حل اسئلة شبكات الحاسوب قسم الحاسوب الجامعة المستنصرية نموذج رقم1

1حل اسئلة شبكات الحاسوب قسم الحاسوب الجامعة المستنصرية نموذج رقم

 Q.1) Chose the correct answer (chose Ten Only)

1- ------- cable can carry signals of higher frequency ranges than --------- cable .

a- Twisted-pair; fiber-optic

b- Coaxial; fiber-optic

c- Coaxial; twisted-pair

d- None of the above

2- Assume six devices are arranged in a mesh topology. How many cables are needed?

a- 15

b- 6

c- 12

d- 18

3- A _______ connection provides a dedicated link between two devices.

a- primary

b- multipoint

c- point-to-point

d- secondary

4- In an optical fiber, the inner core is --------- the calling

a- Denser than

b- Less dense than

c- The same density as

d- Another name for

5- The ____ address uniquely defines a host on the Internet.

a- IP

b- port

c- specific

d- physical

6- Physical layer is responsible for

a- Mode to node communication

b- Peer to peer communication

c- Hop to hop communication

d- Both a and c

e- Both b and c

7- What is the netid of the address 227.78.19.21?

a- 227

b- 227.78

c- 227.78.19

d- None of the above

8- What is the result of ANDing 192 and 65?

a- 192

b- 65

c- 64

d- none of the above

9- You have two computers connected by an Ethernet hub at home. This network is

a- LAN

b- WAN

c- MAN

d- None of the above

10- Frequency of failure and network recovery time after a failure are measures of the _______of a network.

a- Performance

b- Security

c- Reliability

d- Feasibility

11- Loss in signal power as light travels down the fiber is called?

a- Attenuation

b- Propagation

c- Scattering

d- Interruption

12- The session, presentation, and application layers are the ____ support layers.

a- user

b- network

c- both (a) and (b)

d- neither (a) nor (b)

Q.2) Answer Two branches Only

a- Draw a hybrid topology with a ring backbone and two bus networks.

Sol//

1. Ring Backbone: The ring backbone connects all the bus networks in a circular manner. Each bus network is connected to the ring backbone at two points.

1. العمود الفقري الدائري: يربط العمود الفقري الدائري جميع شبكات الحافلات بشكل دائري. ترتبط كل شبكة حافلات بالعمود الفقري الدائري عند نقطتين.

2. Bus Networks: There are two bus networks connected to the ring backbone. Each bus network consists of multiple nodes connected in a linear fashion. Nodes on the same bus network can communicate directly with each other.

2. شبكات الحافلات: هناك شبكتان من الحافلات متصلة بالعمود الفقري الدائري. تتكون كل شبكة حافلات من عقد متعددة متصلة بطريقة خطية. يمكن للعقد الموجودة على نفس شبكة الحافلات التواصل مباشرة مع بعضها البعض.

Here's a simple ASCII diagram to illustrate:

```

       /---[Node]--[Node]--[Node]---\

      /                              \

[RING]----[Node]--[Node]--[Node]---[RING]

      \                              /

       \---[Node]--[Node]--[Node]---/

```

In this diagram:

- `[RING]` represents the ring backbone.

- `[Node]` represents a node connected to the bus network.

- The lines connecting nodes represent the bus network connections.

- The ring backbone connects all the bus networks in a circular manner.

b- A 100-byte message is sent through a private internet using the TCP/IP protocol suite. If the

protocol adds a 10-byte header at each layer, what is the efficiency of the system (the ratio of

the number of useful bytes to the number of total bytes)?

Sol//

Given:

• Original message size: 100 bytes
• Header size added by each layer: 10 bytes

There are typically 4 layers in the TCP/IP protocol suite where headers are added:

1. Data Link Layer
2. Network Layer
3. Transport Layer
4. Application Layer

So, the total header size added to the original message is $4\times 10\text{bytes}=40\text{bytes}$.

