The Elevator Problem - NCSSM Links

The Elevator Problem

By Helen Compton and Dan Teague

Introduction

The elevator problem can be used with many different classes. Naturally, the
more advanced the class, the more that can be expected in the solution. This problem can
be used as the setting for developing a simple linear function of two variables, for
investigating aspects of proportional reasoning, or as an application of combinatorial
probabilities. It focuses attention to the importance of making reasonable, simplifying
assumptions. The problem can be made as simple or complex as the teacher desires by
altering a few of the parameters in the problem. Each teacher can determine how
difficult the problem should be and how far into the problem they want their students to
go.

Problem Statement

In some buildings, all of the elevators can travel to all of the floors, while in
others, the elevators are restricted to stopping only on certain floors. Why? What is the
advantage of having elevators which travel only between certain floors?

Suppose a building has 5 floors (1-5) that are occupied. The ground floor (0) is
not used for business purposes. Each floor has 60 people working on it and there are 3
elevators (A, B, and C) available to take these employees to their offices in the morning.
Each elevator holds 10 people and takes approximately 25 seconds to fill on the ground
floor. The elevators then take 5 seconds to travel between floors and 15 seconds on each
floor on which it stops. If everyone arrives at approximately the same time and enters the
elevators on the ground (0) floor, how long will it take everyone to get to work? How
should the elevators be organized to get the employees to their offices as quickly as
possible?

A Basic Solution

One way to approach this problem is
to consider the worst case scenario. The
worst that can happen is for each elevator
to always have at least one person from
each floor on it. This means the elevator
will have to make the longest possible trip
each time. Since there are 60 people on
5 floors, there are 300 employees to take to
their offices. On average, we would expect
that each of the elevators will carry 100
people. Since the capacity of each elevator
is ten people, each elevator will make 10
trips. How long will each of these trips
take? The diagram below illustrate one
complete trip. Since no one is being picked
up or let off, the elevators do not stop on
the way down.

Figure 1: Graphical Model of an Elevator Transit

The total time of the trip is given by

T  25  510  515 150 seconds per trip.

Since each elevator makes 10 trips, the total time is 1500 seconds, or about 25 minutes.

Rerouting the Elevators

What would happen if one elevator went only to floors 1, 2, and 3, while the other
two elevators went to floors 4, and 5? How long would this configuration take to get
everyone to the proper floor?

Rather than draw a diagram to model the situation, we can create an algebraic
expression whose value gives the time of the trip. The transit time depends upon two
things, the number (N) of floors at which the elevator stops, and the highest floor (F) to
which the elevator travels. The transit time is

T  2510F 15N seconds per trip.
In the example given above, the trip to the first, second and third floors takes

T  25 10(3) 15(3)  100 seconds per trip,

while the trip to the fourth and fifth floors takes
T  25 10(5) 15(2)  105 seconds per trip.

As shown in Table 1, with this configuration, the slowest elevator takes 1260 seconds, a
reduction in total transit time of 4 minutes over having all elevators travel to all floors.

Elevators Floors People Trips Time/Trip Total Travel
Selected Carried Required (secs) Time (secs)

A&B 1, 2, 3 180 9 each 100 900

C 4, 5 120 12 105 1260

Table 1: Total Transit Time with Elevator C to Floors 4 and 5

If we let one elevator travel to floors 1 and 2 and the other two travel to floors 3, 4
and 5, then we can reduce the total transit time to 1080 seconds, or 7 minutes less than
having the elevators travel to all floors.

Elevators Floors People Trips Time/Trip Total Travel
Selected Carried Required (secs) Time (secs)

A 1, 2 120 12 75 900

B&C 3, 4, 5 180 9 each 120 1080

Table 2: Total Transit Time with Elevator A to Floors 1 and 2

Naturally, other floor combinations are possible. Remember, we base our calculations on
the worst case assumption that every elevator stops at every floor to which it can travel.

