Tutorial 1

Q1 1- A company has an internet address 172.30.12.192/22 and want to divide this block of addresses to 8 equal blocks ( subnets). 172.30.12.192/22 172.30.00001100.11000000

In the subnetting, the number of bits added to the mask could be calculated as: #of bits = log2 ( #of subnets) = log2 ( 8) = 3 172.30.00001100.11000000

Subnet1 172.30.00001100.00000000 172.30.12.0/25 27= 256 Subnet2 172.30.00001100.10000000 172.30.12.128/25 Subnet3 172.30.00001101.00000000 172.30.13.0/25 Subnet4 172.30.00001101.10000000 172.30.13.128/25 Subnet5 172.30.00001110.00000000 172.30.14.0/25 Subnet6 172.30.00001110.10000000 172.30.14.128/25 Subnet7 172.30.00001111.00000000 172.30.15.0/25 Subnet8 172.30.00001111.10000000 172.30.15.128/25

Q2 2-Channel with BW = 3 Mbps Calculate the utilization in a go-back-in protocol(with window size =31 ) with a 10KByte packet. Utilization=(total data in bits/bandwidth) *100 =((31*10*1000*8)/3*1000000)*100

How could you increase the utilization in this system By increasing the packet size or increasing the window size or both

Q3 3-In a ring topology : What is the difference between ring and dual ring if one of the links damaged ? Damage in ring topology stops the entire network Link damage in dual ring redirects the traffic in to the opposite side

Show the fault management steps in ring topology? Detecting the damaged link Isolating the fault by removing the damaged link and reconnect the rest of network Correcting the fault by replacing the link and reconnect the network Documenting the fault

Q4 4- A component with a MTBF of 8 days and a MTTR of 3 hours . Calculate the availability. Availability = MTBF / (MTBF + MTTR) = 8*24/(8*24+3) MTBF= mean time between failures MTBF= MTTF+MTTR MTTR Mean time to repair MTTF mean time to failure

How could you increase the availability. prove that How could you increase the availability ?prove that. By decreasing the MTTR

5- Calculate the channel capacity if the Bandwidth is 5KHz and the SNR is 100 and find Two different ways to increase the capacity. C=B * log(1+SNR) =5*1000 log(1+100) bps SNR signal to noise ratio (compared the level of desired signal to the level of background noise)

Q6 6- What is the transmission time if the distance between the two points is 24,000 km, the propagation speed is 2.4 × 108 m/s, queuing time and processing time is zero and latency is 1s? Latency = propagation time + transmission time+ queuing time + processing time Propagation time = distance/propagation speed Transmission time=packet size/ BW propagation time وقت النشر amount of time takes for the head of signal to travel from sender to reciver

7- A system with 5 replica managers : Draw the system if an active replication is applied? Draw the system if a passive replication is applied? Calculate the availability if there is 7% probability of a replica manager failure. Calculate the availability if we use just a single server. Replication is used to provide fault tolerance Active replication :each client request is processed by all servers Passive replication : only one server(called primary) process client request after processing request the primary update the state on the other backup servers and and send back respond to the client

Draw the system if an active replication is applied? RM FE C RM FE means Front ends ===When data is replicated it should be done in a transparent manner: the client should not normally be aware that multiple physical copies of the data exist, either in terms of submitted requests or returned values. This is achieved by interposing a front end between the client and the service. RM

Draw the system if a passive replication is applied? FE C RM Primary Backup RM RM

Calculate the availability if there is 7% probability of a replica manager failure. A=1-Probability(all managers failed or unreachable)=1-pn P=0.07 N=5 A= 1-pn = 1-(0.07)5 =1-0.0000016807=0.999=99.9%

Calculate the availability if we use just a single server. P=0.07 N=1 A=1-Probability(all managers failed or unreachable)=1-pn P=0.07 N=1 A= 1-pn = 1-(0.07)1 =0.93=93%

8- If the time required to open a certain page from the browser cache is 500ms , proxy cache is 2555ms and web server cache is 5000ms fill in the table below Action The proximate time required PC1 requests the page for the first time 5000ms PC2 requests the page for the first time 2555ms PC1 requests the page for the second time 500ms PC2 requests the page for the second time

2- If the time required to open a certain page from the browser cache is 500ms , proxy cache is 2555ms and web server cache is 5000ms fill in the table below