QUEUING MODELS Economic Analyses of Queuing Models

Script

QUEUING MODELS – Economic Analyses of Queuing Models

Slide 1

Welcome back.

In this module we look at 5 situations where we can build rather straightforward queuing models to aid in making proper choices for maximizing profits or minimizing costs.

Slide 2

In queuing
Every model is different and requires different analytic approaches
In this module we’ll look at two basic situations where economic analyses can be employed to determine optimal policies.
The first one is a model that will be used to determine the minimum number of servers that will keep the average time a customer spends in the queue below some threshold level.
The other 4 examples offer two or more possibilities for the queuing staffing
For each we will determine the performance measures and use them to determine which model is the best in the sense that it has the lowest overall expected total cost or the highest overall expected profit.

Slide 3

In this first example
Suppose customers arrive to a store according to a Poisson process at an average rate of 100 per hour.
Service times are exponential and average 5 minutes.
Management would like to staff its checkout area with the minimum number of clerks so that the average wait time in the queue for service by customers is less than 30 seconds.
This means less than .5 minutes or .00833 hours

Slide 4

So here is the situation
Customers arrive at an average rate of 100 per hour
Each clerk serves in an average time of 5 minutes, meaning that each clerk’s service rate is 60 divide by 5 or 12 per hour
The goal is to keep the average time a customer spends in the queue to less than .00833 hours
And the question is, how many servers do we need?

Slide 5

We go to the queuing template and
Select the M M k worksheet
We enter lambda and mu in the same time units --- here customers per hour
Then we look down the W Q column until we find the first instance where this number is less than .00833. This is a W Q of .003999 which is associated with k equal to 12 servers.
Thus a minimum 12 servers will be needed to meet management’s goal.

Slide 6

Our second example involves trying to decide which clerk to assign to minimize total costs.
Suppose customers arrive according to a Poisson process to a store at night at an average rate of 8 per hour.
The cost for security, insurance and other factors amount to an average of $4 per hour for each customer that is in the store.
There will be one clerk,
and although service times can be assumed to follow an exponential distribution
if Ann is hired, her average service time is 6 minutes and will cost us $6 per hour in salary; if Bill is hired, his average service time is 5 minutes and will cost us $10 per hour in salary; and if Charlie is hired, his average service time is 4 minutes and he will cost us $14 per hour in salary
The question is “which clerk should be hired?”

Slide 7

Here is the picture for Ann
Customers arrive at an average rate of 8 per hour and stand in line until Ann is free for service
Then, since her average service time is 6 minutes, she serves at a rate of 10 per hour.
Her hourly cost then is her salary of $6 plus the $4 security cost times the average number of customers in the system, L.
Thus to calculate the total hourly cost, we must find L for Ann

Slide 8

We go to the M M k worksheet and enter lambda equal to 8 and mu equal to 10 for Ann; and since Ann is one server, L for Ann is 4
Thus Ann’s total hourly cost is $6 plus $4 times 4 or $22 per hour.

Slide 9

For Bill
Again customers arrive at an average rate of 8 per hour and stand in line until Bill is free for service
Then, since his average service time is 5 minutes, he serves at a rate of 12 per hour.
His hourly cost then is his salary of $10 plus the $4 security cost times his average number of customers in the system, L.
Thus to calculate the total hourly cost, we must find L for Bill

Slide 10

We go to the M M k worksheet and enter lambda equal to 8 and mu equal to 12 for Bill and since Bill is one server, L for Bill is 2
Thus Bill’s total hourly cost is $10 plus $4 times 2 or $18 per hour.

Slide 11

For Charlie
Again customers arrive at an average rate of 8 per hour and stand in line until Charlie is free for service
Then, since his average service time is 4 minutes, he serves at a rate of 15 per hour.
His hourly cost then is his salary of $14 plus the $4 security cost times his average number of customers in the system, L.
Thus to calculate the total hourly cost, we must find L for Charlie

Slide 12

We go to the M M k worksheet and enter lambda equal to 8 and mu equal to 15 for Charlie and since Charlie is one server, L for Charlie is 1.14
Thus Charlie’s total hourly cost is $14 plus $4 times 1.14 or $18.56 per hour.

