1. What is queuing theory?
In simple terms queuing theory is the study of waiting in lines or queue. Furthermore, Queuing theory is a field of study that deals with Queuing or waiting lines that are analyzed with a set of mathematical formulas. Different queuing models and mathematical models exist to deal with different types of waiting line systems.
2. What is finite calling population and give an example of it?
Finite calling population is when one knows exactly the population that is in a queue system. E.g. a mine engineer knows the exact number of trucks (i.e. 20) which will be served by a number of shovels (i.e. 2).
Or finite calling population has a specific , countable number of potential customers that planning is often easy, for example fixed production equipment at mining sites, trucking terminals, trains and airplanes.
3. What is the expected logic between the arrival and service rates? Why is this important?
The expected logic is that the Arrival Rate must be less than the Service Rate. In other words, the service rate must be greater than the arrival rate. So it is logic to assume that the rate at which services are completed must exceed the arrival rate of customers to increase productivity. This logic is very important because if this is not the case then, the waiting line will continue to grow and there will be no average solution. Hence, it is generally assumed that the service rate must exceed the arrival rate, λ<µ.
4. What is the difference between the average number of customers in the queue system and average number of customers in the waiting line?
Average number of customers in queue system is when customers waiting in a single or multiple line to be served in a Bank, dinner at the mess etc. not only people but trucks/machines/ships parts and products queue to be loaded/unloaded and manufacturing operation to be worked on.
Whereas
Average waiting line system comprises of arrivals, servers and waiting line structure. Waiting lines are based on averages of customers or trucks/machines/ships/plane/train arrivals and service times.
5. Explain the cost relationships between service cost and level of service, and waiting cost and level of service. Explain with an example of each.
Generally the service cost is inversely related to waiting cost. For example, as the level of service is improve by increasing the number of servers, the cost of services will increase whereas waiting cost decreases.
Cost of providing the service is reflected in the cost of servers like loaders/shovels, bank clerk or repair crew in maintenance plant. As the number of servers is increased to reduce service time, service cost rises. The major effect of waiting cost is the loss of business because customers might be tired of waiting and leave, or loss production due to time waste (e.g. in mining) but this loss is temporary.
The curve below shows these relationships.
Part B
1. Given the information, calculate the operating characteristics and determine if there is any improvement required, including the possibility of increasing number of mechanics.
Number of customers in the queue is exactly known so this is a finite calling population.
Data;
Population size N (no. of trucks) = 10 trucks
Number of servers = 1
Arrival rate (λ) per customer = 1/140 = 0.00714 per hour
Service rate (µ) is =1/4.5 = 0.2222 hour
i. Po = 
= 
= 0.692162 n=0, 1, 2, 3……….10
Po is the probability of no trucks in the system.
n
|
0
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
Po
|
1
|
0.3213
|
0.0929
|
0.0239
|
0.0054
|
0.001
|
0.00017
|
2.14E-05
|
2.1E-06
|
1.3E-07
|
4E-09
|
sum(1/sum)
|
0.692162
| | | | | | | | | | |
ii. Lq = 
=
= 0.1121 waiting for maintenance.
iii. L = Lq + (1-Po) = 0.1121 + (1 – 0.692162) = 0.419938 trucks in the queue system.
iv. Wq = 
= 
=1.64 hours waiting for repair.
v. W = Wq + 
= 1.64 + 
= 6.14 hours’ time in the maintenance workshop.
From the results above, it is seen that the mechanics are busy for about 51% of 12 hour shift each day repairing one machine on the queue system. Out of the 10 haul trucks, an average of 0.42 machines are in the queue for maintenance or4.2 % trucks are broken down waiting for repair or under repair. Each broken down truck is idle which means in 12 hour shift, about two tucks are repaired (12hrs/6.14). Therefore there is no need to increase the number of mechanics. But if the fleet size is increased to 20 or so, then there is a need for couple of mechanics at the workshop. But otherwise, no need.
2. Given the information, calculate the operating characteristics and recommend if there is anything to be done to improve truck – shovel productivity and reduce queue time.
