Our client is a ski company. At one of their ski parks, they are facing long queues at the ski lifts. They would like us to reduce the amount of time their customers spend waiting. In order to do so, the company would like us to investigate two ski lift alternatives:
1) Installing a Faster Ski Lift
2) Installing a 2nd Ski Lift, which is exactly like the current ski lift
We can recommend no change if the existing setup is most efficient.
Info provided:
1) There are three components to the ski park: the lift, the slope, and the queue.
2) All customers must be in one of these 3 locations.
3) Customers go up the lift, ski down the slope, wait in the queue, and keep repeating this cycle. (Lift, Slope, Queue, Lift, Slope, Queue, Lift, Slope, Queue, etc.)
Current ski lift:
Entrance/Exit Rate: 5 customers/min
Lift time: 10 min
Ski slope time: 5 min
Faster ski lift:
Entrance/Exit Rate: 5 customers/min
Lift time: 5 min
Ski slope time: 5 min
2nd ski lift:
Entrance/Exit Rate: 10 customers/min
Lift time: 10 min
Ski slope time: 5 min
Any ideas? Seems much more of a math problem than an actual case.
Do you know how many customers there are and how fast they're coming? It seems like an optimization problem, but its not at all clear what the problem is. If it means there is one lift that travels up taking X lift time and it holds Y customers, then the faster ski lift and 2nd add the same capacity. They can both transport 1 customer per minute. But that could be completely wrong if the definitions are wrong - I think you (or they) really need to clarify problem.
Oh, and I don't ski, which might help..
Tell them to steal Whistler's patented queuing system.....easy
Combine 1 and 2 and install a second, faster lift. Because fuck you arbitrary limitations
You need to investigate gains and losses associated with the current queue. How much does each option cost, if they're losing money for the long queues (i.e. competing slopes are taking their customers, etc.), and any externalities that could affect the decision (expanding the number of slopes, increasing the size of the lodge, etc.). These outstanding questions are what makes it a "case" as opposed to just a math problem.
This was the solution posted on glassdoor.
We first begin with this logic: If the entrance and exit rate of each location are the same, then the quantity of customers at a given location will be the entrance or exit rate multiplied by the time spent at location.
Here it is in formula form: "Entrance or Exit Rate" x "Time at Location" = "Quanitity of Customers at Location"
With this formula, we can still derive how much time a customer spends in the queue.
Current Ski Lift
Using the information provided, we can derive the following information for the current park:
and etc.
How does the formula make any sense? I simply don't understand that logic.
Where did the 200 total customers come from?
That explanation doesn't make sense.
Faster ski lift: Entrance/Exit Rate: 5 customers/min Lift time: 5 min Ski slope time: 5 min
By those parameters:
Lift - 5x5 = 25 Slope 5x5 = 25
Queue = 200 - 25 - 25 = 150
150 / 5 = 30
The wait time gets longer with the faster lift?
Precisely. You can just do this out logically too I think, without having to plug numbers in. Skiiers either spend time in line, on the lift, or skiing down the slope. They will spend the same amount of time on the slopes regardless of the type of lift, meaning that the rest of their time will be split between waiting in line or being on the lift. So it's inversely related: the faster the lift runs, the quicker they will get down and the longer time they will have to wait in line. That scratches out the Faster Lift as a potential answer. Then it comes down to extra capacity with the 2nd lift, and I guess you would have to come up with an arbitrary volume (i.e. people=200), because per person wait times wouldn't help you figure out that it best solves the excess demand problem.
This seems to be a pure math case, or more likely just the beginning or middle of a broader case. It's about how you add capacity in an efficient way to meet the surplus demand.
It doesn't make sense because it doesn't answer the question, "Should we buy a new, faster lift, add another lift, or stick with what we have".
See Brady Chiu's answer at http://www.quora.com/Management-Consulting-and-Management-Consulting-Fi… for a discussion of the case.
Can somebody explain to me this formula?
"Entrance or Exit Rate" x "Time at Location" = "Quanitity of Customers at Location"
Doesn't make much logical sense to me.
Try thinking about it a different way
In the current system:
5 people get on the lift every minute and 5 people get off every minute, and the lift takes 10 minutes. Suppose 5 people get on. After one minute, another 5 get on behind them. After another minute, another 5 people get on behind those. By the time those initial 5 people get to the end of the lift, there are 9 lots of 5 people behind them (since the lift journey is 10 minutes long). Thus the entire lift system can hold 50 people at any one time.
