Issue 3/2004
05/02/04
Up-peak traffic calculation – an improved Procedure
Dr.-Ing. Julian Jimenez
The conventional procedure to calculate the expected number of stops and the highest reversal floor of lift systems in up-peak condition with floors equally populated and passenger arrivals uniformly distributed on time, implies that the number of stops made by the lift is a random variable with a binomial distribution. This assumption is not true. The improved procedure described below to do this calculation allows us not only to find the correct expected values of these variables but also the detailed probability distribution of them.
Category: Issue 3/2004
Posted by: Editor
Conventional calculation
The conventional procedure to calculate the expected number of stops (S*) and the highest reversal floor (H*) of lift systems for the up-peak traffic situation, for floors with equal populations and assuming that passengers arrive uniformly in time, leads to the well known expressions:
where N is the number of floors served by the lift and m the number of passengers in the car.
This calculation procedure implies the assumption that the number of stops made by the lift is a random variable with a binomial distribution. In this case, the probabilities that the lift stops k times, from k = 0 to k = N, are given by the terms of the binomial development [3]
This calculation procedure implies the assumption that the number of stops made by the lift is a random variable with a binomial distribution. In this case, the probabilities that the lift stops k times, from k = 0 to k = N, are given by the terms of the binomial development [3]
where p and N are the parameters of the distribution and q = 1 – p. The parameter p is the probability that the lift stops in a particular floor. As all floors have the same population, the probability that one of them will be the destination floor of one of the m car passengers is 1/N. The probability that this passenger will have another floor destination is (1-1/N) as both events are mutually exclusive. Hence, the probability that this floor do not be the destination of any passenger is
In this case, the lift does not stop in this particular floor. The probability of the opposite event, that is to say, the lift stops in this floor will be
Evidently, the probability that the lift performs a number of stops between o and k is
As p + q = 1, the sum of the terms of [3] must be also equal to 1, in accordance with the binomial formula.
The probability distribution of the number of stops (S) being a binomial one, of parameters p and N, the mean value of S will be S* = N p, that is the same expression [1]. The variance V(S) will be equal to N p q.
Review of conventional calculation
In fact, the assumption made by the conventional up-peak traffic calculation procedure exposed above in the sense that de the number of stops performed by a lift follows a binomial probability distribution in not true. This can be seen because this assumption leads to some erroneous conclusions:
i) The lift can make zero stops, which corresponds to the first term of [3]. This conclusion is obviously erroneous as a lift in up-peak motion must stop at least one time.
ii) The number of stops can be greater than the number of passengers in the car, which corresponds to the terms of [3] for m < k ≤ N. This conclusion is also erroneous as it is clear that the greatest number of stops that can be made by a lift in up-peak condition is equal to the number of the passengers carried.
As a consequence of these facts, the sum of the binomial expression [3] cannot be equal to 1 and hence the binomial distribution is not the actual distribution of the lift stops.
Combinatorial calculation
We can obtain the true probability distribution of the number of stops made by a lift in up-peak traffic situation for a population likely distributed in the floors served by the lift by means of the combinatorial analysis, in other words evaluating the probability that the lift stops a certain number of times that is the quotient between the number of the possibilities of this event by the number of all the possibilities. In the same way we can also obtain the distribution of the highest reversal floor. The calculations are easy to carry out using a simple spreadsheet.
Distribution of the number of stops
This approach consists in fact in evaluating all arrangements of car passengers depending on their floor destination. Each of these arrangements can be represented by a set of m integer numbers with variable values from 1 to N, m being the number of passengers carried by the car and N the number of served floors.
One of these arrangements, for m = 8 and N = 10 could be the following:
One lift leaving the main floor with this arrangement will stop 4 times, as many as the different values existing in the arrangement and precisely in the floors (5), (8), (9) and (10). Among the 8 passengers, two will leave the car in the 5th floor, one in the 8th floor, two in the 9th and three in the 10th floor.
One conclusion is that one lift in this traffic condition with 8 passengers can stop a minimum of one time and a maximum of eight times. The first case will occur when all passengers have the same floor destination and the second when each of them has a different one.
One conclusion is that one lift in this traffic condition with 8 passengers can stop a minimum of one time and a maximum of eight times. The first case will occur when all passengers have the same floor destination and the second when each of them has a different one.
Consequently, the number of stops made by the lift will be equal to the number of different groups that we can form wit them attending to their floor destination. The fact that the lift stops s times implies that there are s different groups of passengers depending on their floor destination which can be occur in
different cases.
In the example mentioned above where N = 10 y s = 4 will exist
passengers orderings with four different floor destinations among the 10 possible ones according to [8].
Now we consider one of the possible car passengers arrangements i.e. the showed in [7] for a car with 8 passengers. Is evident that if we permute the numbers, each of these permutations will be a possible arrangement of car passengers but in all of them the lift will stop the same number of times and in the same floors. The total number of these arrangements will be the permutations of 8 numbers in which the (5) appears two times, the (8) one time, the (9) two times and the (10) three times.