Data:
λ = 4 trucks per hour (poison distribution)
µ = 3 trucks per hour
s = 2 shovels
sµ = 2*3 = 6, (>λ = 4)
Po =
= 
= 0.2 This is the probability that
No trucks are in the waiting line or queue.
L = 
*Po + 
= 
*0.2 + 
= 2.4, so 2 trucks arrive for loading at the pit.
W = 
= 
= 0.5 hours or 30 minutes in the queue (waiting and served).
Lq = L - = 2.4 - = 1.0667 trucks in queue to be loaded by the two shovels.
Wq = 
= 1.0667/4 = 0.267 hours or 16 minutes in the queue.
Pw = 
= 
= 0.533333
Now, there is 0.533 probability that trucks must wait or queue for loading which means that there is one or two trucks in the open pit queue system to be loaded by the two shovels.
Result Discussion
It can be seen from the results above which shows that the truck – shovel combination productivity is not attractive. There is more truck capacity and less number of shovels causing low truck productivity. The queuing time of 16 minutes for one or two trucks in queue is not attractive. There are 2 trucks entering the pit for loading at any time. By loading only one of the total 2, it consumes 16 minutes of production time where the truck(s) remain idle. It is now a need to reduce queuing time at all cost as it is recommended here. It is also recommended that, one shovel must be added to the existing fleet to increase productivity but keeping the fleet of trucks constant. With this recommendation the following are the results of improvements made in the queue system:
Recommended data: s = 3; λ = 4; µ = 3
Po = 0.254 probability that no trucks in the queue or loading
L = 1.48, so 1 truck enter the pit for loading.
W = 0.25 hour or 15 minutes queue time (average waiting time to load)
Lq = 0.15 waiting to be served.
Wq = 0.0375 hour or 2.25 minutes waiting in line before one truck is being loaded.
Pw =0.18 probability that one truck must wait for loading or all 4 shovels are busy.
Upon the above recommendation, it is seen that productivity y is improved when one truck’s average queuing time is reduced to 2.25 minutes from 16 minutes. In the future the cost of adding one shovel to existing fleet will be compensated by the higher production which is objective of the mine management.
3. Given the data, the manager wants to determine the average length of waiting line and average waiting time at counter. If the office space can only accommodate 10 clients, what do you recommend?
Data: λ = 5 rate per hour; µ = 60/10 = 6 clients per hour.
Now, Lq = 
= 
= 2.08 clients waiting
Wq = 
= 
= 0.416 hour or 24.96 minutes waiting in line or queue.
The results above shows that the client service provided by the law firm is not effective as there is almost half an hour consumes while the clients are waiting to be served. Therefore, it is recommended that the constant service time must be reduced 8 minutes and by increasing one server to the exiting but clients arriving rate will be constant as it can’t be controlled. That means the law firm must accommodate two clients at a time with two servers serving the clients at all cost. Upon these changes, the following results are obtained;
Λ = 5 rate per hour; µ = 60/8 = 7.5 clients per hour.
Lq = 0.667 clients waiting
Wq = 0.133 hours or 7.98 minutes waiting in line or queue.
Now this effective since the waiting line is reduced to 7.98 minutes from 24.96 minutes. It is assumed that there will be no or one client waiting to be served. The cost of the additional server will be covered the more clients paying for the business as they are being served.
Note also that the office capacity to accommodate only ten clients doesn’t mean that all of them are going to be served at the same time but, will sit comfortably waiting to be served. That doesn’t have an effect on the queuing time but queuing time depends on serving time. Therefore, I think the purpose of office capacity to accommodate ten clients is just for the firm’s good reputation and attractiveness of the business.
NB; D=demand per period, Co = ordering cost, Ch = unit holding cost over the period.
= 1095.45 tonnes size of order.
= 
= 0.11 years
= 
= 9 /year NB: N=number of orders. So 9 orders in a year.
+ 
+ D*C
+ 
+ 10000t*K25/t
= 
= 692.82 tonnes
+ 
+ D*C
+ 
+ 10000t*K25/t
