5 · Throughput & bandwidth-delay product ★

HW1 P6 · slides 1-52 → 1-54

Throughput

Throughput = the rate (bits/s) at which bits actually get from sender to receiver.
Instantaneous = the rate at one moment. Average = the rate over a longer period (e.g. a whole file).
Picture fluid in pipes: the flow is limited by the narrowest pipe.

Two links: server rate $R_s$, client rate $R_c$

$$\text{throughput} = \min(R_s,\ R_c)$$

If $R_s < R_c$, the server's link is the limit. If $R_s > R_c$, the client's link is the limit. The slowest link on the path is called the bottleneck link.

Network scenario: $n$ connections share a backbone link $R$

The backbone is split fairly, so each connection gets $R/n$ of it:

$$\text{throughput per connection} = \min\!\left(R_s,\ R_c,\ \frac{R}{n}\right)$$

Slide example: 10 connections → $\min(R_c, R_s, R/10)$. In practice $R_c$ or $R_s$ (the access links) is often the bottleneck, because backbones are fast.

File transfer time (ignoring propagation and queuing): $$T = \frac{F}{\text{throughput}}$$

Worked example: throughput

Q: A server sends a 4 MB file to a client. $R_1 = 500$ kbps, $R_2 = 2$ Mbps, $R_3 = 1$ Mbps on the path. Throughput? Transfer time?

$$\text{throughput} = \min(500 \text{ kbps},\ 2 \text{ Mbps},\ 1 \text{ Mbps}) = \mathbf{500 \text{ kbps}}$$

Convert the file to bits first: 4 MB = 4×10⁶ bytes × 8 = 32×10⁶ bits.

$$T = \frac{32 \times 10^6}{5 \times 10^5} = \mathbf{64 \text{ s}}$$

Worked example: shared backbone

Q: $R_s = 2$ Mbps, $R_c = 1$ Mbps, and 10 connections share a 5 Mbps backbone. Throughput per connection?

$$\min\!\left(2,\ 1,\ \frac{5}{10}\right) = \min(2,\ 1,\ 0.5) = \mathbf{0.5 \text{ Mbps}}$$

Here the backbone is the bottleneck, not the access links.

Bandwidth-delay product

$$\text{BDP} = R \cdot d_{prop} \qquad d_{prop} = \frac{m}{s}$$

Meaning: the maximum number of bits that can be in the link at once. Think "how many bits fit in the pipe." The sender pushes out $R$ bits every second, and each bit stays in the wire for $d_{prop}$ seconds.

Width of a bit

If the link (length $m$) is full with BDP bits, each bit takes up:

$$\text{width} = \frac{m}{R \cdot d_{prop}} = \frac{m}{R \cdot m/s} = \frac{s}{R}$$

So a faster link has thinner bits (more of them squeezed into the same wire).

Worked example: HW1 P6

Q: A and B are 20,000 km apart on a direct link with $R = 2$ Mbps and $s = 2.5 \times 10^8$ m/s.

(a) Bandwidth-delay product

Convert units first: 20,000 km = 2×10⁷ m.

$$d_{prop} = \frac{2 \times 10^7}{2.5 \times 10^8} = 0.08 \text{ s}$$ $$R \cdot d_{prop} = 2 \times 10^6 \times 0.08 = \mathbf{160{,}000 \text{ bits}}$$

(b) 800,000-bit file sent as one message. Max bits in the link at once?

160,000 bits. The link can't hold more than the BDP, and the file is bigger than that, so the link fills up completely.

If the file is smaller than the BDP, the answer is the file size. In general: max bits in link $= \min(F,\ R \cdot d_{prop})$.

(c) Interpretation

The BDP is the maximum number of bits that can be in the link at any time.

(d) Width of a bit. Longer than a football field?

$$\text{width} = \frac{m}{\text{BDP}} = \frac{2 \times 10^7}{1.6 \times 10^5} = \mathbf{125 \text{ m}}$$

Same answer from $s/R = 2.5 \times 10^8 / 2 \times 10^6 = 125$ m. Yes, longer than a football field (~100 m).

(e) General expression

$$\text{width} = \frac{s}{R}$$

By hand: divide powers of 10 separately. $\frac{2 \times 10^7}{2.5 \times 10^8} = \frac{2}{2.5} \times 10^{-1} = 0.8 \times 0.1 = 0.08$.

Quick check

1. $R_s$ = 10 Mbps, $R_c$ = 3 Mbps, nothing else on the path. Throughput?

$\min(10, 3) = \mathbf{3 \text{ Mbps}}$. The client link is the bottleneck.

2. $R_s$ = 2 Mbps, $R_c$ = 4 Mbps, 4 connections share a 10 Mbps backbone. Throughput per connection?

$\min(2,\ 4,\ 10/4) = \min(2,\ 4,\ 2.5) = \mathbf{2 \text{ Mbps}}$. The server link is the bottleneck.

3. Same as #2, but the backbone is 6 Mbps. Throughput now? How long for a 30 Mbit file?

$\min(2,\ 4,\ 6/4) = \min(2,\ 4,\ 1.5) = \mathbf{1.5 \text{ Mbps}}$. Time = 30 / 1.5 = 20 s.

4. A 1000 km link, $s = 2 \times 10^8$ m/s, $R = 10$ Mbps. BDP? Width of a bit?

$$d_{prop} = \frac{10^6}{2 \times 10^8} = 5 \text{ ms} \qquad \text{BDP} = 10^7 \times 0.005 = \mathbf{50{,}000 \text{ bits}}$$ $$\text{width} = \frac{s}{R} = \frac{2 \times 10^8}{10^7} = \mathbf{20 \text{ m}}$$

5. In #4, a 20,000-bit file is sent. Max bits in the link at once?

20,000 bits. The file is smaller than the BDP (50,000), so the whole file fits in the link.

6. You double R. What happens to the BDP and to the width of a bit?

BDP doubles ($R \cdot d_{prop}$). Width of a bit halves ($s/R$).

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