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soldier1979 [14.2K]
3 years ago
7

Which of the following is true of how the Internet has responded to the increasing number of devices now using the network?

Computers and Technology
1 answer:
Rasek [7]3 years ago
3 0

Answer:

C: The protocols of the Internet were designed to scale as new devices were added.

Explanation:

The internet protocols are changed every year to adapt to the new devices that have been connected to the network. Back in the 1990s, most traffic used a few protocols.  Pv4 routed packets, TCP turned those packets into connections, SSL (later TLS) encrypted those connections, DNS named hosts to connect to, and HTTP was often the application protocol using it all.

For many years, there were negligible changes to these core Internet protocols; HTTP added a few new headers and methods, TLS slowly went through minor revisions, TCP adapted congestion control, and DNS introduced features like DNSSEC. The protocols themselves looked about the same ‘on the wire’ for a very long time (excepting IPv6, which already gets its fair amount of attention in the network operator community.)

As a result, network operators, vendors, and policymakers that want to understand (and sometimes, control) the Internet have adopted a number of practices based upon these protocols’ wire ‘footprint’ — whether intended to debug issues, improve quality of service, or impose policy.

Now, significant changes to the core Internet protocols are underway. While they are intended to be compatible with the Internet at large (since they won’t get adoption otherwise), they might be disruptive to those who have taken liberties with undocumented aspects of protocols or made an assumption that things won’t change.

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Explanation:

<u>1) How many bits is needed to address a 2M * 32 memory </u>

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hence number of bits =  log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

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b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

<u>3) How many bits are required to address a 1M * 8 main memory </u>

1M = 1^1 * 1^20 ,  item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

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