The problem can be stopped here if desired. The focus of the problem may be on
setting the simplifying assumptions, generating the time equation T  2510F 15N ,
and using it in an interesting, problem solving investigation. This problem lends itself
well to group work, as students working in groups can quickly investigate other
promising combinations of elevators. Perhaps the class can take a "field trip" to some
nearby elevators and use their time measurements and elevator capacity to set the
constants in the problem.

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Proportional Reasoning Component

Proportional reasoning is difficult for most students. The elevator problem can be
used to reinforce the concept of proportional reasoning. There are two ways in which
proportional reasoning can enter the problem. The first is in the manner in which the
problem is presented. Consider the presentation of the elevator problem in the form of a
series of memos, as suggested by Walton and Davidson in the Spode Group's wonderful
volume Solving Real Problems with Mathematics, Volume 2.

Memo #1

From: Your Boss
To: You
Re: Late Arrivals

It has come to my attention that many of our employees are arriving in the offices well
after the 9:00 a.m. starting time. They complain that the three elevators cannot handle the rush
at the start of the day. We cannot, at this time, install any extra elevators or increase the
capacity of existing ones above the current ten persons. Please investigate and let me have
some possible solutions to the problem.

Memo #2

From: You
To: Your Assistant
Re: Late Arrivals

Can you find out:
1. How long the elevators take to move between floors and how long they stop for

people to exit?
2. How many people from each floor use the lift in the morning?
3. How many people were late this morning?

Memo #3

From: Your Assistant
To: You
Re: Answers to your questions

1. The elevators appear to take 5 seconds between each floor, and extra 15 seconds for

each stop. It also seems to take about 25 seconds for the elevator to fill on the ground

floor.

2. The number of workers on each floor are:

Floor G 1st 2nd 3rd 4th 5th

Number 0 60 60 60 60 60

3. About 80 people were late today.

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From: Memo #4
To:
Re: You
Your Boss
Solutions to the lateness problem

?

In this problem statement, the students are required to write a short memo to their
boss explaining their decision. We have already seen how to reduce the time needed to
get everyone to work. Will our solution get everyone there on time? To determine if the
solution is good enough to solve the lateness problem, the students must first determine
the arrival time of the employees.

Earlier, we found that it took approximately 1500 seconds for everyone to be
taken to their offices. Our assistant found that 220 (73%) of the employees arrived on
time. Evidently, people begin arriving at a time that allows only 73% of the required
1500 seconds to occur before nine o'clock. That is, they begin to arrive approximately
1095 seconds, or about 18 minutes to 9:00. We now have a target time of 18 minutes if
we are to get everyone to their offices on time. We have already found one configuration
that accomplishes this. Is it the best solution?

A more sophisticated problem in proportional reasoning can arise in the elevator
distribution process. Consider the following elevator assignment: Elevator A goes to
floors 1, 2, and 3, while elevators B and C go to floors 3, 4 and 5. With our time
equation, it is easy to determine the time of transit, but how many trips does each elevator
make? The number of trips required of each elevator is determined by the number of
people transported. We know that elevator A will take everyone from floors 1 and 2 and
some from floor 3, while elevators B and C will take everyone from floors 4 and 5 and
some from floor 3. Since all three elevators can stop at the third floor, does each carry
one-third of the occupants of floor three?

The employees who work on the third floor will get on the first elevator that
arrives. Elevators that arrive more often will take proportionally more employees to the
third floor. Elevator A makes one trip in 100 seconds while elevators B and C require
120 seconds. Elevator A will carry a greater percentage of the employees on the third
floor than either of the other two elevators. How can we compute that percentage?

One way to have students think about this problem is to consider the number of
trips made by each elevator in 600 seconds, the lowest common multiple of 100 and 120.
Elevator A makes 6 trips while Elevators B and C make 5 trips each. In the 16 total trips,
Elevator A has made 6, or 6  0.375 of the total. So, we expect that Elevator A

556
will carry approximately 38% of the employees of the 3rd floor, while Elevators B and C
will each carry approximately 31% of the employees. This configuration is not an
improvement over earlier efforts.