Now, let's calculate the total number of bytes transmitted:

• Original message size: 100 bytes
• Total header size added: 40 bytes

Total number of bytes transmitted = Original message size + Total header size added $=100\text{bytes}+40\text{bytes}=140\text{bytes}$

Now, the efficiency of the system is given by the ratio of useful bytes to total bytes: $\text{Efficiency}=\frac{\text{Useful bytes}}{\text{Total bytes}}$

Useful bytes = Original message size = 100 bytes Total bytes = Total number of bytes transmitted = 140 bytes

$\text{Efficiency}=\frac{100\text{bytes}}{140\text{bytes}}$

$\text{Efficiency}\approx 0.714$

So, the efficiency of the system is approximately $71.4\mathrm{\%}$.

c- What are the different among physical address, logical address, and port address?

Sol/

Physical Address:

- A physical address, also known as a MAC (Media Access Control) address, is a unique identifier assigned to a network interface controller (NIC) for communication on a physical network.

- It is typically expressed as a hexadecimal number and is used at the Data Link Layer (Layer 2) of the OSI model.

- Physical addresses are used to identify devices at the hardware level and are used for local communication within a network segment.

العنوان الفعلي:

- العنوان الفعلي، المعروف أيضًا باسم عنوان MAC (التحكم في الوصول إلى الوسائط)، هو معرف فريد يتم تعيينه لوحدة تحكم واجهة الشبكة (NIC) للاتصال على شبكة فعلية.

- يتم التعبير عنه عادةً كرقم سداسي عشري ويتم استخدامه في طبقة ارتباط البيانات (الطبقة 2) لنموذج OSI.

- تُستخدم العناوين الفعلية لتحديد الأجهزة على مستوى الأجهزة وتُستخدم للاتصالات المحلية داخل مقطع الشبكة.

Logical Address:

- A logical address, also known as an IP (Internet Protocol) address, is a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication.

- It is used to identify a device on a network and facilitate routing of data packets between devices.

- Logical addresses are typically expressed as a series of four numbers separated by dots (e.g., 192.168.1.1) and are used at the Network Layer (Layer 3) of the OSI model.

العنوان المنطقي:

- العنوان المنطقي، المعروف أيضًا باسم عنوان IP (بروتوكول الإنترنت)، عبارة عن تسمية رقمية يتم تعيينها لكل جهاز متصل بشبكة كمبيوتر تستخدم بروتوكول الإنترنت للاتصال.

- يتم استخدامه لتحديد جهاز على الشبكة وتسهيل توجيه حزم البيانات بين الأجهزة.

- يتم التعبير عن العناوين المنطقية عادةً كسلسلة من أربعة أرقام مفصولة بنقاط (على سبيل المثال، 192.168.1.1) ويتم استخدامها في طبقة الشبكة (الطبقة 3) لنموذج OSI.

Port Address:

- A port address is used in the context of TCP (Transmission Control Protocol) and UDP (User Datagram Protocol) communications.

- Ports are logical endpoints used by applications and services for communication over a network.

- Each port is associated with a specific protocol (TCP or UDP) and a unique number ranging from 0 to 65535.

- Port addresses are used to distinguish between different services running on the same device.

- In TCP/IP networking, ports are used in conjunction with IP addresses to enable communication between processes running on different devices.

عنوان المنفذ:

- يتم استخدام عنوان المنفذ في سياق اتصالات TCP (بروتوكول التحكم في الإرسال) وUDP (بروتوكول مخطط بيانات المستخدم).

- المنافذ هي نقاط نهاية منطقية تستخدمها التطبيقات والخدمات للاتصال عبر الشبكة.

- يرتبط كل منفذ ببروتوكول معين (TCP أو UDP) ورقم فريد يتراوح من 0 إلى 65535.

- تُستخدم عناوين المنافذ للتمييز بين الخدمات المختلفة التي تعمل على نفس الجهاز.

- في شبكات TCP/IP، يتم استخدام المنافذ جنبًا إلى جنب مع عناوين IP لتمكين الاتصال بين العمليات التي يتم تشغيلها على أجهزة مختلفة.

باختصار، يتم استخدام العناوين الفعلية في طبقة ارتباط البيانات لتحديد الأجهزة على مستوى الأجهزة، ويتم استخدام العناوين المنطقية في طبقة الشبكة لتحديد الأجهزة الموجودة على الشبكة، ويتم استخدام عناوين المنافذ لتحديد خدمات أو عمليات معينة تعمل على الجهاز في طبقة النقل.

