CN102521723A - Discrete workshop logistics information online obtaining method based on radio frequency identification technology - Google Patents

Abstract

The invention discloses a discrete workshop logistics information online obtaining method based on radio frequency identification technology, which includes: 1, assigning operation task, and binding the operation task-material-radio frequency identification (RFID) label and station-process-RFID reader-writer; 2, initializing a logistics status matrix and a logistics status change matrix; 3, obtaining RFID initial data at the moment of tn; 4, converting the RFID initial data into logistics simple events; 5, collecting the logistics simple events into logistics complicated events; 6, obtaining logistics status change matrix at the moment of tn through the logistics complicated events; 7, generating logistics status matrix at the moment of tn; 8, obtaining logistics information at the moment of tn according to rank-sum elements of each block matrix in the logistics status matrix; and 9, setting the logistics status change matrix into a null matrix, going to the step 3, and obtaining logistics information at the moment of t(n+1). The discrete workshop logistics information online obtaining method based on the radio frequency identification technology achieves real-time synchronization of material flow and information flow in a discrete workshop.

Description

A kind of online acquisition methods of discrete workshop logistics information based on REID

Technical field

The present invention relates to the online acquisition methods of a kind of discrete workshop logistics information, especially a kind of based on RF identification (Radio Frequency IDentification, RFID) the online acquisition methods of discrete workshop logistics information of technology.

Background technology

Exist comparatively general " information tomography " phenomenon in the discrete workshop, wherein the tomography problem of logistics information is particularly outstanding.For realizing the synchronous in real time of streams and information flow in the discrete workshop; More and more researchers begins to pay close attention to the online problem of obtaining of logistics information in the discrete workshop at present, and main mode or the technological means obtained have: papery card, bar code and RFID technology etc.Wherein the papery card exist fragile or lose, shortcomings such as typing is time-consuming, easy error, accuracy is low, the data non real-time obtains; Bar code exist reading/writing distance limited, need visually to read, data capacity is little, only support shortcoming such as tracking in batches; The RFID technology has then overcome the deficiency of papery card, bar code; Have contactless with read in batches, fouling resistance (like oil rub resistance, dust etc.), adaptive capacity to environment strong (like high temperature resistant, high humidity etc.), support that single article are followed the tracks of, recyclable utilization once more, can write outstanding advantage such as data, thereby increasing discrete manufacturing enterprise is applied to the RFID technology in the middle of the workshop logistics information obtains.

The RFID technology is applied to the main difficulty that discrete workshop logistics information faces in obtaining is: problems such as the time interval that the raw data that rfid interrogator produces automatically on each station has enormous amount, produce data is short, semantic hierarchies are low; Be difficult to by upper system (like enterprise resource planning (Enterprise ResourcePlanning; ERP), supply-chain management system (Supply Chain Management; SCM), warehouse management system (Warehouse Management System; WMS) etc.) directly be applied, also be difficult to understood, bring the phenomenon of " data are had a surplus, information is not enough (or deficient) " thus by the general management personnel; More stubborn problem is; If a large amount of RFID raw data can not be handled effectively, not only can't realize the online target of obtaining of logistics information, and will take the lot of data storage space; Frequent data write also will increase the I/O of data base management system (DBMS) (I/O) expense greatly, bring extra system to handle burden.

In a word; The introducing of RFID technology helps to realize better obtaining in real time of logistics information in the discrete workshop; On the other hand, the RFID raw data has that data volume is huge, data shortcomings such as the time interval is short each other, semantic hierarchies are low, therefore must it effectively be analyzed and handle; The target of can the finishing department logistics information obtaining in real time, thus realize the synchronous in real time of workshop streams and information flow.

Summary of the invention

For in time, obtain the logistics information in discrete workshop effectively; And overcome huge, the data typical problems such as the time interval is short each other, semantic hierarchies are low of original data volume that exist when the RFID technology is applied to discrete online the obtaining of workshop logistics information; The object of the present invention is to provide a kind of online acquisition methods of logistics information towards discrete workshop based on the RFID technology; Through the RFID raw data is effectively handled, thereby extract the logistics information that wherein contains.

The technical solution adopted for the present invention to solve the technical problems is:

1, a kind of online acquisition methods of discrete workshop logistics information based on REID is characterized in that the step of this method is following:

(1) assign job task, bind " job task-material-RFID tag " and " station-operation-radio-frequency identification reader/writer ":

At first need to bind " job task-material-RFID tag ", its relationship description is following:

Wherein:

1) ID _TExpression job task numbering;

2) M _kExpression thing item number;

3) o _xThe sign of expression radio frequency identification marking;

Between expression job task and the material is many-one relationship, and job task ID is described _TIn K material arranged;

Between

expression material and the RFID tag is one-one relationship; So k=x=1; 2; ..., K explains and has pasted a RFID tag on each material;

Need once more to bind " station-operation-radio-frequency identification reader/writer ", its relationship description is following:

Wherein:

1) E _iExpression radio-frequency identification reader/writer r _yLocation type, i=1,2,3,4,5,6, and:

1. if i=1 has E _i=E ₁, expression read write line r _yBe positioned at neck material outbound station;

2. if i=2 has E _i=E ₂, expression read write line r _yBe positioned at the arrival district of processing or assembly station;

3. if i=3 has E _i=E ₃, expression read write line r _yBe positioned at the district of doing over again of processing or assembly station;

4. if i=4 has E _i=E ₄, expression read write line r _yBe positioned at the district of scrapping of processing or assembly station;

5. if i=5 has E _i=E ₅, expression read write line r _yBe positioned at inspection station eventually;

6. if i=6 has E _i=E ₆, expression read write line r _yBe positioned at finished product warehouse-in station;

2) P _jThe expression operation number, j=1,2 ..., J, J are the total number of operation:

1. when j=1, corresponding operation is neck material outbound operation;

2. work as j=2,3 ..., during J-2, corresponding operation is processing or assembly process;

3. when j=J-1, corresponding operation is examined operation eventually for completion;

4. when j=J, corresponding operation is a finished product warehouse-in operation;

3) W _lExpression worker item, l=1,2 ..., J,

4) r _yExpression radio-frequency identification reader/writer numbering, y=1,2 ..., Y, Y=3 (J-3)+3=3J-6, the concrete value of y is relevant with j, and computing method are:

1. when j=1, y=1;

2. work as j=2,3 ..., during J-2, $y=\left\{\begin{array}{c}3j-4\mathrm{if\; PlaceType}\left(r_y\right)={\mathrm{<figure-callout\; id="2"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_2\\ 3j-3\mathrm{if\; PlaceType}\left(r_y\right)={\mathrm{<figure-callout\; id="3"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_3\\ 3j-2\mathrm{if\; PlaceType}\left(r_y\right)=E_4\end{array}\right.$ (the read write line location type is got in " PlaceType " expression);

3. when j=J-1, y=3J-7;

4. when j=J, y=3J-6;

5) be many-one relationship between

expression radio-frequency identification reader/writer location type and the radio-frequency identification reader/writer; Promptly to any one radio-frequency identification reader/writer; Its location type is unique; And identical location type can be to there being a plurality of radio-frequency identification reader/writers

Between

expression operation, station and the radio-frequency identification reader/writer is one-one relationship

(2) initialization logistics state matrix and logistics state variation matrix:

If the logistics state matrix is M _L(t), initialized logistics state matrix is:

M _L(t ₀)＝O _(6×J) (3)

If logistics state variation matrix is Δ M _L(t), initialized logistics state variation matrix is:

ΔM _L(t ₀)＝O _(6×J) (4)

(3) obtain t _nRF identification raw data constantly:

If RFID (t) expression RF identification raw data, then t _nA RF identification raw data constantly is:

RFID(t _n)＝{{o ₁，r ₁，t _n}，{o ₂，r ₂，t _n}，...，{o _x，r _y，t _n}，...} (5)

(4) convert the RF identification raw data into the logistics simple event:

By t _nA RF identification raw data { o constantly _x, r _y, t _n, obtain t _nA logistics simple event E constantly _r(t _n):

E _r(t _n)＝e(o _x，r _y，t _n) (6)

(5) the logistics simple event is polymerized to the logistics complicated event:

By t _nA logistics simple event e (o constantly _x, r _y, t _n), polymerizable generates a t _nLogistics complicated event E (t constantly _n):

E(t _n)＝e(id，{ID _T，E _i，P _j}，{E _r(t _n)}，t _n) (7)

Wherein:

1) id is a natural number, obtains through the mode that increases certainly;

2) ID _T=(o _x→ ID _T), expression is passed through formula (1) by o _xObtain ID _T

3) E _i=(r _y→ E _i), expression is passed through formula (2) by r _yObtain E _i, be specially:

If 1. radio-frequency identification reader/writer r _yBe positioned at neck material outbound station, then E _i=E ₁

If 2. radio-frequency identification reader/writer r _yBe positioned at processing or assembly station and arrive district, then E _i=E ₂

If 3. radio-frequency identification reader/writer r _yBe positioned at processing or assembly station do over again district, then E _i=E ₃

If 4. radio-frequency identification reader/writer r _yBe positioned at processing or assembly station and scrap the district, then E _i=E ₄

If 5. radio-frequency identification reader/writer r _yBe positioned at inspection station, then E eventually _i=E ₅

If 6. radio-frequency identification reader/writer r _yBe positioned at finished product warehouse-in station, then E _i=E ₆

4) P _j=(r _y→ P _j), expression is passed through formula (3) by r _yObtain P _j,

(6) obtain t by the logistics complicated event _nLogistics state variation matrix constantly:

For t _nLogistics state variation matrix constantly has:

${\mathrm{\&Delta;M}}_L\left(t_n\right)=\left[\begin{array}{ccc}\mathrm{\&Delta;}{M}_L^{\left(1\right)}\left(t_n\right)& {\mathrm{\&Delta;M}}_L^{\left(3\right)}\left(t_n\right)& O\\ O& \mathrm{\&Delta;}{M}_L^{\left(2\right)}\left(t_n\right)& O\\ O& O& \mathrm{\&Delta;}{M}_L^{\left(4\right)}\left(t_n\right)\end{array}\right],$

Wherein:

The outbound of

expression neck material changes submatrix, and have:

is the element in the matrix

;

expression operation arrival and the inventory disposal and the variation submatrix of doing over again have:

<maths num="0003"> <! [CDATA [<math> <mrow> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 2 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub > <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> <mo> = </ mo> <mfenced open = '[' close = ']'> <mtable> <mtr> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 22 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi > n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. < / mo> </ mtd> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> < mn> 2 </ mn> <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t < / mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> <mo> [</ mo> <mn> 2 </ mn> <mrow> <mo> (</ mo> <mi> J </ mi> <mo> - </ mo> <mn> 2 </ mn> <mo>) </ mo> </ mrow> <mo>] </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> </ mtr> <mtr> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi > L </ mi> <mrow> <mo> (</ mo> <mn> 32 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> < mo> (</ mo> <mn> 3 </ mn> <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> <mo> [</ mo> <mn> 3 </ mn> <mrow> <mo> (</ mo> <mi> J </ mi> <mo> - </ mo> <mn> 2 </ mn> <mo >) </ mo> </ mrow> <mo>] </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi > n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> </ mtr> <mtr> <mtd> <msubsup> <mi> ΔM </ mi> < mi> L </ mi> <mrow> <mo> (</ mo> <mn> 42 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (< / mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. < / mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 4 </ mn> <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> <mo> [</ mo> <mn> 4 </ mn> <mrow> <mo> (</ mo> <mi> J </ mi> <mo> - </ mo> <mn > 2 </ mn> <mo>) </ mo> </ mrow> <mo>] </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi > t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> </ mtr> </ mtable> </ mfenced> <mo>, </ mo> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 2 </ mn > <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> <mo>, </ mo> </ mrow> </ math>]]> </maths> <img file =" BDA0000115410870000051.GIF " he =" 72 " img-content =" drawing " img-format =" tif " inline =" yes " orientation =" portrait " wi =" 562 "/> is the matrix <img file = "BDA0000115410870000052.GIF" he = "65" img-content = "drawing" img-format = "tif" inline = "yes" orientation = "portrait" wi = "211" /> elements;

Lead material to change submatrix after

expression is scrapped again, have:

<maths num="0004"> <! [CDATA [<math> <mrow> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> <mo > (</ mo> <mn> 3 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> <mo> = </ mo> <mo> [</ mo> <mi> Δ </ mi > <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 12 </ mn> <mo>) </ mo> </ mrow > </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow > <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L < / mi> <mrow> <mo> (</ mo> <mn> 1 </ mn> <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> < mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> [</ mo> <mn> 1 </ mn> <mrow> <mo> (</ mo> <mi> J </ mi> <mo> - </ mo> <mn> 2 </ mn> <mo >) </ mo> </ mrow> <mo>] </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi > n </ mi> </ msub> <mo>) </ mo> </ mrow> <mo>] </ mo> <mo>, </ mo> </ mrow> </ math>]]> </maths> <img file = "BDA0000115410870000055.GIF" he = "71" img-content = "drawing" img-format = "tif" inline = "yes" orientation = "portrait" wi = "237 "/> is the matrix <img file =" BDA0000115410870000056.GIF " he =" 64 " img-content =" drawing " img-format =" tif " inline =" yes " orientation =" portrait " wi = "209" /> elements;

expression material completion and warehouse-in change submatrix, have:

< maths num= " 0005 " > <! [CDATA[< math > < mrow > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 4 </ mn > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > < mo >=</ mo > mfenced open='['close='] ' < mtable > < mtr > < mtd > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > [</ mo > < mn > 5 </ mn > < mrow > < mo > (</ mo > < mi > J </ mi > < mo >-</ mo > < mn > 1 </ mn > < mo >) </ mo > </ mrow > < mo >] </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > < mtd > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 5 </ mn > < mi > J </ mi > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > </ mtr > < mtr > < mtd > < msubsup > < mi > &Delta; M </ mi > < mi > L </ mi > < mrow > < mo > [</ mo > < mn > 6 </ mn > < mrow > < mo > (</ mo > < mi > J </ mi > < mo >-</ mo > < mn > 1 </ mn > < mo >) </ mo > </ mrow > < mo >] </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > < mtd > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 6 </ mn > < mi > J </ mi > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > </ mtr > </ mtable > </ mfenced > < mo >; </ mo > < msubsup > < mi > &Delta; M </ mi > < mi > L </ mi > < mrow > < mo > [</ mo > < mn > 5 </ mn > < mrow > < mo > (</ mo > < mi > J </ mi > < mo >-</ mo > < mn > 1 </ mn > < mo >) </ mo > </ mrow > < mo >] </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > < mo >; </ mo > < msubsup > < mi > &Delta; M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 5 </ mn > < mi > J </ mi > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > < mo >; </ mo > < msubsup > < mi > &Delta; M </ mi > < mi > L </ mi > < mrow > < mo > [</ mo > < mn > 6 </ mn > < mrow > < mo > (</ mo > < mi > J </ mi > < mo >-</ mo > < mn > 1 </ mn > < mo >) </ mo > </ mrow > < mo >] </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > < mo >; </ mo > </ mrow > </ math >]] > </maths> <img file= " BDA0000115410870000059.GIF " he= " 72 " img-content= " drawing " img-format= " tif " inline= " yes " orientation= " portrait " wi= " 249 " /> be the element in the matrix <img file= " BDA00001154108700000510.GIF " he= " 65 " img-content= " drawing " img-format= " tif " inline= " yes " orientation= " portrait " wi= " 211 " />