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Elevators Floors People Trips Time/Trip Total Travel
Selected Carried Required (secs) Time (secs)
A
B&C 1, 2, 3 120 + 23 15 100 1500

3, 4, 5 120 + 37 8 each 120 960

Table 3: Overlapping Elevators to the Third Floor

One interesting solution involving overlapping floors is to have elevator A travel
to floors 1 and 3, elevator B travel to 2 and 4, and elevator C travel to 4 and 5. This
configuration takes only 17 minutes for everyone to arrive at work.

Elevators Floors People Trips Time/Trip Total Travel
Selected Carried Required (secs) Time (secs)
A
B 1, 3 120 12 85 1020
C 855
2, 4 60 + 29 9 95 900

4, 5 31 + 60 10 90

Table 4: Overlapping Elevators to the Fourth Floor

Probability Considerations

How realistic is our worst case assumption? For example, how likely is it that an

elevator with 10 passengers carries none of the 60 employees working on the fifth floor?

 240 60
  
This probability is given by  10  0   0.10 . Of the ten trips an elevator makes, we

 300
 
 10 

would expect that 1 of them would go only to the first 4 floors and take only 125 seconds.

In a similar manner, we find that 10% of the transits don't stop on floor 1, 10% don't stop

on floor 2, 10% don't stop on floor 3, and 10% don't stop on floor 4. Each of these

transits takes 135 seconds Of course, some of the 10% that don't stop on the 5th floor

also don't stop on the 4th. The probability that an elevator with 10 people has no one

180  120 
  
from either the 4th or 5th floors is  10  0   0.005 . This means that of the 10 trips,

 300
 
 10 

we would expect none of them to miss the top two floors. Indeed, all two floor

combinations have this same probability, since all floors have the same number of

people. Three floor combinations are even less likely. In 10 trips, it is reasonable to

think that some of the elevator transits will miss one floor, but none will miss more than

one. Of the 10 trips, we expect 5 go to all floors, while the remaining 5 each miss one

floor. A more realistic total travel time is

5150 1125  4135 1415 seconds,

or only about 24 minutes. Similar arguments can be given for elevators that are restricted
to travel only to certain floors. Having an unequal number of employees on the floors
greatly increases the importance of the probabilistic considerations.

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For many classes, these probability considerations are well beyond the abilities of
the students and the interest of the teacher. However, it is a nice application of
combinatorics and in a course in which combinatorial probabilities are discussed, it offers
an interesting and challenging extension to the problem.

Problem Variations

Teachers can increase or decrease the difficulty of the problem by changing the
number of elevators and floors. Having unequal populations on the floors significantly
increases the challenge offered by the problem. Adding in a time constraint for doors
having to reopen when the elevator is crowded changes the time equation. Students need
to determine in what situations will the door likely reopen.

For an advanced class, I suggest the following employee distribution:

Memo #3

From: Your Assistant

To: You

Re: Answers to your questions

1. The elevators appear to take 5 seconds between each floor, and extra 15 seconds for

each stop, and another 9 seconds if the doors have to reopen. It also seems to take about 25

seconds for the elevator to fill on the ground floor.

2. The number of workers on each floor are:

Floor G 1st 2nd 3rd 4th 5th

Number 0 100 100 20 100 20

3. About 80 people were late today.

Have fun!

References

Walton, John, and Bob Davidson, 1982. An Uplifting Problem. Solving Real
Problems with Mathematics, Volume 2: 37-43. Cranfield, United Kingdom: The Spode
Group, Cranfield Press.

The North Carolina School of Science and Mathematics, 1991. Precalculus
Through Applications. Providence, Rhode Island: Janson Publications.

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