Combinatory method of definition of discrete decisions of the problem of rational distribution of vehicles on radial routes Текст научной статьи по специальности «Математика»

Shermukhamedov Abdulaziz Adilkhakovich, professor of Department of Road construction machines and equipment Tashkent Institute of design, construction and maintenance of automotive roads E-mail: sheraziz@mail.ru Juraev Mukhiddin Nortojievich, doctorant of Department of Road construction machines and equipment Tashkent Institute of design, construction and maintenance of automotive roads

COMBINATORY METHOD OF DEFINITION OF DISCRETE DECISIONS OF THE PROBLEM OF RATIONAL DISTRIBUTION OF VEHICLES ON RADIAL ROUTES

Abstract: In article three properties of the combinatorial method that allows determining discrete solutions of the problem of rational distribution of vehicles on radial routes are considered and justified. On the basis of the properties considered, the problem of effective distribution of vehicles for transportation of cotton on radial routes with application of a combinatory method of discrete programming is solved. One of decisions of a problem is received and the number the trips of each of considered five radial directions are defined.

By the proposed method it is possible to find and other decisions of a problem. As a result there will be a possibility of effective distribution carriages in a current of week, month and year.

Keywords: motor vehicles, discrete programming, cotton, radial routes, rational distribution, shipping, loading/ unloading mechanism, nodal point, cotton, working time, from the beginning forming, satisfy need, index number, property.

Analysis of the research work on the management of the The scheme of a route which connects points of the sender and process of transportation of goods by motor vehicles showed the addressee is the scheme of delivery and they can gather in that, in essence, the problem under consideration uses the uniform points or disperse from uniform points. Such schemes following modeling methods: linear programming, discrete usually name assembly or distributive radial routes, and assem-programming, dynamic programming and stochastic linear bly (or distributive) point name nodal point (NP) [4; 6]. programming. The satisfaction of requirement NP to transportation of

If it is required to determine the objective function with con- cargoes is defined by parameters intensity of goods traffics straint equations in the tasks of cargo transportation planning, which are provided vehicles, routes from a point of depar-then linear programming methods with widely used computa- ture to reception point. RR, rendering transport services NP, tional schemes and computer algorithms [1; 2; 3; 4] are used. consist of routes with various distances. The distance of a A separate method for solving optimization problems in route of transportation of cargo and a type of cargo, on the transportation is dynamic programming, according to which, one hand, defines number of vehicles which can effectively to find the optimal solution, the problem is divided into a se- be used in these directions, and, on the other hand, defines ries of steps (stages) and is solved sequentially from stage to number of flights which should be delivered the addressee or stage [1; 5]. is taken out from the sender, depending on distribution of the

Some problems require solutions in the form of discrete limited quantity of vehicles. Vehicles, distributed in separate values. These solutions require the use or development of meth- directions RR, cannot be big or small as we want. Because ods for formulating and analyzing discrete solutions that meet the quantity vehicles, allocated for routes, should not exceed all requirements. Such issues include the task of efficiently dis- possibility of points of sending (loading process) and accep-tributing vehicles along routes when transporting raw cotton. tance (unloading process) cargo, and also their degree more

Clap-raw transportation is carried out on radial routes (RR), an effective utilisation. i.e. cargoes are transported from one sender to several address- That is why the problem park distribution vehicles on RR

ees or, on the contrary, from several senders to one addressee. demands the fresh wording and workings out of methods of

its decision on the basis of formation and the analysis of various combinations of variants of effective loading working on routes of final number vehicles within the working day.

Such a problem can be solved by means of discrete programming [7; 8]. The analysis of existing methods of discrete programming (a method of branches and borders, a method of casual search and combinatory methods) has shown that combinatory methods correspond essence of a considered problem. At the same time, for an effective utilisation of this method has arisen necessity of improvement of this method. Theoretical properties of a combinatory method are more low considered.

a, - the possible number of working 1 oading (unloading) mechanisms in point of departure (acceptance) of cargo .-the route; y - number of daily rides that can be performed as part of effective work with a. = 1; Z{ - number of trips to perform the daily volume of cargo transportation in NP RR; Zp. - a set of possible number of rides on radial routes; ^U^ ■ -amount of time required to complete all rides.

