Mathematical and Computer Modelling 52 (2010) 545–555

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Mathematical and Computer Modelling

journal homepage: www.elsevier.com/locate/mcm

The effect of multiple entrances on the elevator round trip time underup-peak trafficLutfi Al-Sharif ∗Mechatronics Engineering Department, University of Jordan, Amman 11942, Jordan

a r t i c l e i n f o

Article history:Received 31 October 2009Received in revised form 27 March 2010Accepted 29 March 2010

Keywords:ElevatorLiftRound trip timeIntervalUp peak trafficBasement

a b s t r a c t

The design of vertical transportation systems relies on the calculation of the interval as anindicator of the quality of service. This in turn involves the accurate calculation of the roundtrip time of a single elevator. The calculation of the round trip time is well established forthe single entrance scenario (main entrance) but not for the case of multiple entrances.This paper presents two methods for evaluating the effect of the multiple entrances

on the value of the round trip time equation. One of the methods is based on the useof a weighted average of the round trip time of all possible scenarios, but it becomesimpractical when the number of basements exceeds two. The second method finds theweighted average of the round trip for two specific scenarios: no basements served andbasements served and uses the basement equivalent of the highest reversal floor (HB) andthe basement equivalent of the probable number of basement stops (SB). Verification of theresult using a Monte-Carlo simulator is carried out.The formula for calculating the interval as a performance indicator at any of the

entrances is also introduced.© 2010 Elsevier Ltd. All rights reserved.

1. Introduction

The traditional method used in the design of vertical transportation systems is the calculation of the round trip timefor an elevator during the up-peak traffic. This assumes that all traffic is entering the building (i.e., incoming) and that theelevator picks up the passengers from the main entrance and delivers them to the upper levels. The round trip time (τ ) isthe cycle time taken by the elevator to pick up the passengers from the main entrance, deliver them to the upper levels andthen return back to the main entrance.The value of the round trip time (τ ) for an elevator during up peak conditions can be calculated as follows [1,2]:

τ = 2 · H ·(dfv

)+ (S + 1) ·

(tf −

dfv+ tdo + tdc + tsd − tao

)+ P

(tpi + tpo

)(1)

where:

τ is the round trip time in sH is the highest reversal floor (where floors are numbered 0, 1, 2 N)S is the probable number of stops (not including the stop at the ground floor)df is the typical height of one floor in mv is the top rated speed in m s−1

∗ Tel.: +962 6 5355000x23025, mobile: +962 796 000 967; fax: +44 207117 1526.E-mail address: lal-sharif@theiet.org.

0895-7177/$ – see front matter© 2010 Elsevier Ltd. All rights reserved.doi:10.1016/j.mcm.2010.03.053

546 L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555

tf is the time taken to complete a one floor journey in s assuming that the elevator attains the top speed vP is the number of passengers in the car when it leaves the ground floortdo is the door opening time in stdc is the door closing time in stsd is the motor start delay in stao is the door advance opening time in s (where the door starts opening before the car comes to a complete standstill)tpi is the passenger boarding time in stpo is the passenger alighting time in s.

The probable number of stops S for equal floor populations was first derived in [3]. It can be calculated as follows for equalfloor populations:

S = N ·

(1−

(1−

1N

)P)(2)

where N is the number of floors above the main entrance (ground floor). It is calculated as follows for unequal floorpopulations (where Ui is the population of floor i and U is the total building population) [1,2]:

S = N −N∑i=1

(1−

UiU

)P. (3)

The highest reversal floor H was first derived in [4]. It can be calculated as follows for equal floor populations [4–7]:

H = N −N−1∑i=1

(iN

)P(4)

. . . and as follows for unequal populations (where Ui is the population of floor i and U is the total building population) [1,2]:

H = N −N−1∑j=1

(j∑i=1

UiU

)P. (5)

A third parameter that can be used in the derivation of the round trip time for an elevator in the up peak conditions is thelowest call express (LCE) [8]. However, it is shown that its effects are so insignificant on the final value of the round triptime [8] and it is thus not used in practice.The formula for the round trip time implicitly makes a number of assumptions:

1. The elevator cabin will fill up to a total number of P passengers (usually assumed to be 80% of the rated car capacityCC [1,2]).

