Switching — circuit, packet, virtual circuit: Worked examples

This page is the drill floor for Switching — circuit, packet, virtual circuit. We take the three delay formulas and push them into every corner: single packet vs many, zero hops, huge setup, the pipelining limit, a real-world word problem, and an exam twist that mixes queueing in. Each example tells you which cell of the scenario matrix it hits, so you can be sure nothing is left uncovered.

Before we start, let us re-anchor the symbols so this page stands on its own — never trust that you remember them.

Definition The symbols, in plain words

$L$ = size of the whole message in bits. Picture a long ribbon; its length is $L$ .
$R$ = link rate in bits per second — how fast one link can push bits out. Picture the width of a doorway: wider door, more bits per second.
$\frac{L}{R}$ = transmission time: ribbon length ÷ push speed = seconds to shove the whole ribbon through the doorway. This is the only place message size and link speed meet.
$d$ = number of links (hops) on the path. Two switches between you and the goal means $d = 3$ links.
$t_{p}$ = propagation delay per link: time for one bit to travel the distance of a wire (limited by speed of light-ish), independent of $L$ or $R$ .
$S$ = setup time: one-time cost to build a circuit or a virtual circuit before any data moves.
$P$ = number of packets we chop $L$ into; each packet is $L / P$ bits (we ignore header bits here).

Recall Which term ignores message size entirely?

Propagation $d t_{p}$ ::: it depends only on distance (hops), not on how many bits you send.

The scenario matrix

Every switching problem lands in one of these cells. The worked examples below each carry a tag like [Cell A2] so you can check coverage.

	Circuit	Datagram	Virtual Circuit
Normal message	A1 · one $L / R$ , add setup	A2 · single packet, $d L / R$	A3 · setup + cheap forward
Chopped into $P$ packets	— (streamed, not chopped)	B2 · pipelining formula	B3 · same pipeline, label swap
Degenerate: $d = 1$ (one hop)	C1	C2 · datagram = circuit here	—
Limiting: $P \to \infty$	—	D2 · transmission $\to L / R$	—
Zero setup / zero propagation	E1 · $S = 0$ or $t_{p} = 0$	E2 · $t_{p} = 0$	—
Real-world word problem	—	F2 · file transfer race	—
Exam twist: add queueing	—	G2 · $+$ per-hop queue delay	—

We now walk cells A1, A2, A3, B2, C2, D2, E1, F2, G2 — that touches every row and every column at least once.

Worked example Example 1 — Circuit switching, normal message

[Cell A1] Send $L = 2 \times 1 0^{6}$ bits over a circuit-switched path. $R = 1 0^{6}$ bps, $d = 4$ links, $t_{p} = 2$ ms per link, setup $S = 50$ ms. Find the total time.

Forecast: guess now — will the answer be near $2$ s, or near $8$ s? Circuit switching sends one copy of the message, so bet on the small one.

Setup phase. $S = 0.050$ s. Why this step? A circuit must be reserved end-to-end before any bit flows; that reservation is not free.
Propagation across all links. $d t_{p} = 4 \times 0.002 = 0.008$ s. Why this step? The first bit still has to physically travel every wire; four hops means four wire-crossings.
Transmission of the message — ONCE. $\frac{L}{R} = \frac{2 \times 1 0 ^{6}}{1 0 ^{6}} = 2$ s. Why this step? After setup the path is a continuous pipe with no store-and-forward, so the source pushes all $L$ bits once; the switches do not re-buffer. Hence no $\times d$ here.
Add. $T = 0.050 + 0.008 + 2 = 2.058$ s.

Verify: Units all seconds. Transmission dominates (2 s), setup and propagation are small ms-scale corrections — sensible for a slow link and a big file. ✓

Worked example Example 2 — Datagram, single packet

[Cell A2] Same numbers: $L = 2 \times 1 0^{6}$ bits, $R = 1 0^{6}$ bps, $d = 4$ , $t_{p} = 2$ ms. Send it as one datagram packet. Find total time.

Forecast: datagram is store-and-forward. Will it beat or lose to the 2.058 s from Example 1?

Transmission at each hop, serially. Each switch must fully receive the packet before forwarding: $d \cdot \frac{L}{R} = 4 \times 2 = 8$ s. Why this step? This is the definition of store-and-forward. The whole ribbon is re-shoved through each of the 4 doorways one after another — the packet is "reborn" at every switch.
Propagation. $d t_{p} = 4 \times 0.002 = 0.008$ s. Why this step? Distance cost is unchanged by store-and-forward.
Add. $T = 8 + 0.008 = 8.008$ s.

