In the posed elevator problem, your assistant notified you that there are 60 wor

Question

In the posed elevator problem, your assistant notified you that there are 60 workers operating on each of 5 floors atop the ground floor in your office building - 300 in total. The elevator occupancy is 10 people, and there are 3 elevators in total. In the worst-case scenario in terms of total transition time, on each elevator ride at least one person is getting off at each floor. Naturally, we can ask: what is the probability that an elevator with 10 passengers has none of the 5th floor workers? For that trip, it should reduce the elevator travel time by t45 + texit + t54 = 5 + 15 + 5 = 25sec. To see how often this might happen, we can perform random draws of elevator occupants, and average over a long time. This is a way to check the validity of certain modeling assumptions.

A back-of-the-napkin calculation for this probability can be done using simple
combinatorics. Assume that the full pool of workers are available to enter this
particular elevator. The probability of there being no 5th floor workers on the
elevator is given by:
Combinations without 5th floor employees/Combinations of all workers =(240C10* 60C0)/300C10

the binomial coefficient stated as "N choose k". Calculate this probability using
the MATLAB command "nchoosek".
Answer the following questions in your lab report:
• 1. How does your value calculated in Part 1 compare to the result of th
calculation shown in Part 2?
• 2. What is probability that a given elevator only contains workers from the
1st, 2nd, and 3rd floors?
• 3. (Tougher) Given your answers to 1. and 2., can you justify and calculate
a more realistic expected transition time than if you were to simply assume
each elevator goes to each floor on its travel?

Explanation / Answer

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 2510F 15N
seconds per trip.
In the example given above, the trip to the first, second and third floors takes
T 2510(3) 15(3) 100
seconds per trip,
while the trip to the fourth and fifth floors takes
T 2510(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
Selected
People
Carried
Trips
Required
Time/Trip
(secs)
Total Travel
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
Selected
People
Carried
Trips
Required
Time/Trip
(secs)
Total Travel
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

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