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本帖最后由 coolyfoot123 于 2009-12-7 20:07 编辑
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, n. `" R8 \, k: d. N. S4 [我今天遇到一个问题，请问你知道如何把来自两种不同的load1和load2在一个机器上加工，合成一个load,该如何使用order命令？我每次运行的时候，给出的报告里面资源机器Massm没有load经过？以下是我的程序：
) X+ l9 u0 I1 ]begin P_in arriving
% E" [( P+ ? t3 V move into Q_in0 _3 u( M; t: \4 a
use oper for 2 min
1 ~# y! l' ]+ h$ K send to oneof(0.95:P_in1,0.05:die)
) D P( a* S! Y" Lend9 }% A2 d" q. i. i7 N
begin P_in1 arriving$ `7 ^1 K4 Y' K8 ?8 r g, m
if load type = load1 then6 g1 [* s/ a1 L$ ]) d
begin
9 B: Q: ^5 L+ G9 e/ @( ^4 J' g move into Q_m1+ t& x" e+ }: b* j
use M1 for u 5,1 min
" H# d+ G: {/ M) G W move into Q_out1
# J6 y+ C# z4 y: Y( Z' I wait to be ordered on orderlist OL_1! c3 \" i* o; ?! U. p* Q1 G8 g2 T
end
' m4 W) u, m# w5 |1 p if load type = load2 then
" M: [9 }* ]! ` t begin2 B3 W- H6 K1 s. C) g
move into Q_m27 q$ J" T+ h. L7 _ b& g5 a
use M2 for n 8,1 min( m% _+ j& e# b: J7 c. M
move into Q_out2 T- b1 l+ Y/ R
wait to be ordered on orderlist OL_2
- x! Z* f+ D3 W+ `: `5 a end
# ]; `1 {8 W- o/ b# Aend$ \9 ]; `2 ?/ J' T" c
begin P_dunmy arriving; I" y' d( r- A
set i = 16 y, U: \4 b3 A/ \/ U7 v
while i > 0 do
6 M: N$ N6 T+ h* l7 G P! b begin
0 ]# V) d, E1 d. A3 y order 1 loads from OL_1 to die+ }: G m+ {/ q* e
in case order not filled backorder on OL_1! S5 J y/ U) T
order 1 loads from OL_2 to continue/ Z' k. k2 Q# O, I! U- Q0 W/ a
in case order not filled backorder on OL_2
; `, }/ k0 G% m- B- q move into Q_13 a3 R; Y" j" a0 U
use Massm for n 5,1 min( W7 \7 |3 S) `1 s( A: R- L6 c
send to die
! s @" j# N6 |0 u9 w- j end
* y8 h# A3 Q+ ]' j0 ^% R6 ]+ Zend
3 N! s | j* w/ ~( {9 v* l# p这是我的报告$ z0 e) j% v W4 ^# Z! _' c* n9 j
*** AutoMod 11.1 ***! q* T* Y4 h+ l8 ?2 s& W
Model q
) D1 F4 G7 ` h3 q) v: |* A# sStatistics at Absolute Clock = 1:00:00:00.00, Relative Clock = 1:00:00:00.00$ o! Z, I% {, y7 e+ C. l' V
CPU time: Absolute: 42.859 sec, Relative: 42.859 sec, g9 z! F( G* W% b) y
Statistics for Process System "q"
* z/ m; R7 {6 T6 ]Process Statistics
+ K2 \$ z4 Q: h8 jName Total Cur Average Capacity Max Min Util Av_Time Av_Wait
) C( n3 B( {6 g& j2 b' |- O+ J========================================================================================0 v. T; T; f; s6 `" R/ Z* S, m
P_in 222 1 0.32 -- 2 0 -- 126.04 --
% S8 U' P0 O" g. n! I( JP_in1 212 210 102.38 -- 210 0 -- 41725.92 --
. p2 s8 A3 A0 C4 d& x! z. RP_in3 0 0 0.00 -- 0 0 -- 0.00 --+ \9 n# k$ [& q$ k
P_in4 0 0 0.00 -- 0 0 -- 0.00 --# \4 D! U% @' i6 A7 r+ r
P_dunmy 1 0 0.00 -- 1 0 -- 277.31 --6 o, o0 `' y, ]% E' S
Process Traffic Limit Statistics