In general, the number of permutations of m elements in which there are s different elements, one repeated n1 times, another n2 times and so on until the element s repeated ns times, is [10]:
This formula can be easily deduced if we consider that the number of permutations of m different elements is m! but the n1!, n2!, ......, ns! permutations we can form permuting the repeated elements of each group are indistinguishable.
For example, consider the sequence [7] in which n1 = 2, n2 = 1, n3 = 2 and n4 = 3, being (n1 + n2 + n3 + n4) = 8. The number of different permutations we can form with these numbers will be:
Consequently, the number of different car passenger arrangements we can form with the numbers 1 to m, representing their floor destinations, in which there are s different numbers and producing hence s stops is:
being (n1 + n2 + …… + ns) = m and n equal to the number of possible breakdowns of the number m in s addends, the highest being (m–s+1) and the smallest the unit.
Thus, the total number of the possible arrangements of car passengers will be:
The probability that the lift stops s times will be:
This expression is hence the probability distribution of the number of stops made by the lift.
The expected or main value of this variable will be evidently:
The expected or main value of this variable will be evidently:
In the above mentioned example, for N = 10 and m = 8 we will get the results of the Table 1.
By the conventional calculation procedure and according to [1] we would obtain for S* a value equal to 5,6953.
Probability distribution of the highest reversal floor
Now we will return to [7] that is one of the possible arrangements of car passengers, formed by 4 different numbers and thus implying 4 lift stops. The reversal floor will be the highest number included in this arrangement, that is to say 10. This will be the reversal floor for all arrangements formed by these four numbers regardless the permutations we can do with them according to [10].
If we consider then the example in which we have N = 10 possible stops, m = 8 passengers and s = 4 different numbers, we can wonder in how many cases the 10th floor will be the reversal floor. Evidently, the number (10) will be the highest floor reached by the lift in the up-peak trip in all cases it appears, precisely in C9³ = 84 cases. The number (9) will be the highest floor in the cases it will appear with the remaining numbers except wit (10), precisely in C8³ = 56 cases. We can repeat the reasoning until the number (4) that can be the highest floor when it will appear with (3), (2) and (1), exactly in C3³ =1 case. The numbers (3), (2) and (1) can not be the highest floor in theses cases in which we have four numbers as the fourth will be necessarily higher. Therefore, the total number of possible combinations of the N = 10 elements we can form with s = 4 different numbers can be expressed with the equation [16]:
The first addend of the second member of this expression represents the only case in which (4) is the highest reversal floor, the second the 4 cases in which the (5) is it as so on until the seventh term that represent the 84 cases in which (10) is the highest floor. In all of these cases, the permutations we can form with the 8 numbers of each of them according to [9] do not modify the fact that the highest reversal floor will be the same that the corresponding to the initial combination.
Therefore, according to [16] the number of combinations in which H (1 ≤ H ≤ N) will be the highest reversal floor for a given number of stops s (1 ≤ s ≤ m) will be:
This value will be zero for H ≤ s-1 as (1) can’t be the highest floor in the combinations consisting in two different numbers, the (2) in the combinations of three different numbers and so on.
The total number of cases in which a given number H will be the highest reversal floor for all the possible values of s will be expressed as
being (n1 + n2 +…..+ ns) = m
For example, according to the preceding considerations and the results showed in Table 1, we can observe that in the case N = 10 and m = 8, the number of passenger arrangements in which H = 7 will be the highest reversal floor for s = 5 is:
For example, according to the preceding considerations and the results showed in Table 1, we can observe that in the case N = 10 and m = 8, the number of passenger arrangements in which H = 7 will be the highest reversal floor for s = 5 is:
The total number of cases in which H will be the highest reversal floor is hence:
On the other hand is evident that
The probability that H be the highest reversal floor will be of course
This last expression represents the distribution of the probability of the highest reversal floor H.
The expected or mean value of the highest reversal floor will be:
The probability distribution of the highest reversal floor and the calculation of the expected value of this variable is showed in Table 2 for the case N = 10 y m = 8.
Each cell of this Table contents the number of cases in which the floor corresponding to its row is the highest reversal floor for a number of stops corresponding to its column. The result is given by [18]. For example we can observe that for H = 7 and s = 5 the result is the same given by [19]. The last but one column contents the total cases in which the row value is the highest reversal floor. We can observe that the total sum of this column is the same that the Qs column in Table 1 as both represent the total possible car passenger arrangements for N = 10 and m = 8. The last column contents the distribution of probability of H according to [22].
Applying [23] we obtain H* = 9,36. With the conventional calculation procedure we would obtain H* = 9,92.
The differences of the results obtained with both procedures increase with the difference between the values of N and m respectively.
Conclusion
The procedure described above for the calculation of the number of stops and the highest reversal floor of a lift in up-peak traffic condition with floors equally populated and passenger arrivals uniformly distributed on time allows us not only to find the correct expected values of these variables but also de detailed probability distribution of them.
3/2004
