High Performance Computing MCQs with answers Page - 27

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Q. A CUDA program is comprised of two primary components: a host and a _____.

  • (A) gpu kernel
  • (B) cpu kernel
  • (C) os
  • (D) none of above

A

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Q. The kernel code is dentified by the ________qualifier with void return type

  • (A) _host_
  • (B) __global__
  • (C) _device_
  • (D) void

A

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Q. In All-to-All Personalized Communication on a Ring, the size of the message reduces by at each step

  • (A) p
  • (B) m-1
  • (C) p-1
  • (D) m

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Q. All-to-All Broadcast and Reduction algorithm on a Ring terminates in steps.

  • (A) p+1
  • (B) p-1
  • (C) p*p
  • (D) p

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Q. In All-to-all Broadcast on a Mesh, operation performs in which sequence?

  • (A) rowwise, columnwise
  • (B) columnwise, rowwise
  • (C) columnwise, columnwise
  • (D) rowwise, rowwise

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Q. Messages get smaller in and stay constant in .

  • (A) gather, broadcast
  • (B) scatter , broadcast
  • (C) scatter, gather
  • (D) broadcast, gather

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Q. The time taken by all-to- all broadcast on a ring is .

  • (A) t= (ts + twm)(p-1)
  • (B) t= ts logp + twm(p-1)
  • (C) t= 2ts(√p – 1) - twm(p-1)
  • (D) t= 2ts(√p – 1) + twm(p-1)

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Q. The time taken by all-to- all broadcast on a mesh is .

  • (A) t= (ts + twm)(p-1)
  • (B) t= ts logp + twm(p-1)
  • (C) t= 2ts(√p – 1) - twm(p-1)
  • (D) t= 2ts(√p – 1) + twm(p-1)

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Q. The time taken by all-to- all broadcast on a hypercube is .

  • (A) t= (ts + twm)(p-1)
  • (B) t= ts logp + twm(p-1)
  • (C) t= 2ts(√p – 1) - twm(p-1)
  • (D) t= 2ts(√p – 1) + twm(p-1)

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Q. The prefix-sum operation can be implemented using the kernel

  • (A) all-to-all broadcast
  • (B) one-to-all broadcast
  • (C) all-to-one broadcast
  • (D) all-to-all reduction

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