According to formula (7), can calculate the value that obtains each element in the above-mentioned partitioned matrix:

$1)\mathrm{\&Delta;}{M}_L^{\left(11\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i=E_1\mathrm{AndE}\left(t_n\right).P_j=P_1\\ 0& \mathrm{Else}\end{array}\right.;$

(annotate: logistics complicated event E (t is got in symbol ". " expression _n) in certain property value, down with)

$2)\mathrm{\&Delta;}{M}_L^{\left(2j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="2"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_2\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$3)\mathrm{\&Delta;}{M}_L^{\left(3j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="3"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_3\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$4)\mathrm{\&Delta;}{M}_L^{\left(4j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i=E_4\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$5)\mathrm{\&Delta;}{M}_L^{\left(1j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i=E_1\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$6)\mathrm{\&Delta;}{M}_L^{\left[5(J-1)\right]}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="5"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_5\mathrm{AndE}\left(t_n\right).P_j=P_{(J-1)}\\ 0& \mathrm{Else}\end{array}\right.;$

$7)\mathrm{\&Delta;}{M}_L^{\left(5J\right)}\left(t_n\right)\&equiv;0;$

$8)\mathrm{\&Delta;}{M}_L^{\left[6(J-1)\right]}\left(t_n\right)\&equiv;0;$

$9)\mathrm{\&Delta;}{M}_L^{\left(6J\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="6"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_6\mathrm{AndE}\left(t_n\right).P_j=P_J\\ 0& \mathrm{Else}\end{array}\right.,$

(7) generate t _nLogistics state matrix constantly:

M _L(t _n)＝M _L(t _(n-1))+ΔM _L(t _n) (8)

In the formula (8), M _L(t _(n-1)) expression t _(n-1)Logistics state matrix constantly,

(8) the sum of ranks element according to each partitioned matrix in the logistics state matrix obtains t _nLogistics information constantly:

To the logistics state matrix $M_L\left(t_n\right)=\left[\begin{array}{ccc}{M}_L^{\left(1\right)}\left(t_n\right)& M_L^{\left(3\right)}\left(t_n\right)& O\\ O& M_L^{\left(2\right)}\left(t_n\right)& O\\ O& O& M_L^{\left(4\right)}\left(t_n\right)\end{array}\right]$ Each partitioned matrix analyze, obtain corresponding logistics information (wherein: lead material for making clothes matrix,

expression material completion again and put submatrix in storage after

expression neck material outbound submatrix,

expression operation arrival and inventory disposal are scrapped with the submatrix of doing over again,

expression):

1) obtains neck and expect library information

During 1. as

, material is not neck material outbound as yet;

(annotate: rank of matrix is got in symbol " Rank () " expression, down together)

During 2. as ; Material has been led the material outbound, and neck material outbound quantity is

2) obtain operation arrival and inventory disposal, the information of doing over again

During 1. as

, job task does not begin to carry out;

2. work as

The time, job task has begun to carry out, when

And

The time, job task implements j procedure P _j

3. work as

The time, there is material to do over again or the generation of inventory disposal situation, if

Then do over again and occur in j procedure P _j, if

Then scrap and occur in j procedure P _jOn;

During 4. as

; Existing material is done over again also has the inventory disposal situation to take place, scrap and the operation localization method of doing over again the same;

3) obtain and lead material information again after scrapping

During 1. as

, lead the material situation to take place after not scrapping again;

During 2. as

; Have and lead after scrapping the material situation to take place again, and if neck material

was individual again after

scrapped;

4) obtain material completion and go into library information

During 1. as

, material is not inspection eventually as yet;

During 2. as

; Material has carried out whole inspection, and inspection quantity is

eventually

During 3. as ; The material warehouse-in that completed, warehousing quantity is

(9) logistics state variation matrix is changed to null matrix, forwarded for (3) step to, obtain t _(n+1)Logistics information constantly.

The beneficial effect that the present invention has is:

The present invention pastes each material to be processed or to be assembled in the job task and goes up the RFID label; On this basis; Set up the logistics state matrix; Sum of ranks element through each partitioned matrix in the analyte stream state matrix obtains logistics information, thus realized streams and information flow in the discrete workshop synchronously in real time.The present invention can obtain the logistics information in the discrete workshop in real time, online, thereby has realized fine-grained logistic track (contain neck and expect links such as outbound, material arrival station, material are done over again, inventory disposal, finished product inspection eventually, finished product completion warehouse-in).

Description of drawings

Fig. 1 is based on the implementation process figure of the online acquisition methods of discrete workshop logistics information of REID.

Embodiment

Below in conjunction with accompanying drawing and embodiment the present invention is further specified.

Fig. 1 is an embodiment of the present invention, may further comprise the steps:

(1) assign job task, bind " job task-material-RFID tag " and " station-operation-radio-frequency identification reader/writer ":

At first need to bind " job task-material-RFID tag ", its relationship description is following:

Wherein:

1) ID _TExpression job task numbering;

2) M _kExpression thing item number;

3) o _xThe sign of expression radio frequency identification marking;

4)

Between expression job task and the material is many-one relationship, and job task ID is described _TIn K material arranged;

5) be one-one relationship between

expression material and the RFID tag; So k=x=1; 2; ..., K explains and has pasted a RFID tag on each material;

Need once more to bind " station-operation-radio-frequency identification reader/writer ", its relationship description is following:

Wherein:

1) E _iExpression radio-frequency identification reader/writer r _yLocation type, i=1,2,3,4,5,6, and:

1. if i=1 has E _i=E ₁, expression read write line r _yBe positioned at neck material outbound station;

2. if i=2 has E _i=E ₂, expression read write line r _yBe positioned at the arrival district of processing or assembly station;

3. if i=3 has E _i=E ₃, expression read write line r _yBe positioned at the district of doing over again of processing or assembly station;

4. if i=4 has E _i=E ₄, expression read write line r _yBe positioned at the district of scrapping of processing or assembly station;

5. if i=5 has E _i=E ₅, expression read write line r _yBe positioned at inspection station eventually;

6. if i=6 has E _i=E ₆, expression read write line r _yBe positioned at finished product warehouse-in station;

2) P _jThe expression operation number, j=1,2 ..., J, J are the total number of operation:

1. when j=1, corresponding operation is neck material outbound operation;

2. work as j=2,3 ..., during J-2, corresponding operation is processing or assembly process;

3. when j=J-1, corresponding operation is examined operation eventually for completion;

4. when j=J, corresponding operation is a finished product warehouse-in operation;

3) W _lExpression worker item, l=1,2 ..., J,

4) r _yExpression radio-frequency identification reader/writer numbering, y=1,2 ..., Y, Y=3 (J-3)+3=3J-6, the concrete value of y is relevant with j, and computing method are:

1. when j=1, y=1;

(the read write line location type is got in " PlaceType " expression);

3. when j=J-1, y=3J-7;

4. when j=J, y=3J-6;

Between

expression radio-frequency identification reader/writer location type and the radio-frequency identification reader/writer is many-one relationship; Promptly to any radio-frequency identification reader/writer; Its location type is unique; And identical location type can be to there being a plurality of radio-frequency identification reader/writers

Between

expression operation, station and the radio-frequency identification reader/writer is one-one relationship

(2) initialization logistics state matrix and logistics state variation matrix:

The logistics state matrix is the register or the register of logistics information during job task is carried out; Be used for storing the different logistics informations constantly of entire job task implementation; Wherein, The row of logistics state matrix has been represented six types of different logistics complicated events, matrix column represented when carrying out job task the set of operation of process, so the element representation in the logistics state matrix accumulative total frequencies of different moment logistics complicated events on different operations.