Q.3) Answer Two branches Only

a- If the signal at the beginning of a cable with -0.3 dB/km has a power of 2mW, what is the power

of the signal at 5 km?

Sol//

To calculate the power of the signal at 5 km given the initial power and attenuation rate, we can use the formula for power loss:

${𝑃}_{\text{final}}={𝑃}_{\text{initial}}\times 10^{\left(\frac{-\text{attenuation rate}\times \text{distance}}{10}\right)}$

Given: Initial power (${𝑃}_{\text{initial}}$) = 2 mW Attenuation rate = -0.3 dB/km Distance = 5 km

We first need to convert the attenuation rate from dB/km to a linear scale:

$-0.3\text{dB/km}=10^{\left(\frac{-0.3}{10}\right)}$

Now, we can calculate the final power using the formula:

${𝑃}_{\text{final}}=2\text{mW}\times 10^{\left(\frac{-0.3\times 5}{10}\right)}$

${𝑃}_{\text{final}}=2\text{mW}\times 10^{-1.5}$

${𝑃}_{\text{final}}=2\text{mW}\times 0.0316$

${𝑃}_{\text{final}}=0.0632\text{mW}$

So, the power of the signal at 5 km is $0.0632\text{mW}$.

b- Match the following to one or more layers of the OSI model:

i. route determination

ii. flow control

iii. interface to transmission media

iv. provides access for the end user

v. defines frames

Sol//

i. Route determination - Network Layer (Layer 3): The network layer is responsible for routing packets through intermediate nodes in the network from source to destination.

ii. Flow control - Transport Layer (Layer 4): Flow control is managed at the transport layer, where it ensures that data is transmitted at an appropriate rate between sender and receiver to prevent overwhelming the receiver.

iii. Interface to transmission media - Physical Layer (Layer 1): The physical layer is responsible for transmitting raw data bits over a communication channel, including the physical connection and characteristics of the transmission medium.

iv. Provides access for the end user - Application Layer (Layer 7): The application layer provides interfaces for end-user applications to interact with the network, including protocols like HTTP, FTP, SMTP, etc.

v. Defines frames - Data Link Layer (Layer 2): The data link layer defines the format and rules for constructing frames, which are the basic units of data transmission over the network.

c- One host of organization have an IP address 25.34.12.56/16. What are the first address (network address) and the last address (limited broadcast address) in this block?

Sol//

To find the network address and the limited broadcast address from the given IP address and subnet mask (/16), we need to perform the following steps:

1. Identify the network portion and the host portion of the IP address based on the subnet mask.

2. Calculate the range of addresses within the subnet.

Given:

IP address: 25.34.12.56

Subnet mask: /16

The subnet mask /16 indicates that the first 16 bits are used for the network portion, and the remaining 16 bits are used for the host portion.

1. Network portion: The first 16 bits are fixed, so the network address is obtained by setting the host portion to all zeros.

2. Host portion: The last 16 bits can vary to represent different hosts within the subnet.

Calculations:

- The first 16 bits of the IP address 25.34.12.56 are 25.34 (in binary: 00011001.00100010).

- The network address is obtained by setting the host portion (last 16 bits) to all zeros: 25.34.0.0.

- To find the limited broadcast address, we set all bits in the host portion to 1 and keep the network portion unchanged.

- The limited broadcast address for the given subnet is obtained by setting the host portion (last 16 bits) to all ones: 25.34.255.255.

Therefore, in the given subnet:

- The network address (first address) is 25.34.0.0.

- The limited broadcast address (last address) is 25.34.255.255.

Q.4) Answer Two branches Only

a- From the figure below, assume that the communication is between a process running at

computer A with port address i and a process running at computer D with port address j. Show

the contents of packets and frames at the network, data link, and transport layer for each hop.

