本帖最后由 GJM 于 2009-12-5 21:43 编辑 % f" N6 s/ B& i: T7 u! |$ x/ n
- L- ?9 F" f* s8 O! t) O3 P底下是小弟做AutoMOD里面PDF练习的(Exercise 5.9)逻辑文件但问题是,程序只Run到Machine A和Machine B就没继续下去
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不知道是哪里出错,另外这题和Exercise7.1的题型类似,请问若要符合Exercise7.1的题意又该如何修改呢?请各位先进指导,谢谢!5 I2 p* M7 G7 T6 f% `# }' ~# J
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begin P_something arriving
2 K- m1 U: z: |" d" I move into Q_wait) t8 G: s+ Y$ A
move into nextof(Q_mA,Q_mB,Q_mC)
+ e: m3 v# O$ h: Q0 }; A use nextof(R_mA,R_mB,R_mC) for normal 48, 5 min6 |# m, M- \! d- p' Y
send to nextof(P_mA_down,P_mB_down,P_mC_down,P_mA_clean,P_mB_clean,P_mC_clean)
: f% n$ f/ Q. z5 s( [/ F# X4 e* M. F send to die( r% O8 {+ [8 I6 E7 I* J
end: b8 m9 h! r* ^0 p
+ a5 {: j( _& o- e( W& ]begin P_mA_down arriving
- i3 Y R) S" G1 W while 1=1 do 1 U. t% X4 _- m' H
begin1 ]+ l8 ~. l7 {0 U8 b" Q
wait for e 110 min
1 s( Y2 I- Q2 R( M6 q4 @& ` take down R_mA. N; f' j7 k# s& U
wait for e 5 min- I# A: y, D2 j. t: U* u
bring up R_mA
8 H6 p8 D2 @. n. I6 ~, c w7 h4 O end
0 e/ k6 i' L. U1 `end' S5 ^4 p+ r$ g m
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begin P_mB_down arriving
5 ]2 U# D- Q P( q2 z while 1=1 do
% `" H3 J9 [/ W# F5 | begin1 E# \" D2 Q& f
wait for e 170 min0 P# O! P8 B1 ]4 S
take down R_mB
, X4 c/ j/ r9 d- X* ^ wait for e 10 min8 U) f+ }) n6 k, \
bring up R_mB. b4 F% @) z& L) \( `8 b
end i3 z0 J9 |2 ]3 M
end* P+ ]- |' A( d' `5 H2 e5 B6 o" [
" q4 x, o, c6 Y) c' Jbegin P_mC_down arriving
" X% [3 h8 u, I, t" Y4 c while 1=1 do 0 C' _3 ^7 M- y* Z u9 b5 y- X
begin" P9 ~8 |& H8 G( u: h8 ]! \( b
wait for e 230 min/ \- i8 C5 H$ E
take down R_mC) J% W! E* v# g) e1 ?
wait for e 10 min
0 {, L1 n: h) E) b bring up R_mC" s) X/ x1 t7 s5 b
end- e8 U, t5 p" N( v
end
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begin P_mA_clean arriving
$ r: a, F0 h- V8 U: Z; z$ L( Y while 1=1 do
0 H, B) C- ? } begin
h' C+ |/ e3 K; ^ wait for 90 min
5 H. c, Q: C& S4 g0 g' x9 E take down R_mA* x* P, J. [' C6 S; { n
wait for 5 min
7 E! @8 D* ]2 n: r, j bring up R_mA
1 L6 ^0 E: c* s1 C end4 ?6 [( _( w# u* J4 P* Q4 d6 g
end
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begin P_mB_clean arriving X$ H3 D7 j' L Q1 Z
while 1=1 do
5 M* R8 n5 N( Z begin
6 j$ C, b; y2 n8 A wait for 90 min8 Y7 F3 p' t4 J" x' P0 x+ y
take down R_mB: s4 Y3 Q) C+ t+ O) p
wait for 5 min1 i$ c3 ?+ p z4 b
bring up R_mB' q- Y; [5 o# j) m7 D
end. G4 R, m. ~% I5 j) U, N/ U
end
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" a. E: m% y* p% F% N6 w# ~begin P_mC_clean arriving
# M9 }1 u- ^0 P3 U while 1=1 do( z' D' w% @* E8 @% \5 L3 J7 r: z
begin, \& f9 P7 ?& E; @. e, V4 ]
wait for 90 min' q9 I, H. Q3 C3 z- z* S
take down R_mC
0 p- z* H, y3 @* v0 f" |8 T wait for 10 min
