Member Avatar for wynn

1.Given that there are 5 nodes in a LAN and a connection between each pair of nodes is
$10. Based solely on the price of connections, how much would a full mesh topology
cost?

2.Given that there are 5 nodes in a LAN and a connection between each pair of nodes is
$10, how much would a ring topology cost?

3.If the 8-port 10BaseT hub is replaced by an 8-port 10BaseT switch, how many
pairs of PCs, in the 6-PCs LAN, can be communicating with each other through
that 8-port 10BaseT switch at a time, under the most ideal condition?

4.How many pairs of PCs, in the 6-PCs LAN, can be communicating with each
other through that 8-port 10BaseT hub at any one time?

5.State a topology that would be cheaper than the ring topology. How much would this
topology cost using the same $10 per connection for each pair of nodes, based on 5
nodes in the LAN?

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Generated

Clear, concise answers and the minimal reasoning needed to check work. Assumptions: link cost is $10 per directly connected pair and device prices are not included; “pairs communicating” means distinct pairs of endpoint hosts (not counting multiple simultaneous flows per host). 's reminder to sketch the topology is useful for counting links; the numeric results below follow from those counts.

Full mesh (complete graph): number of links = N*(N-1)/2. With N = 5 that is 5*4/2 = 10 links → 10 * $10 = $100.
Ring: number of links = N → 5 * $10 = $50.

6-PC LAN on an 8-port 10Base-T switch (ideal, non‑blocking, full‑duplex): each port is its own collision domain and the switch can forward independent conversations simultaneously. The maximum number of disjoint communicating pairs among 6 hosts is floor(6/2) = 3 pairs. If hosts initiate multiple conversations at once the switch can carry many flows, but the count of independent two‑host pairs (no shared endpoints) is three.
Same 6‑PC LAN on an 8‑port hub (shared half‑duplex medium): the hub is one collision domain, so at any instant only one station can transmit successfully — effectively only one communicating pair at a time.

A cheaper topology than a ring is any spanning tree (common example: a star). A tree needs N-1 links, so for N = 5 that is 4 * $10 = $40. Note: trees minimize link cost but trade off redundancy (single point of failure at the central node in a star). ’s instinct on the full‑mesh cost was correct.

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Member Avatar for Suspishio

Is this your exam queastion?

Member Avatar for wynn

No this is no my exam question but tutorial question

Member Avatar for wynn

Just explain to me how to calculate it thanks very much

Member Avatar for Suspishio

It's not what the forum is for. Can you imagine the amount of work you're trying to freeload off forum members when YOU should be doing the research?

It's not fair, is it?

Member Avatar for wynn

First question izzit 10 * 10 = $ 100?

Member Avatar for Suspishio

You don't give up even though nobody else has bothered to answer your freeloading question.

To get you started out of the kindness of my heart, here is the key tip.

Draw the fully meshed topology; multiply the item cost by the number of items and you have your answer. Apply similar thinking to the other questions.

Then go away please.

Member Avatar for wynn

You dont know how to answer my question?

Member Avatar for Suspishio

I did answer your question. I told YOU how to do it. It is wrong of you to freeload in the way you are trying.

You have to be a complete fool not to be able to be able to draw a meshed diagram and multiply the number of components by their cost.

Member Avatar for wynn

I mean the rest you dont know how to do is ok thanks for reply

Member Avatar for Suspishio

Of course I know how to answer the other questions. The answer to the first question is the principle behind the answers to the other questions.

Now YOU can do the research on the different topologies and produce the answers.

Freeloading in this way is not honourable. If you have a technical problem and need help, we give it. if you want exam or tutorial answers, study the subject as you're supposed to do.

Now go away.

Member Avatar for wynn

Can you explain question 3 and 4. But if you don't know how to do is really OK THANKS

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