Sol//

Hop 1: Computer A to Router R1

• Transport Layer:
  • Segment containing the data or message from the application process on computer A.
  • Destination port address: j (port of the process on computer D).
  • Source port address: i (port of the process on computer A).
• Network Layer:
  • Datagram (IP packet) containing the segment from the transport layer.
  • Destination IP address: IP address of Router R1 (connected to the LAN where computer A is located).
  • Source IP address: IP address of computer A.
• Data Link Layer:
  • Frame containing the datagram (IP packet) from the network layer.
  • Destination MAC address: MAC address of Router R1's interface connected to the LAN where computer A is located.
  • Source MAC address: MAC address of computer A's network interface card (NIC).

Hop 2: Router R1 to Router R2

• Transport Layer:
  • Segment remains unchanged (encapsulated within the datagram).
• Network Layer:
  • Datagram (IP packet) may be modified by Router R1.
    • Destination IP address: IP address of Router R2 (connected to the network where computer D is located).
    • Source IP address: IP address of Router R1.
• Data Link Layer:
  • New frame is created containing the modified datagram (IP packet).
  • Destination MAC address: MAC address of Router R2's interface connected to the network where computer D is located.
  • Source MAC address: MAC address of Router R1's interface connected to the network where computer A is located.

Hop 3: Router R2 to Computer D

• Transport Layer:
  • Segment remains unchanged (encapsulated within the datagram).
• Network Layer:
  • Datagram (IP packet) may be modified by Router R2.
    • Destination IP address: IP address of computer D.
    • Source IP address: IP address of Router R2 (depending on routing configuration).
• Data Link Layer:
  • New frame is created containing the modified datagram (IP packet).
  • Destination MAC address: MAC address of computer D's NIC.
  • Source MAC address: MAC address of Router R2's interface connected to the network where computer D is located.

b- i) Change the following IP addresses from dotted-decimal notation to binary notation.

a. 114.34.2.8

b. 129.14.6.8

c. 208.34.54.12

ii) Find the class of the following IP addresses.

a. 11110111 11110011 10000111 11011101

b. 10101111 11000000 11110000 00011101

Sol//

i) Changing IP addresses from dotted-decimal notation to binary notation:

a. IP address: 114.34.2.8

In binary notation:

- 114 = 01110010

- 34 = 00100010

- 2 = 00000010

- 8 = 00001000

So, the IP address 114.34.2.8 in binary notation is: 01110010.00100010.00000010.00001000

b. IP address: 129.14.6.8

In binary notation:

- 129 = 10000001

- 14 = 00001110

- 6 = 00000110

- 8 = 00001000

So, the IP address 129.14.6.8 in binary notation is: 10000001.00001110.00000110.00001000

c. IP address: 208.34.54.12

In binary notation:

- 208 = 11010000

- 34 = 00100010

- 54 = 00110110

- 12 = 00001100

So, the IP address 208.34.54.12 in binary notation is: 11010000.00100010.00110110.00001100

ii) Finding the class of the following IP addresses:

a. IP address: 11110111 11110011 10000111 11011101

This IP address starts with binary '1111', which falls in the range of Class E addresses. Class E addresses are reserved for experimental use and are not intended for public use on the Internet.

b. IP address: 10101111 11000000 11110000 00011101

This IP address starts with binary '10', which falls in the range of Class B addresses. Class B addresses are used for medium-sized networks and have a default subnet mask of 255.255.0.0.

c- What is the differences between OSI and TCP/IP model?

Sol//

The OSI (Open Systems Interconnection) model and the TCP/IP (Transmission Control Protocol/Internet Protocol) model are both conceptual frameworks used to understand and describe the functions of network protocols and the communication process in computer networks. While they share similarities in that they both describe the communication process in layers, they also have several key differences:

يعد نموذج OSI (ربط الأنظمة المفتوحة) ونموذج TCP/IP (بروتوكول التحكم في الإرسال/بروتوكول الإنترنت) إطارين مفاهيميين يستخدمان لفهم ووصف وظائف بروتوكولات الشبكة وعملية الاتصال في شبكات الكمبيوتر. في حين أنهما يشتركان في أوجه التشابه حيث أنهما يصفان عملية الاتصال في طبقات، إلا أن لديهما أيضًا العديد من الاختلافات الرئيسية:

1. Number of Layers:

   - OSI Model: The OSI model consists of seven layers: Physical, Data Link, Network, Transport, Session, Presentation, and Application.

