DSST MIS study materials
DSST MIS study resources

DSST MIS: Telecommunications: Media, Multiplexing, Switching and Capacity Math

DSST MIS · Work through your exam, one topic at a time.

All guide sections

Telecommunications: Media, Multiplexing, Switching and Capacity Math

Telecommunications

The Telecommunications domain of the DSST MIS exam tests how data moves between systems: transmission media and their limits, analog versus digital signalling, multiplexing and switching, LAN/WAN technologies, the OSI and TCP/IP models, and basic capacity arithmetic. Expect definitional items plus one small calculation.

Core concepts

  • Bandwidth vs throughput. Bandwidth is theoretical channel capacity; throughput is what a user actually receives after overhead, contention and errors. A 100 Mbps LAN rarely delivers 100 Mbps of file data.
  • Media. Twisted pair (Cat5e/Cat6) is cheap and distance-limited to roughly 100 m per run; coaxial cable carries more bandwidth per run; fibre offers huge bandwidth, immunity to EMI and long spans; microwave and satellite suit difficult terrain but add latency and weather fade.
  • Multiplexing. FDM splits the band into sub-channels; TDM assigns fixed time slots; statistical TDM allocates slots on demand and improves utilisation.
  • Switching. Circuit switching reserves a path (predictable delay, wasteful of capacity); packet switching shares links and underpins IP.
  • Models. OSI's seven layers are a reference model; TCP/IP's four layers are what is implemented. Each layer adds a header as data descends the stack (encapsulation).
  • Addressing. IPv4 uses 32-bit dotted-decimal addresses plus a mask separating network bits from host bits.

Worked example

A 12 MB file crosses a 10 Mbps link whose protocol overhead is 20%, over a 3,000 km path where signals travel at 2 x 10^8 m/s. How long?

  1. Convert: 12 MB x 8 = 96 megabits.
  2. Usable rate = 10 Mbps x 0.80 = 8 Mbps.
  3. Transmission time = 96 / 8 = 12 seconds.
  4. Propagation delay = 3,000,000 m / (2 x 10^8 m/s) = 0.015 s = 15 ms.
  5. Total is about 12.015 s. Serialisation dominates here; on a short LAN the propagation term would be negligible.

Traps

  • Mixing bytes and bits: MB x 8 = megabits. Answers off by a factor of eight are the most common error.
  • Quoting bandwidth where the question asks for delivered throughput.
  • Treating OSI layers as implemented protocols rather than a teaching model.
  • Forgetting that statistical TDM, not fixed TDM, exploits idle slots.