Class 1 MGMT 370 with Lonnie Turpin McNeese State Operations Management (Part 2)

this is part two for class number one management 370 with Dr Turpin right now we are going to be talking about the mm1 queuing model this was a super fascinating class um I'll put the problems in the link in the description um so that might be the easiest way to follow along and I'll read them and let's get into it all right so the first problem is um setting up this entire problem that we're going to go through it says a movie theater complex has a single ticket booth cashier who can maintain an average service rate of two movie goers per minute so what we just learned is there's a movie theater and the cashier can service two people per minute arrivals on a typically active day average 1.6 per minute the diagram below stylizes the queing system so you have this diagram um I won't draw it out but if you want to look at it the Lambda 1.6 is the average rate per minute then you have that waiting period Then the me is the service rate which is 2 per minute so that is like the maximum that um our cashier can do we are told that both the interval times and service times follow an exponential distribution which we just talked about in part one to determine efficiency of the current ticket operation the manager wishes to examine several Q operating characteristics all right we are on problem six and it says based on the above information is the system effective AKA is it stable and so what this is just basically asking is um is capacity um big enough to support the average rate per minute so the formula for this is uh if the Lambda is less than the MU then it's stable and it sounds a little more complicated than it is I'll write it out and you'll be like a easy so that's our formula to determine if a system is stable and if you look at your um little little diagram you see that our Lambda is 1.6 because that is how many come per minute usually and then our me our capacity or our service rate is two two so obviously this is correct one or this is stable 1.6 is less than two and this um formula is basically saying that the movie theater cashier can service two people per minute and 1.6 people come per minute so it's stable of course so we're already done with the first one okay number seven says what percentage of the time is the cashier busy so if the cashier has a capacity of two people but 1.6 come every minute what percentage of time are they busy super simple formula again p is what we're going to be finding we're going equal Lambda over mu and before we solve this which you could probably even do in your head I thought of like a good a fun analogy for this um basically if you think about like your stomach capacity I find that to be like helpful so let's just say you have a stomach capacity of three Donuts you can eat three donuts and then you can't eat anymore um that would be your Mew and if you eat two Donuts uh that would be the um average I guess your average rate so if you wanted to think and say hm how would I find how much time does cashier is busy it would basically for the doughnut analogy it would say how full is my stomach so if you can eat three donuts and you do eat two of course like comment knowledge would be like just divide 2 by three and then you can find out how much room um is taken up by the donuts in your stomach so you would just divide 2x3 that'd be like 666 six 66 repeating repeating um and that would be how often your stomach is busy how often your stomach is full when you eat two Donuts when you have the capacity for three so this is the exact same um the cashier at this rate sees 1.6 people per minute and they can see two so all you have to do is divide it and that is 80% the cashier is busy 80% of the time there we go all right only three more questions this is going by fast and I think these problems are pretty fun our next question is what is the probability that the cashier is Idle or not busy I shouldn't have maybe erased that it's all right we remember this is uh a super simple formula again but basically all you have to do is subtract what we just found our P um from one because we we're assuming the percentage is like 100% um so it'd be like 100% - 80% they're idle 20% of the time I guess I'll write it out so you go 1 -8 gives You2 they are idle 20% of the time and that'll be a on your sheet awesome all right number nine what is the average number of movie goers waiting in line to purchase a ticket these next to get a little more complic complicated in the formulas but they're still pretty interesting and we're given a formula sheet I believe so it's going to be too hard either all right here's the formula to see how many um people are waiting in line so we have our P again which we just found which was 8 and that's all it is so we're just going to substitute um8 in there and we'll get our answer s and this is something pretty interesting um that you might want to be careful of cuz I tried to like Mental Math this in class and then I got it wrong and I was like oh shoot um basically when you see 64 /. 2 you might be tempted to say okay it's 32 um cuz you just like I did maybe do it so fast and you think like oh cross off the decimal you're just dividing by two but actually if you want to um a little hack if you want to see the numbers a little better you can just add a zero onto here so they have the same amount of decimal points and so of course um 20 goes into 64 about three times so you'll know your answer will be 3 point something rather than um like a decimal place like I did like 32 for my answer um but this should give you 3.2 people waiting in line awesome and I'll keep this up for our last problem this has gone by so fast what is the average time or minutes spent waiting in line to get a ticket out the window okay I'll write our formula and then we'll solve it all right so as you can see um we just found this so all we have to do is plug 3.2 up top and then we'll go back to our first page and the Lambda if you guys remember was 1.66 and that should give us um 2 minutes everybody's waiting about 2 minutes in line all right there we go hope this was helpful um good luck bye