刚接触R语言菜鸟，问题小白不要见怪。

题目：

A car takes 40 hours of labor and 1 ton of steel

A truck takes 60 hours and 3 hours of steel

Resources: 1600 hours of labor and 70 tons of steel eachweek.

建立一个R代码，使过高计划产量（例如30辆轿车与20辆卡车），或者过低的计划产量逐步调整到最优化水平（最大量生产+资源剩余也不多）.

起始代码：

factory <- matrix(c(40,1,60,3),nrow=2,
   dimnames=list(c("labor","steel"),c("cars","trucks")))
available <- c(1600,70); names(available) <- rownames(factory)
slack <- c(8,1); names(slack) <- rownames(factory)
output <-c(30,20); names(output) <- colnames(factory)

passes <- 0 # How many times have we  been around the loop?
repeat {   
     passes <- passes + 1   
     needed <- factory %*% output # What do we need for that output level?   # If we're not using too much, and are within the slack, we're done   
if (all(needed <= available) &&       all((available - needed) <= slack)) {     
break()   
}   
# If we're using too much of everything, cut back by 10%   if (all(needed > available)) { output <- output * 0.9     next()   
}   

# If we're using too little of everything, increase by 10%   if (all(needed < available)) {  output <- output * 1.1     next()   
}   

# If we're using too much of some resources but not others, randomly   # tweak the plan by up to 10%   
output <- output * (1+runif(length(output),min=-0.1,max=0.1))
}

要求：以for loop 代替上面代码里的repeat loop.

问题来了：在建立 for loop() 时候，（）里面循环的次数passes不是预先知道的，所以无法设定。那如何让R在函数达到最优化的时候，自动放弃执行for loop 命令里面剩余的循环次数。（例如假设最大的循环次数设定为2000，达到最优化的时候，循环次数是1200，如何让R放弃执行剩余的800次？）