If the logistics state matrix is M _L(t), initialized logistics state matrix is:

M _L(t ₀)＝O _(6×J) (3)

Logistics state variation matrix is the deposit unit of logistics information sometime during job task is carried out, sometime logistics information in carrying out in order to the storage operation task.

If logistics state variation matrix is Δ M _L(t), initialized logistics state variation matrix is:

ΔM _L(t ₀)＝O _(6×J) (4)

(3) obtain t _nRF identification raw data constantly:

The RF identification raw data is meant that the RFID tag that sticks on sometime on the material is read by the radio-frequency identification reader/writer on certain station and the data that form; Be a tlv triple, comprise attributes such as radio frequency identification marking sign, radio-frequency identification reader/writer numbering, time for reading.

If RFID (t) expression RF identification raw data, then t _nA RF identification raw data constantly is:

RFID(t _n)＝{{o ₁，r ₁，t _n}，{o ₂，r ₂，t _n}，...，{o _x，r _y，t _n}，...} (5)

(4) convert the RF identification raw data into the logistics simple event:

Article one, the generation of RF identification raw data can be considered the generation of a logistics simple event; Therefore can directly convert the RF identification raw data into the logistics simple event; Wherein the radio frequency identification marking sign in the RF identification raw data corresponds to the radio frequency identification marking sign in the logistics simple event; The radio-frequency identification reader/writer numbering corresponds to the radio-frequency identification reader/writer numbering in the logistics simple event; Time for reading corresponds to logistics simple event time of origin, thereby, comprised ingredients such as radio frequency identification marking sign, radio-frequency identification reader/writer numbering, logistics simple event time of origin in the logistics simple event of acquisition.

By t _nA RF identification raw data { o constantly _x, r _y, t _n, obtain t _nA logistics simple event E constantly _r(t _n):

E _r(t _n)＝e(o _x，r _y，t _n) (6)

(5) the logistics simple event is polymerized to the logistics complicated event:

A logistics complicated event can obtain according to the polymerization of a logistics simple event.According to the radio frequency identification marking sign in the logistics simple event,, can obtain the job task numbering in the logistics complicated event in conjunction with " job task-material-RFID tag " corresponding relation of setting up in (1) step; According to the numbering of the radio-frequency identification reader/writer in the logistics simple event,, can obtain the operation numbering in the logistics complicated event in conjunction with " station-operation-radio-frequency identification reader/writer " corresponding relation of setting up in (1) step; According to the numbering of the radio-frequency identification reader/writer in the logistics simple event, " radio-frequency identification reader/writer-radio-frequency identification reader/writer location type " corresponding relation in conjunction with setting up in (1) step can obtain the event type in the logistics complicated event; Logistics simple event time of origin according in the logistics simple event can obtain logistics complicated event time of origin.Thereby can obtain a complete logistics complicated event; Comprise ingredients such as logistics complicated event numbering (be a natural number, obtain), job task numbering, operation numbering, event type, Event origin (promptly corresponding logistics simple event), logistics complicated event time of origin according to the mode that increases certainly.

By t _nA logistics simple event e (o constantly _x, r _y, t _n), polymerizable generates a t _nLogistics complicated event E (t constantly _n):

E(t _n)＝e(id，{ID _T，E _i，P _j}，{E _r(t _n)}，t _n) (7)

Wherein:

1) id is a natural number, obtains through the mode that increases certainly;

2) ID _T=(o _x→ ID _T), expression is passed through formula (1) by o _xObtain ID _T

3) E _i=(r _y→ E _i), expression is passed through formula (2) by r _yObtain E _i, be specially:

If 1. radio-frequency identification reader/writer r _yBe positioned at neck material outbound station, then E _i=E ₁

If 2. radio-frequency identification reader/writer r _yBe positioned at processing or assembly station and arrive district, then E _i=E ₂

If 3. radio-frequency identification reader/writer r _yBe positioned at processing or assembly station do over again district, then E _i=E ₃

If 4. radio-frequency identification reader/writer r _yBe positioned at processing or assembly station and scrap the district, then E _i=E ₄

If 5. radio-frequency identification reader/writer r _yBe positioned at inspection station, then E eventually _i=E ₅

If 6. radio-frequency identification reader/writer r _yBe positioned at finished product warehouse-in station, then E _i=E ₆

4) P _j=(r _y→ P _j), expression is passed through formula (3) by r _yObtain P _j,

(6) obtain t by the logistics complicated event _nLogistics state variation matrix constantly:

According to the attribute in the logistics complicated event; Comprise event type and operation numbering; Can locate certain respective element in the logistics state variation matrix (wherein the row of incident homologue stream mode transformation matrices, the row of operation homologue stream mode transformation matrices), thereby upgrade the value of this element.

For t _nLogistics state variation matrix constantly has:

${\mathrm{\&Delta;M}}_L\left(t_n\right)=\left[\begin{array}{ccc}\mathrm{\&Delta;}{M}_L^{\left(1\right)}\left(t_n\right)& {\mathrm{\&Delta;M}}_L^{\left(3\right)}\left(t_n\right)& O\\ O& \mathrm{\&Delta;}{M}_L^{\left(2\right)}\left(t_n\right)& O\\ O& O& \mathrm{\&Delta;}{M}_L^{\left(4\right)}\left(t_n\right)\end{array}\right],$

Wherein:

The outbound of

expression neck material changes submatrix, and have:

is the element in the matrix

;

expression operation arrival and the inventory disposal and the variation submatrix of doing over again have:

