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?)
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