1. ECE 358 Examples #1 Xuemin (Sherman) Shen Office: EIT 4155
Phone: x 32691

2. Problem 1. Customers arrive at a fast food restaurant at a rate of five per minute and wait to receive their order for an average of 5 minutes. With probability 0.5 customers eat in the restaurant and carry out their food without eating with probability 0.5. A meal requires an average of 20 minutes to finish eating. What is the average time a customer spends in the restaurant? What is the expected number of customers in the restaurant?
Method 1: For those customers carrying out their food, they stay in the restaurant with an average of 5 minutes (for waiting). This situation happens with the probability of 0.5. For those customers eating in, they stay with an average of 25 minutes (for waiting and eating), which also happens with the probability of 0.5. Two situations considered together, the average customer time in the restaurant is
We know that customers arrive at a rate of = 5. By Little’s Theorem, the average number in the restaurant is

3. M/M/1 queue
The time a customer must wait in the queue is There are n customers in the system when a new customer arrives
Let W denote the mean time that a customer has to wait from the moment he arrives until he departures then
The mean number of customers in the system is

4. Method 2:

5. M/M/1 queue Find P0: Since , we have Then, a geometric distribution
The average number of customers in the system is:
The average number of customers in the queue is:

6. M/M/1/N queue Arrival rate: packet/sec; Blocking probability: ;
Departure rate: packet/sec;
Throughput:

7. Problem 2. A person enters a bank and finds all of the four tellers busy serving customers. There are no other customers in the bank, so the person will start receiving service as soon as one of the customers in service leaves. Customers have independent, identical, exponential distribution of service time with mean
a). What is the probability that the person will be the last to leave the bank assuming no other customers arrive?
b). If the average service time is 1 minute, what is the average time the person spend in the bank?
a). The probability that the person will be the last to leave is 1/4 because the exponential distribution is memoryless, and all customers have identical service time distribution. In particular, at the instant the customer enters service there are 4 persons at service, and the remaining service time of each of the other three customers served has the same distribution as the service time of the customer.

8. b). The average time in the bank is the average time (1 minutes) plus the average waiting time before being served. The average waiting time equals to the expected time for the first customer to finish service, which is 1/4 minute since the departure process is statistically identical to that of a single server facility with 4 times larger service rate. More precisely, we have
The above equation use the fact that the 4 service times are 4 independent exponential distributed random variables. Therefore,
just meaning the first departure time is exponential distributed and the expected time for the first departure is 1/4. Now, we can get the average time the person will spend in the bank is 1 + 1/4 = 5/4 minutes.

9. Problem 3. (Regenerative method) A packet has to be sent from node A to node D via nodes B and C. The transmission proceeds as follows. First, A transmits the packet to B. This transmission is always successful and takes T unit of time. Second, B sends the packet to C. This is successful with probability and takes T time units. If the transmission from B to C is unsuccessful, then B finds out that the transmission is incorrect after 2T time units. It then repeats the transmission until the fist success. Third, C sends the packet to D. This takes T time units and is successful with probability If it is not successful, then A finds out after 3T time units, and A must then repeat the whole process. Find the mean time needed until D first gets a successful packet.

10. Network of M/M/1 Queues l1 = g1 + g2 l2 = g1 + g2 + g3 l3 = g1 + g3 m2

11. l1 = g1 + g2
l2 = g1 + g2 + g3
l3 = g1 + g3 m2 m3 m1

12. l1 = g1 + g2
l2 = g1 + g2 + g3
l3 = g1 + g3 m2 m3 m1
The time through the system is the sum of the time through each queuing component.

13. ECE710 Wireless Communications Networks
Network of M/M/1 Queues
Three streams of packets go through this network. Assume that the arrival streams are Poisson processes with rate , and respectively. Assume also that the service times at the three buffers are independent and exponentially distributed with rates , , and , respectively.
It can be shown that the average number of packets in each buffer is equal to
2018/11/18
ECE710 Wireless Communications Networks

14. ECE710 Wireless Communications Networks
Network of M/M/1 Queues
The average delay T per packet is equal to
The assumptions required are
The arrival streams from outside into the network form independent Poisson process.
The packet transmission times at all the queues are independent and exponentially distributed.
In practice, assumption1 may be verified. Assumption2 cannot be since the transmission times of a given packet into the various nodes are all proportional to the packet length, and so they can’t be independent.
Note: Little’s result holds for systems that are not necessarily first come- first served.
2018/11/18
ECE710 Wireless Communications Networks

15. Queues are represented via the notation: A/S/C/K
A: The arrival process of packets. M stands for Markovian (Poisson) process; λ is the arrival rate (the average number of packets arriving by unit time). The interarrival time is exponentially distributed with mean 1/λ.
S: The packet departure process. In case of M, the interdeparture time (the service time) is exponentially distributed with average service time 1/μ, where μ is the service rate.
- C: the number of parallel servers in the system.
- K: max. number of packets in the queue (the max. number of packets that can be accommodated in the buffer plus the number of servers). If K is missing, K = infinity.
Ex: M/M/1 or M/M/1/N queue (Poisson arrivals, Exponential service time)