AutoMOD内附练习问题
本帖最后由 GJM 于 2009-12-5 21:43 编辑底下是小弟做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
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!
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begin P_something arriving
move into Q_wait
move into nextof(Q_mA,Q_mB,Q_mC)
use nextof(R_mA,R_mB,R_mC) for normal 48, 5 min
send to nextof(P_mA_down,P_mB_down,P_mC_down,P_mA_clean,P_mB_clean,P_mC_clean)
send to die
end
begin P_mA_down arriving
while 1=1 do
begin
wait for e 110 min
take down R_mA
wait for e 5 min
bring up R_mA
end
end
begin P_mB_down arriving
while 1=1 do
begin
wait for e 170 min
take down R_mB
wait for e 10 min
bring up R_mB
end
end
begin P_mC_down arriving
while 1=1 do
begin
wait for e 230 min
take down R_mC
wait for e 10 min
bring up R_mC
end
end
begin P_mA_clean arriving
while 1=1 do
begin
wait for 90 min
take down R_mA
wait for 5 min
bring up R_mA
end
end
begin P_mB_clean arriving
while 1=1 do
begin
wait for 90 min
take down R_mB
wait for 5 min
bring up R_mB
end
end
begin P_mC_clean arriving
while 1=1 do
begin
wait for 90 min
take down R_mC
wait for 10 min
bring up R_mC
end
end
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Exercise 5.9
Create a new model to simulate the following system:
Loads are created with an interarrival time that is exponentially
distributed with a mean of 20 minutes. Loads wait in an infinite-
capacity queue to be processed by one of three single-capacity,
arrayed machines. Each machine has its own single-capacity queue
where loads are processed. Waiting loads move into one of the three
queues in round-robin order. Each machine has a normally
distributed processing time with a mean of 48 minutes and a standard
deviation of 5 minutes.
The three machines were purchased at different times and have
different failure rates. The failure and repair times are exponentially
distributed with means as shown in the following table:
Note The solution for this assignment is required to complete
exercise 7.1 (see “Exercise 7.1” on page263); be sure to save a copy of
your model.
MachineMean time to failMean time to repair
A110 minutes 5 minutes
B 170 minutes 10 minutes
C230 minutes 10 minutes
The machines also must be cleaned according to the following
schedule. All times are constant:
MachineTime between cleanings Time to clean
A90 minutes 5 minutes
B 90 minutes 5 minutes
C90 minutes 10 minutes
Place the graphics for the queues and the resources.
Run the simulation for 100 days.
Define all failure and cleaning times using logic (rather than resource
cycles). Answer the following questions:
a.What was the average number of loads in the waiting queue?
b.What were the current and average number of loads in Space?
How do you explain these values?
如果把逻辑改成底下这样，直觉
send to nextof(P_mA_down,P_mB_down,P_mC_down,P_mA_clean,P_mB_clean,P_mC_clean)
这行好像也不太对，各位先进觉得如何呢?
begin P_something arriving
move into Q_wait
move into nextof(Q_mA,Q_mB,Q_mC)
use nextof(R_mA,R_mB,R_mC) for n 48,5 min
send to nextof(P_mA_down,P_mB_down,P_mC_down,P_mA_clean,P_mB_clean,P_mC_clean)
end
begin P_mA_down arriving
wait for e 110 min
take down R_mA
wait for e 5 min
bring up R_mA
end
begin P_mB_down arriving
wait for e 170 min
take down R_mB
wait for e 10 min
bring up R_mB
end
begin P_mC_down arriving
wait for e 230 min
take down R_mC
wait for e 10 min
bring up R_mC
end
begin P_mA_clean arriving
wait for 90 min
take down R_mA
wait for 5 min
bring up R_mA
end
begin P_mB_clean arriving
wait for 90 min
take down R_mB
wait for 5 min
bring up R_mB
end
begin P_mC_clean arriving
wait for 90 min
take down R_mC
wait for 10 min
bring up R_mC
end begin P_something arriving
move into Q_wait
set A_q to nextof(Q_m1,Q_m2,Q_m3)
move into A_ｗ
if A_ｗ=Q_ｍ1 then use R_m1 for u 48,5 min
else if ．．．．．then use ．．．．for u 48,5 min
else if ．．．．．then use ．．．．for u 48,5 min
send to die
end
供您參考看看～
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