Verify: $8.008 > 2.058$ — datagram with a single fat packet is far worse, exactly because of the $\times d$ replay. This is the pain that pipelining (Example 4) cures. ✓ See Store-and-Forward and Network Delays.

Figure — Switching — circuit, packet, virtual circuit

Worked example Example 3 — Virtual circuit, normal message

[Cell A3] A VC is set up ( $S = 30$ ms) over $d = 4$ links, then $L = 2 \times 1 0^{6}$ bits are sent as one packet with a small label. $R = 1 0^{6}$ bps, $t_{p} = 2$ ms. Find total time.

Forecast: VC does a setup like circuit, but forwards like a packet (store-and-forward). Guess: closer to Example 1 or Example 2?

Setup. $S = 0.030$ s — build the VC tables (label mappings) hop by hop. Why this step? VC installs per-VC state once; but note it reserves no bandwidth, only a logical path.
Store-and-forward transmission. VC packets are still buffered fully at each hop: $d \cdot \frac{L}{R} = 4 \times 2 = 8$ s. Why this step? Label swapping does not remove store-and-forward; it only makes the per-hop lookup cheap, not the buffering. A common trap.
Propagation. $d t_{p} = 0.008$ s.
Add. $T = 0.030 + 8 + 0.008 = 8.038$ s.

Verify: VC with one fat packet ≈ datagram with one fat packet ( $8.008$ s) plus setup — as expected, since both store-and-forward. The win for VC comes with many small packets (Example 5) and cheap label lookup, not from the delay formula alone. ✓

Worked example Example 4 — Datagram,

$P$ packets, pipelining [Cell B2] Chop the same $L = 2 \times 1 0^{6}$ bits into $P = 100$ packets (ignore headers). $R = 1 0^{6}$ bps, $d = 4$ , $t_{p} = 2$ ms. Find total time.

Forecast: more packets should help. Guess: will it drop below Example 2's 8.008 s? By how much?

First packet crosses all links. One packet is $L / P = 2 \times 1 0^{4}$ bits, so per-hop transmission is $\frac{L / P}{R} = \frac{2 \times 1 0 ^{4}}{1 0 ^{6}} = 0.02$ s. Across $d$ links: $d \cdot \frac{L / P}{R} = 4 \times 0.02 = 0.08$ s. Why this step? The lead packet still store-and-forwards through 4 hops, but it is 100× smaller, so each hop costs 100× less.
Remaining $P - 1$ packets pipeline behind it. Once packet 1 clears link 1, packet 2 uses link 1 while packet 1 uses link 2 — parallel occupancy. The tail arrives $(P - 1) \cdot \frac{L / P}{R} = 99 \times 0.02 = 1.98$ s after the lead. Why this step? Pipelining: only the last packet's extra queue behind the first matters; the middle overlaps.
Combine transmission + propagation. Use the compact formula: $T = \frac{d + P - 1}{P} \cdot \frac{L}{R} + d t_{p} = \frac{4 + 99}{100} \times 2 + 0.008 = 1.03 \times 2 + 0.008 = 2.068 s .$ Why this step? $\frac{d + P - 1}{P}$ folds steps 1 and 2 into one factor times $L / R$ .

Verify: Check the pieces: $0.08 + 1.98 + 0.008 = 2.068$ s — matches the compact formula. And $2.068 ≪ 8.008$ : pipelining slashed the delay ~4×, roughly the number of hops. ✓

Worked example Example 5 — Limiting case

$P \to \infty$ [Cell D2] Keep everything from Example 4 but let $P$ grow without bound. What transmission cost do we approach?

Forecast: guess the "best possible" transmission time. Can it be smaller than $L / R = 2$ s?

Take the limit of the factor. $P \to \infty lim \frac{d + P - 1}{P} = P \to \infty lim (1 + \frac{d - 1}{P}) = 1$ . Why this step? As $P \to \infty$ the $\frac{d - 1}{P}$ term vanishes; only the " $1$ " survives.
So transmission $\to \frac{L}{R} = 2$ s and total $\to 2 + d t_{p} = 2.008$ s. Why this step? With infinitely fine pipelining, the multi-hop path behaves like a single hop for transmission — the $\times d$ penalty of store-and-forward disappears entirely.

Verify: Compare with $P = 100$ (2.068 s) and $P = 1$ (8.008 s): the sequence $8.008 \to 2.068 \to 2.008$ is monotonically approaching $2.008$ from above. The limit is a floor you can approach but never beat (headers would eventually push it back up in reality). ✓ Related idea: Multiplexing — TDM, FDM shares links; here we time-share by pipelining.

