Network Performance Metrics: Delay, Throughput and BDP

Link and path performance is described by a few quantities: how long a packet takes to cross a link or a path (delay or latency), how many bits per second actually arrive (throughput), and how many bits are "in flight" on a path at once (the bandwidth-delay product). Efficiency formulas for stop-and-wait, sliding windows, CSMA/CD and TCP are all built from these.

Two delays that sound alike must be kept apart. Transmission delay is how long it takes to push a packet's bits onto the link; it depends on the packet size and the link rate. Propagation delay is how long one bit takes to travel the length of the link; it depends on distance and signal speed. Which one dominates decides almost every design choice in the later notes.

The nodal delay at one hop is $d_{\text{proc}} + d_{\text{queue}} + T_t + T_p$.

Transmission versus propagation. $T_t$ has nothing to do with distance: a 1000-byte packet takes the same time to put onto a 10 m link or a 10,000 km link of the same rate. $T_p$ has nothing to do with packet size or link rate: a bit takes the same time to cross the link whether the link is 1 kbps or 100 Gbps.

Units convention used in all CN notes. Link rates use powers of ten: 1 kbps = $10^3$, 1 Mbps = $10^6$, 1 Gbps = $10^9$ bit/s. Sizes in bytes are multiplied by 8. Unless a question says otherwise, 1 KB = 1000 bytes; when it says $2^{10}$, use that, because the answer changes (Example 4).

The round-trip time (RTT) is the time from sending a bit until a reply can come back; on a single link with negligible reply size and processing it is $2T_p$. The dimensionless ratio

compares the two delays. If $a \ll 1$ (short links, long frames) the sender finishes a frame long after its first bit has arrived; if $a \gg 1$ (long or fast links) the sender has finished before the first bit arrives, and the link holds many frames at once. The efficiency of stop-and-wait is $1/(1+2a)$ and that of CSMA/CD falls as $a$ grows (Sliding Window Protocols, Medium Access Control).

The bandwidth-delay product is the number of bits a link or path holds when the sender transmits continuously:

The round-trip form is the amount a sender must be allowed to have outstanding (sent but not acknowledged) to keep the link fully busy. A window smaller than $R \cdot \text{RTT}$ caps throughput at $\text{window}/\text{RTT}$. The physical length of one bit on the wire is $v/R$ metres.

Throughput is the rate at which bits arrive at the receiver, as opposed to the link rate (often called bandwidth). For a path of links in series, the end-to-end throughput of one flow is the bottleneck rate:

where a link shared equally by $n$ flows contributes $R/n$. The time to move a file of $F$ bits is about $F/\text{throughput}$ plus the path latency, and for large files the first term dominates.

Utilisation of a link is the fraction of time it is busy sending useful bits; useful throughput is utilisation times $R$.

If packets of $L$ bits arrive at a router output at an average $\lambda$ packets per second, the traffic intensity is $\lambda L / R$. If it exceeds 1, bits arrive faster than they can leave and the queue grows without bound (in practice, packets are dropped when the buffer is full). As it approaches 1 from below, the average queueing delay grows sharply. Queueing delay is the only component that varies from packet to packet.

End-to-end delay over $N$ store-and-forward links. With equal rates, no queueing and processing $d_{\text{proc}}$ at each of the $N - 1$ routers, one packet needs

and $P$ back-to-back packets need $(N + P - 1)L/R$ plus the propagation and processing terms (Circuit, Packet and Virtual-Circuit Switching).

Burst queueing. If $N$ packets of $L$ bits arrive at an idle link at the same instant, the $k$-th one (counting from 1) waits $(k - 1)L/R$, so the average queueing delay is

Window-limited throughput. A sender allowed at most $W$ bits outstanding sends $W$ bits per RTT at best, so

A 1000-byte packet (8000 bits) on a 10 Mbps link, signal speed $2 \times 10^8$ m/s.

On a 1 Gbps link 3000 km long, a 1 MB file (8,000,000 bits) has $T_t = 8$ ms and $T_p = 15$ ms: even a large transfer is propagation-dominated on a fast long-haul link.

Host to host through two routers: three links of 1 Mbps, each with $T_p = 1$ ms, packets of 1500 bytes (12,000 bits, so $T_t = 12$ ms), no queueing.

A 100 Mbps path with RTT 40 ms: $R \cdot \text{RTT} = 4 \times 10^6$ bits $= 500{,}000$ bytes must be outstanding to keep it busy. One way (20 ms) the path holds $2 \times 10^6$ bits. At $v = 2 \times 10^8$ m/s each bit is $2 \times 10^8 / 10^8 = 2$ m long.

A geostationary satellite link with $T_p = 270$ ms each way (RTT 540 ms) at 10 Mbps has a round-trip BDP of $5.4 \times 10^6$ bits $= 675{,}000$ bytes. A sender limited to a 65,535-byte window gets at most $65{,}535 \times 8 / 0.54 \approx 970{,}889$ bit/s, under 1 Mbps on a 10 Mbps link. This is why TCP needs window scaling on such paths (TCP Flow Control).

A server with a 2 Mbps access link sends to a client with a 1 Mbps access link through a 50 Mbps core link shared equally by 10 flows (5 Mbps each). Throughput is $\min(2, 5, 1) = 1$ Mbps. A 4 MB file takes $32 \times 10^6 / 10^6 = 32$ s with 1 MB = $10^6$ bytes, but $4 \times 2^{20} \times 8 / 10^6 = 33.554432$ s with 1 MB = $2^{20}$ bytes.

Packets of 12,000 bits arrive at 100 packets/s at a 1.5 Mbps link: $\lambda L/R = 1.2 \times 10^6 / 1.5 \times 10^6 = 0.8$. At 150 packets/s the intensity is 1.2, and the queue grows without bound until the buffer overflows.

Five 12,000-bit packets arrive together at an idle 1 Mbps link. Their queueing delays are 0, 12, 24, 36 and 48 ms; the average is 24 ms, which matches $(5 - 1) \times 12/2$.

Common misconceptions.