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For a stipulated population of amplifiable copies, let $N_n$ be the amount after cycle $n$ and define PCR efficiency by the fractional new yieldHere $E_n=1$ means one additional copy per existing copy, hence ideal doubling. If every cycle has the same PCR efficiency $E$, the exponential growth model gives $N_n=N_0(1+E{)}^n$. Starting with $N_0=100$ copies, twenty ideal cycles give $104857600$ copies; the stipulated value $E=0.8$ gives about $12748236$. An efficiency of 80 percent therefore means multiplication by 1.8 each cycle, not retention of 80 percent of the final ideal yield.

In the ideal strand-counting example, all duplex-equivalent material doubles, but exact-length amplicons lag initially. Starting from one longer duplex and retaining its two original strands, cycle $n\ge 1$ contains $2(2^n-n-1)$ exact-length strands out of $2^{n+1}$ total strands. Thus their fraction is $1-(n+1)/2^n$: zero after cycle 1, one quarter after cycle 2 and one half after cycle 3. This counts strands rather than assuming every strand immediately finds an exact-length partner. The generic yield model above ignores this early distinction.

A PCR plateau is the late phase in which further cycles add little product. Depletion of substrates and accumulation of reaction by-products can reduce PCR efficiency. Consequently, similar final amplicon amounts need not mean similar initial template amounts. In the stipulated green model, $0<N_n<K$ implies $N_n<N_{n+1}<K$, and growth approaches the imposed ceiling. That illustrates loss of starting-amount information without claiming to identify the limiting chemical process in an actual tube.^[ref]