PCB Populating Challenge

I hate these.

Working separately - 4 hours, since that's the minimum time it takes for Pam to do two.

As a "team":

Let's see between them they can do 5 boards in 6 hours, right? So that means they can do 3 boards in 3/5*6 = 216 minutes.

Oops, I forgot to also ask: since only one person at a time can work on any PCB, what will be their optimal work share plan?

Jorrie

Four hours. If they both start at the same time Pam will start her second board while what's his name is still working on his first. There would be no point in slowing down Pam's second board by involving the slower worker. In fact, send the slow one home when he is done so he doesn't bother Pam while she finishes her second board.

I had a "Pam" and "Jim" working on my assembly line once. They were working late and I happened to come in to discover they were watching a movie on their laptop while assembling boards. Needless to say, I fired them. I don't mind them listening to music but visual distraction while doing a visual task? Please, what were they thinking.

6 hours. Have Pam populate all the boards (3 boards x 2 hours) and task Jim with caring for Pam (gathering and providing water, food, etc) while she works.

Jim's labour is not compensated, so your production costs are materials, labour (Pam's only) and operational overhead.

That has been the model used in most capitalist countries the world over for the last 100 years.

Together they will be able to populate 5 boards in 6 hours. (1min 12sec per board)

Total time therefore 3 min 36 sec.

Additional time may have to be allowed for swapping the boards.

It would mean a lot for the economy if an assistant is given to each of them. The job can be then be completed in 8 hours flat.

Hi Hendrik, wow! I guess you meant 3 hours and 36 minutes? And the 8 hours??

Assuming it takes a few seconds to swap boards, what will their best "swap plan" be?

Jorrie

Yes hours & min - Must have aimed at increasing production rates.

The 8 hours is from measured production statistics. (only if you start with 4 boards)

The slowest person must do the swapping. less production will be lost.

Can i recommend a supplier for pick and place equipment?

Share plan = 1 min 12 sec each per board

Jim works on board A for 36 minutes then starts on board C for three hours. Pam does board C in 2 hours then picks up board A, and finishes it in 1 hour 36 minutes.

But maybe Hendrik's scheme would be more efficient in reality.

...........................Jim..........Pam

1hr. 12 minutes...... A ........... B

1hr. 12 minutes...... C ........... A

1hr. 12 minutes...... B ........... C

This way they could each fit the same subset of components to all three boards.

your scheme may be better because less swaps implies less time lost. But time would be required to set up the stations and train both for placing all parts.

On the older pick and place machines we have a led shining on the bin and then on the position. It is really hard work to be controlled by a red dot.

If they have to work as a team, it implies that there is only one work station.

From the times given, it could be assumed that Pam can pick faster than Jim, as taking longer to solder would lead to many overheated components.

So get Pam to pick while Jim is placing and soldering, and time for each board will be minimised.....but by how much??

Tricky question . . .

May we assume that the project of populating 3 PCB's starts with both Jim and Pam at the same time?

If so, Pam can finish one in 2 hours and start on another. 3 hours from the start, Jim can help Pam finish the third PCB which has been half-populated by her in the hour she worked on it by herself, but that's the hard way . . .

Pam's rate is 1/2 PCB/hour (or 12/day) while Jim's is 1/3 PCB/hour (or 8/day). Their combined rate is 20/day. Since only 3 PCB's are needed, only 3/20 days are needed, or 3 hrs and 36 minutes.

What technology are they using? If they are using GSM placement machines, Pam can program the PCB placement in two hours using the first PCB as a template. The next two boards will only take a few minutes. Jim gets a cup of coffee and goes back to his cube to check out CR4 on the internet.

I would guess just over 4 hours. Pam finishes the first one then takes a few moments to get set up on her second board. Jim will complete his board and Pam is not going to let him put his clumsy hands in her way. She will have her system all set up and Jim would just be in the way.

This is not the same as if Jill starts here and Joe starts there and one walks one speed and the other walks at another, and how far will they walk in x-time.

