0
$\begingroup$

I have run a batch of a few thousand random 2-qubit QuantumCircuits on multiple different IBMQ Backends, and the execution times seem to significantly vary across the backends (obviously excluding time in queue). The general trend seems to be that increased:

  1. Qubits or
  2. Quantum Volume(QV)

results in faster execution. As per my knowledge, this could be explained either by

  1. increased parallel execution of the batch with more qubits or
  2. systems with higher QV operate at proportionately higher clock speeds (I am more inclined to this explanation as the fastest job executed on ibmq_montreal despite having fewer qubits than ibmq_rochester)

Are any of these explanations correct or is there an entirely different explanation for this observation?

I have used the following backends:

  • ibmq_montreal - 27 qubits with QV 32
  • ibmq_vigo - 5 qubits with QV 16
  • ibmq_rochester - 53 qubits QV 8
$\endgroup$

1 Answer 1

0
$\begingroup$

With a higher Quantum Volume, your circuits will run faster, partly because a backend with a higher quantum volume can run complex circuits with a greater width and depth than backends with lower quantum volume. However, the number of qubits on a backend does not affect the speed of a circuit, because each logical qubit in your circuit is mapped to one physical qubit on the backend, so having more physical qubits than needed does not affect your running time, since they are not involved in the circuit. There are a lot of factors that go into determining the quantum volume of a backend, so if you are interested I would read more about it at the following links: https://qiskit.org/textbook/ch-quantum-hardware/measuring-quantum-volume.html and https://qiskit.org/documentation/tutorials/noise/5_quantum_volume.html

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.