Study on Establishment and Improvement Strategy of Aviation Equipment

Assume to enter the tree Ti(t)(I =1,...... ,n) regard R(t) as the root, and Lj1(t),...... Ljk(t),...... Ljm(t) is regarded as the end, then the included branch vectors are all components, and will be sorted one by one according to the size of the subscript jk contained in the mid-flow potential variable Ljk(t), and thus formed, defined as (Ri (t), Aij1 (t), Lj1 (t)... , (Ri (t), Aijk (t), Ljk (t)... Ti (t) tree vector of, (Ri (t), Aijm (t), Ljm (t)), can be denoted as (Ri (t), Lj1-... - jk -... Jm (t).

This content will transform flow graph into vector, and can use the calculation method of advanced mathematics to deal with the vector relationship, such as determinant, matrix, etc., which can simplify the calculation process and build a more standardized processing process. Assume that using (Ri (t), Lj1-... - jk -... Jm (t)) to indicate that all roots are R (t), ending is Lj1 (t),... Ljk (t)... The branch vector of LJM of T, then that means multiplication of the tree vector.

(Ri(t),Lj1 − ... − jk − 1 − jk − jk + 1 − ... jm(t)) × (Rt(t),Lp1 − ... − pq − 1 − pq − pq + 1 − ... pn(t)) {=((Ri(t),Aijk(t),Lt(t),Rt(t),Atpq(t),Lpq(t))),(Ri(t),Lj1−jk−1−...jk+1−...−jm(t)),(Rt(t),Lp1−...−pq−1−...−pq+1−...−pn(t))when.Ljk(t).is.the.Rt(t).stream.bit.and.Aijk(t).and.Atpq(t).have.no.same.variable;=0,other \left\{{\matrix{{= \left({\left({{\rm{Ri}}\left(t \right),Aijk\left(t \right),Lt\left(t \right),Rt\left(t \right),Atpq\left(t \right),Lpq\left(t \right)} \right)} \right),} \hfill \cr {\left({{\rm{Ri}}\left(t \right),Lj1 - {\rm{jk}} - 1 -...jk + 1 -... - {\rm{jm}}\left(t \right)} \right),} \hfill \cr {\left({{\rm{Rt}}\left(t \right),Lp1 -... - {\rm{pq}} - 1 -... - pq + 1 -... - {\rm{pn}}\left(t \right)} \right)} \hfill \cr {when.Ljk\left(t \right).is.the.Rt\left(t \right).stream.bit.and.Aijk\left(t \right).and.Atpq\left(t \right).have.no.same.variable;} \hfill \cr {= 0,other} \hfill \cr {} \hfill \cr}} \right.\;

In combination with the above formula analysis, it can be seen that when the tree vector is multiplied by the tree vector, in order to make the final result not to be 0, the middle shunt of the two must be guaranteed to be the same. In other words, the tree vector long hair and nested operation are equivalent procedures.