Let's spend theoretical preconditions for application of a combinatory method of discrete decisions of a problem of rational distribution of vehicles on radial routes.

Property 1. For all admissible sets on ¡-the sum ofvalues of indexes a, of a set or indexes f 1 ... f 'e ... f C - in number equally also is equal to a version of trailer Z;/y.

Justice of this property follows from a condition

Zj = UP then Z X j =Yj'a,=fYfs,, whence

M M ih ¡¿h

X = Ui/7.

jh

As Z{ it is identical to all admissible sets on i - the version of trailer and j - number of daily rides that can be performed as part of effective work with a. = 1, that for all admissible sets of elements Up ■ value - is identical and equal Z/j.

jh

It is known, that number of an admissible set consists of set of indexes , (e e {1,2,., e,..., c}), where e - a serial number of element Up, in a set. Value of each index

We in

number is defined by value of corresponding index a. in an admissible set of elements Up.. From here also follows, that

X U=Yy-« = YYO , whence Xctj = Ui/Y.

Mi j£Jt jsJi jh

This property will be a basis at formation of every possible variants of admissible sets: their numbers are formed by assignment to indexes y/'e e e E = {1,2,...,e,...c}, value from set

e U

Z = {0,1,...,c} so that took place TV' = —-, i.e. distribution

e=i e y

of values Z/j on indexes taking into account that value of each index belongs to set Z= {0,1,...,a}.

Consider the formation of the number of the first admissible set "from the beginning". Assign the first index (^ ) its maximum possible value, check y/[oax = U, /y and if

y/^max < Zj / y , then we assign the maximum possible value to the 2nd index (.2').Check again whether the ratio holds max + max = Z / y and if (wjmax + w2max) < ^ / y, then go to the 3rd index and etc. Continues it until we will find out the such e-th index, for which the ratio holds (1max... + 1max) > Z / y . If thus ^imax... + ^2max = Z /7 , that number . ma\... ma\...+1j min .... min, where ¥e+1j min..¥cj min take zero values, is the number of the first valid number "from the beginning". Otherwise, i.e. if (i^1max...+ylemax) > Z /y,index . must take on the value determined by the difference Z;/y and y'emax +... + y'e-1max . The first admissible set "from the beginning" will look like thus: y/'™ —y/'e = { / y -+... ... + wlmax) } 7 ~vlmin. When forming the number ofthe first allowable dialing in order "from the end", the maximum possible values of the indices are assigned not starting from the first but from the last index. The rest of the process of formation does not differ from the above.

Determining the number of the first admissible set in the manner described above can be quite a long process, since before identifying the desired number gets a significant number of options.

Let's enter concept about a trial set and its number. The set is named trial "from the beginning" ("from the end") and

/ \fbeg ft .A fbeg , .fend

designated - H(j ) I £ I (or H(Z ) ), if the

V j )samp I V j J samp

V e-1 / samp

first (last) element in it takes the maximum possible, while the remaining elements take the minimum possible values.

In the general case, the definition of the number of the first admissible set "from the beginning" or "from the end" is represented as follows. The number of the test set "from the beginning" is formed (l ma>2min... i.min) or "from the end" (.Cmm...y'c_1min^imax) . For the trial dial number, the value is

f c Afbeg ( ] fend

determined I X Ve I or I ^ y/'e I and compared with

V e=1 J samp V e=1

Z/y There may be three cases:

samp

case

fend

1 IX tfl > Ze / Y or (X^e) > Ze / 7 . In this

V e=1 J samp ^ e=1 J samp

, the following actions are taken:

r e ^ fbeg

• the difference is calculated A1 = I ^ y/'e I - Ze / y

V e=1 J samp

and the first index of the number is determined Vi = vimax - . As a result, get the number of trial dialing H(Zja ) : {{ min....^CmiD};

j samp fend

• to determine the test dial number H(Z) calcu-

e Z jaj .samp

lated A1' = Yw '--^ and = y/'cmax - A]. As a ree e=1 e 7 .....

sult, its number is set: {i//1'mm ...^-TVe}.