   - TCP/IP Model: The TCP/IP model consists of four layers: Link, Internet, Transport, and Application. Sometimes, a fifth layer, the Network Interface layer, is added to align it more closely with the OSI model.

1. عدد الطبقات:

    - نموذج OSI: يتكون نموذج OSI من سبع طبقات: المادية، ورابط البيانات، والشبكة، والنقل، والجلسة، والعرض التقديمي، والتطبيق.

    - نموذج TCP/IP: يتكون نموذج TCP/IP من أربع طبقات: الارتباط والإنترنت والنقل والتطبيق. في بعض الأحيان، تتم إضافة طبقة خامسة، وهي طبقة واجهة الشبكة، لمواءمتها بشكل وثيق مع نموذج OSI.

2. Layer Names and Functions:

   - OSI Model: Each layer in the OSI model has its own specific name and function, and each layer interacts with adjacent layers using well-defined protocols and services. For example, the Transport layer ensures end-to-end communication between hosts.

   - TCP/IP Model: The TCP/IP model combines the OSI's Physical and Data Link layers into the Link layer and the Session and Presentation layers into the Application layer. The Transport layer in the TCP/IP model performs similar functions to the OSI Transport layer.

2. أسماء الطبقات ووظائفها:

    - نموذج OSI: كل طبقة في نموذج OSI لها اسمها ووظيفتها المحددة، وتتفاعل كل طبقة مع الطبقات المجاورة باستخدام بروتوكولات وخدمات محددة جيدًا. على سبيل المثال، تضمن طبقة النقل الاتصال من طرف إلى طرف بين المضيفين.

    - نموذج TCP/IP: يجمع نموذج TCP/IP بين الطبقات المادية وطبقات ارتباط البيانات الخاصة بـ OSI في طبقة الارتباط وطبقات الجلسة والعرض التقديمي في طبقة التطبيق. تؤدي طبقة النقل في نموذج TCP/IP وظائف مشابهة لطبقة النقل OSI.

3. Standardization:

   - OSI Model: The OSI model is a theoretical framework developed by the International Organization for Standardization (ISO). While widely used as a conceptual model, it has not been as widely implemented in practice.

   - TCP/IP Model: The TCP/IP model was developed based on real-world protocols used in the development of the Internet. It is more closely aligned with the actual protocols and technologies used in networking, making it the de facto standard for internet communication.

3. التقييس:

    - نموذج OSI: نموذج OSI هو إطار نظري تم تطويره من قبل المنظمة الدولية للمعايير (ISO). وعلى الرغم من استخدامه على نطاق واسع كنموذج مفاهيمي، إلا أنه لم يتم تنفيذه على نطاق واسع في الممارسة العملية.

    - نموذج TCP/IP: تم تطوير نموذج TCP/IP بناءً على بروتوكولات العالم الحقيقي المستخدمة في تطوير الإنترنت. وهو أكثر توافقًا مع البروتوكولات والتقنيات الفعلية المستخدمة في الشبكات، مما يجعله المعيار الفعلي للاتصالات عبر الإنترنت.

4. Flexibility and Scalability:

   - OSI Model: The OSI model offers a more modular and flexible approach to networking, allowing for easier understanding and design of complex systems. However, its complexity can sometimes make it less practical for real-world implementation.

   - TCP/IP Model: The TCP/IP model is simpler and more streamlined, making it easier to implement and troubleshoot. Its scalability and adaptability have contributed to its widespread adoption as the foundation of the modern internet.

4. المرونة وقابلية التوسع:

    - نموذج OSI: يقدم نموذج OSI أسلوبًا معياريًا ومرونًا أكثر للاتصال بالشبكات، مما يتيح فهمًا وتصميمًا أسهل للأنظمة المعقدة. ومع ذلك، فإن تعقيدها قد يجعلها في بعض الأحيان أقل عملية للتنفيذ في العالم الحقيقي.