{(Ri(t),Lj1−...−jk−1−jk−jk+1−...−jm(t))×(Rt(t),Lp1−...−pf−1−pf−pf+1−...pq−1−pq−pq+1−...−pn(t))×(Rr(t),Ls1−...−sd−1−sd−sd+1−...−jm(t)) {=((Ri(t),Aijk(t),Lt(t),Rt(t),Atpq(t),Lr(t),Rr(t),Arsd(t),Lsd(t))),(Ri(t),Lj1−...−jk−1−jk+1−...jm(t)),(Rt(t),Lp1−...−pq−1−pq+1−...−pn(t)),(Rr(t),Ls1−...−sd−1−sd+1−...sm(t)),when.Ljk(t).and.Lpq(t).are.streams.of.Rt(t).and.Rr(t).respectively.and.there.is.no.same.variable.in.Aijk(t).and.Atpq(t),Atps(t).and.Arsd(t)=(Ri(t),Aijk(t),Lt(t),Ri(t),Atpq(t),Lpq(t)),(Ri(t),Atpf(t),Lr(t),Rr(t),Arsd(t),Lsd(t)),(Ri(t),Lj1−...−jk−1−jk+1−...jm(t)),(Rt(t,Lp1−...−pf−1−...pf+1−...−pq−1−pq+1−pn(t))),(Rr(t),Ls1−...sd−1−sd+1−su(t)),when.Ljk(t).and.Lpf(t).are.streams.of.Rt(t).and.Rr(t).respectively,and.there.is.no.same.variable.in.Aijk(t).and.Atpq(t).Atpf(t).and.Arsd(t);=0, other. \eqalign{& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{{\matrix{{\left({{\rm{Ri}}\left(t \right),Lj1 -... - {\rm{jk}} - 1 - {\rm{jk}} - jk + 1 -... - {\rm{jm}}\left(t \right)} \right) \times} \hfill \cr {\left({{\rm{Rt}}\left(t \right),Lp1 -... - {\rm{pf}} - 1 - {\rm{pf}} - pf + 1 -...{\rm{pq}} - 1 - {\rm{pq}} - pq + 1 -... - {\rm{pn}}\left(t \right)} \right) \times} \hfill \cr {\left({{\rm{Rr}}\left(t \right),Ls1 -... - {\rm{sd}} - 1 - {\rm{sd}} - sd + 1 -... - {\rm{jm}}\left(t \right)} \right)} \hfill \cr}} \right.\; \cr & \left\{{\matrix{{= \left({\left({{\rm{Ri}}\left(t \right),Aijk\left(t \right),Lt\left(t \right),Rt\left(t \right),Atpq\left(t \right),Lr\left(t \right),Rr\left(t \right),Arsd\left(t \right),Lsd\left(t \right)} \right)} \right),} \hfill \cr {\left({{\rm{Ri}}\left(t \right),Lj1 -... - {\rm{jk}} - 1 - jk + 1 -...{\rm{jm}}\left(t \right)} \right),} \hfill \cr {\left({{\rm{Rt}}\left(t \right),Lp1 -... - {\rm{pq}} - 1 - pq + 1 -... - {\rm{pn}}\left(t \right)} \right),} \hfill \cr {\left({{\rm{Rr}}\left(t \right),Ls1 -... - {\rm{sd}} - 1 - sd + 1 -...{\rm{sm}}\left(t \right)} \right),} \hfill \cr {when.{\rm{Ljk}}\left(t \right).{\rm{and}}.{\rm{Lpq}}\left(t \right).{\rm{are}}.{\rm{streams}}.{\rm{of}}.{\rm{Rt}}\left(t \right).{\rm{and}}.{\rm{Rr}}\left(t \right).{\rm{respectively}}.{\rm{and}}.} \hfill \cr {there.{\rm{is}}.{\rm{no}}.{\rm{same}}.{\rm{variable}}.{\rm{in}}.{\rm{Aijk}}\left(t \right).{\rm{and}}.{\rm{Atpq}}\left(t \right),Atps\left(t \right).{\rm{and}}.{\rm{Arsd}}\left(t \right)} \hfill \cr {= \left({{\rm{Ri}}\left(t \right),Aijk\left(t \right),Lt\left(t \right),Ri\left(t \right),Atpq\left(t \right),Lpq\left(t \right)} \right),} \hfill \cr {\left({{\rm{Ri}}\left(t \right),Atpf\left(t \right),Lr\left(t \right),Rr\left(t \right),Arsd\left(t \right),Lsd\left(t \right)} \right),} \hfill \cr {\left({{\rm{Ri}}\left(t \right),Lj1 -... - {\rm{jk}} - 1 - jk + 1 -...{\rm{jm}}\left(t \right)} \right),} \hfill \cr {\left({{\rm{Rt}}\left({t,Lp1 -... - {\rm{pf}} - 1 -...pf + 1 -... - {\rm{pq}} - 1 - pq + 1 - {\rm{pn}}\left(t \right)} \right)} \right),} \hfill \cr {\left({{\rm{Rr}}\left(t \right),Ls1 -...{\rm{sd}} - 1 - sd + 1 - {\rm{su}}\left(t \right)} \right),} \hfill \cr {when.{\rm{Ljk}}\left(t \right).{\rm{and}}.{\rm{Lpf}}\left(t \right).{\rm{are}}.{\rm{streams}}.{\rm{of}}.{\rm{Rt}}\left(t \right).{\rm{and}}.{\rm{Rr}}\left(t \right).respectively,} \hfill \cr {and.there.{\rm{is}}.{\rm{no}}.{\rm{same}}.{\rm{variable}}.{\rm{in}}.{\rm{Aijk}}\left(t \right).{\rm{and}}.{\rm{Atpq}}\left(t \right).{\rm{Atpf}}\left(t \right).{\rm{and}}.{\rm{Arsd}}\left(t \right);} \hfill \cr {= 0,\,other.} \hfill \cr}} \right.\; \cr}