Slide 13

Summarizing
Ann’s total hourly cost is estimated to be $22
Bill’s $18
And Charlie’s $18.56
So we would recommend hiring Bill

Slide 14

The next example involves what kind of line to have.
A fast food restaurant will be installing a new drive-thru window service. It is willing to assume that service distributions are exponential….
And that customers arrive according to a Poisson process at an average rate of 24 per hour.
Customer time, waiting to be served, is valued at $25 per hour.
Each clerk makes $6.50 per hour
And each drive-thru lane that is built costs $20 per hour to operate.
Management is considering three possibilities for the queuing configuration and must decide which one to choose.

Slide 15

The first configuration looks like this. A single line and a single clerk (the red circle) handling the orders and the dispensing of the food at the window shown in black. The line of cars will be to the right of the restaurant.
Now cars arrive at an average rate of 24 per hour.
The average service time of this one clerk is 2 minutes, so that his average service rate is 30 per hour; and since 24 is less than 30, steady state will be reached.
The total hourly cost of this system is the $6.50 clerk cost plus the $20 lane cost plus $25 times the average number of cars in line (not being served) at the restaurant, L sub Q.
(double click)

Slide 16

We go to the M M k worksheet and enter the arrival rate of 24 for lambda and the service rate of 30 for mu. Since there is only one server, L sub Q is the value for k equal to 1 and is 3.2.
Substituting L sub Q equals to 3.2 into our total hourly cost equation gives an hourly cost of $106.50 for this configuration.

Slide 17

The second configuration looks like this, which is the way most McDonald’s operates now. There is a single line and but two clerks (the red circles) working together, one handling the orders and collecting the money, and the other dispensing of the food at the window shown in black. The line of cars will be to the right of the restaurant.
Now cars arrive at an average rate of 24 per hour.
But because there are two clerks working together, the average service time decreases to 1.25 minutes, so that the average service rate is now 48 per hour.
The total hourly cost of this system is the $6.50 times 2 or $13 the salaries of the 2 clerks cost plus the $20 lane cost plus $25 times the new average number of cars in line (not being served) at the restaurant, L sub Q.
(double click)

Slide 18

This is still an M M 1 system because although we have two clerks, they work together on each customer. So again we go to the M M k worksheet and enter the arrival rate of 24 for lambda and the service rate of 48 for mu. Since there is only one service operation, L sub Q is the value for k equal to 1 and is .5.
Substituting L sub Q equals to .5 into our total hourly cost equation gives an hourly cost for this configuration of $45.50.

Slide 19

The third configuration looks like this. There are two service windows and one long tine of cars with the lead car going to the first available window. This is an M M 2 queuing situation.
Cars arrive at an average rate of 24 per hour, join the queue and wait for the first available server.
Each server has a two minute average service time, so each server serves at an average rate of 30 per hour.
The total hourly cost of this system is the $6.50 times 2 or $13 the salaries of the 2 clerks cost plus a 2 times $20 or $40 lane cost, since there are two lanes, plus $25 times the new average number of cars in line (not being served) at the restaurant, L sub Q.
(double click)

Slide 20

This is an M M 2 system because the two clerks serve different customers. So again we go to the M M k worksheet and enter the arrival rate of 24 for lambda and the service rate of 30 for mu for each clerk. Since there are two servers acting independently, L sub Q is the value for k equal to 2 and is .152.
Substituting L sub Q equals to .152 into our total hourly cost equation gives an hourly cost for this configuration of $56.80.

Slide 21

Summarizing. The total hour cost for the single server system, option 1 is
$106.50
For the second speeded up single server system is$45.50
and for the third two server system$58.80
Thus the best option is the second configuration.

Slide 22

Thus far we have been concerned with minimizing expected costs. In this example we will be concerned about maximizing expected profits.
Suppose customers are expected to arrive to a particular store location at an average rate of 30 per hour according to a Poisson process.
The store will be open 10 hours per day
Each sale grosses the company $25.
Store clerks are paid $20 per hour including all benefits.
The cost estimate of having a customer in the store for security, insurance, and so forth is $8 per hour per customer
A clerk’s service time follows an exponential distribution and averages about 6 minutes, meaning that each clerk’s service rate is 10 customers per hour.
The company has two options. One is to lease a large store for $1000 per day, with virtually limitless space for all customers and staff it with 6 clerks. The other is to lease a small store which has room for only three customers, for $200 per day, and staff it with 2 clerks. The question is which one appears more profitable?

