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| COMPARTMENT INITIALIZATION BLOCK |
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MEANING: Abbreviated code for compartment initialization
CONTEXT: $PK abbreviated code
SAMPLE:
$PK
IF (A_0FLG.EQ.1) THEN
... compartment initialization block ...
ENDIF
DISCUSSION:
A "compartment initialization block" is a block of abbreviated code
that sets the initial state of the kinetic system. It is to be exe-
cuted only when A_0FLG=1. Such a block may be present only in $PK.
PREDPP sets A_0FLG to 1 at a call to PK with the first event record of
an individual record (if the data are population data), with the first
event record of the data set (if the data are single-subject data),
and with a reset record. The flag A_0FLAG may be found in NONMEM-
PREDPP common PROCMC.
Special rules apply to such a block.
1) Values may be assigned to reserved variables A_0(n), but only in
a compartment initialization block. The value of the amount in
the nth compartment (the nth element of the state vector) is set
to the value assigned to A_0(n). If PK is called with a dose
record or a dose-reset record where the dose is input into the
nth compartment, this amount is then increased by the amount of
the (bioavailable) dose. If a value is assigned to A_0(n), then
it is not necessary that values be assigned to any of the remain-
ing variables A_0(m). A value to the output compartment cannot
be assigned. A_INITIAL(n) is a synonym for A_0(n).
2) The statement "IF (A_0FLG.EQ.1)" and the corresponding "ENDIF"
statement may be included explicitly in abbreviated code, thus
defining an explicit compartment initialization block. (See
example 1 below.)
3) A_0(n) may be assigned a value with an unconditional statement.
This defines an implicit compartment initialization block; NMTRAN
inserts "IF (A_0FLG.EQ.1) ..." before the statement and "ENDIF"
after it. (See example 2 below.)
4) An IF statement testing ICALL and A_0FLG together is not permit-
ted. Instead, two separate nested IF statements must be used:
an IF statement testing ICALL must occur as the outermost state-
ment, and an IF statement testing A_0FLG must occur as the inner-
most statement. The latter may be supplied by NM-TRAN as a
result of using an implicit initialization block. (See example 4
below.)
5) Within an explicit compartment initialization block, A_0(n) may
be assigned conditionally. The usual rules apply if A_0(n) is a
random variable. E.g., A_0(n) cannot be assigned within a nested
IF, and it defaults to 0 if it is assigned conditionally but
incompletely. However, in checking for a nested IF, tests of
A_0FLG and of ICALL are ignored. (See example 3 below.)
6) User-defined variables may be defined in compartment initializa-
tion blocks, but not reserved variables such as basic or addi-
tional PK parameters.
EXAMPLES OF USAGE:
The following two fragments of code yield identical results:
(1) Explicit compartment initialization block:
IF (A_0FLG.EQ.1) THEN
A_0(1)=THETA(1)*(1+ETA(1))
ENDIF
(2) Implicit compartment initialization block:
A_0(1)=THETA(1)*(1+ETA(1))
(3) This is an example of conditional assignment of A_0(n) (X is a
data item or user-defined variable):
IF (A_0FLG.EQ.1) THEN
IF (X.EQ.1) THEN
A_0(1)=THETA(1)*(1+ETA(1))
ELSE
A_0(1)=THETA(2)*(1+ETA(2))
ENDIF
ENDIF
(4) Suppose compartment initialization should occur only during the
Simulation step. The following is not permitted:
IF (ICALL.EQ.4.AND.A_0FLG.EQ.1) A_0(1)=THETA(1)*(1+ETA(1))
Instead, use:
IF (ICALL.EQ.4) THEN
IF (A_0FLG.EQ.1) A_0(1)=THETA(1)*(1+ETA(1))
ENDIF
or simply
IF (ICALL.EQ.4) THEN
A_0(1)=THETA(1)*(1+ETA(1))
ENDIF
or even more simply
IF (ICALL.EQ.4) A_0(1)=THETA(1)*(1+ETA(1))
(See prdpk3, procmc, pk).
REFERENCES: None.
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