2. The floor heights are equal.3. The top speed is attained in one floor to floor journey.4. All passenger arrivals are from the main entrance. It is assumed that this is the only entrance (i.e., no basement or otherentrances).

The first assumption is not significant and can be addressed by adjusting the number of passengers P . The effect of the secondassumption can be accounted for by taking the average of the height of all floors without a major loss of accuracy. However,the third assumption is very significant especially at high rated speeds. For speeds above 2.5 m/s and typical floor heightsof around 4 m, the error can be become very significant. This issue is addressed by the work in [9] that amends the roundtrip time equation to account for the fact that the top speed is not attained during one floor jump.The fourth assumption is also very significant especially with large numbers of basements having a high percentage of

the arrivals.This paper addresses the case of multiple entrances and develops a methodology for calculating its effect on the round

trip time during up peak traffic conditions.An approach for dealing with basements during up peak conditions has been introduced in [2,10–12]. The method uses

the values for the highest reversal floor and the probable number of stops for the basement. No explanation however isgiven as to the basis for arriving at the values of H and S for the basement and the values of both parameters are estimated.The method also incorrectly assumes that the elevator will go to the basement in every elevator round trip journey.A general set of equations dealingwith up peak traffic from one entrance are developed in [13,14] and then verified using

simulation methods in [13].This paper develops two methods for calculating the effect of multiple entrances on the value of the round trip time

equation. The formula developed for the second method is simple and can be carried out manually without the need forsoftware. The paper also examines how the interval (as a performance indicator) varies on the different entrances. Theinterval is the time cycle between consecutive elevator arrivals at an entrance (usually the main entrance).

L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555 547

Fig. 1. Demand and performance model of an elevator system.

2. Multiple entrances

Multiple entrances are said to exist where passenger arrivals take place frommore than one entrance. The most obviouscase is the case of basements (e.g., basement car parks) where a significant number of passengers board the elevator fromthe basement floors. The ground floor represents the main entrance (ME) and the other entrances (e.g., the basements)represent the sub-entrances (SE).The sub-entrances will not necessarily always be basement floors. Some buildings have two main entrances at two

different street levels and can be analysed using the same methodology presented here. However, for simplicity the sub-entrances will be referred to as basements. The terms ground and main-entrance will be used synonymously.The condition under which all traffic is entering the building is referred to as the up peak scenario or incoming traffic

only. This is shown graphically in Fig. 1 for the case of multiple entrances. Passengers board the elevator from the basementfloors and the ground floor and are then delivered to the upper floors based on their destination selections.Each of the entrances has a percentage arrival probability thatwill be referred to as Parr (i)which stands for the probability

of a passenger arriving from entrance floor i. The entrance floors range from 0, B1 to NB. The number of basements is NB, inaddition to the ground floor denoted as 0. Each of the upper floors has a destination probability Pdes (i) that will be referredto as which stands for the probability of a passenger heading to the destination floor i. The destination floors range from 1to N .So one complete cycle for the elevator round trip comprises the following stages:

(a) Picking up a total of P passengers (some from the sub entrances and then the rest from the main entrance).(b) The P passengers will select their destinations from 1 to N .(c) The elevator travels to the upper floors and delivers all the P passengers to their destinations.(d) The probability of a passenger selecting a floor as a destination depends on the percentage population on that floor.(e) Once the elevator finishes at the topmost served floor it then reverses direction, and expresses back to the basements(sub-entrances) bypassing the ground floor (main entrance) without stopping.

(f) It stops at the ‘furthest’ basement floor from the main entrance at which there is a call.(g) It then reverses direction and travels in the up direction picking up any other passengers originating at the basements(sub-entrances).

(h) It then stops at the ground floor (main entrance) and picks up the remaining passengers (usually the majority ofpassengers will originate at the ground floor (main entrance).

(i) The probability of a passenger originating at an entrance floor depends on the percentage population arrival from thatentrance.

(j) The cycle then repeats again.(k) The elapsed time between the doors opening at the lowest-most entrance floor (basement) until the door re-open againat the lowest-most entrance floor in the next cycle is the round trip time (τ ).