Jorrie , I love a challenge but PCB's are not my thing. Can you give a definition of 'populating a PCB' for us lesser mortals. It may not change the nature of your poser , but I'm curious. Thanks , Kris

'populating a PCB'

Taking a circuit board (etched and drilled) and adding/soldering in the components.

Jims rate = .33333_r boards/hr

Pams rate = .5 boards/hr

Maximum total rate = .83333_r boards/hr

Time at above rate = 3/.8333r _boards/hr = 3.6 Hr

Time = 1.2 hr/board

Pam is 1.5 x the speed of jim.

Jim starts on board 2 , Pam on board 1. They follow each other , changing boards at the same time .Jim will have done 2/5 of a board when Pam has completed 3/5 of a board.Pam will complete her 3/5 of a board in (120 Min's x3/5) 72 Min's.

If they are quick with the 3 changes , they can meet 3.6 h (3x72 Min's)

OK so maybe we can do better by not wasting so much of Jim's available time. As soon as Pam finishes her first board she takes Jim's first board (2/3's done) and finishes it while he starts the third board. When she finishes the second board she takes Jim's second partially complete board (the third and final board) and finishes it. This way, the fastest worker gets a head start on finishing the second and third board but at no point is the faster worker impeded. Jim is still idle at the end but for not as long. This assumes that only one person can work on a board at a time. Also, from a practical standpoint I don't believe that a complex switching arrangement is a good idea though in theory you could squeeze a few minutes out of the total by doing so. Probably not a good plan with regular humans.

I can see no reason for Jim to be 'idle'. Get them to work in proportion to their output rate , simply balance changeover time in proportion to the rate.Pay rate - well thats a different kettle of fish.

Having owned and managed an assembly shop, I have to take a practical approach. As stated, there are only three boards to be assembled. You will spend more time trying to figure out how to swap things around and do so at a supervisory rate of pay than you could possibly save to just pay Jim to be idle for a few minutes. Also, and I speak from experience, when it becomes difficult to determine who did what on what board, more errors will occur and you won't be able to attribute them to anyone. If Jorrie had said that there were 300 boards to assemble then you have a situation that is easier to manage accross different rates. I had a three station assembly line and everybody had different work rates. It was always a juggling act to balance the work flow depending on who was working. Usually, the person who got ahead would jump back and forth between some preparatory task but again this assumes a constant flow of work and not just three boards.

Above post is me. I'm on a different computer and didn't notice I wasn't logged in.

I've done exactly the same myself , so I can't fault you . I'm waiting 'till Jorrie looks at my posting and tells me I've missed something !

Hi Kris, you wrote: "I'm waiting 'till Jorrie looks at my posting and tells me I've missed something"

I like your idea, originally proposed by Hendrik in post #6 and commented on by Randall in #8, I think!

I will reserve judgement for a few more days to see if a better scheme does not pop up...

Regards, Jorrie

Apologies to Hendrik and Randell - I jumped back into this without careful re-reading of previous answers .. '...a better scheme..' ? Now I'm really curious .

Well numbers and time calculations all look well and good on paper but how about a more practical approach to get an idea of how long it will ACTUALLY take. Here is how I used to do it when I populated pcbs for a living.

Place the three pcbs side by side on a jig and populate all three at the same time, fitting the same component to all three of the boards in succession and then switching to the next component. Repeat until all boards have been fully populated. If we assume that all components are ready to be placed (i.e.- you don't have to spend an hour trying to find that box of damn R13 resistors that you KNOW you were using yesterday but cannot remember where you put it), and Pam is doing the work, the approximate minimum time (based on my experience) will be around 4-5 hours. Having Jim help assemble the boards is likely going to increase the time necessary to complete all 3 boards due to crossover inefficiency when Pam takes over finishing Jim's board (which is possible). Best to let Pam do all the boards and get Jim to track down where those R13 resistors went!

That's how we do it down here .

The penny has dropped (I think)

They only need to change twice.

Pam spends 2 hours completing board a , whilst Jim starts on board b

At t+36 mins , Jim moves to board c

At t+120 min P moves to board b where she finishes it off in 80 mins

Jim will have done board c in his remaining 120 mins.

