RAND(<array>) / <,stream> /);This is in reality a procedure. It is used to generate random variates taken from multivariate random functions. The n-dimensional random functions are given by an n-dimensional <array> whose values are an n-dimensional frequency table. It must be declared of FREQ type and initialized in the INIT section.
Example 1:
Test RAND procedure for multivariate random values. This example generates random numbers from a distribution given by a three variable frequency table given by an array Mul of three indexes. The multivariant frequency table is: 15 18 15 18 ------ ------ 2.0 | 1 3 2.0 | 3 2 0.1 5.0 | 4 3 0.4 5.0 | 2 1 10.0 | 3 1 10.0 | 3 6 So, Mul[0.1,5.0,15]=4 is the number of cases in which the values of the variables were (0.1, 5.0, 15); Mul[0.4,10.0,18]=6 The sum of all values is 32. The first matrix sum up 15, the second one, 17. The RAND algorithm selects one vector, for example (0.1, 5.0, 15) with a probability according to the frequency table. Each value is selected according to its marginal probability distribution conditional to the previous values. It is decided between 0.1 and 0.4 according its marginal probabilities: 15/32 for 0.1 and 17/32 for 0.4. If the selected is, for instance 0.1, the second value is selected according to the marginal probabilities: (3+1)/17 for the value 2.0,(4+3)/17 for the value 5.0, (3+1)/17 for the value 10.0. If the selected is, for instance 5.0, the third value is selected according to the probabilities 4/7 for 15.0 and 3/7 for 18.0. See that the probability of the value (0.1, 5.0, 15) is then 17/32*7/17*4/7=4/32, as may be see directly. The following program computes 32000 random numbers with the frequencies in the array Mul, counts the different results and put them in a table Zzz similar to Mul. The values in Zzz must be approximately those in Mul. NETWORK A:: IT := 1; RAND(Mul,2); IF RANDV_Mul[1] = 0.1 THEN I := 1 ELSE I := 2; IF RANDV_Mul[2] = 2.0 THEN J := 1 ELSE IF RANDV_Mul[2] = 5.0 THEN J := 2 ELSE J := 3; IF RANDV_Mul¤[3] = 15.0 THEN K := 1 ELSE K := 2; Zzz[I, J, K] := Zzz[I, J, K] + 1; Result (A):: WRITELN; FOR I := 1 TO 2 DO BEGIN WRITELN; FOR J := 1 TO 3 DO BEGIN WRITELN; FOR K := 1 TO 2 DO WRITE(Zzz[I, J, K] / 1000: 7: 4); END END; PAUSE; ENDSIMUL; INIT ACT(A, 0); ASSI Mul[1..2, 1..3, 1..2] := ( ( ( 1, 3 ),( 4, 3 ),( 3, 1 ) ) , ( ( 3, 2 ),( 2, 1 ),( 3, 6 ) ) ) ; ASSI VAL_Mul[1..3, 1..3]:=( (0.1, 0.4, 0.0), (2, 5, 10), (15, 18, 0) ) ; ASSI Zzz[1..2, 1..3, 1..2] := ( ( ( 0, 0 ), ( 0, 0 ), ( 0, 0 ) ) , ( ( 0, 0 ), ( 0, 0 ), ( 0, 0 ) ) ) ; ACT(Result, 32000); DECL VAR Mul: ARRAY[1..2, 1..3, 1..2] OF FREQ; Zzz: ARRAY[1..2, 1..3, 1..2] OF REAL; VAL_Mul: ARRAY[1..3, 1..3] OF REAL; I, J, K: INTEGER; END.Example 2:
Logs are processed by a set of machines. The processing time
was adjusted by regression to a linear function of the Length L
and the Diameter D of the logs. Logs arrive each 10 m. with random
sizes. A frequency table aggregated in 3 diameters (32, 36, 40
inches) and 4 lengths (15, 25, 35, 45 feet), obtained by sampling,
gives the quantity in each class.
NETWORK
Arrivals (I) :: IT := 10; RAND(Fld);
L := RANDV_Fld[1]; D := RANDV_Fld[2];
Machines (R) :: STAY := 8.5 + 0.32 * L + 0.21 * D;
Dep (E) ::
INIT
ACT(Arrivals, 0); TSIM := 2000; Machines := 3;
(*Multivariate frequency array D: rows L: columns: *)
ASSI Fld[1..3, 1..4] :=
{ Length feet }
{ 15 25 35 45 }
{32} ( ( 123, 108, 47, 12 ),
{Diameter inches} {36} ( 89, 104, 99, 43 ),
{40} ( 22, 87, 106, 172) ) ;
(*values of the variables (columns) : *)
ASSI VAL_Fld[1..2, 1..4]:=
( (32, 36, 40, 0), (15, 25, 35, 45)) ;
DECL VAR_Fld: ARRAY[1..3, 1..4] OF FREQ;
VAL_Fld: ARRAY[1..2, 1..4] OF REAL;
MESSAGES Arrivals(L, D: REAL);
STATISTICS ALLNODES;
END.