<maths num="0017"> <! [CDATA [<math> <mrow> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 2 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub > <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> <mo> = </ mo> <mfenced open = '[' close = ']'> <mtable> <mtr> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 22 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi > n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. < / mo> </ mtd> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> < mn> 2 </ mn> <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t < / mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> <mo> [</ mo> <mn> 2 </ mn> <mrow> <mo> (</ mo> <mi> J </ mi> <mo> - </ mo> <mn> 2 </ mn> <mo>) </ mo> </ mrow> <mo>] </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> </ mtr> <mtr> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi > L </ mi> <mrow> <mo> (</ mo> <mn> 32 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> < mo> (</ mo> <mn> 3 </ mn> <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> <mo> [</ mo> <mn> 3 </ mn> <mrow> <mo> (</ mo> <mi> J </ mi> <mo> - </ mo> <mn> 2 </ mn> <mo >) </ mo> </ mrow> <mo>] </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi > n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> </ mtr> <mtr> <mtd> <msubsup> <mi> ΔM </ mi> < mi> L </ mi> <mrow> <mo> (</ mo> <mn> 42 </ mn> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (< / mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. < / mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 4 </ mn> <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> <mtd> <mo>. </ mo> <mo>. </ mo> <mo>. </ mo> </ mtd> <mtd> <msubsup> <mi> ΔM </ mi> <mi> L </ mi> <mrow> <mo> [</ mo> <mn> 4 </ mn> <mrow> <mo> (</ mo> <mi> J </ mi> <mo> - </ mo> <mn > 2 </ mn> <mo>) </ mo> </ mrow> <mo>] </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi > t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> </ mtd> </ mtr> </ mtable> </ mfenced> <mo>, </ mo> <mi> Δ </ mi> <msubsup> <mi> M </ mi> <mi> L </ mi> <mrow> <mo> (</ mo> <mn> 2 </ mn > <mi> j </ mi> <mo>) </ mo> </ mrow> </ msubsup> <mrow> <mo> (</ mo> <msub> <mi> t </ mi> <mi> n </ mi> </ msub> <mo>) </ mo> </ mrow> <mo>, </ mo> </ mrow> </ math>]]> </maths> <img file =" BDA0000115410870000112.GIF " he =" 71 " img-content =" drawing " img-format =" tif " inline =" yes " orientation =" portrait " wi =" 562 "/> is the matrix <img file = "BDA0000115410870000113.GIF" he = "65" img-content = "drawing" img-format = "tif" inline = "yes" orientation = "portrait" wi = "211" /> elements;

Lead material to change submatrix after

expression is scrapped again, have:

${\mathrm{\&Delta;M}}_L^{\left(3\right)}\left(t_n\right)=[\mathrm{\&Delta;}{M}_L^{\left(12\right)}\left(t_n\right)...\mathrm{\&Delta;}{M}_L^{\left(1j\right)}\left(t_n\right)...\mathrm{\&Delta;}{M}_L^{\left[1(J-2)\right]}\left(t_n\right)],\mathrm{\&Delta;}{M}_L^{\left(1j\right)}\left(t_n\right)$ Be the element in the matrix

;

expression material completion and warehouse-in change submatrix, have:

< maths num= " 0019 " > <! [CDATA[< math > < mrow > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 4 </ mn > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > < mo >=</ mo > mfenced open='['close='] ' < mtable > < mtr > < mtd > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > [</ mo > < mn > 5 </ mn > < mrow > < mo > (</ mo > < mi > J </ mi > < mo >-</ mo > < mn > 1 </ mn > < mo >) </ mo > </ mrow > < mo >] </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > < mtd > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 5 </ mn > < mi > J </ mi > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > </ mtr > < mtr > < mtd > < msubsup > < mi > &Delta; M </ mi > < mi > L </ mi > < mrow > < mo > [</ mo > < mn > 6 </ mn > < mrow > < mo > (</ mo > < mi > J </ mi > < mo >-</ mo > < mn > 1 </ mn > < mo >) </ mo > </ mrow > < mo >] </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > < mtd > < mi > &Delta; </ mi > < msubsup > < mi > M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 6 </ mn > < mi > J </ mi > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > </ mtd > </ mtr > </ mtable > </ mfenced > < mo >; </ mo > < msubsup > < mi > &Delta; M </ mi > < mi > L </ mi > < mrow > < mo > [</ mo > < mn > 5 </ mn > < mrow > < mo > (</ mo > < mi > J </ mi > < mo >-</ mo > < mn > 1 </ mn > < mo >) </ mo > </ mrow > < mo >] </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > < mo >; </ mo > < msubsup > < mi > &Delta; M </ mi > < mi > L </ mi > < mrow > < mo > (</ mo > < mn > 5 </ mn > < mi > J </ mi > < mo >) </ mo > </ mrow > </ msubsup > < mrow > < mo > (</ mo > < msub > < mi > t </ mi > < mi > n </ mi > </ msub > < mo >) </ mo > </ mrow > < mo >; </ mo > </ mrow > </ math >]] > </maths> <img file= " BDA0000115410870000119.GIF " he= " 72 " img-content= " drawing " img-format= " tif " inline= " yes " orientation= " portrait " wi= " 632 " /> be the element in the matrix <img file= " BDA00001154108700001110.GIF " he= " 65 " img-content= " drawing " img-format= " tif " inline= " yes " orientation= " portrait " wi= " 211 " />

According to formula (7), can calculate the value that obtains each element in the above-mentioned partitioned matrix:

$1)\mathrm{\&Delta;}{M}_L^{\left(11\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i=E_1\mathrm{AndE}\left(t_n\right).P_j=P_1\\ 0& \mathrm{Else}\end{array}\right.;$

(annotate: logistics complicated event E (t is got in symbol ". " expression _n) in certain property value, down with)

$2)\mathrm{\&Delta;}{M}_L^{\left(2j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="2"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_2\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$3)\mathrm{\&Delta;}{M}_L^{\left(3j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="3"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_3\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$4)\mathrm{\&Delta;}{M}_L^{\left(4j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i=E_4\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$5)\mathrm{\&Delta;}{M}_L^{\left(1j\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i=E_1\mathrm{AndE}\left(t_n\right).P_j=P_j\\ 0& \mathrm{Else}\end{array}\right.;$

$6)\mathrm{\&Delta;}{M}_L^{\left[5(J-1)\right]}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="5"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_5\mathrm{AndE}\left(t_n\right).P_j=P_{(J-1)}\\ 0& \mathrm{Else}\end{array}\right.;$

$7)\mathrm{\&Delta;}{M}_L^{\left(5J\right)}\left(t_n\right)\&equiv;0;$

$8)\mathrm{\&Delta;}{M}_L^{\left[6(J-1)\right]}\left(t_n\right)\&equiv;0;$

$9)\mathrm{\&Delta;}{M}_L^{\left(6J\right)}\left(t_n\right)=\left\{\begin{array}{cc}1& \mathrm{IfE}\left(t_n\right).E_i={\mathrm{<figure-callout\; id="6"\; label="E"\; filenames="HDA0000115410880000011.png"\; state="\{\{state\}\}">E</figure-callout>}}_6\mathrm{AndE}\left(t_n\right).P_j=P_J\\ 0& \mathrm{Else}\end{array}\right.,$

(7) generate t _nLogistics state matrix constantly:

t _nLogistics state matrix constantly equals t _(n-1)Logistics state matrix and t constantly _nLogistics state variation matrix sum constantly, so have:

M _L(t _n)＝M _L(t _(n-1))+ΔM _L(t _n) (8)

In the formula (8), M _L(t _(n-1)) expression t _(n-1)Logistics state matrix constantly,

(8) the sum of ranks element according to each partitioned matrix in the logistics state matrix obtains t _nLogistics information constantly:

With t _nLogistics state matrix constantly carries out piecemeal; Obtain neck material outbound submatrix, operation arrival and the inventory disposal and the submatrix of doing over again, lead material for making clothes matrix, material completion and warehouse-in submatrix again after scrapping; Through analyzing the sum of ranks element of these four partitioned matrix, can obtain job task at t _nLogistics information constantly.

