For purposes of licensing its software running on Linux, IBM assigns a metric called "PVUs" (Processor Value Units) to each type of processor core. Most IBM software products for Linux are licensed according to the number of PVUs at a price per PVU. Likewise, annual software subscription and support renewals for those products are based on the same metric.
So here's an example. Let's say you're going to run WebSphere Portal on Linux. One zEC12 IFL has the same PVU metric as one Intel Xeon Nehalem EX (X86) core, provided the latter is part of a server with 5 or more processor sockets. They both are rated at 120 PVUs in that case. So you can purchase WebSphere Portal licenses for, say, 6 IFLs (720 PVUs), and the same licensing will be exactly what you need to run the same IBM software on the same number of Nehalem EX cores (with that caveat above). Sometimes core vary in their PVUs. Much less powerful X86 cores, such as older pre-Nehalem cores, are rated at 50 PVUs per core, for example. In that case 1 zEC12 IFL would require as much software licensing as 2.4 X86 cores. Well, you can't actually license 2.4 cores -- it has to be a whole number. So 1 zEC12 IFL (120 PVUs) would require software licensing that's in between the licensing for 2 (100 PVUs) and 3 (150 PVUs) older X86 cores -- closer to 2 than 3. The biggest possible gap in IBM's current PVU table between zEC12 and something else is with Oracle/Sun UltraSPARC T1. Those SPARC cores are...well, let's just say they're not very good. So IBM rates them at 30 PVUs each. Oracle uses "core factors," and it's very similar conceptually at first glance. An IFL has an Oracle core factor of 1, and (usually) an X86 core has a core factor of 0.5, to pick a couple examples. So 2 X86 cores has the same Oracle licensing as 1 IFL.... ....Well, not exactly. I don't speak for Oracle (or for IBM), but my understanding is that Oracle requires that you round up per X86 socket, and Oracle isn't so disposed to sub-capacity licensing as IBM is. IFLs are simple: count the number of IFLs running Oracle software, and that's that -- no rounding required since the core factor is 1. Capacity Backup (CBU) IFLs don't count. But for other core types, such as X86, it could get "weird." Let's use this slightly artificial (but illustrative) configuration as an example: Server #1: 2 sockets, 5 X86 cores each. One chip (5 cores) runs IBM WebSphere Portal, and the other runs Oracle Database. Server #2: DR server, 2 sockets, 3 X86 cores each. One chip would run IBM WebSphere Portal in a disaster, the other Oracle Database -- at diminished capacity/performance because there are fewer cores. The server is installed but on "cold standby." OK, first let's consider IBM software licensing. If the servers are running a valid virtualization product (according to IBM's list), and if you're following the other sub-capacity licensing rules (which aren't difficult to follow), then you would need 5 cores of WebSphere Portal. IBM does not require software licensing for DR servers if "cold" -- and in certain other cases. A 2 socket multi-core Intel X86 server typically has 70 PVUs per core, so you'd need to license 5 times 70 = 350 PVUs of WebSphere Portal. Now let's turn to Oracle. Oracle uses a core multiplier of 0.5 for this type of server, but that's applied per socket. And sub-capacity licensing is at least harder to get. Oracle does not typically waive licensing for DR servers either -- if it's installed and ready, even powered off, it still needs a license. So in this case the math would be: 5 * 0.5 = 2.5 rounded up to 3 3 * 2 = 6 full licenses required for Server #1 3 * 0.5 = 1.5 rounded up to 2 2 * 2 = 4 full licenses required for Server #2 6 + 4 = 10 full licenses required for both servers Oracle licensing = 10 * license price = Total License Amount Oracle maintenance = 10 * maintenance price = Total Maintenance Amount Anyway, I think you can see the problem. In this case you'd need 10 full licenses (at core factor 1) when the Oracle software is actually only running productively on 5 X86 cores -- and undoubtedly not at 90% average utilization. So you've got an effective Oracle core factor of 2 in this example instead of 0.5. -------------------------------------------------------------------------------------------------------- Timothy Sipples Consulting Enterprise IT Architect (Based in Singapore) E-Mail: sipp...@sg.ibm.com ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to lists...@listserv.ua.edu with the message: INFO IBM-MAIN
