JMS Performance

Question

          Hi All,
          i read the BEA WebLogicTM JMS Performance Guide (July 31, 2003) and have some
          questioon:
          http://dev2dev.bea.com/products/wlserver/whitepapers/WL_JMS_Perform_GD.jsp
          The only thing that indicates performance in the whole document is page 60 with
          the 3 examples.
          the problem there is that the arithmetic just doesn't work.
          All examples use a 100 Mbit link at 80% efficiency, which gives 10 mbyte/sec.
          Example 1: Messages of length 1Mbyte, 2 operations: (Calculating in IN Kbytes)
          10000/(1000*2) = 5. Bravo !!!
          Example 2: Message length 50kbyte, 10 operations : 10000/(50*10) = 20. (Am i wrong
          ? not 160 as the document says). Out of each 10 operations, 1 is write (send)
          and 9 are receive (read) , so each subscriber gets 2 messages per second, and
          not 18 as the document says.
          Example 3: Message size 50kByte, 2 Operations (because of the Multicast). 10000/(50*2)
          = 100 , and not 800. each subscriber gets 50 per second, and not 89.
          The arithmetic is simple, you can do it on your fingers - no need for excel.
          Can you, Please
          1. Explain the calculations
          2. Supply real life, Measured statistics that DO take into consideration server
          power, Memory constraints and other trivial, but necessary, facts of life...
          Thanks yuval
          

Answer 1

Well, I must admit I was the primary author, so you will have
          trouble getting additional info. But, heck, I'm game:
          (1) We have literally had dozens of customers not
          realize that the network can be a bottleneck, thus the
          network calculations. Congrats on being one of the majority
          that didn't need this bit of guidance, but please don't look
          down on those that do - they are pretty bright people in
          their own right.
          (2) Memory usage is a function of number of messages,
          usage of queues or topics, and message size. Assuming
          paging is enabled, I think it is fair to be conservative
          and estimate for queues:
          number of messages * .5K = memory used
          This assumes .5K overhead per message - as the guide
          states, WL JMS does not page out message-headers, so
          messages have overhead even when they are paged out.
          (3) This is a performance guide, not a capacity guide, the
          guide was not intended to address capacity.
          As the paper states, there is a reason for this.
          Hard performance statistics are VERY application and
          hardware dependent. To get anything even close to the statistics
          that map to a particular app, run the app. That said, the performance
          book referenced in the guide DOES contain statistics, pretty
          graphs, and even a tool. Keep in mind that the first edition of the
          book never tests using multiple producers - which limits
          scalability, and kind of limits the applicability of some
          of the graphs - but the authors realize this, and are addressing
          the issue in the next addition.
          Tom
          Yuval Avni wrote:
          > Hi All,
          >
          > i read the BEA WebLogicTM JMS Performance Guide (July 31, 2003) and have some
          > questioon:
          > http://dev2dev.bea.com/products/wlserver/whitepapers/WL_JMS_Perform_GD.jsp
          > The only thing that indicates performance in the whole document is page 60 with
          > the 3 examples.
          > the problem there is that the arithmetic just doesn't work.
          >
          > All examples use a 100 Mbit link at 80% efficiency, which gives 10 mbyte/sec.
          >
          > Example 1: Messages of length 1Mbyte, 2 operations: (Calculating in IN Kbytes)
          > 10000/(1000*2) = 5. Bravo !!!
          >
          > Example 2: Message length 50kbyte, 10 operations : 10000/(50*10) = 20. (Am i wrong
          > ? not 160 as the document says). Out of each 10 operations, 1 is write (send)
          > and 9 are receive (read) , so each subscriber gets 2 messages per second, and
          > not 18 as the document says.
          >
          > Example 3: Message size 50kByte, 2 Operations (because of the Multicast). 10000/(50*2)
          > = 100 , and not 800. each subscriber gets 50 per second, and not 89.
          >
          > The arithmetic is simple, you can do it on your fingers - no need for excel.
          >
          > Can you, Please
          >
          > 1. Explain the calculations
          > 2. Supply real life, Measured statistics that DO take into consideration server
          > power, Memory constraints and other trivial, but necessary, facts of life...
          >
          > Thanks yuval