Worked example Example 6 — Degenerate: one hop,

$d = 1$ [Cell C2] A source connects directly to the destination: $d = 1$ . $L = 2 \times 1 0^{6}$ bits, $R = 1 0^{6}$ bps, $t_{p} = 2$ ms. Compare datagram (1 packet) with circuit switching (setup $S = 0$ ).

Forecast: with only one link and no middle switches, is there any difference left between the two schemes?

Datagram, 1 packet, $d = 1$ . $T = d \cdot \frac{L}{R} + d t_{p} = 1 \times 2 + 1 \times 0.002 = 2.002$ s. Why this step? Store-and-forward across one link means exactly one $L / R$ — the $\times d$ replay has nothing to replay.
Circuit, $S = 0$ , $d = 1$ . $T = S + d t_{p} + \frac{L}{R} = 0 + 0.002 + 2 = 2.002$ s. Why this step? With no setup and one hop, circuit's formula collapses to the same terms.

Verify: Both give $2.002$ s — identical. The whole distinction between switching schemes lives in the intermediate switches; remove them and all methods agree. A clean sanity check on the formulas. ✓

Worked example Example 8 — Real-world word problem: the file-transfer race

[Cell F2] Aisha uploads a $1 MB = 8 \times 1 0^{6}$ bit file across $d = 3$ hops, $R = 1 0^{6}$ bps, $t_{p} = 1$ ms. Her app can send it either as 1 packet or as 1000 packets (ignore headers). Which finishes first, and by how much?

Forecast: intuition says "smaller packets = faster because pipelining." Predict roughly how many seconds each takes.

One packet. $T_{1} = d \cdot \frac{L}{R} + d t_{p} = 3 \times 8 + 3 \times 0.001 = 24 + 0.003 = 24.003$ s. Why this step? Baseline: full store-and-forward replay of 8 Mbit three times.
1000 packets, pipelined. $T_{1000} = \frac{d + P - 1}{P} \cdot \frac{L}{R} + d t_{p} = \frac{3 + 999}{1000} \times 8 + 0.003 = 1.002 \times 8 + 0.003 = 8.019$ s. Why this step? The factor $\frac{1002}{1000} = 1.002$ is barely above 1, so transmission ≈ $L / R = 8$ s.
Difference. $24.003 - 8.019 = 15.984$ s saved. Why this step? Quantifies the pipelining win in the reader's own units — nearly 16 seconds off a 24-second transfer.

Verify: $8.019$ s ≈ single-hop transmission $8$ s plus tiny propagation — physically the best a 1 Mbps link can do for 8 Mbit. The 1-packet answer is ~3× larger, matching $d = 3$ . This is precisely why the Internet chops files into packets and why TCP — reliable ordered delivery rides on packet switching. ✓

Worked example Example 9 — Exam twist: add per-hop queueing delay

[Cell G2] The clean formulas ignore congestion. Now suppose each of the $d = 3$ switches adds a queueing delay $t_{q} = 5$ ms because other traffic is waiting ahead of our packet. Send $L = 2 \times 1 0^{6}$ bits as 1 packet, $R = 1 0^{6}$ bps, $t_{p} = 1$ ms. Find the realistic total.

Forecast: how much does 3 queues of 5 ms each add — noticeable, or lost in the noise?

Base store-and-forward delay. $d \cdot \frac{L}{R} + d t_{p} = 3 \times 2 + 3 \times 0.001 = 6.003$ s. Why this step? Start from the known datagram formula, then bolt on the new effect.
Add queueing at each hop. Extra $= d \cdot t_{q} = 3 \times 0.005 = 0.015$ s. Why this step? Queueing delay is a per-hop wait, like store-and-forward, so it multiplies by $d$ . See Queueing Delay.
Total. $T = 6.003 + 0.015 = 6.018$ s. Why this step? Full realistic delay = transmission + propagation + queueing, summed over hops.

Verify: Queueing added only $15$ ms to a $6$ s transfer here — negligible for a fat packet on a slow link, but on fast links where $L / R$ is tiny, queueing can dominate (its whole reason for mattering in real networks). Units: all seconds; the extra term $d t_{q}$ has the same shape as $d t_{p}$ , a good consistency check. ✓

Recall In one line: why does packet switching win the real-world race?

Pipelining drives the $\times d$ store-and-forward penalty down toward $L / R$ , and there is no setup cost — so datagram beats circuit for bursty file transfer (Examples 4, 8). ::: This is why the Internet is packet-switched.

See also: Routing Algorithms (how datagram switches pick the outgoing link), MPLS and ATM (real virtual-circuit systems using label swapping), Computer Networks (the parent subject).