7 s+ {. n# N D% P) R( B/ | bring up R_mC0 \4 B+ S6 G) X6 I, S! G* o
end7 \$ v8 |$ Q# N( {8 o/ G6 s9 ^; x
end. w6 s9 p# \* ~2 Z! k: z9 m
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Exercise 5.99 [3 h6 g1 H8 o8 `
' }8 ~; v( P0 o! x, U6 ]# A
8 b+ j& l8 r- u* v* n: y# ^Create a new model to simulate the following system:* p3 o! r1 M* b
Loads are created with an interarrival time that is exponentially
4 T1 [' B5 e s* f" L" r* N+ v: Rdistributed with a mean of 20 minutes. Loads wait in an infinite-4 o, V" z- u1 a" O7 t: H2 k+ p
capacity queue to be processed by one of three single-capacity, 1 [: d \' s9 g( m8 E* o
arrayed machines. Each machine has its own single-capacity queue
4 x' n( M. J+ k5 b5 i& c Iwhere loads are processed. Waiting loads move into one of the three
4 q, F6 D0 |9 ]1 Wqueues in round-robin order. Each machine has a normally 6 T3 j, { N# w6 |& G. h
distributed processing time with a mean of 48 minutes and a standard 4 I" |. W: U! c2 G! U( a
deviation of 5 minutes.
9 n2 D3 [( e {) l0 I' Z6 i4 k, K2 aThe three machines were purchased at different times and have 7 `6 b, J! d/ _, G& r7 T
different failure rates. The failure and repair times are exponentially ; A3 s7 f6 y9 P3 L1 k0 i6 F
distributed with means as shown in the following table:
. x! h( G3 \, p5 nNote The solution for this assignment is required to complete
9 c+ i6 Z3 O6 q, O/ V) A/ W; Cexercise 7.1 (see “Exercise 7.1” on page263); be sure to save a copy of
+ m! s |6 U8 O: oyour model.
: B+ u, K% F( J m' V. z9 t
+ b5 m8 F( G7 s* e% ~2 MMachineMean time to failMean time to repair
( P1 U0 @1 o# ] ]$ yA110 minutes 5 minutes
9 D8 {8 c1 ` ~! L9 l L9 h; _B 170 minutes 10 minutes# Z4 s) l4 g* Y3 W4 H+ o
C230 minutes 10 minutes5 [; R/ {8 u* h5 \) L; t1 a" b
: A) K4 g/ e4 u6 T6 L1 R( S8 \
The machines also must be cleaned according to the following
3 |* M% A- Y- e4 n9 m' {6 Yschedule. All times are constant: * n; [/ N7 b. ]1 Z
3 q4 j! ?2 J( c9 u7 H* @MachineTime between cleanings Time to clean
4 p& I' l8 M# O+ ~( T9 jA90 minutes 5 minutes" R1 U" d) m5 u k" `0 W+ A' i7 D
B 90 minutes 5 minutes
3 D) n- ~) O8 l# F/ [) oC90 minutes 10 minutes
5 L: c a& b/ j9 u; A
& c( J6 j! x7 `. i! k- qPlace the graphics for the queues and the resources.
2 ~% D( D; o' URun the simulation for 100 days.
* a, U$ x" g/ i. dDefine all failure and cleaning times using logic (rather than resource
/ E6 O t$ i- x0 b2 ~$ k. \cycles). Answer the following questions:9 J k+ A1 v/ l& _, Q2 l/ v9 C
a.What was the average number of loads in the waiting queue?+ E) l' M- R; n, `. t* I
b.What were the current and average number of loads in Space? : M' r& w0 O8 R
How do you explain these values? - T- B3 F2 z4 |2 p$ M
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