This cycle is shown graphically in Fig. 2, where the x-axis represents time, the y-axis represents vertical position within thebuilding and the solid line represents the position of the elevator in the building at any point in time.

548 L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555

Fig. 2. Round trip time timeline where the elevator goes down to the basement.

Under these conditions where the elevator serves basements, the round trip time is amended as follows (assuming thatthe elevator goes to the basement in every journey):

τB = 2 · H ·(dfv

)+ (S + 1) ·

(tf −

dfv+ tdo + tdc + tsd − tao

)+ P

(tpi + tpo

)+ 2 · HB ·

(dBv

)+ (SB) ·

(tB −

dBv+ tdo + tdc + tsd − tao

)(6)

where:τB is the average round trip time of the elevator where the elevator goes to the basement in every journeyHB is the ‘highest’ (furthest from ground) reversal floor in the basementSB is the probable number of stops in the basementdB is the basement floor height in mtB is the time taken to travel from one basement floor to the next in s.

It is worth noting that a subscript of either G, B or GB will be added to the symbol of the round trip time (τG, τB, τGB) inorder to distinguish between the three possible round trip journeys. The second of those has been introduced in Eq. (6). τGis identical to τ shown in Eq. (1). τGB will be introduced later in the paper.

3. Sub-entrance (basement) equation derivation

Before introducing the methods for calculating the effect of the multiple entrances on the round trip time two importantformulae are first derived.

3.1. Calculation of the probability of stopping at an entrance

The probability of stopping at any of the entrances (whether the main entrance or the sub-entrance) can be derived asfollows:The probability of any passenger originating (entering) at one of the basement floors i is equal to:Parr(Bi). (7)

The probability of any passenger j not stopping at that basement i is:1− Parr(Bi). (8)

The probability that all P passengers will not originate at that entrance is (assuming that their decisions are independent):

(1− Parr(Bi))P . (9)Thus the probability that at least one passenger will originate from a basement i is (which is the probability that a stop willtake place on that basement):

Pstop (Bi) =(1− (1− Parr (Bi))P

)(10)

where:Pstop (Bi) is the probability that the elevator will make a stop at the basement i

L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555 549

Parr (Bi) is the probability that a passenger jwill originate at the basement floor iP is the number of passengers boarding the elevator form all the entrances in one journey.

3.2. Calculation of the probability of going to the one of the basements

In this sub-section, the probability that the elevator will go to the basement in any one journey is derived. This is theprobability that the elevator will make at least one basement stop in any round trip journey.The probability that all P passengers will not originate at basement i is equivalent to the probability that a stop will not

take place at that basement and is equal to:

(1− Parr(Bi))P . (11)The probability that the elevator will not go to any of the sub-entrances (i.e., basements) will be equal to the product of theprobability of not stopping at each of the sub-entrances:

(1− Parr (B1))P · (1− Parr (B2))P · · · · (1− Parr (BNB))P =NB∏i=1

(1− Parr (Bi))P . (12)

The probability that the elevator will go to at least one of the basements (sub-entrances) in a round trip journey is equal tothe probability of the elevator going to the basement in any one round trip journey:

Pstop (B) =

(1−

NB∏i=1

(1− Parr (Bi))P)

(13)

where:Pstop (B) is the probability that the elevator will go to the basement in any one round trip journey

NB is the number of basement floorsParr (Bi) is the probability that a passenger will originate at the basement floor i

P is the number of passengers boarding the elevator from all entrances in one round trip journey.

3.3. Notes on the practical values of the basement probabilities

In this sub-section, practical values of percentage arrivals from the ground floor (main entrance) and the basements (sub-entrances) are used to draw general conclusions regarding the probability of stopping at the ground floor (main entrance).In practice within real life buildings, the percentage of arrivals from the ground floor (main entrance) which is equal to

the probability of a passenger originating (arriving) on that floor is rarely less than 50% (and in most cases is as high as 80%).For example a typical office buildingmight have four car park basementswith 5% arrival from each basement (e.g., car parks)and 80% arrival from the ground floor (e.g., public transport). A typical elevator rated car capacity for such an office buildingwould be 1600 kg/21 persons. P will be equal to 80% of that value which equates to an average of 16.8 persons. Under anup-peak traffic scenario within this building, the probability of the elevator stopping at the ground floor (main entrance) inany one round trip journey can be calculated from (10) as follows:

Pstop (G) =(1− (1− Parr (G))P

)=(1− (1− 0.8)16.8

)=(1− 1.8× 10−12

)≈ 1.

