = 8(1)($7) = $56

Total daily cost of the queuing system = $106.67 + $56 = $162.67
Suppose that Jimmy Smith, an expert mechanic, applies for Blank’s position, claiming that he is able to
install 4 mufflers per hour. Should Arnold hire him (and fire Blank) if his asking pay is $9 per hour?
Recomputing all the operating characteristics:
l = 2 cars arriving per hour

µ = 4 cars serviced per hour
2
L= = 1 cars in the system on the average
4-2
1 1
W = = hour that an average car spends in the system
4-2 2
4 1
Lq = = cars waiting on line on the average
4(4 - 2) 2
2 1
Wq = = hour = average waiting time per car
4(4 - 2) 4
l 1
r = = = percent of time mechanic is busy
µ 2
1 1
Po = 1 - = = probability that there are 0 cars in the system
2 2
total daily waiting cost = $40 per day

service cost of Smith = $72
total expected cost = $112

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Identifying Models Using Kendall Notation

• Arrivals are served on a FIFO basis.

Let λ = mean number of arrivals per time period

4. Average time a customer spends waiting in the queue:

l = 2 cars arriving per hour

Let m = number of channels

Total service cost:

If the waiting time is based on time in the system

total waiting cost = (lWq )C w

If the waiting time is based on time in the system

total cost = total service cost + total waiting cost

If the waiting time is based on time in the queue

total cost = mC s + lWq C w

total daily waiting cost = (8 hours per day)(2)(2/3)($10) = $106.67

Total daily service cost = (8 hours per day) mC s = 8(1)($7) = $56

l = 2 cars arriving per hour

total daily waiting cost = $40 per day

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