1.
Explain differences between asynchronous and
synchronous transmission.
The primary differences between synchronous and asynchronous transmission is in overhead, error detection, and message termination.
2.
Suppose that a file of 10,000 characters is to be
sent over a line at 2400 bps
a. calculate
the overhead in bits and time when using an asynchronous transmission. Assume 1 start bit, 1 stop bit and 8 bits
for each character
10,000 characters * 8 bits = 80,000 bits for the message
10,000 characters * (1stop bit + 1start bit) = 20,000 bits of overhead
20,000 overhead / 80,000 message = 25% overhead
20,000 bits / 2400 (bits/sec) = 8.33 seconds
b. Calculate
the overhead in bits and time when using a synchronous transmission. Each frame is 1000 characters long with 48
bits of overhead.
10,000 characters * 8 bits = 80,000 bits for the message
10 frames * 48 bits = 480 bits for the overhead
480 bits / 80,000 bits = .6% overhead
480 bits / 2400 (bits/sec) = .2 seconds
c. What would the answers to parts a and b be
for a file of 10,00 characters with a rate of 9600 bps
The overhead does not change
20,000 bits / 9600 (bits/sec) = 2.08 seconds
480 bits / 9600 (bits/sec) = 0.05 seconds
Synchronous protocols conduct their synchronization at the beginning of a transmission through a synchronization pattern. This pattern is present at the beginning of a message, and periodically throughout long message.
In character-based protocols, transparency is often provided by either a byte count or by a signal drop (with fixed length headers).
In a bit-based protocol, transparency is the function of a special and unique sequence of bits is used to mark off the beginning and ending of a transmission, called the flag. The headers are fixed in size and the message body is variable. Since the frame is not ended until the unique bit sequence is received, all data that isn’t part of the header (or optional footer?) is considered part of the message body. A technique called bit stuffing is used in the case that a user wants to transmit the unique bit sequence as a part of the message.
4. What is forward error correction?
Forward error detection not only allows a protocol to detect errors, but often to also correct them, saving the overhead of retransmiting the entire packet.
One old method is called a Hamming code, which works by appending additional parity bits on different sets of the data, such that each set is unique by only one datum from another set.
6. List three
major multiplexing techniques and briefly explain their characteristics
FDM: Frequency Division Multiplexing transfers several signals “at the same time” by transmitting them at different frequencies. Similar to asynchronous transmissions, small parts of data are transmitted at a time, which requires more overhead. The overhead in this case isn’t a header, or checksum; rather it is a time delay that is inserted to ensure that signals don’t overlap. This delay is referred to as a guard band.
TDM: Time Division Multiplexing allocates a more lengthy duration of time for a signal to transmit (say 100 milliseconds). By this method, costly time delays when switching signals can be minimized. Unfortunately, since TDM assumes that each signal will have data, when its time comes, bandwidth may be wasted.
STDM: Statistical Time Division Multiplexing solves this problem and does not waste it’s time when a signal has no data to transmit. The benefit of allowing other signals to go in a free time slot comes with some overhead, however. Since the receiver has no sure way of determining which signal it is receiving, a header, which identifies the sender, is necessary. This small amount of overhead is more than compensated for the by additional speed STDM provides.
7. Explain differences
between x.25 and frame relay
One of the biggest differences between X.25 and frame relay is in the method of error checking (Acknowledgments). In X.25, not only is an Ack sent at each receiving node along a path, but a final ACK is send when the packet reaches its destination. Frame relay waits until the packet reaches its destination before sending (only one) ACK frame.
Confirmed Information rate, variable length packets
8. Explain at
least four differences between ATM and X.25
ATM provides quality of service levels for each packet, allowing unimportant packets to be “dropped” when the network becomes congested.
An ATM frame is fixed at 53 bytes long. This allows vendors to easily develop cheap and fast hardware implementations, as well as allowing video and voice data to quickly move to its destination.
No link-by-link error and flow control
Different type of addressing
9. Why do
HDLC packets need an address and SLIP packets do not?
HDLC is a routed protocol and it is necessary to include a destination address so that hosts along the way can forward it in the appropriate direction. Slip is a point-to-point protocol; there are only two hosts on a network link so it is not needed to have a destination.
10. How does
disk caching work?
Disk caching works in two ways
One is by storing recently accessed data in high-speed RAM. When another application requests the same data, it is already stored in local memory and there is no need to read from the slow hard drive.
Secondly, when requesting data, the disk cache assumes that you are reading data sequentially and grabs the next few sectors of data at the same time—saving the cost of making another trip back to the hard drive.
18 A’a
10 B’s
2 C’s
2 D’s
1 E, 1 F, 1
G, 1 H
Message length will end up being 74 bits minimum. I’m not drawing a tree in ms-paint for you