Even where the elevator car is only rated at 1000 kg/13 persons and the percentage of arrivals from the ground floor is aslow as 50%, the probability of the elevator stopping at the ground floor (main entrance) is:

Pstop (G) =(1− (1− 0.5)10.4

)=(1− 7.4× 10−4

)≈ 1.

So from a practical point of view, it can be safety assumed that elevator will make a stop at the ground floor in every roundtrip journey.

4. The weighted average round trip time method

The firstmethod examines the possible combination of stops at the entrances (ground and basements). Each combinationis referred to as a scenario. Each scenario has a probability of taking place based on the probability of a stop at each entrance.The probability of the elevator stopping at an entrance in any one round trip journey can be found using (10).For each scenario the change in the value of the highest reversal floor (H) and the probable number of stops (S) for that

scenario is then evaluated. These new amended values of H and S are then used to calculate the relevant round trip time(τ ) for that scenario. The effective round trip time (τ ) is then found by evaluating the weighted average of all the round triptimes for all the scenarios.The number of possible scenarios SC can be found as follows:

SC = 2(NB+1) (14)where:SC is the number of possible scenariosNB is the number of basement floors.

550 L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555

The effective round trip timewhere the elevator serves the basements can then be found using theweighted average of allthe scenarios (sum of the products of the round trip time for each scenario and the probability of that scenario taking place):

τGB =

SC∑i=1

(τsc (i) · Psc (i)) (15)

where:τGB is the average round trip time of the elevator where it serves the ground and basements (but will not necessarily goto the basement in every round trip journey)

Psc(i) is the probability that a certain scenario i (combination of entrance stops) will take place in any one round trip journeyτsc(i) is the round trip time for the journey of a certain scenario i (combination of entrance stops)SC is the total number of possible scenarios.

The probable number of stops in each journey is adjusted based on the number of basement stops. The ‘highest’ (i.e., furthestfrom the ground floor) floor is adjusted based on the furthest basement at which the elevator stops.

5. The HBSB method

The previous method becomes impractical as the number of basement floors increases (e.g., 32 scenarios for fourbasement floors).An alternative simpler method is based on the H and S approach to the basement. It is adapted from the approach for

dealing with basements during up peak conditions that has been introduced in [2,10–12]. This method finds the ‘highest’reversal floor in the basements and the number of stops in the basement and denotes them asHB and SB respectively, so thatthe subscript B refers to basement variables. ‘Highest’ in this context can be translated to the term ‘furthest’ from the mainentrance within the basement. There are two improvements in this method compared to the original method:1. The original method assumes approximated values. The method introduced here calculates exact values for HB and SB.2. The original method assumes that the elevator goes to the basement in every round trip. The method introduced heredeals with the fact that the elevator will not go to the basement in every round trip.

Themain difference betweenH and S compared toHB and SB is the fact that all P passengers contribute the values ofH and S,while only a fraction of the P passengers contribute to the values of HB and SB. Only passengers boarding from the basementfloors will contribute to the value of HB and SB and these passengers will be denoted as PB.Two different round trip times are then derived: one is the original round trip time where the elevator does not go to the

basement. This will be referred to here as τG. The other round trip time is the onewhere the elevator does go to the basementin every round trip journey, making the probable number of basement stops and travelling as far as the ‘highest’ basementreversal floor. This will be referred to as τB. In reality the elevator will not go to the basement in every journey, based onthe probability of going to the basement. The difference between the two values of the round trip time will be referred to as1τGB. The ultimate aim is to calculate the true round trip time with the presence of the basement. It is based on the premisethat the elevatorwill not go to the basement in every round trip. Thus it is the value of the round trip timewhere the elevatorgoes to the basement depending on the probability of arrival from the basement floors. This will be denoted as τGB and thisis the round trip time that needs to be calculated.A new probability is defined P(basement)which is the probability of the elevator going to the basement in any one round

trip. The probability of not going to the basement in any one round trip is the complement, (1-P(basement)).Thus the actual round trip time (referred to as τGB) can be found by finding theweighted average of both round trip times,

as follows.τGB = ((1− P(basement)) · τG)+ (P(basement) · τB) (16)τGB = τG + P(basement) · (τB − τG) (17)τGB = τG + P(basement) ·1τGB. (18)