Under the condition that the product of three tree vectors is not equal to 0, and the product of two tree vectors is also not 0, it is necessary to calculate how many flow rate variables of the first two tree vectors are contained in the third tree vector flow bit, and make clear the flow bit where one of the flow rate variables is located. In combination with the above definition formula analysis, it can be seen that in each tree vector of k (k≥3), the result obtained by multiplication among k−1 tree vectors will either add chain vectors or the final result will be 0. Therefore, when constructing an evaluation model of aviation equipment supply chain based on differential equations, two conditions must be considered in order to apply tree vector multiplication. On the one hand, it must be guaranteed to meet the exchange rate; on the other hand, the addition operator and distribution rate should be guaranteed in the operation process [1].

|X1,1,1,...,1,10,X2,1,...,1,1...,...,...,...,...,...0,0,0,...,Xn,11,1,1,1,1,1| \left| {\matrix{{X1,1,1,...,1,1} \cr {0,X2,1,...,1,1} \cr {...,...,...,...,...,...} \cr {0,0,0,...,Xn,1} \cr {1,1,1,1,1,1} \cr}} \right|

The above is defined as the determinant of x-0-1, where xi (I =1, 2... , n) belongs to vector form, and the corresponding calculation rule is as follows: |a11,a12,…,a1na21,a22,…,a2n…,…,…,…an1,an2,…,ann|=∑j1,j2,…jkaij 1,a2i2,…,anjn \left| {\matrix{{a11,a12, \ldots,a1n} \hfill \cr {a21,a22, \ldots,a2n} \hfill \cr {\ldots, \ldots, \ldots, \ldots} \hfill \cr {an1,an2, \ldots,ann} \hfill \cr}} \right| = \sum\limits_{j1,j2, \ldots jk} {\rm{aij}}1,a2i2, \ldots,{\rm{anjn}}

And ∑_{j1, j2,... jk} is for J1, J2... Sum of permutations of jn. According to the above definition and column analysis, it can be known that all of them are plus sign properties, so the following theorem can be obtained.

If the above determinants are as follows: 1,2... In order of n, the values of the two subexpressions will be the same every time. In the case of n=1, the above determinant is expanded according to column 1, which can be obtained: |X2,1,...,1,1...,...,...,...,...0,0,...,Xn,11,1,1,1,1,1|+|1,1,...,1,1X2,1,1,1,1...,...,...,...,...0,0,...,Xn,1| \left| {\matrix{{X2,1,...,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr {1,1,1,1,1,1} \cr}} \right| + \left| {\matrix{{1,1,...,1,1} \cr {X2,1,1,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr}} \right|

Since all the terms after the expansion of this determinant are plus signs, swapping them does not affect the final result. By converting the last line of the to the previous line, the final line is obtained. Proof. Since all the terms after the expansion of this determinant are plus signs, swapping them does not affect the final result. By converting the last line of the |X2,1,...,1,1...,...,...,...,...0,0,...,Xn,11,1,1,1,1| \left| {\matrix{{X2,1,...,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr {1,1,1,1,1} \cr}} \right| to the previous line, the final line is obtained |X2,1,...,1,1...,...,...,...,...0,0,...,Xn,11,1,1,1,1| \left| {\matrix{{X2,1,...,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr {1,1,1,1,1} \cr}} \right| .