/ e \ fbeg r e \fend

2. (X^e I < Ze / 7 or r£¥e j < Ze / Y . In this

case, the whole part of the relationship is taken

a; =-

Z / y , Z / y ,

e or A'k =-—5—-¡—¡-A',, and based on it is

Z ^e

fbeg

fend

samp

determined by the index y',) = Ze / y -A'k \ X V

(I+A; )

fend

e=1

fb;g

or

samp

<-4 ) = Ze / 7 - 4 ^ I • As a result for H (z^ )

fbeg

samp

get a number { ^

(I+a; -i)r (1+a;, v I+a;„+I

min ,andfor

HI

(j )) number {min ....

min, w' , w'maf ...w'max (1+A-1)r(c-A'; ri+Ai+i Yc

e Z ( \jbeg

3. = —. In this case for a set H(Z a ) number

e Y 'samp

will have the form: { y/[max... v2min... v'm™ } ,and for a set

(\fend

7 I f / min i min .../max \

^Lp : }.

A union of sets is called as trial if, it consists the first admissible sets formed rather: 1) "from the beginning" or; 2) "from the end". In the first case we name its trial association "from the beginning" and value ( ^ Zja ■ t . )samp for it it is

designated Z ^ -t. ^^, and in'lthe second - "from the

m '

end" - (^Zja ■ t. The following property of the trial set

m ' union is valid.

Property 2. Among the values of the sum ^ Z-t j for

m '

all possible combinations of permissible sets of value (X'-f is the greatest and the value £Z. ■t])ffmf -

je] js]

the smallest. Let us prove the validity of this property. The elements Z jaj in a set are located in ascending order t,, that corresponds to a decrease order Z.. In the first valid set "from the beginning," the elements Z jaj, reaching at the beginning of the set, take their maximum possible values, they correspond to the highest values Z., because Z{ for the elements Zpi in the set has a reducing character and therefore the value Z{ for the elements, reaching at the beginning, the greatest and for the subsequent ones - gradually decreases. As a result for (£' -tj.)^ value £ Z ^^ -t j is greatest: of ah to

possible values corresponding every possible other admissible

fend

' jGj * j ) samp

set. The trial set "from the beginning" consists of (Z Z ja ■ t ■)

„ M

for all ' e J, hence the value ^ Zja t will be the greatest of

m '

all. From property 2 the following consequence follows. Consequence 1. If the test association satisfies conditions:

i) (Zv t } tmP > (a3Thu or; 2) (Xit:P <

jsJ jeJ

(A3TH )min, then the task under the given conditions has no solution. Here, A3 - total number ofvehicles used in transport, T - the daily operating time of each vehicle. the total number of cars.

In view of property 2, the validity of consequence 1 is almost obvious. Really meaning Zja ■ t j)fmp is small of all

m '

^ tj fgp or (X j ■tj )Samdp and compared with the val-

and (AT)

possible and if it is less than (ATH)mn, then for any other association the ratio is fair V Z -t. < (AT) ., i.e. the task

/ t j&j j y 3 H min7

has no solution.

Let's continue the problem decision as follows. After forming all the first valid sets, the value is calculated

(Z Z

ues (AT) . and (AT) . If at the same time the ratio pro

3H min 3 h max L

vided by consequence 1, that concludes that the problem has the decision, and if not provided in it we come to a conclusion about necessity of moving "to the right" or "to the left" in sets of initial association.

The choice of directions of movement is made while taking into account the following property.

Property 3. In the movement in the number of the first allowable set "to the right", if possible, allows you to get the number of the set for which the value Z Zja • tj less and vice

m j

versa - "left" - more than the value of this sum in the first

admissible set.