The difference between the two round trip times (i.e., the round trip time where the elevator goes to the basement and theround trip time where the elevator does not go to the basement) can be found by examining the difference between thetwo Eqs. (1) and (6). The difference between the two round trip times is represented by the two extra terms involving thebasement as follows:

τGB = τG +

(P(basement) ·

((2 · HB ·

dBv

)+ (SB · tSB)

))(19)

where:tSB is the time delay caused by a stop in the basement in sτG is the round trip where the elevator does not go to the basement in sτB is the round trip time where the elevator goes to the basement in every trip in sτGB is the round trip time where the elevator goes to the basement in some of the journeys depending on the probabilityof arrival from the basements compared to the probability of arrival from the main entrance (ground floor)

1τGB is the difference between τB and τG (i.e., extra incurred when going to the basement in a round trip).

L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555 551

The equivalent ‘highest’ reversal floor (HB) and the probable number of stops for the basement (SB) can be found by usingthe same formulae used for the floors above ground. HB is substituted for H , SB is substituted for S, PB is substituted for Pand NB is substituted for N . For example, for equal percentage arrival rates from the basements, the probable number ofbasement stops can be evaluated as follows, using (2):

SB = NB ·

(1−

(1−

1NB

)PB)(20)

. . . and the ‘highest’ reversal floor for the basement (i.e., furthest from the ground floor) can be found as follows for equalpercentage arrival rates from the basements using (4):

HB = NB−NB−1∑i=1

(iNB

)PB(21)

where percentage arrivals from the basement floors are unequal, then formulae (3) and (5) can be used for SB and HBrespectively.The critical parameter that decides the value of SB and HB is PB which is the number of passengers that are picked up

from the basements as shown in (20) and (21). This is decided by the total percentage arrivals from the basement and canbe evaluated as follows:

PB = P ·NB∑i=1

Parr (Bi) . (22)

6. Case study

An example from an office building is now used to apply the two methods. The parameters for a typical office buildingare shown below.(a) Number of floors above ground is 14 floors.(b) Number of basement floors below the ground floor is 3 (denoted as B1, B2 and B3).(c) Car capacity is 21 person (1600 kg).(d) Floor height is 4.5 m (finished floor level to finished floor level).(e) Basement floor height is 4.5 m.(f) Top value of speed, v, is 1.6 m s−1.(g) Top value of acceleration, a, is 1 m s−2.(h) Top value of jerk, j, is 1 m s−3.(i) Passenger transfer time out of the car is 1.2 s.(j) Passenger transfer time into the car is 1.2 s.(k) Door opening time is 2 s.(l) Door closing time is 3 s.(m) Advanced door opening is 0.5 s.(n) Start delay is 1 s.(o) Total building population of 1100 persons.(p) Equal floor populations.(q) Probability percentage arrivals from the basements is 5% from each of the three basements.

The first step is to calculate the round trip time without the basement (τG). The equation for the round trip time where theonly entrance is the ground floor (τG) can be written as follows (adapted from (1) with the subscript G added):

τG = 2 · H ·(dfv

)+ (S + 1) ·

(tf −

dfv+ tdo + tdc + tsd − tao

)+ P

(tpi + tpo

). (23)

A check needs to be carried out to ensure that the elevator will attain top speed in a one floor journey, as follows [15]:

df = 4.5 ≥(a2v + v2jaj

)=

(1 · 1.6+ 1.62 · 1

1

)= 4.16 m. (24)

So the top speed of 1.6 m/s will be attained in a one floor jump. In this case, the time taken to complete a one floor journey,tf will be [15]:

tf =dv+v

a+aj=4.51.6+1.61+11= 5.4 s. (25)

The value of P based on 80% car rated capacity is 16.8 passengers (on average). Moving on to calculate H and S as follows(assuming equal floor population), using (2) for S:

S = N ·

(1−

(1−

1N

)P)= 14 ·

(1−

(1−

114

)16.8)= 9.97 (26)

552 L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555

Table 1Probabilities of stopping at each entrance in any one round trip journey (calculated using (10)).