    - نموذج TCP/IP: يعتبر نموذج TCP/IP أبسط وأكثر انسيابية، مما يسهل تنفيذه واستكشاف الأخطاء وإصلاحها. وقد ساهمت قابليتها للتوسع وقابليتها للتكيف في اعتمادها على نطاق واسع كأساس للإنترنت الحديث.

5. Protocol Independence:

   - OSI Model: The OSI model is independent of any specific protocols and can accommodate a wide range of networking technologies and protocols.

   - TCP/IP Model: The TCP/IP model is closely associated with the TCP/IP suite of protocols, which includes protocols such as TCP, IP, UDP, and ICMP. While TCP/IP is the dominant protocol suite for internet communication, the model itself can support other protocols as well.

5. استقلال البروتوكول:

    - نموذج OSI: نموذج OSI مستقل عن أي بروتوكولات محددة ويمكنه استيعاب مجموعة واسعة من تقنيات وبروتوكولات الشبكات.

    - نموذج TCP/IP: يرتبط نموذج TCP/IP ارتباطًا وثيقًا بمجموعة بروتوكولات TCP/IP، والتي تتضمن بروتوكولات مثل TCP وIP وUDP وICMP. في حين أن TCP/IP هو مجموعة البروتوكولات السائدة للاتصالات عبر الإنترنت، إلا أن النموذج نفسه يمكنه دعم البروتوكولات الأخرى أيضًا.

باختصار، في حين أن كلا النموذجين يوفران إطارًا لفهم مفاهيم الشبكات، فإن نموذج OSI أكثر نظرية وشمولاً، في حين أن نموذج TCP/IP أكثر عملية ومتوافق بشكل وثيق مع تطبيقات الشبكات في العالم الحقيقي، لا سيما في سياق الإنترنت.

PRACTICAL EXAM

Q1: Answer two branches:

A- Find the subnet mask and the first host, the last host and the broadcast of 255.255.255.128/C (6mark)

Sol//

1. Subnet Mask:

   The "/C" notation indicates the number of bits used for the network portion of the address. In this case, "C" represents 24 bits (from the 255.255.255 part of the address), and the remaining 7 bits are used for the subnet.

   Since each octet of the subnet mask has 8 bits, and the first 7 bits of the fourth octet are used for the subnet, the subnet mask can be represented as follows:

   ```

   11111111.11111111.11111111.10000000

   ```

   Converting this binary representation to decimal gives us the subnet mask:

   ```

   255.255.255.128

   ```

2. First Host, Last Host, and Broadcast Address:

   - First Host: To find the first host address, we increment the last octet of the network address by 1.

   - Last Host: To find the last host address, we subtract 1 from the broadcast address.

   - Broadcast Address: The broadcast address is the highest address in the subnet. To find it, we set all the host bits to 1 in the subnet and then convert it to decimal.

   With a subnet mask of 255.255.255.128, the network address is the same as the given address: 255.255.255.128. Therefore:

   - First Host: 255.255.255.129

   - Last Host: 255.255.255.254

   - Broadcast Address: 255.255.255.255

So, for the subnet 255.255.255.128/C:

- Subnet Mask: 255.255.255.128

- First Host: 255.255.255.129

- Last Host: 255.255.255.254

- Broadcast Address: 255.255.255.255

B- What is the first valid host on the subnetwork that the 192.168.10.0/29? (6mark)

Sol//

1. Identify the Subnet Mask:

   The "/29" notation indicates that the subnet uses 29 bits for the network portion and 3 bits for the host portion. This results in a subnet mask of 255.255.255.248.

2. Identify the Network Address:

   The given network address is 192.168.10.0.

3. Calculate the First Valid Host:

   - Since the subnet mask is /29, the last octet of the network address is divided into 8 - 3 = 5 bits for hosts.

   - The binary representation of the last octet of the network address is:

     ```

     0000 0000

     ```

   - Incrementing the last octet by 1 gives us the first valid host:

     ```

     0000 0001

     ```

   - Converting this binary representation back to decimal, we get the first valid host:

     ```

     192.168.10.1

     ```

So, the first valid host on the subnetwork 192.168.10.0/29 is 192.168.10.1.

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