To the logistics state matrix $M_L\left(t_n\right)=\left[\begin{array}{ccc}{M}_L^{\left(1\right)}\left(t_n\right)& M_L^{\left(3\right)}\left(t_n\right)& O\\ O& M_L^{\left(2\right)}\left(t_n\right)& O\\ O& O& M_L^{\left(4\right)}\left(t_n\right)\end{array}\right]$ Each partitioned matrix analyze, obtain corresponding logistics information (wherein: lead material for making clothes matrix,

expression material completion again and put submatrix in storage after

expression neck material outbound submatrix,

expression operation arrival and inventory disposal are scrapped with the submatrix of doing over again,

expression):

1) obtains neck and expect library information

During 1. as

, material is not neck material outbound as yet;

(annotate: rank of matrix is got in symbol " Rank () " expression, down together)

During 2. as

; Material has been led the material outbound, and neck material outbound quantity is

2) obtain operation arrival and inventory disposal, the information of doing over again

During 1. as

, job task does not begin to carry out;

2. work as

The time, job task has begun to carry out, when And

The time, job task implements j procedure P _j

3. work as

The time, there is material to do over again or the generation of inventory disposal situation, if

Then do over again and occur in j procedure P _j, if

Then scrap and occur in j procedure P _jOn;

During 4. as ; Existing material is done over again also has the inventory disposal situation to take place, scrap and the operation localization method of doing over again the same;

3) obtain and lead material information again after scrapping

During 1. as

, lead the material situation to take place after not scrapping again;

During 2. as

; Have and lead after scrapping the material situation to take place again, and if neck material

was individual again after

scrapped;

4) obtain material completion and go into library information

During 1. as

, material is not inspection eventually as yet;

During 2. as

; Material has carried out whole inspection, and inspection quantity is

eventually

During 3. as ; The material warehouse-in that completed, warehousing quantity is

(9) logistics state variation matrix is changed to null matrix, forwarded for (3) step to, obtain t _(n+1)Logistics information constantly.

Setting up t _nLogistics state matrix M constantly _L(t _n) and obtain t _nAfter the logistics information constantly, with t _nLogistics state variation matrix Δ M constantly _L(t _n) be changed to null matrix, next forwarded for (3) step to, begin to obtain t _(n+1)Logistics information constantly is up to the entire job task ID _TTill all being finished.

According to above-mentioned embodiment, below will be through the validity of concrete instance illustration method.

(1) assign job task, bind " job task-material-RFID tag " and " station-operation-radio-frequency identification reader/writer ":

Job task to be numbered " SC2010000562 " is an example, and material is all identical in the job task, and material quantity is 5, need be through 7 procedures, and corresponding station has 7.At first bind " job task-material-RFID tag " and " station-operation-radio-frequency identification reader/writer ", respectively shown in table 1, table 2 and table 3.

Table 1 job task numbering/thing item number/tag number corresponding instance

The job task numbering	The thing item number	Tag number
			SC2010000562	M 1	o 1
SC2010000562	M 2	o 2
			SC2010000562	M 3	o 3
SC2010000562	M 4	o 4
			SC2010000562	M 5	o 5

Table 2 radio-frequency identification reader/writer numbering/radio-frequency identification reader/writer location type corresponding instance

The radio-frequency identification reader/writer numbering	The radio-frequency identification reader/writer location type
		r 1	E 1
r 2	E 2

r 3	E 3
		r 4	E 4
r 5	E 2
		r 6	E 3
r 7	E 4
		r 8	E 2
r 9	E 3
		r 10	E 4
r 11	E 2
		r 12	E 3

Continuous table

The radio-frequency identification reader/writer numbering	The radio-frequency identification reader/writer location type
		r 13	E 4
r 14	E 5
		r 15	E 6

Table 3 operation number/worker's item/radio-frequency identification reader/writer numbering corresponding instance

Operation number	Worker's item	The read write line numbering
			P 1	W 1	r 1
P 2	W 2	r 2
			P 2	W 2	r 3
P 2	W 2	r 4
			P 3	W 3	r 5
P 3	W 3	r 6
			P 3	W 3	r 7
P 4	W 4	r 8
			P 4	W 4	r 9
P 4	W 4	r 10
			P 5	W 5	r 11
P 5	W 5	r 12
			P 5	W 5	r 13
P 6	W 6	r 14
			P 7	W 7	r 15

(2) initialization logistics state matrix and logistics state variation matrix:

Initialized logistics state matrix is: $M_L\left(t_0\right)={\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]}_{(6\&times;7)},$

Initialized logistics transformation matrices is: $M_L\left(t_0\right)={\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]}_{(6\&times;7)}.$

(3) obtain t _nRF identification raw data constantly:

RFID(t ₁₁)＝{o ₂，r ₁₀，t ₁₁}。

(4) convert the RF identification raw data into the logistics simple event:

E _r(t ₁₁)＝e(o ₂，r ₁₀，t ₁₁)。

(5) the logistics simple event is polymerized to the logistics complicated event:

According to formula (7), have: E (t _n)=e (id, { ID _T, E _i, P _j, { E _r(t _n), t _n),

Wherein:

(a) id is assumed to 25 here from increasing;

(b) ID _T=(o _x→ ID _T), o here _x=o ₂,, ID is arranged therefore according to table 1 _T=SC2010000562;

(c) E _i=(r _y→ E _i), r here _y=r ₁₀,, E is arranged therefore according to table 2 _i=E ₄

(d) P _j=(r _y→ P _j), r here _y=r ₁₀,, P is arranged therefore according to table 3 _j=P ₄

Thereby have:

E(t ₁₁)＝e(25，{SC2010000562，E ₄，P ₄}，{E _r(t ₁₁)}，t ₁₁)。

(6) obtain t by the logistics complicated event _nLogistics state variation matrix constantly:

Because E is (t ₁₁) E _i=E ₄, E (t ₁₁) P _j=P ₄So, have only

And all the other variation parts are 0.So t ₁₁The logistics transformation matrices that generates constantly does $\mathrm{\&Delta;}{M}_L\left(t_{11}\right)=\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right].$

(7) generate t _nLogistics state matrix constantly:

According to formula (8), M _L(t ₁₁)=M _L(t ₁₀)+Δ M _L(t ₁₁), M wherein _L(t ₁₀) be t ₁₀Logistics state matrix constantly is specially: $M_L\left(t_{10}\right)=\left[\begin{array}{ccccccc}5& 0& 0& 0& 0& 0& 0\\ 0& 5& 5& 4& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right],$ So t ₁₁Logistics state matrix constantly is:

$M_L\left(t_{11}\right)=\left[\begin{array}{ccccccc}5& 0& 0& 0& 0& 0& 0\\ 0& 5& 5& 4& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]+\left[\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right]=\left[\begin{array}{ccccccc}5& 0& 0& 0& 0& 0& 0\\ 0& 5& 5& 4& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right].$

(8) the sum of ranks element according to each partitioned matrix in the logistics state matrix obtains t _nLogistics information constantly:

The partitioning of matrix is had: $M_L\left(t_{11}\right)=\left[\begin{array}{ccc}{M}_L^{\left(1\right)}\left(t_{11}\right)& M_L^{\left(3\right)}\left(t_{11}\right)& O\\ O& M_L^{\left(2\right)}\left(t_{11}\right)& O\\ O& O& M_L^{\left(4\right)}\left(t_{11}\right)\end{array}\right],$ Wherein:

$M_L^{\left(1\right)}\left(t_{11}\right)=\left[5\right],M_L^{\left(2\right)}\left(t_{11}\right)=\left[\begin{array}{cccc}5& 5& 4& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right],M_L^{\left(3\right)}\left(t_{11}\right)=\left[\begin{array}{cccc}0& 0& 0& 0\end{array}\right],M_L^{\left(4\right)}\left(t_{11}\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right].$

Analyzing the sum of ranks element of each partitioned matrix can know:

is so material has been led the material outbound, and neck material outbound quantity is

So doing over again, existing material also have the inventory disposal situation to take place,

Scrap and occur in the 4th procedure P ₄On, and the situation of doing over again is at t ₁₁Took place constantly;

is so lead the material situation to take place again after not scrapping;

be not so material begins whole inspection as yet.