The same logic can be applied to the slope to see that it can hold 25 people at any one time
Since a skier is either on the lift, on the slope or in the queue, if there are N skiers in total then N - 75 of them are in the queue in the current system. Since 5 people leave the queue to enter the lift each minute, the waiting time in the queue is thus (N-75)/5 (rounded up to the nearest integer)
In the faster lift system:
The lift can hold 25 people and the slope can hold 25, so there are N - 50 people in the queue giving a waiting time of (N - 50)/5 (again, you'd have to round up to the nearest integer)
In the double lift system:
The lifts can hold 50 people each, i.e. 100 in total. The slope can hold 50 people in this instance (since 10 skiers can start skiing every minute, rather than 5), so there are N - 150 waiting at the bottom. This is divided into two queues, so there are (N-150)/2 people in each queue, with a waiting time of (N-150)/10 in each
So we see that the double lift system is best. Intuitively, this is what one might have expected
If you've taken an operations course, that's just another way to express the standard inventory process flow formula I = R x T, so Inventory = Flow Rate x Flow Time
To me this is a waiting line problem in Queuing Theory. Basically, the main components are the arrival rate (no. of skiers that enter the queue) and service times (how long it takes to serve the skier). More specifically, the problem follows a "M/M/1" queue, written in Kendall's notation representing a single server and single line queue, think Customer A; Cust B; Cust C, etc ----> Service point ----> End .
Here, the arrival rate is determined from the ski slope time - 5 min/cust ( or 1/5 cust/min) The service TIME is determined from the lift time and entrance/exit time - 5 cust/min take 10min to go up.
In the M/M/1 model, a queue is stationary only if the arrival rate service rate = 1/10 cust/min, then the system is not stable and will grow indefinitely, which is why we are tasked with finding a solution.
[Service TIME = 1/service RATE, so service RATE = 1/10 customers/min. ]
Now, the way to solve this is to compare a single metric of choice against the alternatives and choose the optimum solution. Here, the objective is to minimise the amount of time waiting. [ the metric that will need to be used here is the average time a customer spends waiting in the system i.e. average time spent waiting in line]. I will just sprout some formulas from queuing theory that you can easily read up on on your own time so bear with me.
If p (rho) = arr/serv = (1/5)/(1/10) = 2, then
number of cust in queue = Q = p^2/(1-p) = 2^2/(1-2) --> the answer is not feasible as it is negative, which further illustrates the unstable condition.
SCENARIO 1 still follows M/M/1 with a faster service time, while SCENARIO 2 follows an M/M/k model, where k is the number of servers (here, k = 2 for the 2 ski lifts).
This is all somewhat from memory from my Queuing Theory course during undergrad, so I'm not going to attempt to solve it until I find my formula sheet somewhere in my pile of old notes :p
Cheers
To me this is a waiting line problem in Queuing Theory. Basically, the main components are the arrival rate (no. of skiers that enter the queue) and service times (how long it takes to serve the skier). More specifically, the problem follows a "M/M/1" queue, written in Kendall's notation representing a single server and single line queue, think Customer A; Cust B; Cust C, etc ----> Service point ----> End .
Here, the arrival rate is determined from the ski slope time - 5 min/cust ( or 1/5 cust/min) The service TIME is determined from the lift time and entrance/exit time - 5 cust/min take 10min to go up.
In the M/M/1 model, a queue is stationary only if the arrival rate service rate = 1/10 cust/min, then the system is not stable and will grow indefinitely, which is why we are tasked with finding a solution.
[Service TIME = 1/service RATE, so service RATE = 1/10 customers/min. ]
Now, the way to solve this is to compare a single metric of choice against the alternatives and choose the optimum solution. Here, the objective is to minimise the amount of time waiting. [ the metric that will need to be used here is the average time a customer spends waiting in the system i.e. average time spent waiting in line]. I will just sprout some formulas from queuing theory that you can easily read up on on your own time so bear with me.
If p (rho) = arr/serv = (1/5)/(1/10) = 2, then
number of cust in queue = Q = p^2/(1-p) = 2^2/(1-2) --> the answer is not feasible as it is negative, which further illustrates the unstable condition.
SCENARIO 1 still follows M/M/1 with a faster service time, while SCENARIO 2 follows an M/M/k model, where k is the number of servers (here, k = 2 for the 2 ski lifts).
This is all somewhat from memory from my Queuing Theory course during undergrad, so I'm not going to attempt to solve it until I find my formula sheet somewhere in my pile of old notes :p
Cheers
Facilis eveniet voluptate numquam nesciunt saepe nostrum perferendis. Debitis nulla voluptas ipsum ut dolores quia sunt. Voluptas vitae sed debitis ut cum.
Magni sit sed voluptas dolorem quae repudiandae. Voluptatem perferendis doloremque nihil est. Non qui minima magnam ratione.
Tempora quisquam quia reprehenderit ullam. Cumque commodi soluta rerum quo id rerum. Dolores sapiente consequuntur sed ut voluptas iusto. Tempora dicta quam earum impedit totam sequi voluptatem. Eligendi magnam et molestiae nemo expedita quod dolores.
See All Comments - 100% Free
WSO depends on everyone being able to pitch in when they know something. Unlock with your email and get bonus: 6 financial modeling lessons free ($199 value)
or Unlock with your social account...