Level Percentage arrival Probability of stopping at the floor

G 85 1.000B1 5 0.578B2 5 0.578B3 5 0.578

Table 2Calculation of the probability of the possible 16 scenarios.

Level G B1 B2 B3 Resultant Probability for floor Product of the fourprobabilities

Prob. of stop at thisfloor (from (10))

1.000 0.578 0.578 0.578 P(0) orP(0)

P(B1) orP(B1)

P(B2) orP(B2)

P(B3) orP(B3)

Probability ofscenario k Psc(k)

Scenario 0 0a 0 0 0 0.000 0.422 0.422 0.422 0.000000Scenario 1 0 0 0 1 0.000 0.422 0.422 0.578 0.000000Scenario 2 0 0 1 0 0.000 0.422 0.578 0.422 0.000000Scenario 3 0 0 1 1 0.000 0.422 0.578 0.578 0.000000Scenario 4 0 1a 0 0 0.000 0.578 0.422 0.422 0.000000Scenario 5 0 1 0 1 0.000 0.578 0.422 0.578 0.000000Scenario 6 0 1 1 0 0.000 0.578 0.578 0.422 0.000000Scenario 7 0 1 1 1 0.000 0.578 0.578 0.578 0.000000Scenario 8 1 0 0 0 1.000 0.422 0.422 0.422 0.075382Scenario 9 1 0 0 1 1.000 0.422 0.422 0.578 0.103066Scenario 10 1 0 1 0 1.000 0.422 0.578 0.422 0.103066Scenario 11 1 0 1 1 1.000 0.422 0.578 0.578 0.140917Scenario 12 1 1 0 0 1.000 0.578 0.422 0.422 0.103066Scenario 13 1 1 0 1 1.000 0.578 0.422 0.578 0.140917Scenario 14 1 1 1 0 1.000 0.578 0.578 0.422 0.140917Scenario 15 1 1 1 1 1.000 0.578 0.578 0.578 0.192668

Check 1.000000a 1 stop at the floor; 0 no stop at the floor.

. . . and using (4) for H:

H = N −N−1∑i=1

(iN

)P= 14−

13∑i=1

(i14

)16.8= 14− 0.288− 0.075− 0.0174 · · · = 13.62. (27)

Substituting in the round trip time equation (23) gives:

τG = 2 · H ·(dfv

)+ (S + 1) ·

(tf −

dfv+ tdo + tdc + tsd − tao

)+ P

(tpi + tpo

)= 2 · 13.62 ·

(4.51.6

)+ (9.97+ 1) ·

(5.4−

4.51.6+ 2+ 3+ 1− 0.5

)+ 16.8 · (1.2+ 1.2)

= 76.5+ 89+ 40.3 = 205.77 s. (28)

The problem can now be solved by the using the weighted average round trip time method. The probabilities of stopping ateach of the entrances are first calculated using (10). The resultant probabilities are shown in Table 1. As discussed earlier,the probability of stopping at the ground floor is effectively 100%.For each scenario the probability of that scenario taking place is calculated based on the individual entrance stopping

probabilities that make up that scenario. As predicted by (14), the number of scenarios is 16 (which is calculated using2(3+1)).Each of the possible combinations are denoted as a scenario. The index k is used to denote scenario k out of the 16 total

possible scenarios. The index for the scenarios runs from 0 to 15 as shown in Table 2. For example, scenario 9 is the scenariowhere the elevator makes a stop at B3, no stop at B2, no stop at B1 and a stop at the ground floor (0). When this patternis spelt out as ones and zeros (a one for a stop and a zero for a non-stop) it produces the binary equivalent of 9 which is1001. This is shown in Table 2 in columns 2 to 5, which are populated by ones and zero depending on whether the scenarioinvolves a stop or a non-stop at the corresponding level.The probability of a scenario taking place is the product of the four probabilities that correspond to the four events at the

four entrance floors. For each floor we use the probability of stopping at that floor if the event is true (i.e., a one representinga stop) or the complement of that probability of stopping at that floor if the event is false (i.e., a zero representing a non-stop). The complement of the probability of a stop at the entrance floor is the probability of stopping at the entrance floorsubtracted from 1.