Slide 23

Here is the situation for the large store. It leases for $1000 per day, which over a 10-hour day is $100 per hour.
It has 6 clerks
and a potentially limitless queue.
Customers arrive to the system at an average rate of 30 per hour
And all customers join the system and get served.

Slide 24

The small store leases for $200 per day, which over a 10-hour day is $20 per hour.
It has 2 clerks
But room for only one other customer other than the two being served.
Customers arrive to the system at an average rate of 30 per hour
A potential arrival finding less than 3 customers in the system will join the system and eventually get served.
But a potential arrival finding 3 customers in the system when he arrives will not join the system and his business is lost.

Slide 25

So let’s set up an hourly profit analysis table…..
For the large and small store.
All customers get through for the large store so its arrival rate is 30 per hour. For the small store, its potential customers only get through if there are two or less customers in the system. So its effective arrival rate is 30 times 1 minus p sub 3.
The hourly revenue is figured at $25 per customer that actually arrives each hour
For the large store, that’s 25 times 30 or $750; for the small store that’s 25 times lambda sub e
We have several hourly costs
First there are the lease costs of $100 for the large store and $20 for the small store
Then there are the $20 per hour clerk costs. The large store has 6 clerks, so that’s $120 and the small store has 2 clerks, so that’s $40.
Then there are the customer wait costs of $8 times the average number of customers in the store; which is $8 time its L for the large store, and $8 times its different value of L for the small store
The net hourly profit then is the difference between the revenue and the costs. So for the large store, we need its value of L, and for the small store we need lambda sub e and its value of L to complete the calculations.

Slide 26

We note that the large store is an M M 6 system, so we go to the M M k worksheet, enter lambda equal to 30 and mu per server equal to10 and look up L for k equal to 6.
We find that it is 3.099143

Slide 27

The small store is an M M 2 3 finite queuing model. So we go to the M M k F worksheet and enter lambda equal to 30 mu equal to 10 per server, that there are k equal to 2 servers and that there is a maximum of F equal to 3 customers in the system, and we look for the value of L.
We find that it is 2.11475
We also note that p sub 3 is .44262….
So that the effective arrival rate is 30 times 1 minus .44262 or 16.7213. So for this system 30 customers per hour try to join the system, but only 16.7213 get through and we lose the others’ business.

Slide 28

So going back to our hourly profit table…..
For the large and small stores
We see that the arrival rates are 30 and 30 and 16.7213 respectively
This gives hourly revenues of
$750 and $418 respectively
The hourly costs
are $100 and $20 for the leases
$120 and $40 for the clerks
and 8 times L or $25 and $17 respectively for the customer wait costs
Subtracting the costs from the revenues shows that the expected hourly profit for the large store is $505 and for the small store is only $341
Thus we would recommend leasing the large store.

Slide 29

Our last example deals with choosing which of two machines to purchase. The approach is less economical, but is based on which seems to give the overall better performance.
Suppose jobs will arrive to the machine according to a Poisson process at an average rate of 5 per hour.
But service times do not follow an exponential distribution.
Two machines are being considered.
The first has a mean service time of 6 minutes (which is a service rate of 10 per hour) and a standard deviation of 3 minutes (which is .05 hours).
The second is a bit slower on the average, a mean service time of 6.25 minutes (which is a service rate of 9.6 per hour), but has a much smaller standard deviation of .6 minutes (which is .01 hours).
The question is, “which of these two M G 1 designs seems preferable?”

Slide 30

For the first machine and we go to the M G 1 worksheet enter everything in terms of hours: lambda is 5 per hour; mu is 10 per hour; and the standard deviation is .05 hours – giving these performance measures.

Slide 31

For the second machine we change mu to 9.6 and the standard deviation to .01 – giving these performance measures.

Slide 32

Let’s compare the results for machine 1 and machine 2.
We see that a job is more likely to wait with machine 2 – advantage machine1.
We see the average time spent to actually do the job is less on machine 1 – again advantage machine 1.
But the average number of jobs in the system
And in the queue
And the average time a job spends in the system
And in the queue are all less with machine 2.
Thus we would probably recommend machine 2.

Slide 33

Let’s review what we’ve done in this module.
For each system, we listed the components we had to calculate
And developed a model
We then used the queuing template to get estimates for various parameters of the models
And selected the optimal design the maximized or minimized something.

That’s it for this module. Do any assigned homework and I’ll be back to talk to you again next time.