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Flura [38]
2 years ago
15

We have an internal webserver, used only for testing purposes, at IP address 5.6.7.8 on our internal corporate network. The pack

et filter is situated at a chokepoint between our internal network and the rest of the Internet. Can such a packet filter block all attempts by outside hosts to initiate a direct TCP connection to this internal webserver? If yes, design suitable packet filter rule sets (similar to those shown in the table below) that provides this functionality; if no, explain why a (stateless) packet filter cannot do it.
Table: Packet-Filtering Examples
Rule, Direction, Src address, Dest addresss, Protocol, Dest port, Action
1, In, External, Internal, TCP, 25, Permit
2, Out, Internal, External, TCP, >1023, Permit
3, Out, Internal, External, TCP, 25, Permit
4, In, External, Internal, TCP, >1023, Permit
5, Either, Any, Any, Any, Any, Deny

Computers and Technology
1 answer:
lukranit [14]2 years ago
3 0

Answer:

Check the explanation

Explanation:

A packet filter firewall is used as a check point between internal corporate network to the outside internet. It blocks all the inbound traffic from the outside hosts trying to initiate a direct TCP connection to the internal corporate webserver. The network design with firewall is shown in the attached image below:

The figures in the attached image below shows an internal corporate network is protected with a packet filter firewall to minimize the inbound traffic from the external network or an internet. Therefore, the packet filter is used as a check point between the network.

The packet filter blocks all attempts by the outside hosts in order to initiate a direct TCP connection to the internal webserver of the internal corporate network.

Going by the second part of the attached image below can can therefore conclude that:

• Rule 1 specifies that, deny any packet with the destination address 5.6.7.8 if the STN flag of TCP header is set.

• Rule 2 specifies that, allow the inbound email traffic from the external source.

• Rule 3 specifies, allows the Outbound TCP traffic from the internal corporate network.

• Rule 4 specifies, allows outbound Email traffic from the internal corporate network to the external network.

• Rule 5 specifies, block any traffic from any source to the any destination.

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<u>The total time elapsed from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B is </u><u>25.11 ms</u>

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Host A first converts the analog signal to a digital 64kbps stream and then groups it into 56-byte packets. The time taken for this can be calculated as:

time taken 1= \frac{Packet Size in Bits}{Bit Rate}

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Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that pac
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Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

Explanation:

Given details are below:

Length of the link = 10 meters

Bandwidth = 150 bits/sec

Size of a data packet = 100,000 bits

Size of a control packet = 200 bits

Size of the downloaded object = 100Kbits

No. of referenced objects = 10

Ler Tp to be the propagation delay between the client and the server, dp be the propagation delay and dt be the transmission delay.

The formula below is used to calculate the total time delay for sending and receiving packets :

d = dp (propagation delay) + dt (transmission delay)

For Parallel downloads through parallel instances of non-persistent HTTP :

Bandwidth = 150 bits/sec

No. of referenced objects = 10

For each parallel download, the bandwith = 150/10

  = 15 bits/sec

10 independent connections are established, during parallel downloads,  and the objects are downloaded simultaneously on these networks. First, a request for the object was sent by a client . Then, the request was processed by the server and once the connection is set, the server sends the object in response.

Therefore, for parallel downloads, the total time required  is calculated as:

(200/150 + Tp + 200/150 + Tp + 200/150 + Tp + 100,000/150 + Tp) + (200/15 + Tp + 200/15 + Tp + 200/150 + Tp + 100,000/15 + Tp)

= ((200+200+200+100,00)/150 + 4Tp) + ((200+200+200+100,00)/15 + 4Tp)

= ((100,600)/150 + 4Tp) + ((100,600)/15 + 4Tp)

= (670 + 4Tp) + (6706 + 4Tp)

= 7377 + 8 Tp seconds

Thus, parallel instances of non-persistent HTTP makes sense in this case.

Let the speed of propogation  of the medium be 300*106 m/sec.

Then, Tp = 10/(300*106)

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The Tp value 0.03 micro seconds as calculated above is negligible. This would lead to a similar value of time delay for both persistent HTTP and non-persistent HTTP. Thus, persistent HTTP is not faster than non-persistent HTTP with parallel downloads.

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