(9) logistics state variation matrix is changed to null matrix, forwarded for (3) step to, obtain t _(n+1)Logistics information constantly:

Obtaining t ₁₁On the basis of logistics information constantly, with logistics state variation matrix Δ M _L(t ₁₁) be changed to null matrix, forwarded for (3) step to, obtain t ₁₂Logistics information constantly, detailed process is the same, repeats no more.

Claims (1)

1. An online acquisition method of logistics information in discrete workshops based on radio frequency identification technology, characterized in that the steps of the method are as follows: (1) Release the job task and bind "job task-material-RFID tag" and "station-process-RFID reader": First, you need to bind "job task-material-RFID tag", and the relationship is described as follows:

in: 1) ID _T represents the job task number; 2) M _k represents the material number; 3) o _x represents the radio frequency identification tag number;

Indicates that there is a one-to-many relationship between the job task and the material, indicating that there are K materials in the job task ID _T ;

Indicates that there is a one-to-one relationship between the material and the RFID tag, so k=x=1, 2, ..., K, indicating that each material is pasted with an RFID tag; Once again, it is necessary to bind the "station-process-RFID reader", and the relationship is described as follows:

in: 1) E _i represents the location type of the radio frequency identification reader r _y , i=1, 2, 3, 4, 5, 6, and: ①If i=1, there is E _i ＝E ₁ , which means that the reader r _y is located at the pick-out station; ②If i=2, E _i =E ₂ means that the reader r _y is located in the arrival area of the processing or assembly station; ③ If i=3, E _i =E ₃ means that the reader r _y is located in the rework area of the processing or assembly station; ④ If i=4, there is E _i =E ₄ , which means that the reader r _y is located in the scrapping area of the processing or assembly station; ⑤If i=5, there is E _i ＝E ₅ , which means that the reader r _y is located at the final inspection station; ⑥If i=6, E _i =E ₆ means that the reader r _y is located at the finished product storage station; 2) P _j represents the process number, j=1, 2,..., J, J is the total number of processes: ① When j = 1, the corresponding process is the material picking and delivery process; ② When j=2, 3, ..., J-2, the corresponding process is a processing or assembly process; ③ When j=J-1, the corresponding process is the final inspection process; ④ When j=J, the corresponding process is the finished product warehousing process; 3) W _l represents the station number, l=1, 2,..., J, 4) r _y indicates the serial number of the radio frequency identification reader, y=1, 2, ..., Y, Y=3(J-3)+3=3J-6, the specific value of y is related to j, the calculation method for: ①When j=1, y=1; ② When j=2, 3, ..., J-2, $\mathrm{the\; y}=\left\{\begin{array}{c}3j-4\mathrm{if\; PlaceType}\left(r_{\mathrm{the\; y}}\right)={\mathrm{E.}}_2\\ 3j-3\mathrm{if\; PlaceType}\left(r_{\mathrm{the\; y}}\right)={\mathrm{E.}}_3\\ 3j-2\mathrm{if\; PlaceType}\left(r_{\mathrm{the\; y}}\right)={\mathrm{E.}}_4\end{array}\right.$ ("PlaceType" indicates the location type of the reader); ③When j=J-1, y=3J-7; ④ When j=J, y=3J-6;

Indicates that there is a one-to-many relationship between the location type of the RFID reader-writer and the RFID reader-writer, that is, for any RFID reader-writer, its location type is unique, and the same location type can correspond to multiple RF recognize the reader,

Indicates that there is a one-to-one relationship between the process, the station and the RFID reader, (2) Initialize the logistics state matrix and the logistics state change matrix: Let the logistics state matrix be M _L (t), the initialized logistics state matrix is: M _L (t ₀ )=O _(6×J) (3) Let the logistics state change matrix be ΔM _L (t), and the initialized logistics state change matrix is: ΔM _L (t ₀ )=O _(6×J) (4) (3) Obtain the raw radio frequency identification data at t _n moment: Let RFID(t) represent the original data of radio frequency identification, then a piece of original data of radio frequency identification at time t _n is: RFID(t _n )={{o ₁ , r ₁ , t _n }, {o ₂ , r ₂ , t _n }, ..., {o _x , r _y , t _n }, ...} (5 ) (4) Convert RFID raw data into logistics simple events: From a piece of radio frequency identification raw data {o _x , _ry , t _n } at time t _n , a simple logistics event E _r (t _n ) at time t _n is obtained: E _r (t _n )=e(o _x , _ry , t _n ) (6) (5) Aggregate simple logistics events into complex logistics events: A simple logistics event e(o _x , _ry , t _n ) at time t _n can be aggregated to generate a complex logistics event E(t _n ) at time t _n : E(t _n )=e(id, {ID _T , E _i , P _j }, {E _r (t _n )}, t _n ) (7) in: 1) id is a natural number, which is obtained by self-increment; 2) ID _T = (o _x → ID _T ), which means that the ID _T is obtained from o _x through the formula (1); 3) E _i = (ry _y → E _i ), which means that E _i is obtained from ry _y through formula (2), specifically: ① If the radio frequency identification reader r _y is located at the pick-out station, then E _i = E ₁ ; ②If the radio frequency identification reader r _y is located in the processing or assembly station arrival area, then E _i =E ₂ ; ③ If the radio frequency identification reader _ry is located in the rework area of the processing or assembly station, then E _i =E ₃ ; ④ If the radio frequency identification reader r _y is located in the scrapping area of the processing or assembly station, then E _i =E ₄ ; ⑤ If the radio frequency identification reader r _y is located at the final inspection station, then E _i = E ₅ ; ⑥If the radio frequency identification reader r _y is located at the finished product storage station, then E _i =E ₆ ; 4) P _j = (ry _y → P _j ), which means that P _j is obtained from ry _y through formula (3), (6) Obtain the logistics state change matrix at time t _n from complex logistics events: For the logistics state change matrix at time t _n , there are: ${\mathrm{<span\; class="notranslate">\; \&Delta;M</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left[\begin{array}{ccc}\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span>& {\mathrm{<span\; class="notranslate">\; \&Delta;M</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 3</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span>& \mathrm{<span\; class="notranslate">\; o</span>}\\ \mathrm{<span\; class="notranslate">\; o</span>}& \mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span>& \mathrm{<span\; class="notranslate">\; o</span>}\\ \mathrm{<span\; class="notranslate">\; o</span>}& \mathrm{<span\; class="notranslate">\; o</span>}& \mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 4</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span>\end{array}\right]<span\; class="notranslate">\; ,</span>$ in:

