# Arrival rate problem

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- This topic has 10 replies, 3 voices, and was last updated 13 years, 6 months ago by chhabra.

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- August 28, 2007 at 12:02 pm #47968
Hi

How to calculate departure rate at the end of a tunnel if the arrival rate ( mean of exponential distribution) at the other end is known and also the time it takes to pass through the tunnel (normal distribution with known mean and sd) is known ?

Thanks0August 28, 2007 at 1:48 pm #160474Amit:Use Monte Carlo simulation (Google “CrystalBall” “Decisioneering” or “@Risk”).Cheers, BTDT

0August 28, 2007 at 2:25 pm #160477Thanks BTDT….but even for simulating, I need to have a formula for creating the “forecast”. While I do have the “assumptions” (arrival rate and passthrough time distribution), I am not sure how to formulate the forecast(departure).

0August 28, 2007 at 3:05 pm #160480Amit:The forecast will have a particular distribution of its own. You can use the “fit distribution” option to find out what the parameters are.If you have Minitab, generate 10,000 values of the arrival times, 10,000 values of the passthrough time, add the two and see what the result looks like.Cheers, BTDT

0August 28, 2007 at 4:26 pm #160489BTDT, I am not sure I understand that the forecast distribution will be a summation of the interarrival distribution and the pass through time distribution…..and how to extract the departure rate from this.

Had this problem been deterministic, say an interarrival time of 0.25 minutes ( 4 arrivals per minute), and a pass through time of 5 minutes, how will the sum (0.25+5 OR 4+1/5) help in calculating departure rate? Intutively we understand that in this case, during the first five minutes, there will be no departures. But once the 5 minutes warm up is over, we will expect the departure rate to equal the arrival rate at the other end…ie…4 per minute.0August 28, 2007 at 4:59 pm #160495The simulation suggested to you would look something like this -Exponential = -Mean*Log(1-Uni, 2.71828)

Arrival Rate Arrival Time Travel Time Departure

Time

Random Uniform Random Exponential Last Arrival + Next

Arrival Random Normal Arrival Time + Travel Time

Average of 0.25 minutes Average of 5 minutes, STDEV

of 0.3

0.903286843 0.58400185 0.58400185 4.937225425

5.521227276

0.027588733 0.006994117 0.590995968 5.036314418

5.627310386

0.567949461 0.209803319 0.800799286 5.05426034

5.855059626

0.660847804 0.270326761 1.071126047 5.015371597

6.086497645

0.301614429 0.089746044 1.160872091 4.67920121

5.840073301

0.401379437 0.128281919 1.289154011 4.524678969

5.81383298

0.649189734 0.261877614 1.551031625 5.326508143

6.877539768

0.820368053 0.429211604 1.980243229 4.957991577

6.938234806

0.043488876 0.011115723 1.991358952 4.711776582

6.703135533

0.663838618 0.272541166 2.263900118 5.024019528

7.287919646

0.869319742 0.508750773 2.77265089 5.059822014

7.832472904

0.498611408 0.172593577 2.945244468 4.540214958

7.485459426

0.278756066 0.081694523 3.02693899 4.910914312

7.937853302

0.355418561 0.1097886 3.136727591 4.943778903

8.080506494

0.279671621 0.082012077 3.218739668 5.091318952

8.31005862

0.31019013 0.09283488 3.311574548 4.523060978

7.834635526

0.202063051 0.056431462 3.36800601 5.615810677

8.983816687

0.598010193 0.22783229 3.5958383 4.95769042

8.55352872

0.965849788 0.844247188 4.440085488 4.814989337

9.255074826

0.554734947 0.202271523 4.642357012 5.248208153

9.890565164

0.001800592 0.000450554 4.642807566 4.746747562

9.389555128

0.305734428 0.091225242 4.734032808 4.755649242

9.48968205

0.588915677 0.22223938 4.956272188 5.007907147

9.964179335

0.540055544 0.194162517 5.150434704 4.67378244

9.824217145

0.142582476 0.038457598 5.188892302 5.007516633

10.19640893

0.800103763 0.402489486 5.591381788 4.914839918

10.50622171

0.995422224 1.34663638 6.938018168 4.556600869

11.49461904

0.356608783 0.110250653 7.048268821 4.504638254

11.55290707

0.825525681 0.436494723 7.484763544 5.150776714

12.63554026

0.798028504 0.399907444 7.884670989 4.93221968

12.81689067

0.48893704 0.167815735 8.052486723 4.927668227

12.98015495

0.821497238 0.430788091 8.483274814 5.400096042

13.88337086

0.970000305 0.876642607 9.359917422 5.247530124

14.60744755

0.241920225 0.06924171 9.429159132 4.858859269

14.2880184

0.858027894 0.488031497 9.917190629 4.901030719

14.81822135

0.811487167 0.417147579 10.33433821 5.294550091

15.6288883

0.560930204 0.205774361 10.54011257 4.950864094

15.49097666

0.94021424 0.704247417 11.24435999 5.381753125

16.62611311

0.753440962 0.350038688 11.59439867 5.222692734

16.81709141

0.703268532 0.303732129 11.8981308 5.358940952

17.25707175

0.70455031 0.304814382 12.20294519 4.927006229

17.12995141

0.487014374 0.166876976 12.36982216 5.112031216

17.48185338

0.508774071 0.177712899 12.54753506 5.012384589

17.55991965

0.139927366 0.037684634 12.58521969 4.948490313

17.53371001

0.237281411 0.06771658 12.65293627 4.533104528

17.1860408

0.430036317 0.140545753 12.79348203 5.143375019

17.93685705

0.998321482 1.597462076 14.3909441 5.084013436

19.47495754

0.842432936 0.461976337 14.85292044 5.110531573

19.96345201

0.231086154 0.065694131 14.91861457 4.971488023

19.8901026

0.155339213 0.04220507 14.96081964 4.761549589

19.722369230August 28, 2007 at 4:59 pm #160497Yikes – send me an email address and I’ll send you the [email protected]

0August 28, 2007 at 10:56 pm #160516Stan:Yikes indeed!I looked more carefully and realized it is a queue management problem. I’d do this with a process simulator to keep track of the queue length, etc.If you’ve got a way of managing the queue in Excel, I’d like to see the spreadsheet too.Cheers, BTDT6Sigmaguru(at)gmail(dot)com

0August 28, 2007 at 10:57 pm #160517Amit:I didn’t see the problem immediately. Ignore what I posted and see what Stan comes up with.Cheers, BTDT

0August 29, 2007 at 11:32 am #160530We use a nice example that is not intuitive because of the queues.I’ll send the problem set up and a solution built in Excel later.

0September 2, 2007 at 6:33 am #160669Thanks a lot Stan. Appreciate yours taking time out to help.

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