6100 ADCCP Programming Manual
Station and Data Capacity on Links
ADCCP Link and Station Management
069225 Tandem Computers Incorporated 2–17
The SYSGEN MAXFRAME parameter determines the size of each input and output
buffer. The size of a buffer is calculated as follows:
size = MAXFRAME bytes (for the information field)
+ 4 bytes (for the address field)
+ 2 bytes (for the control field)
+ 12 bytes (for ADCCP internal information)
The address length and control field length that you define at system generation do
not affect the size of the buffer that the ADCCP protocol module allocates. The size of
an actual outgoing or incoming frame does not affect the size of the buffer either. The
protocol module always allocates space for the largest possible buffer.
The WINDOW and STATIONS parameters determine the total space allowed for
input/output buffers. Each station can have as many output buffers as the window
size allows. All stations together can have one window of buffers. Thus, the total
space allowed for buffers is calculated as follows:
(size x STATIONS x WINDOW) + (size x WINDOW)
! !
! !
output buffers input buffers
Of this space allowed for buffers, ADCCP permanently allocates only a small portion
to the buffers as follows:
size x (WINDOW + 3)
The rest of the space is allocated in response to application traffic.
Note The criteria of size x (WINDOW + 3) must be met; otherwise, the line will not start.
The STATIONS parameter (defined at system generation or with a SET
CONFIGURATION request) determines the maximum space used for station control
blocks. The largest station list is calculated as follows:
STATIONS x 26 bytes
ADCCP allocates two trace buffers while a trace is in progress , each large enough to
contain a trace block (as defined in the CMI TRACE command).
Without proper planning, you could possibly to set up a link whose memory
requirements vastly exceed the memory available on the LIU. For example, if you
define 255 stations, a window size of 127, and a maximum I-field size of 2046, the
memory requirement would (excluding possible trace buffers) exceed the total amount
of memory available on the LIU, which is 16K words. The calculation for this link
would be:
((2046 + 4 + 2 + 12) x 255 x 127) + 127(2046 + 4 + 2 + 12)
= 67104768 bytes
= 65532K bytes
= 32766K words