Let's prove this property. Assume that ^ ...y'd -K) ...y'c -the number of the first valid set, formatted in the order "from the beginning": Zjaj m ...j (d) ...j (C-k)...zJCTj (c ). The elements of

1 t t t

this set of corresponding values t: t. Known that t.,,,< t ,„, <... < t ,,, <... < tY n < ^,. In this case, the first allowable set corresponds to the value of the sum

Πz

i^ji

t )fbeg = z' t . li hsN iai(1) i(1)

"j hSN

Z

t;rj ^ +

(d) i(d)

yZ -t. :

¿—l JVj J j£]

+ ... + Z'- t; ,+ ... + Z'- ■ t;

J°-j(«) j (c-k) J°-j(c) j (c)

Suppose that in this set, moving to the right is possible (moving to the left is impossible here, which is proved by consequence 2, if moving to the right is also impossible, then the problem has no solution). We make an arbitrary unit move in the number of this set: let us assume that we move the unit from the d-th index to (c-k)-th. It is clear that d < (c-k) and moving from the d-th index to (c-k)-th is moving "to the right". As a result of this move, we get the number: Vi'...(wd - 1)...(^lc-k) +1)..-¥'c. This number corresponds to the set Z „ ...(Z' -Yv)...(Z „ (c K) +y)...Z „ ,). It is clear that the

j aj(i) (d) v j aj (c-k) > f j((c)

movement of the unit in the number corresponds to the movement of the number y in the set.

Amount value Z Z ^ -tj for the result set is determined by:

N

■t ) = Z'

j 'ww jaj (1)

Π.

ij

+(.Zi«,(c—K) +7) ■ti(c—K)

Obviously,

ti (1)+...+(ZL,— /)■ t(d ) +

i^i (d )

Z

if

iai (c)

A =

(Z Z

. ieJi

i&i i / new -Z Z

. t. )

i&i i f new

Πz;

property 3

ij

> (Z Zi

iJ

t \fbeg li hSN

n (c )• the

> 0

i(d )

difference that

t which proves the validity of

iaj " i >1SN

e=1

e=1

samp

For check of possibility of such case will define A:

A = j (1)tj (1) + ... + (j (d) -7) j + ... + (j (c-k) +7) txe-K) +

+... + Z,

■t - Zl

■ t - - Zl

■ t - - Zl

(e) j(e) jaj (1) j(1) jaj (d) j(d) jaj (c-k)

' tj(e-K) - Zja, (e) 'tj (e) = tj(d)(Zja, (d) ~7 - Zja, (d))+ tj (e-K) ( j (c-k) ) +

+ 7 - j (C-k)) = 7(tj (e-k) - tj(d)).

Because t ,(-k) > t j(dy then A = yj - t.(d)) > 0, as required. It should be noted that the movement "right" must be carried out starting from the sets of H ,where the index i has the highest value with the decrement order, and moving "left" is the smallest with the increment order.

The effective distribution of vehicles by route considers the example of transportation and raw cotton.

Let the volume of homogeneous goods with an index J, ( J = 1) cotton of cleaning factory satisfying requirement makes ZJ=1 = 130 carriage. Radial routes of transportation of cargoes

are numbered as increase of full time of the turn (t) one vehicle on the route, t- {2,3,4,5,6}.

Set of numbers carriage (Zp^, that can be performed on each route (j) present in the form of an ordered matrix ||zJ| (Table 1).

II HI;v

Table 1. Matrix possible the carriage \\Z.

"> S,

Shippers (j) Quantity Oî^^^^^ loading mechanisms($j)^^^ Radial route numbers

1 2 3 4 5

0 0 0 0 0 0

1 10 10 10 10 10

2 20 20 20 20 20

3 30 30 30 30 30

4 40 40 40 40 40

Problem statement:

It is necessary to determine the combination and possible aggregations that are formed from the elements of the matrix ||UjCT. || , representing a multitude of carriages running on radial routes so that a given number of carriages can be completed by destination Z{ = 130 with available car park ( A3 = 60 units) within minimum (A3 -TM)oin = 600 hours and maximum (A3 -TM)oax = 660 hours of their congestion starting with minimum conditions (TM oil] = 10 hours) and maximum ( TM oill = 11 hours) time of performance of works.