L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555 553

Table 3Calculation of the weighted average round trip time for basement service.

Psc(k) Change in S Change in H Sk Hk τk(s) Weighted τk(s)

0.000000 0 0 10 13.6 205.92 2.24038E−130.000000 1 3 11 16.6 222.795 3.31418E−130.000000 1 2 11 15.6 217.17 3.2305E−130.000000 2 3 12 16.6 230.895 4.69603E−130.000000 1 1 11 14.6 211.545 3.14683E−130.000000 2 3 12 16.6 230.895 4.69603E−130.000000 2 2 12 15.6 225.27 4.58163E−130.000000 3 3 13 16.6 238.995 6.64588E−130.075382 1 0 11 13.6 205.92 15.522734740.103066 2 3 12 16.6 230.895 23.797479720.103066 2 2 12 15.6 225.27 23.217732110.140917 3 3 13 16.6 238.995 33.678441230.103066 2 1 12 14.6 219.645 22.63798450.140917 3 3 13 16.6 238.995 33.678441230.140917 3 2 13 15.6 233.37 32.885783510.192668 4 3 14 16.6 247.095 47.60733326

τGB 233.03

So taking scenario 9 again, we can calculate the probability of it taking place by multiplying the probably of its fourelements taking place. The four elements are:

(a) A stop at the ground floor. The probability of this happening based on is 1.0.(b) No stop at B1: The probability of this happening is the complement of the stop. The probability of the stop is 0.578. Thusthe probability of a stop not happening at B1 is 1− 0.578 = 0.422.

(c) No stop at B2: The probability of this happening is the complement of the stop. The probability of the stop is 0.578. Thusthe probability of a stop not happening at B1 is 1− 0.578 = 0.422

(d) A stop at B3: The probability of a stop at this entrance is 0.578.

The product of these four probabilities is the probability of this scenario taking place. This equals 1.0 × 0.422 × 0.422 ×0.578 = 0.103, as shown in Table 2.For each scenario the change in H and S is found and thus the new values of H and S are found. The corresponding round

trip time is then calculated using the basic round trip time equation (1). This is shown in Table 3. The weighted average ofall the round trip times for all the scenarios is then calculated using (15). This gives a round trip time with basement serviceof 233.03 s. Index k is used to denote the scenario number.The problem can now be also solved using the HBSB method as follows. There are three basements so NB is 3. PB is

the number of passengers arriving from the basement in each elevator round trip journey. It can be calculated as followsusing (22):

PB = P ·NB∑i=1

Parr (Bi) = 16.8 · (0.05+ 0.05+ 0.05) = 2.52. (29)

So on average the elevator will pick up 2.52 passengers from the basements, when it does go to the basement in a round tripjourney. SB and HB can be found as follows using (20) and (21) respectively:

SB = NB ·

(1−

(1−

1NB

)PB)= 3 ·

(1−

(1−

13

)2.52)= 1.92 (30)

HB = NB −NB−1∑i=1

(iNB

)PB= 3−

2∑i=1

(i3

)2.52= 3− 0.063− 0.36 = 2.577. (31)

Thus the extra term (τGB) becomes using (19):

1τGB = 2 · HB ·(dfBv

)+ (SB) ·

(tf −

dfBv+ tdo + tdc + tsd − tao

)= 2 · 2.577 ·

(4.51.6

)+ (1.92) ·

(5.4−

4.51.6+ 2+ 3+ 1− 0.5

)= 14.5+ 15.53 = 30 s. (32)

554 L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555

The probability of the elevator going to the basement in any one round trip journey can be found as follows using (13):

Pstop (B) =

(1−

NB∏i=1

(1− Parr (Bi))P)

= 1−((0.95)16.8

)·((0.95)16.8

)·((0.95)16.8

)= 0.9246. (33)

Substituting the result for the difference in (18) gives the value of the round trip time where the elevator serves the groundand the basements as follows:

