[HW] HW#7S 2020 (Updated)

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Prof. Hsiao

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Jan 4, 2021, 11:19:45 PM1/4/21
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True (select A in the homework system) or false (B). 


Consider the following forwarding table. 

Destination IP Address Range  ->  Link Interface

11100000 00000000 00000000 00000000 through 11100000 00111111 11111111 11111111   -> 0

11100000 01000000 00000000 00000000 through 11100000 01000000 11111111 11111111  -> 1

11100000 01000001 00000000 00000000 through 11100001 01111111 11111111 11111111  -> 2

otherwise  -> 3

 

0: 224.0.0.0 – 224.63.255.255

1: 224.64.0.0 – 224.64.255.255

2: 224.65.0.0 – 225.127.255.255

3: others

 

Q1. A datagram whose destination IP is 11001000 10010001 01010001 01010101, and this datagram will be forwarded to interface 1? 

B. False. 200.145.81.85 goes to interface 3.

 

Q2. 11100001 01000000 11000011 00111100 will goes to interface 2?

A. True. 225.64.195.60 goes to interface 2.

 

Q3. 11100001 10000000 00010001 01110111 will goes to interface 3?

B. False. 225.128.17.119 goes to interface 3.

 

Q4. Give an IP 192.168.0.1 and its netmask is /24. The subnet ID of this LAN is 11000000 10101000 00000000 00000000.

A. True.

IP:    11000000 10101000 00000000 00000001

MASK: 11111111 11111111 11111111 00000000

ID:    11000000 10101000 00000000 00000000

 

Q5. Give an IP 192.168.0.1 and its netmask is /22. The subnet ID of this LAN is 11000000 10101000 00000000 00000000.

A. True.

IP:    11000000 10101000 00000000 00000001

MASK: 11111111 11111111 11111100 00000000

ID:    11000000 10101000 00000000 00000000

 

Q6. A /25 subnet has a mask 255.255.255.128 and it contains 128 addresses.

A. True.

MASK: 11111111 11111111 11111111 10000000

 => 255.255.255.128

It has 7 bits for differentiating hosts. 2^7 = 128.

 

Q7. Give a host IP address 192.168.1.123 in a subnet with a netmask 255.255.255.192. The last address of this subnet is 192.168.1.127.

A. True.

IP:    11000000 10101000 00000001 01111011

MASK: 11111111 11111111 11111111 11000000

ID:    11000000 10101000 00000001 01000000

Last IP: 11000000 10101000 00000001 01111111 -> 192.168.1.127

 

Q8. Given a 172.16.32.1 IP in a /20 subnet, the subnet’s last IP address is 172.16.47.255.

A. True.

IP:    10101100 00010000 00100000 00000001

MASK: 11111111 11111111 11110000 00000000

ID:    10101100 00010000 00100000 00000000

Last IP: 10101100 00010000 00101111 111111111 -> 172.16.47.255

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Ying-yin Chu

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Jan 5, 2021, 10:36:25 AM1/5/21
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Dear professor:
Why question 3 is false?


Prof. Hsiao 在 2021年1月5日 星期二下午12:19:45 [UTC+8] 的信中寫道：

Prof. Hsiao

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Jan 6, 2021, 10:03:00 AM1/6/21
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255.128.x.x 是落在 interface 2 之外的。
如果你有答案或我算錯請貼上你的答案。

蕭舜文

Ying-yin Chu 在 2021年1月5日 星期二下午11:36:25 [UTC+8] 的信中寫道：

匿名濃厚奶茶

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Jan 7, 2021, 6:27:44 AM1/7/21
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老師好:
想請問關於編號的定義問題

0: 224.0.0.0 – 224.63.255.255

1: 224.64.0.0 – 224.64.255.255

2: 224.65.0.0 – 225.127.255.255

3: others

這邊的interface由上至下編號是interface0,1,2,3還是 interface1,2,3,4?

若是0,1,2,3的話
225.128.17.119應該不在 224.65.0.0 – 225.127.255.255的範圍內
應使用 3: others->因此我們覺得這題應該是true
謝謝
Prof. Hsiao 在 2021年1月6日 星期三下午11:03:00 [UTC+8] 的信中寫道：

Prof. Hsiao

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Jan 7, 2021, 6:35:07 AM1/7/21
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Hi,

對，你說的是正確的。
我後頭答案明明寫的是 goes to interface 3. (但我前面寫錯)
所以第三題答案是 A. True.

Thanks.
Hsiao

raychi...@gmail.com 在 2021年1月7日 星期四下午7:27:44 [UTC+8] 的信中寫道：
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