& P6 T' s+ j" O( h1 DName Total Cur Average Capacity Max Min Util Av_Time Av_Wait5 `( F) z9 f+ y: E
========================================================================================
# n+ k5 y% Q, G" ZP_in 222 1 0.32 Infinite 2 0 -- 126.04 --9 l E K m6 \' K
P_in1 212 210 102.38 Infinite 210 0 -- 41725.92 --
7 x! z: j' S) H7 N: F8 }# i7 BP_in3 0 0 0.00 Infinite 0 0 -- 0.00 --5 Z/ ~( g3 b, W
P_in4 0 0 0.00 Infinite 0 0 -- 0.00 -- Y; j5 s% K% T
P_dunmy 1 0 0.00 Infinite 1 0 -- 277.31 --* s3 K3 w- b# s; o
Queue Statistics
5 Q0 ^4 b; i) G* [Name Total Cur Average Capacity Max Min Util Av_Time Av_Wait7 q( v/ C8 s! e! t
========================================================================================+ T+ t! _& t$ v" _6 g+ \. m5 b
Space 223 0 0.00 Infinite 1 0 -- 0.00 --. x+ S) v6 A* `9 p
Q_in 222 1 0.32 Infinite 2 0 -- 126.04 --9 R/ h6 M. |* X5 k* W( [) y
Q_m1 140 0 0.49 Infinite 2 0 -- 299.88 --( n6 p; c' R* l" U8 k9 r, W! b7 r
Q_m2 72 0 0.40 Infinite 2 0 -- 485.61 --! W R7 y0 G1 \$ C
Q_out1 140 139 67.23 Infinite 139 0 -- 41490.93 --
% `$ N" x0 \' A, f' g* \! IQ_out2 72 71 34.26 Infinite 71 0 -- 41114.13 --
! Q# h$ [% Z, {7 R$ B" D* ?' UQ_1 1 0 0.00 1 1 0 0.003 277.31 0.00) A1 f/ F/ q3 m( c% u
Order List Statistics
6 ~! O$ }3 [$ m1 `3 jName Total Cur Average Max Min Av_Time Tot_Back_Ordered, C, A. J% h' v/ z0 \
================================================================================0 Q" B. N# j( J. ^/ C7 I1 o3 U
OL_1 139 139 67.23 139 0 41789.43 1
+ W- F3 O# x. K; ]" u+ N5 e5 TOL_2 71 71 34.26 71 0 41693.20 1, e( |* L8 `, n3 a* ~
Resource Statistics# g1 S* o; {2 N% l( k5 v o( e% Z/ n6 x. f
Name Total Cur Average Capacity Max Min Util Av_Time Av_Wait State
* s* \$ u3 h1 a5 \1 }0 M====================================================================================================, J2 d5 e: ~6 ~ b" y- m$ E5 R
M1 140 0 0.49 1 1 0 0.486 299.67 0.21 Up ---- $ ?$ {0 W; K1 }- s, n
M2 72 0 0.40 1 1 0 0.404 484.99 0.63 Up ---- $ b3 N. Y1 | }0 w: k
Massm 1 0 0.00 1 1 0 0.003 277.31 0.00 Up ----
8 V( o" ?9 B6 Z. Loper 222 1 0.31 1 1 0 0.307 119.62 6.42 Up ----
" a: r3 H/ {( }* V; d/ F' WRandom Number Streams
" X& F$ R+ x/ ?4 C: o# HName Total
! Q9 k) V# i0 R" U# j n# L============================
3 k6 U( c1 P, c0 _3 ?stream0 455
8 O0 F9 s0 _( _( \4 E" h Lstream_M1_1 06 z* |/ H- y& h; j7 R
stream_M2_1 0, |+ R, e' Q+ n: [
stream_Massm_1 02 I' [$ N w0 \9 @
stream_oper_1 0
- K; I& O3 R0 G# @* O' \stream_load1_1 1843 _ {1 u# j5 R Q
stream_load2_1 76' S! b6 ]$ F2 f$ X- ^* E
stream_dummy_1 03 F, C6 o- Y5 R k% k* h. n; b
4 N/ _' X% V, M请问我的错误在那里？我用的版本是11.1的学生版 |
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发表于 2009-12-7 19:14:35