τGB = τG + P (basement) ·1τGB = 205.77+ 0.9246 · 30 = 233.5 s. (34)

There is a difference of around 0.5 s between the two methods (around 0.2%).In order to verify the two methods, the Monte Carlo simulation method is used. This method generates random arrivals

from the basements and random destinations to the destination floors based on the probabilities of arrival and destination.This random generation specifies the stops at the basements and the stops at the destination floors. For each randomcombination the round trip time is calculated using Eq. (6). This value of the round trip time is stored. Another randomcombination is then generated and a new round trip time is calculated. By repeating the above combinations a large numberof times (say 10000) and finding the average of all runs of the round trip timewe can get the true value of the round trip time.Using Monte Carlo simulation results in a value for the round trip time of 233.0974± 0.308 s (with a certainty of 95.4%)

based on 10000 runs. This is nearer to the weighted average method.

7. The interval as an indicator of the quality of service

The performance indicator for the quality of service is the interval. The interval is the average time between the arrivalsof the elevators at an entrance. This is found by dividing the round trip time by the number of elevators in a group. If theprobability of the elevator stopping at an entrance is less than 1, the interval has to be further adjusted to reflect this fact.Thus the interval at any of the entrance floors can be calculated as follows:

Ii =τGB

L · Pstop (i)(35)

where:τGB is the round trip time for the elevator when it serves the ground and the basements (but not necessarily going tothe basement in every round trip journey)L is the number of the elevators in the groupIi is the interval at the entrance i in s

Pstop (i) is the probability that the elevator will stop at the entrance i in a round trip journey.

The probability of stopping at any one of the three basements in the example discussed earlier in a round trip journeycan be calculated as follows using (13):

Pstop (Bi) =(1− (1− Parr (Bi))P

)= 1− 0.9516.8 = 0.577. (36)

Thus the interval at the ground floor will be (assuming that 8 elevators are used in order to meet the de facto standard 30 sinterval requirement for the ground floor) using (35):

IG =τGB

L · Pstop (G)=2338 · 1= 29.13 s. (37)

The interval at any of the basements will be using (35):

IB =τGB

L · Pstop (Bi)=

2338 · 0.577

= 50.47 s. (38)

Thus the quality of service at the basement entrances is worse than that achieved at the main entrance (ground floor) andthis will have to be taken into consideration during the design process. Based on the case study analysed earlier, 85% of thepassengers will witness an interval of 29 s while 15% of the passengers will witness an interval of 51 s. In effect, the averageinterval is around 32.3 s (using a weighted average calculation).

8. Conclusions

The design methodology for vertical transportation systems relies on the derivation of the round trip time during theup peak traffic. The simplest case involves one single entrance, for which the derivation formulae are well established. Themultiple entrance case (e.g., with more than one main entrance and basements) is not very well established and usuallyrequires simulation to evaluate it.

L. Al-Sharif / Mathematical and Computer Modelling 52 (2010) 545–555 555

This paper outlines and details two suggested methods for evaluating the effect of multiple entrances on the value ofthe elevator round trip time under up-peak traffic conditions. The first method denoted as the weighted average round triptimemethod evaluates the possiblemultiple entrance stopping scenarios, finds the probability of each scenario taking place,finds the resultant new round trip time for each scenario and then finds the weighted average of all the scenarios.The second method relies on the use of adapted highest reversal floor, H , and probable number of stops, S, for the

basement. It then looks at two possible scenarios: going to the basement in any one round trip journey and not going tothe basement in any one round trip journey. The round trip time for the basement case is calculated using the developed Hand S formulae for the basement case. The probability of going to the basement and not going to the basement in any onejourney are then both calculated. A weighted average of the two scenarios is then calculated.A case study is used to find the value of the round trip time from both methods. Both methods gave very similar results

(within 0.2%). The result is also verified using a Monte Carlo simulator using 10000 runs.The interval is used as a performance indicator within a multicar elevator group. A general formula for calculating the

interval at any of the entrances based on the probability of stopping at that entrance in any one round trip time is derived.The advantage of the second method is that it uses a simple formula that can be evaluated manually without the need

for software.

References

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