Resource Estimates for Simulation Workloads

A resource estimate is a conditional calculation: given an algorithm, accuracy target, architecture, and error model, what memory, operations, qubits, and time are required? A count without these assumptions is not reproducible. This page explains stable accounting methods, not forecasts for a particular device.

Accuracy defines the workload

Begin with the physical output and its tolerance. Split error into contributions such as

$$\epsilon_{\rm total} \lesssim\epsilon_{\rm model}+\epsilon_{\rm discretization} +\epsilon_{\rm algorithm}+\epsilon_{\rm synthesis} +\epsilon_{\rm sampling}.$$

This schematic sum is justified only when each term controls a compatible error measure. Do not add a trace distance, energy error, and standard deviation without converting them to the requested observable.

For a bounded observable with variance $\sigma^2$, independent sampling to standard error $\epsilon$ requires approximately $\sigma^2/\epsilon^2$ samples. Tightening tolerance tenfold can therefore increase sampling cost one hundredfold, independently of gate optimization.

Classical memory and operation counts

A dense statevector uses $b\,2^n$ bytes; a dense density matrix uses $b\,4^n$, with $b=16$ for complex double precision. Include work vectors, basis tables, communication buffers, and temporary contraction copies in peak memory.

An iterative eigensolver with $m$ stored Krylov vectors may need roughly $bmD$ bytes for those vectors alone. A sparse Hamiltonian’s memory depends on nonzero count and integer-index widths. “Matrix-free” removes matrix storage but not the state or Krylov basis.

For a concrete budget, suppose 32 GiB is available, reserve 8 GiB for the operating environment, basis maps, and other allocations, and require three simultaneous complex-double state arrays. The array budget is $B=24\cdot2^{30}$ bytes, so

$$3(16)2^n\leq B \quad\Longrightarrow\quad n\leq29.$$

At 29 qubits the three vectors occupy exactly 24 GiB; at 30 they need 48 GiB. If the same three-array assumption is applied to dense density operators, $3(16)4^n\leq B$ gives only 14 qubits. These are conditional storage ceilings: an unbudgeted transpose copy or larger solver workspace lowers them. For a fixed-magnetization spin problem replace $2^n$ by the actual sector dimension, not by a fractional “effective qubit” count when allocating arrays.

Wall time is not operation count divided by a processor’s advertised peak. Memory bandwidth, communication, setup, and serial work can dominate. Calibrate a representative kernel and state the extrapolation range.

Logical and physical quantum resources

Record logical qubits, total gates by type, circuit depth, non-Clifford count, non-Clifford depth, measurements, and classical feedforward latency. Gate count and depth answer different scheduling questions.

For $N_{\rm loc}$ logical fault locations and total failure budget $\delta$, a union-bound allocation gives

$$p_{\rm logical}\lesssim\delta/N_{\rm loc}.$$

The union bound does not require independent faults, provided each location’s marginal failure probability is bounded by the assigned value. It can be conservative. Conversely, using a fitted per-round failure probability as though it were a per-gate location probability mixes two different counting models. Define what one location and one fitted error event mean before multiplying them.

Suppose a hypothetical fitted model is $p_{\rm logical}(d)=A(p/p_{\rm th})^{(d+1)/2}$ below threshold. It can be inverted to select a code distance $d$, but its constants and validity range must come from the chosen code, decoder, circuit, and physical noise model. This expression is not universal.

Physical qubits then include encoded data, syndrome ancillas, routing space, and state-distillation factories. Runtime depends on code cycles and whether factories supply non-Clifford resources fast enough. Increasing factories can reduce time while increasing qubits.

A transparent toy estimate

Assume $N_{\rm loc}=10^8$, $\delta=10^{-2}$, $A=0.1$, and $p/p_{\rm th}=0.1$. Then require $p_{\rm logical}\leq10^{-10}$ and

$$0.1\,10^{-(d+1)/2}\leq10^{-10},$$

so odd distance $d=17$ reaches the bound in this toy model. If $N_{\rm loc}$ grows by a factor 100, the same algebra gives $d=21$. This shows how a large workload change can increase distance modestly while physical area, proportional to $d^2$ in many layouts, still increases substantially.

The example deliberately does not turn distance into a device qubit count: that requires a specified layout and factory design.

Sensitivity and verification

Vary physical error rate, target failure probability, cycle time, synthesis precision, parallelism, and factory throughput. Report ranges when assumptions are uncertain. Check all units and distinguish decimal GB from binary GiB.

A useful estimate contains the formulas or executable accounting steps, not only final numbers. Validate the compilation and resource counts on a small instance, then separate measured scaling from extrapolated scaling. No numerical fit to a small range proves asymptotic advantage.

Exercise

An estimator’s statistical tolerance is halved while all deterministic errors stay fixed. What changes in the sample budget, and why might the total accuracy barely improve?

Solution

The independent sample count increases by a factor four at fixed variance. If discretization or model bias dominates, reducing statistical error has little effect on the total error. Resource allocation should first identify the largest controlled contribution.

References

• Beverland, M. E., et al., “Assessing requirements to scale to practical quantum advantage,” (2022), arXiv:2211.07629 — explicit architectural assumptions and resource accounting.
• Häner, T., and Steiger, D. S., “0.5 petabyte simulation of a 45-qubit quantum circuit,” Proceedings of SC17, article 33 (2017), arXiv:1704.01127 — classical memory and communication costs.