Layering & switching formulas

Layering and switching: circuit, packet and virtual circuit. 6 formulas for GATE CS Computer Networks, each with when to use it. Each is explained in the topic notes.

All 6 Layering & switching formulas

HEADER EFFICIENCY

η = \frac{payload bits}{payload bits + header (and trailer) bits}

Fraction of transmitted bits that is useful data; with several layers, add up the headers of all layers that wrap the payload.

MESSAGE DELAY IN A CIRCUIT-SWITCHED NETWORK

T_{circuit} = T_{setup} + \frac{L}{R} + T_{prop,path}

L = message size in bits, R = rate of the reserved circuit in bits per second (R/n of the link rate when n circuits share it); T_setup is the set-up time given in the question (teardown ignored); T_prop,path is the sum of the propagation delays of all links. Switch processing is ignored unless stated.

ONE PACKET OVER K LINKS (STORE-AND-FORWARD)

T = k \cdot \frac{L}{R} + i = 1 \sum k \frac{d _{i}}{s}

The whole packet is transmitted again on every link, so transmission delay is counted k times; all links at rate R, no queueing or processing. If the links have different propagation delays, add the delays given.

A MESSAGE SENT AS N EQUAL PACKETS OVER K LINKS

T = (k + n - 1) \cdot \frac{p}{R} + i = 1 \sum k \frac{d _{i}}{s}

Each packet has p bits including its header, all links at rate R, no queueing, and the packets go back to back. The first packet needs k transmission times; each later packet arrives one transmission time after the one before it (pipelining). T is the time from the first bit leaving the source until the last bit of the last packet reaches the destination.

N EQUAL PACKETS OVER LINKS OF DIFFERENT RATES

T = i = 1 \sum k \frac{p}{R _{i}} + (n - 1) \frac{p}{R _{m i n}} + i = 1 \sum k \frac{d _{i}}{s}

R_i = rate of link i, R_min = rate of the slowest link (the bottleneck). The first packet crosses every link once; the later packets leave the bottleneck one p/R_min apart and keep that spacing. With all rates equal this reduces to the previous formula. No other traffic, no queueing apart from waiting for the packet ahead.

PAYLOAD SIZE THAT MINIMISES TOTAL DELAY

x_{opt} = \frac{M h}{k - 1}

M = message bits, h = header bits per packet, k >= 2 links of equal rate, ignoring the rounding of M/x to a whole number. Derivation: delay times R is (k-1+M/x)(x+h) = (k-1)x + (k-1)h + M + Mh/x; setting its derivative with respect to x to zero gives (k-1) = Mh/x^2. Propagation does not depend on x.

Where this was asked: 2018 (1), 2025 (1). See the 2 previous-year questions on Layering & switching. This shows what was asked, not what will be asked.

KEEP GOING

Layering & switching: the notes and 2 solved previous-year questions