Equation 3 · Part 1 · Model Systems in 2035: Four Scenarios, Their Signals, and What Would Falsify Them

Symbol I^*

I^{*}(t) = \frac{F(t)}{B(t)} = I^{*}_0 \left(\frac{g_F}{g_B}\right)^{t},

I^{*}

What this part means

$I^*$ is part of the quantity the equation computes from the expression on the right.

Its job in the formula

$I^*$ is part of the quantity the equation computes from the expression on the right.

Full expressionSymbol I^*Article meaning

The passage around this formula

Write F(t) for peak arithmetic throughput and B(t) for memory bandwidth. The quantity that matters for a serving system is their ratio, because it sets the arithmetic intensity — operations per byte moved — at which a machine becomes compute-bound rather than bandwidth-bound: $I^{*}(t) = \frac{F(t)}{B(t)} = I^{*}_0 \left(\frac{g_F}{g_B}\right)^{t}$ . with $g_F$ and $g_B$ the annual growth factors. When $g_F$ > $g_B$ , $I^{*}$ grows without bound, and the batch size required to keep the arithmetic units busy grows with it. Autoregressive decoding sits on the wrong side of this: generating one token requires streaming the weights and the accumulated key–value cache, so decode time is bounded below by

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the article section

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