Pure Aloha and Slotted Aloha Examples in Computer Network

Pure Aloha and Slotted Aloha Examples in Computer Network are explained with the following Timestamps:
0:00 - Examples of Aloha - Computer Network
0:32 - Example 1
2:55 - Example 2
5:23 - Example 3

Pure Aloha and Slotted Aloha Examples in Computer Network are explained with the following outlines:
0. Computer Network
1. Random Access Protocols
2. Aloha
3. Pure Aloha
4. Slotted Aloha
5. Examples of Pure Aloha and Slotted Aloha
6. Efficiency of Pure Aloha
7. Throughput of Pure Aloha
8. Efficiency of Slotted Aloha
9. Throughput of Slotted Aloha

Chapter-wise detailed Syllabus of the Computer Network Course is as follows:

Computer Network Basics, Types of Data Communication in Computer Network, Types of Casting in Computer Network, Peer to Peer Network Vs Client Server Network, Types of Computer Network, Topologies of Computer Network.

Basics of OSI Model, OSI Model Details, Comparison of TCP and UDP, Comparison of OSI Model and TCP/IP Model, Transmission Medium in Computer Network, IP Address, MAC Address and Port Address.

Data Link Layer Basics, Transmission Delay & Propagation Delay, Stop & Wait ARQ Protocol, Capacity of Channel in Computer Network, Sliding Window Protocol, Go Back N ARQ Protocol, Selective Repeat ARQ Protocol, Comparison of Sliding Window Protocol, Go Back N ARQ Protocol and Selective Repeat ARQ Protocol, Media Access Protocols, Channelization Protocols: FDMA, TDMA, CDMA, Polling Protocol, Reservation Protocol, Token Passing Protocol, Pure ALOHA, Slotted ALOHA, CSMA - Carrier Sense Multiple Access, CSMA/CD, Back off Algorithm for CSMA/CD, CSMA/CA.

VRC - Vertical Redundancy Check & LRP - Longitudinal Redundancy Parity, Checksum, CRC - Cyclic Redundancy Check, Hamming Code for Error Correction and Error Detection, Minimum Hamming Distance, Error Detection Capabilities & Error Correction Capabilities.

Framing, Ethernet LAN Protocol - IEEE 802.3, Wireless LAN Protocol - IEEE 802.11, Token Ring LAN Protocol - IEEE 802.5, Examples on CSMA Protocols, Examples on Flow Control Protocols.

Repeater, Hub, Bridge, Switch, Router and Brouter, Comparison of Repeater Hub Bridge Switch Router and Brouter, Network Segment, Collision Domain & Broadcast Domain, Spanning Tree Protocol.

Network Layer Basics, logical Addressing, Classful Addressing of IP Address, Subnetting, Classes Inter-Domain Routing - CIDR, Subnet Masking, Supernetting, IPV4 Frame Format, Fragmentation, IPV4 Broadcasting, Localhost and Loopback, ARP, RARP, BOOTP , DHCP, Public IP and Privet IP, NAT - Network Address Translation, ICMP, Routing, Non Adaptive Routing, Distance Vector Routing, Link State Routing, Circuit Switching, Packet Switching and Message Switching, IPV6, IPV6 Casting, IPV6 Frame Format.

Transport Layer Basics, TCP Frame Format, Port Number & Network Socket, Sequence Number & Acknowledgment Number in TCP, Wrap Around Time and Life Time in TCP, Header Length and Checksum in TCP, TCP Connections, PSH URG RST Urgent Pointer and Option Field in TCP, Flow Control in TCP Protocol, Sliding Window Protocol in TCP, Retransmission in TCP, Congestion Control in TCP, Timers in TCP, Basic Algorithm for Timeout Timer in TCP, Jacobson's Algorithm for Timeout Timer in TCP, Karn's Modification of Timeout Timer in TCP, Silly Window Syndrome and It's Solution in TCP, Congestion Control using Traffic Shaping in TCP, Leaky Bucket Algorithm, Token Bucket Algorithm, UDP - User Datagram Protocol, UDP Segment Header, UDP Frame Format.

Application Layer Basics, DNS - Domain Name System, SMTP POP and IMAP, HTTP/HTTPs , FTP SFTP and TFTP - File Transfer Protocols.

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[kamelkamatchu]

sir thanks for the tutorial but the throughput for 500 trame /s is 18.4 not 36.8 because 0.183 is 18.3% of 500 received so 500*0.183 = 92 packet of 500 received so 92 * 200 bits is 18.4 bpbs

[zziikiincrse]

it's a wrong process perspective as book solution sir. why did u use different formula?

[debjitdey]

for 250 frames/sec - throughput = 30.32 kbps

[shubhrambharti]

Sir doubt about calculation of G it is no of host but you calculate by frame/sec.

[benasabu]

Sir i have a doubt since g represent the average number of packet that can be transmitted during the vulnerability time here in caseof pure aloha its 2transmisssion time is it, so here 1000 frames are sen in 1 sec implies for 1 frame 1 milli second is needed, here transmission time is 1 millisecond and vulnerability time is 2 milli sec. so g must be 2 right sir,

[benasabu]

but sir you told g is number of host want to transmit . how come it be number of frames the host transmit

[jayden]

3rd problem 3rd question answer pls explain sir
30.32 getting

[ProgrammingJam]

You are teaching rubbish. You teaching all wrong about Aloha throughput.