The decision of the given problem can be found formation of all possible variants, allocation of variants "satisfying requirements", their check on degree of congestion and the performance analysis. However in this case it is required to consider 3125 variants and to check up to satisfaction to a condition of a problem of each variant. It is difficult enough work and demands performance of great volume of calculations. The following order of formation and the analysis of combinations the shipping therefore is recommended Up..

Possible set of elements Up. represents sequence of indexes f 1... f e . - f C, each of which takes the value of the index element c. from the aggregate.

In aggregate shipping Upi, satisfying the requirements of Z{ each consumer by type of cargo I, indexes c, or the sum of the indexes of the series f 1... f 'e ... f 'c are identical and

equal U / y, and therefore ^ aj = U / y ; where j - quantity

jh

of the articulated lorries, under condition of presence of one loading mechanism oj = 1.

Proceeding from the above mentioned, will form numbers of the first skilled set the carriage Ujaj. For this initial few elements ¥'c assign possible maximum values ( = aoai = 4),oth-

er indexes - minimum values (ct- = om = 0 ).Then numbers of

fend

U1 ^ ) carriage Ulp will be 4141410101.

jai / samp j

For the generated numbers define the sum Y^è :

5 eeE

^y1 = 41 + 41 + 41 + 01 + 01 = 121.

e= U 1301 Calculates value of a parity ( UJy ): —1 =-= 131.

y 10 U

Value sum Wei compare with value — :

tE 5 Y

U=131 >t =121. 7 t1

cU

That is the condition is available V¥'e < —, thereby the . e=1 7

totality numbers H (Up fN are determined by:

A'k - value of parameters for sets generated on-order "about the beginning", count a choice of the whole part of the

Z,

following fraction =L fip =

, =,

121 ,6

6ym

Proceeding from value indexes are defined :

.+Ak ) = U, / y-Aj • £¥e' : = ¥il = 131 -1 • 121 = 11.

e=1

e=1

Thus, numbers of the first of sets "satisfying requirements" articulated lorries, written as - 414i411101 and set in a kind:

<7 1 ^ \ ^ \ ry\ ry\

Z14 Z24 Z34 Z41Z50 .

• t yg

^ Jaj j samP

fmr s;xper^

Calculate the value of the sum (^ Zjc

«

enced associations of the first set of "satisfying needs" carriages:

Afbeg =

samp

Z Z

= Z Zj°¡ ' tj = Z14 t1 + Z24 12 + Z34 13 +

samP

m

Z411t4 + Z501t5 = 40 • 2 + 40 • 3 + 40 • 4 +10 • 5 + 0 • 6 = 410.

If to compare this result with congestion of the car park in a current of day (A3TH )min and (A3TH)max whether it is possible to define satisfy the planned articulated lorries to following conditions:

(A3TH )min - (Z Zja- 'tj )amp — (A3 TH )max .

From here it is visible, that skilled association does not satisfy to the put condition, i.e. Afbmp = 410 < (A3 TH )min = 600 hours.

If for the first associations j • tj )ab,eg < (A3 • TJmin ,

j£]

then, in aggregate, calculate the value "satisfying needs" by moving "to the right":

3N = (A3 Th)min - AfZf = 600 - 410 = 190 hours;

3N = ( A3 • Th )max - Am = 660 - 410 = 250 hours.

