- Type
- Quiz
- Course
- CS601 Data Communication
- Size
- PDF 915 KB
- Updated
- 23 Oct 2026
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Datagram approach and virtual circuit approach are two popular approaches which lead us to two different types of packet-switched networks. What is the primary function of a checksum in digital communication?Select the correct option Error correction Data compression Encryption Error detection Error detection.
The time in the Reservation method is divided ________________________.Select the correct option in intervals with a reservation frame preceding data frames randomly allocated intervalsi nto equal intervals for all stations based on station priority
into equal intervals for all stations
TOKEN possession time management primarily controls the __________________________.Select the correct option frequency of token generation within the network encryption applied to the TOKEN packet speed at which data packets are transmitted
he main purpose of inserting guard times in TDMA systems is to___________________.Select the correct option reduce the number of available time slots account for propagation delays and ensure proper synchronization compensate for variations in signal strength extend the duration of each time slot
account for propagation delays and ensure proper synchronization
CHAP is an Authentication Protocol, which uses ___________ process to authenticate user information.s an authentication protocol that uses a challenge-response
Among the following cables given below, __________cable has the highest cost as compared to others. In random access method, when two or more than two nodes transmit frames at the same time then _________________. All the frames in collision are lost Collision has no effect on frames All the frames in collision survives none All the frames in collision are lost.
In even parity check code, the value of syndrome is ______ if the number of 1s is even.
In optical fiber a glass or plastic core is surrounded by a __________. cladding
_________ uses thick coaxial cable. 10BASE5
he network address "47:20:1B:2E:08:EE” is consisted of ________bytes. Select the correct option812106 6 bytes.
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