If you're willing to use some more resources, throughput can be improved further (to 1 load/0.75 hours) with one more dryer. With a second dryer, there is no wait for the first dryer to finish to start drying another load.

eknight7

The discussion in the lecture said that the throughput in this case is 1 load per hour. I believe its 4 loads/5 hours which is 0.8 load per hour.

mallocanswer

@eknight7, if you have n loads, throughput is approximate to 1 load/hour ((3/4 + n + 1/4)/n).

ote

But to clarify, under this system with one washer, dryer, and the time to load, you could never actually get a throughput of 1 load/hour, even as n approached infinity, the throughput would approach 1, but never actually hit it?

maxdecmeridius

I was thinking the same thing, we are approximating the throughput at 1 load/hour. Looking at the middle of the pipeline, without the initial stage where we don't have any parallelism, the throughput is 1 load/hour, right?

yangwu

@maxdecmeridius what do you mean by middle of the pipeline? form what i see, if we pick any random 1-hour interval, all of them only finish part of one load..

update: i think throughput is considered as how many jobs finishes at a end of every cycle. in this case, the throughput is 1 load/hour if we look at middle of pipeline

CC

@ote @maxdecmeridius Yes I think the throughput can be considered as an approximation, but since the instruction sequence is usually long, the first few cycles of 'less' parallelism do not hurt much for the overall throughput.

efficiens

@ mallocanswer how exactly is it ((3/4 +n + 1/4)/n)?

bojianh

@efficiens: I think it is more like (n/(3/4+n+1/4))

3/4: washing
1: drying
1/4: folding

Since bottleneck is at drying, hence it is the total processing time is comprised of 1 washing + n drying + 1 folding after the nth drying.

If you're willing to use some more resources, throughput can be improved further (to 1 load/0.75 hours) with one more dryer. With a second dryer, there is no wait for the first dryer to finish to start drying another load.

The discussion in the lecture said that the throughput in this case is 1 load per hour. I believe its 4 loads/5 hours which is 0.8 load per hour.

@eknight7, if you have n loads, throughput is approximate to 1 load/hour ((3/4 + n + 1/4)/n).

But to clarify, under this system with one washer, dryer, and the time to load, you could never actually get a throughput of 1 load/hour, even as n approached infinity, the throughput would approach 1, but never actually hit it?

I was thinking the same thing, we are approximating the throughput at 1 load/hour. Looking at the middle of the pipeline, without the initial stage where we don't have any parallelism, the throughput is 1 load/hour, right?

@maxdecmeridius what do you mean by

middleof the pipeline? form what i see, if we pick any random 1-hour interval, all of them only finishpartof one load..update: i think throughput is considered as how many jobs finishes at a end of every cycle. in this case, the throughput is 1 load/hour if we look atmiddleof pipeline@ote @maxdecmeridius Yes I think the throughput can be considered as an approximation, but since the instruction sequence is usually long, the first few cycles of 'less' parallelism do not hurt much for the overall throughput.

@ mallocanswer how exactly is it ((3/4 +n + 1/4)/n)?

@efficiens: I think it is more like (n/(3/4+n+1/4))

3/4: washing 1: drying 1/4: folding

Since bottleneck is at drying, hence it is the total processing time is comprised of 1 washing + n drying + 1 folding after the nth drying.