Let’s assume that the network has just been powered on and there hasn’t been any flow of traffic occurred over the network yet.
a) What might be in the MAC address table of the switch?
b) If PC-1 sends a packet to PC-4, what action(s) will be performed by the switch? Solution:
a) As the network is just switched on, so is the switch. Therefore, there will be no entry in the Address table (forwarding table) of the switch.
b) Switch will perform two operations. First it will check its MAC address table which is empty, and it will populate its address table with the MAC entry of the incoming packet. Secondly, it will repeat the incoming packet on all its ports other than the source port.
Now assume that some data transfer has been occurred between the following devices:
i. PC-1 and PC-4
ii. PC-7 and PC-5
iii. PC-3 and Server
iv. PC-5 and PC-8
What might be in the forwarding table of the switch? Give your answer in the >following format:
Let’s assume that PC-1, PC-2, PC-3 and PC-4 belong to accounts department and PC-5, PC-6, PC-7 and PC-8 belong to admin department of an organization. Both departments have some concerns regarding security of their relevant departmental data, and they want to restrict unnecessary transfer of irrelevant traffic from each other. What approach can be implemented at the switch level to separate out network traffic of both the departments without changing the physical network configuration?
Both networks can be separated out by creating two VLANs on the switch and registering appropriate interfaces on the VLANs. This network behaves just like it would with the two physical switches. Switch does learn all the MAC addresses, but the switch does not forward broadcasts or unicasts from one VLAN to the other. It keeps a separate address table for each VLAN. This way both departments will be separated out.
a) As total number of customers are 3500 in the area, so the ISP will need 3500 hundred IP addresses for assigning unique IP address to the customers. But IP addresses can only be purchased in a range according to multiple of 2 i.e., number of IP addresses should be according to 2n. So,
Minimum number of static addresses required by the ISP = 2n (for minimum ‘n’ such that 2n >=3500)
= 212 = 4096
So, minimum number of static IP addresses purchased by the ISP = 4096
b) Number of remaining IP address = 4096 – 3500 = 596
A total of 596 IP addresses are available to be used in the extended area for providing services. So, maximum number of users in the extended area are 596 that can be provided service using the same static IP addresses.
a) As active number of users are much lesser, so we don’t need to assign static IP address to each customer. Dynamic address allocation technique can be used with a relatively small number of available addresses to be assigned to active users only. As the hosts join and leave, the DHCP (Dynamic Host Configuration Protocol) server allocates an arbitrary address from its current pool of available addresses. Each time a host leaves, its address is returned to the pool. DHCP is ideally suited to these situations, as there are many users coming and going, and addresses are needed for only a limited amount of time
b) As total number of active customers are no more than 950 at a time. The ISP does not need 4,096 IP addresses. Only a block of 1024 IP addresses are sufficient along with the DHCP protocol to provide seamless services to its customers.
Minimum number of IP addresses required by the ISP = 2n (for minimum ‘n’ such that 2n >=950)
= 210 = 1024 > 950
So, many minimum number of IP addresses purchased by the ISP = 1024