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In hierarchical routing with 4800 routers, what region and cluster sizes should be chosen to minimize the size of the routing table for the three-layer hierarchy?

1. 10 clusters, 24 regions and 20 routers
2. 12 clusters, 20 regions and 20 routers
3. 16 clusters, 12 regions and 25 routers
4. 15 clusters, 16 regions and 20 routers
in Others | 3.5k views
+1
I see the same question without options being asked. Is there a way to solve it directly , ans being 51. cube(17)=4913 but we have 4800 routers only. So how the ans is 51 , didn't understand that  part

$Clusters \times regions \times routers =4800 \text{ for all options}$

so we use following

$(clusters- 1) + (regions - 1) + routers$ , which option gives minimum is the ans...

Option 1 gives us 52.

Option 2 gives us 50

Option 3 gives us 51

Option 4 gives us 49

by Boss (25.6k points)
edited
0
How the following formula came?

(clusters- 1) + (regions - 1) + routers , which option gives minimum is the ans...
+2

that's how

(clusters- 1) + (regions - 1) + routers

0

just watch in the order:

routers minimum, then regions minimum, then clusters minimum

so in the first go option C is out of the race

and in the second go, you will get option D with minimum regions.

Suppose if regions also would have been same in 2 choices, we can consider those two choices further and then compare their clusters to get the best choice for minimum size of the routing table.

by (191 points)

+1 vote