Represents the change sub-matrix of picking and delivery, including: for the matrix

elements in

Represents the sub-matrix of process arrival and material scrap and rework change, including: $\mathrm{\&Delta;}{m}_L^{\left(2\right)}\left(t_{\mathrm{no}}\right)=\left[\begin{array}{ccccc}\mathrm{\&Delta;}{m}_L^{\left(\mathrm{twenty\; two}\right)}\left(t_{\mathrm{no}}\right)& ...& \mathrm{\&Delta;}{m}_L^{\left(2j\right)}\left(t_{\mathrm{no}}\right)& ...& {\mathrm{\&Delta;M}}_L^{\left[2(J-2)\right]}\left(t_{\mathrm{no}}\right)\\ \mathrm{\&Delta;}{m}_L^{\left(32\right)}\left(t_{\mathrm{no}}\right)& ...& {\mathrm{\&Delta;M}}_L^{\left(3j\right)}\left(t_{\mathrm{no}}\right)& ...& {\mathrm{\&Delta;M}}_L^{\left[3(J-2)\right]}\left(t_{\mathrm{no}}\right)\\ {\mathrm{\&Delta;M}}_L^{\left(42\right)}\left(t_{\mathrm{no}}\right)& ...& \mathrm{\&Delta;}{m}_L^{\left(4j\right)}\left(t_{\mathrm{no}}\right)& ...& {\mathrm{\&Delta;M}}_L^{\left[4(J-2)\right]}\left(t_{\mathrm{no}}\right)\end{array}\right],\mathrm{\&Delta;}{m}_L^{\left(2j\right)}\left(t_{\mathrm{no}}\right),$

for the matrix

elements in

Represents the sub-matrix of material change after scrapping, including: ${\mathrm{\&Delta;M}}_L^{\left(3\right)}\left(t_{\mathrm{no}}\right)=[\mathrm{\&Delta;}{m}_L^{\left(12\right)}\left(t_{\mathrm{no}}\right)...\mathrm{\&Delta;}{m}_L^{\left(1j\right)}\left(t_{\mathrm{no}}\right)...\mathrm{\&Delta;}{m}_L^{\left[1(J-2)\right]}\left(t_{\mathrm{no}}\right)],\mathrm{\&Delta;}{m}_L^{\left(1j\right)}\left(t_{\mathrm{no}}\right)$ for the matrix

elements in

Represents the material completion and storage change submatrix, including: $\mathrm{\&Delta;}{m}_L^{\left(4\right)}\left(t_{\mathrm{no}}\right)=\left[\begin{array}{cc}\mathrm{\&Delta;}{m}_L^{\left[5(J-1)\right]}\left(t_{\mathrm{no}}\right)& \mathrm{\&Delta;}{m}_L^{\left(5J\right)}\left(t_{\mathrm{no}}\right)\\ {\mathrm{\&Delta;M}}_L^{\left[6(J-1)\right]}\left(t_{\mathrm{no}}\right)& \mathrm{\&Delta;}{m}_L^{\left(6J\right)}\left(t_{\mathrm{no}}\right)\end{array}\right],{\mathrm{\&Delta;M}}_L^{\left[5(J-1)\right]}\left(t_{\mathrm{no}}\right),{\mathrm{\&Delta;M}}_L^{\left(5J\right)}\left(t_{\mathrm{no}}\right),{\mathrm{\&Delta;M}}_L^{\left[6(J-1)\right]}\left(t_{\mathrm{no}}\right),$

for the matrix

elements in the According to formula (7), the value of each element in the above block matrix can be calculated and obtained: $\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 11</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; IfE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; i</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}\mathrm{<span\; class="notranslate">\; AndE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; Else</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ (Note: The symbol "." means to take an attribute value in the logistics complex event E(t _n ), the same below) $\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 2</span>}\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; IfE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; i</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}\mathrm{<span\; class="notranslate">\; AndE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; Else</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ $\mathrm{<span\; class="notranslate">\; 3</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 3</span>}\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; IfE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; i</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; 3</span>}}\mathrm{<span\; class="notranslate">\; AndE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; Else</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ $\mathrm{<span\; class="notranslate">\; 4</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 4</span>}\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; IfE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; i</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; 4</span>}}\mathrm{<span\; class="notranslate">\; AndE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; Else</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ $\mathrm{<span\; class="notranslate">\; 5</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 1</span>}\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; IfE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; i</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}\mathrm{<span\; class="notranslate">\; AndE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; Else</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ $\mathrm{<span\; class="notranslate">\; 6</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; 5</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; J</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; ]</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; IfE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; i</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; 5</span>}}\mathrm{<span\; class="notranslate">\; AndE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; J</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; )</span>}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; Else</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ $\mathrm{<span\; class="notranslate">\; 7</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 5</span>}\mathrm{<span\; class="notranslate">\; J</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&equiv;</span>\mathrm{<span\; class="notranslate">\; 0</span>}<span\; class="notranslate">\; ;</span>$ $\mathrm{<span\; class="notranslate">\; 8</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; 6</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; J</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; ]</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; \&equiv;</span>\mathrm{<span\; class="notranslate">\; 0</span>}<span\; class="notranslate">\; ;</span>$ $\mathrm{<span\; class="notranslate">\; 9</span>}<span\; class="notranslate">\; )</span>\mathrm{<span\; class="notranslate">\; \&Delta;</span>}{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; L</span>}}^{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 6</span>}\mathrm{<span\; class="notranslate">\; J</span>}<span\; class="notranslate">\; )</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; 1</span>}& \mathrm{<span\; class="notranslate">\; IfE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; i</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; E.</span>}}_{\mathrm{<span\; class="notranslate">\; 6</span>}}\mathrm{<span\; class="notranslate">\; AndE</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; t</span>}}_{\mathrm{<span\; class="notranslate">\; no</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; .</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; j</span>}}<span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; P</span>}}_{\mathrm{<span\; class="notranslate">\; J</span>}}\\ \mathrm{<span\; class="notranslate">\; 0</span>}& \mathrm{<span\; class="notranslate">\; Else</span>}\end{array}\right.<span\; class="notranslate">\; ,</span>$ (7) Generate the logistics state matrix at time t _n : M _L (t _n )＝M _L (t _(n-1) )+ΔM _L (t _n ) (8) In formula (8), M _L (t _(n-1) ) represents the logistics state matrix at time t _(n-1) , (8) Obtain the logistics information at time t _n according to the rank and elements of each block matrix in the logistics state matrix: For logistics status matrix $m_L\left(t_{\mathrm{no}}\right)=\left[\begin{array}{ccc}{m}_L^{\left(1\right)}\left(t_{\mathrm{no}}\right)& m_L^{\left(3\right)}\left(t_{\mathrm{no}}\right)& o\\ o& m_L^{\left(2\right)}\left(t_{\mathrm{no}}\right)& o\\ o& o& m_L^{\left(4\right)}\left(t_{\mathrm{no}}\right)\end{array}\right]$ Analyze each of the block matrices to obtain the corresponding logistics information (where:

Represents the sub-matrix of pick-out warehouse,

Indicates the process arrival and material scrap and rework submatrix,

Indicates scrapped and reclaimed sub-matrix,

Indicates material completion and storage sub-matrix): 1) Obtain the picking and delivery information ①When

When , the material has not been picked out of the warehouse; (Note: The symbol "Rank()" means to take the rank of the matrix, the same below) ② when

When , the material has been picked out of the warehouse, and the number of picked out of the warehouse is

2) Obtain process arrival and material scrapping and rework information ①When , the job task does not start to execute; ② when

When , the job task has started to execute, when

and

When , the job task is executed to the jth process P _j ; ③ when

When there is material rework or material scrapping, if

Then rework occurs in the jth process P _j , if

Then scrapping occurs in the jth process P _j ; ④ when

, both material rework and material scrap occur, and the positioning method of scrap and rework process is the same as above; 3) Obtain information on picking materials after scrapping ① when

, no re-picking after scrapping occurs; ② when

, there is a situation of picking after scrapping, and if

Pick up after scrapping

indivual; 4) Obtain material completion and storage information ① when

, the material has not yet been finalized; ② when

When , the material has undergone final inspection, and the final inspection quantity is ③ when

When , the materials have been completed and put into storage, and the put-in quantity is (9) Set the logistics state change matrix to zero matrix, go to step (3) to obtain the logistics information at time t _(n+1) .

Images