Accounting on Snellius

Last updated on 2026-10-27 | Edit this page

Overview

Questions

  • What is a system billing unit, and which row of the table sets the price?
  • How many SBU does 16 cores for 2 hours cost on rome?
  • What does the course cluster charge instead?

Objectives

  • Read the Snellius system-details table: nodes, cores, memory, SBU per hour, smallest allocation.
  • Compute an SBU cost from cores, hours, and the SBU rate of the partition.
  • Say what a full CPU allocation does to idle GPUs, and what a one-core request is rounded up to.

The table you are billed from

Snellius charges system billing units (SBU). The accounting portal is https://portal.cua.surf.nl/.

A rome node is the row most jobs in this course start from: 521 nodes, 128 cores, 256 GB, AMD Rome 7H12 at 2.6 GHz, 128 SBU per node-hour. The smallest allocation is 1/8 of a node, which is 16 cores.

The same table has more partitions. GPU counts are per node. gpu_a100 is 4 NVIDIA A100 40 GB. gpu_h100 is 4 NVIDIA H100 (SXM5) 94 GB.

Partition Nodes Cores Memory SBU/hour Smallest
rome 521 128 256 GB 128 1/8
fat_rome 72 128 1,024 GB 192 1/8
himem_4tb 2 128 4,096 GB 256 1/8
himem_8tb 2 128 8,192 GB 384 1/8
genoa 737 192 384 GB 192 1/8
fat_genoa 48 192 1,536 GB 288 1/8
gpu_a100 63 72 512 GB 512 1/4
gpu_h100 88 64 768 GB 768 1/4
staging 10 16 256 GB 32 1/32

Login nodes are a separate row of 3 machines. Do not run a heavy task there. Their hardware is not one of the rows above.

The whole machine in the deck’s summary line is 1,258 thin rome/genoa nodes, plus the fat and high-memory nodes, plus 604 GPUs. Home is 720 TB. Scratch and project together are 12.4 PB, on GPFS.

The formula

Documentation talks about cores. The charge uses the partition’s SBU per node-hour and the number of cores on that node:

cores x hours x (SBU per node-hour / cores per node)

16 cores for 2 hours on rome:

16 x 2 x (128 / 128) = 32

16 cores is 1/8 of 128, so it is the smallest rome job. At 128 SBU per node-hour that slice is 16 SBU per hour. Two hours is 32 SBU.

A full rome node for one hour is 128 cores, which is 128 SBU. Asking for fewer cores than the smallest unit does not make a cheaper job. The system rounds the request up to that unit, and the job gets every core in the unit.

The tightest resource wins

Each node has cores, memory, accelerators, and local disk. The allocation follows the most restrictive thing you asked for.

If you request every CPU core and no GPU, the node is fully allocated. The GPUs sit idle, and nobody else can use them, because the CPUs are taken.

If you request one core, the system rounds up to the smallest permitted unit. On rome that unit is 1/8 of the node. You then have every core in that unit, not the single core you typed.

Read the tab for the machine in the welcome note. The Snellius tab is the SBU table above. The course-cluster tab has no project budget.

Challenge

Cost a rome job

A job asks for 16 cores on rome and runs for 2 hours. How many SBU is that? Why is 16 the interesting number, rather than 8?

32 SBU.

16 x 2 x (128 / 128) = 32

16 cores is 1/8 of a 128-core rome node, and 1/8 is the smallest allocation on that row. A request for 8 cores is rounded up to that same unit, so it does not cost half as much.

Challenge

Cost a genoa slice

genoa is 192 cores and 192 SBU per node-hour. The smallest allocation is still 1/8 of a node. A job uses 24 cores for 1 hour. How many SBU?

24 SBU.

24 x 1 x (192 / 192) = 24

24 cores is 192/8, so this is one smallest unit for one hour. One full genoa node-hour is 192 SBU.