Математическое моделирование при планировании экспериментов на трех, четырех, пяти уровнях фактора и при неодинаковом количестве уровней первого и второго фактора. Черный А.А. - 39 стр.

1260 IF X = 15 GOTO 1360
1280 IF X = 20 GOTO 1380
1310 GOSUB 4150: GOTO 1410
1320 GOSUB 4210: GOTO 1410
1330 GOSUB 4290: GOTO 1410
1350 GOSUB 4150: GOSUB 4250: GOTO 1410
1360 GOSUB 4150: GOSUB 4340: GOTO 1410
1380 GOSUB 4210: GOSUB 4340: GOTO 1410
1410 S = 0: O(1) = 0: FOR J = 1 TO X: S = S + Y(J): O(1) = O(1) + 1: NEXT J
1420 B(1) = S / O(1): S = 0: O(2) = 0: FOR J = 1 TO X: S = S + I(J) * Y(J)
1430 O(2) = O(2) + I(J) ^ 2: NEXT J: B(2) = S / O(2): S = 0: O(3) = 0
1440 FOR J = 1 TO X: S = S + K(J) * Y(J): O(3) = O(3) + K(J) ^ 2: NEXT J
1450 B(3) = S / O(3): IF X = 3 GOTO 2390
1460 IF X = 4 GOTO 2000
1470 IF X = 5 GOTO 2000
1480 S = 0: O(4) = 0: FOR J = 1 TO X: S = S + P(J) * Y(J)
1490 O(4) = O(4) + P(J) ^ 2: NEXT J: B(4) = S / O(4): S = 0: O(5) = 0
1500 FOR J = 1 TO X: S = S + I(J) * P(J) * Y(J): O(5) = O(5) + (I(J) * P(J)) ^ 2
1510 NEXT J: B(5) = S / O(5): S = 0: O(6) = 0: FOR J = 1 TO X
1520 S = S + Q(J) * Y(J): O(6) = O(6) + Q(J) ^ 2: NEXT J: B(6) = S / O(6)
1530 S = 0: O(7) = 0: FOR J = 1 TO X: S = S + I(J) * Q(J) * Y(J)
1540 O(7) = O(7) + (I(J) * Q(J)) ^ 2: NEXT J: B(7) = S / O(7): S = 0
1550 O(8) = 0: FOR J = 1 TO X: S = S + P(J) * K(J) * Y(J)
1560 O(8) = O(8) + (P(J) * K(J)) ^ 2: NEXT J: B(8) = S / O(8): S = 0: O(9) = 0
1570 FOR J = 1 TO X: S = S + K(J) * Q(J) * Y(J): O(9) = O(9) + (K(J) * Q(J)) ^ 2
1580 NEXT J: B(9) = S / O(9)
1600 S = 0: O(10) = 0: FOR J = 1 TO X: S = S + U(J) * Y(J): O(10) = O(10) + U(J) ^ 2
1610 NEXT J: B(10) = S / O(10): S = 0: O(11) = 0: FOR J = 1 TO X
1620 S = S + I(J) * U(J) * Y(J): O(11) = O(11) + (I(J) * U(J)) ^ 2: NEXT J
1630 B(11) = S / O(11): S = 0: O(12) = 0: FOR J = 1 TO X: S = S + K(J) * U(J) * Y(J)
1640 O(12) = O(12) + (K(J) * U(J)) ^ 2: NEXT J: B(12) = S / O(12)
1650 IF X = 12 GOTO 2390
1670 S = 0: O(13) = 0: FOR J = 1 TO X: S = S + V(J) * Y(J)
1680 O(13) = O(13) + V(J) ^ 2: NEXT J: B(13) = S / O(13): S = 0: O(14) = 0
1690 FOR J = 1 TO X: S = S + I(J) * V(J) * Y(J): O(14) = O(14) + (I(J) * V(J)) ^ 2
1700 NEXT J: B(14) = S / O(14): S = 0: O(15) = 0: FOR J = 1 TO X
1710 S = S + K(J) * V(J) * Y(J): O(15) = O(15) + (K(J) * V(J)) ^ 2: NEXT J
1720 B(15) = S / O(15): IF X = 15 GOTO 2390
1730 S = 0: O(16) = 0: FOR J = 1 TO X: S = S + L(J) * Y(J): O(16) = O(16) + L(J) ^ 2
1740 NEXT J: B(16) = S / O(16): S = 0: O(17) = 0: FOR J = 1 TO X
1750 S = S + P(J) * L(J) * Y(J): O(17) = O(17) + (P(J) * L(J)) ^ 2: NEXT J
1760 B(17) = S / O(17): S = 0: O(18) = 0: FOR J = 1 TO X
1770 S = S + Q(J) * L(J) * Y(J): O(18) = O(18) + (Q(J) * L(J)) ^ 2: NEXT J
1780 B(18) = S / O(18): S = 0: O(19) = 0: FOR J = 1 TO X
1790 S = S + L(J) * U(J) * Y(J): O(19) = O(19) + (L(J) * U(J)) ^ 2: NEXT J
1800 B(19) = S / O(19): S = 0: O(20) = 0: FOR J = 1 TO X
1810 S = S + L(J) * V(J) * Y(J): O(20) = O(20) + (L(J) * V(J)) ^ 2: NEXT J
1820 B(20) = S / O(20): IF X = 20 GOTO 2390
2000 S = 0: O(4) = 0: FOR J = 1 TO X: S = S + L(J) * Y(J): O(4) = O(4) + L(J) ^ 2
2010 NEXT J: B(4) = S / O(4): IF X = 4 GOTO 2390
2020 S = 0: O(5) = 0: FOR J = 1 TO X: S = S + M(J) * Y(J)
2030 O(5) = O(5) + M(J) ^ 2: NEXT J: B(5) = S / O(5): GOTO 2390
2390 PRINT #1, "B(J) ДО АНАЛИЗА": FOR J = 1 TO X
2395 PRINT #1, "B("; J; ")="; B(J)
2397 NEXT J: PRINT
2400 PRINT "ВВОД N0-КОЛИЧЕСТВО ОПЫТОВ НА СРЕДНЕМ УРОВНЕ ФАКТОРОВ"
2407 INPUT N0




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