Here calculate the maximum value on which it is possible

/ 1 \fberg

to move to the right in aggregate H (Z . ) :

\ jaj /1SN

olSN _ ^ Z1 efbeg _ Z efbed _

Max (right ) ^ ^.j ja j '1SN ja. j '1SN

j] js]1

= z 1 • t + Z 1 -t + Z 1 -t + Z1 -t + Z1 -t - Z 1 -t -

10 t1T',1'24 t2T',1'34 t3T',1'41 t4T',1'54 i5 ^14 '1

-Z241 -t2 -Z341 -t3 -Z11 -t4 -Z10-t5 = 0-2 + 40-3 + 40-4 + +10-5 + 40-6 - 40-2 - 40-3 - 40-4 -10-5 - 0-6 = = 570 - 410 = 160 hours.

Comparing value 3maxNu=160 ^ c 3Nhm *=190 hours

and determined them = 160 hours < 3^ = 190

hours.

Let's calculate the rest which is necessary for moving on the right part.

0 oright = o

V^SN min J SN min

. - y 31(sNe hfe = 190 -160 = 30 hours;

in / , Max (right )

l=1 l-1

a oright _ qnght

V^SN max JSN max

-ZomiNghte = 250 -160 = 90 hours.

Thus, will designate = V2 departure index and = receiving index. Let the possible is known m-

(c-Km e

variant of moving of indexes. Identify opportunities for the index am sending and receiving:

K-Te e-¥lr p, = 4-0 = 4

=wp =v!= 4.

Compare the parameters y°'P and ¥t

e = 4 and räP = 4.

(c-rm e

and define:

For a certain variant m will calculate value on which it is possible to move:

V, m, = t- - V ) • V • = (i4 - h) -10 . Vl =

" = (5-3)-10• 4 = 80 hours. For a considered case, that m =

dm (c-k- )

m=1

This value is comparable to the parameters 03^^ and define: Z 3, ,, m = 80 hours > 03NL = 30 .

d (c-k )

m=1

Check the condition > 1, vT = 4 > 1. Conditions

it is carried out.

Now define values AO^T and AO^T :

m-1

A3ri2ht (m e = 03right _ V 3 = 03right = 30 hours

SN min SN min mY m ^°SNmin J W hours,

d (c_K e

m=1 m-1

A3N(m) = 03N - T 3 = 03N = 90 hours.

SN max SN max / , y my m SN max

Because at the above-stated calculations quantity of vari-

m-1

ants of displacement m=1, ¥ = 0.

dm (c-Km )

m=1

The minimum and maximum values of displacements

(nmh(m), nmf(m)), providing individual displacements to an

interval ^Nmin ^ ^Nmax ) can be determined by the expression:

03rlght right (me _ w ^SN m^ _ _

m^ 3I

30

3

(t4-t2e-10 5-3

= 1.5,

03right

right (m e _ w SN max _ _ max 31

90

■ = 4.5.

10•(5 - 3)

dm (c-k"1 )

Now define quantity of individual movings. It is thus checked and define integer presence in an interval

nmn' - n"^(m). In a considered case there are three units of

min max

integers - 2, 3, 4.

Thus, from the possible maximum value of the digits of the aggregate articulated lorries - 0141411141, from the second index to the fourth will execute two displacements and will receive 0121413141 or in aggregate Z^Z^Z^Z^Z^. Let's check up congestion of the car park:

ZZfi, ■ tj = XZ■ tj = Zm\ + Z221t2 + Z34 13 + Z4^t4 + Z541t5 = j] jJ

= 0 • 2 + 20 • 3 + 40 • 4 + 30 • 5 + 40 • 6 = 610 hours.

The reached congestion makes 610 hours and it is between demanded values of600 and 660 hours. This means that one of the solutions to the problem has the form Z^ Z^ Z134 Z ^ Z154, According to which in a current of day are carried out: on 2nd route 20 rider, 3rd route 40 rider, 4th route 30 rider and 5th route 40 rider. At the same time, the need of enterprises for raw materials is satisfied, the efficient use of loading mechanisms is ensured, and the cars are optimally distributed along routes.

By the proposed method it is possible to find and other decisions of a problem. As a result there will be a possibility of effective distribution carriages in a current ofweek, month and year.

m=1

l=1

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