Well I think you got it but didn't quite relate it as perhaps you were thinking. Jim's 36 minutes is equal to 24 minutes of Pam's time so it takes her 120-24=96 minutes to complete board B and that takes her up to 216 minutes. 216 - Jims original 36 leaves him 180 minutes remaining which is what it takes him to do board C by himself.

I'm not sure I get your point ;

At t +36 mins Jim has done 20% of board b (36/180).....Jim takes 3hrs/board. He does board c on his own which takes 180 mins. total time 216 mins

Pam needs to complete 80% of board b = 120 mins/board x 0.8 = 96 mins (on top of the 2 hrs she spent at board a )=216 mins

Their total working time is the same , but they change at different times . Have I explained clearly enough - sometimes I get tangled in my own 'logical' thought train

This is where Jorrie comes in with the 'gotcha' bit !

The statement: "At t+120 min P moves to board b where she finishes it off in 80 mins " seemed unclear. Seems like, and you do state in your last post she finishes it off in 96 minutes which works out. I'm not clear on your reference to 80 minutes.

Sorry rcapper ,in my haste (probably Jims flaw as well ) I made some typos though the principle is sound.

Pam works 120 mins on board a, then 96 mins on board b .Total =216

Jims already done 36 mins work on board b (before Pam moves) when he leaves to do 180 mins on board c. Total =216

It's early morning here , so my brain is duller than normal. Hope this clarifies my meaning , since I'll have to log off soon and catch up tomorrow. Kris

The "official answer", whatever that may mean!

Since management time is "free", I'll work out the theoretical minimum assembly time and get, as many posters did, 3 hours and 36 minutes. Then I'll estimate which 40% of components (by assembly time) to give to Jim and the rest must go to Pam.

I'll instruct the store to issue 6 kits. Pam gets 3 "Pam kits", two of them with bare boards in them. Jim gets 3 "Jim kits", with only one containing a bare board (B in the schedule below).

They than work more or less like in Randall's post 8 (first hinted upon by Hendrik), where in the first 1 hour 12 minutes, Pam populates 60% of A and Jim 40% of B and so on for three sessions as follows:

Session..…................Pam................Jim

1hr. 12 minutes...... A(60%) ........... B(40%)

1hr. 12 minutes...... C(60%) .......... A(40%)

1hr. 12 minutes...... B(60%)........... C(40%)

The rationale is that, although there are more "swaps" than in the alternative mentioned, the efficiency of doing the same subset three times will outweigh the small swap time. This will probably also improve quality, because regular breaks will have to be taken by both people in any case.

BTW, why would Jim be so much slower than Pam?

Regards, Jorrie

Part but perhaps not all of Jim's slower pace is because Jim's fingers are typically larger than Pam's and so his dexterity with small parts is somewhat less. Another reason I have observed is that, and I don't know what this means in other terms, but women seem to tolerate boring repetitive tasks better than men, especially if there are other women to talk to.

I was thinking something similar, with the one who has the smaller fingers fitting all components before trimming, and the other trimming as they went, to make access easier. This may take longer, but result in a neater finish.

Jims rate may be a reflection of better quality. I concur with the answer you've posted , but enjoyed the pedantry of reducing swap time for a theoretical world . The goings-on of a work place can be a bit surreal anyway !

Well, Pam has this ability to work and talk at the same time whilst Jim has difficulty working and listening at the same time. Occassionally, he'd offer a sentence or two or maybe a grunt (like "yes dear").

Or maybe Jim is like me, thinking over what the answer would be while everyone is already giving theirs.

Better late than never....

This could get dangerous , over-talkative women, blokes focused on their work ...!

Still , your remarks have put the smile back on my face (having seen todays Challenge Question) .

Women are alleged to be able to multitask. I think they might be able to look at the next part while placing the current one. As for chatting the constant yap yap will deteriorate poor Jim tempo even further.

I would definitely not get rid of Jim. He cannot become pregnant.

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