If this statement is true for n=k and k≥1, then we get: |Xk+1,1,...,1,1...,...,...,...,...0,0,...,Xn,11,1,1,1,1|=|1,1,...,1,1Xk+1,1,1,1,1...,...,...,...,...0,0,...,Xn,1| \left| {\matrix{{Xk + 1,1,...,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr {1,1,1,1,1} \cr}} \right| = \left| {\matrix{{1,1,...,1,1} \cr {Xk + 1,1,1,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr}} \right|

In the case where n is equal to k plus 1 |Xk+1,1,...,1,1...,...,...,...,...0,0,...,Xn,11,1,1,1,1| \left| {\matrix{{Xk + 1,1,...,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr {1,1,1,1,1} \cr}} \right| , will expand in the first column order, then the submatrix of Xk plus 1 can be expressed as |Xk+1,1,...,1,1...,...,...,...,...0,0,...,Xn,11,1,1,1,1| \left| {\matrix{{Xk + 1,1,...,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr {1,1,1,1,1} \cr}} \right| And the submatrix formula of 1 is |1,1,...,1,1Xk+2,1,1,1,1...,...,...,...,...0,0,...,Xn,1| \left| {\matrix{{1,1,...,1,1} \cr {Xk + 2,1,1,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr}} \right| , which proves that the e|Xk+2,1,...,1,1...,...,...,...,...0,0,...,Xn,11,1,1,1,1|=|1,1,...,1,1Xk+2,1,1,1,1...,...,...,...,...0,0,...,Xn,1| e\left| {\matrix{{Xk + 2,1,...,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr {1,1,1,1,1} \cr}} \right| = \left| {\matrix{{1,1,...,1,1} \cr {Xk + 2,1,1,1,1} \cr {...,...,...,...,...} \cr {0,0,...,Xn,1} \cr}} \right| quation is true.

If the determinant is: |X1,1,1,…,1,10,X2,1,…,1,1…,…,…,…,…0,0,0,…,Xn,11,1,1,1,1,1|=1+X1+…+Xn+X1X2+…+X1Xn+…+Xn−1Xn+X1X2X3+…+Xn−2Xn−1Xn+…+X1X2X3…Xn \matrix{{\left| {\matrix{{X1,1,1, \ldots,1,1} \hfill \cr {0,X2,1, \ldots,1,1} \hfill \cr {\ldots, \ldots, \ldots, \ldots, \ldots} \hfill \cr {0,0,0, \ldots,Xn,1} \hfill \cr {1,1,1,1,1,1} \hfill \cr}} \right|} \cr {= 1 + X1 + \ldots + Xn + X1X2 + \ldots + X1Xn + \ldots + Xn - 1Xn} \cr {+ X1X2X3 + \ldots + Xn - 2Xn - 1Xn + \ldots + X1X2X3 \ldots Xn} \cr}

The following equation can then be proved to be true: |X1,1,1,...,1,10,X2,1,...,1,1...,...,...,...,...,...0,0,0,...,Xn,11,1,1,1,1,1|=(Xn+1)|X1,1,1,...,1,10,X2,1,...,1,1...,...,...,...,...,...0,0,0,...,Xn−1,11,1,1,1,1,1|=...=(Xn+1)(Xn−1+1)...(X3+1)|X1,1,10,X2,11,1,1|=1+X1+...+Xn+X1X2+...+X1Xn+...+X1X2X3+...+Xn−2Xn−1Xn+...+X1X2X3...Xn \matrix{{\matrix{{\left| {\matrix{{X1,1,1,...,1,1} \cr {0,X2,1,...,1,1} \cr {...,...,...,...,...,...} \cr {0,0,0,...,Xn,1} \cr {1,1,1,1,1,1} \cr}} \right| = \left({Xn + 1} \right)\left| {\matrix{{X1,1,1,...,1,1} \cr {0,X2,1,...,1,1} \cr {...,...,...,...,...,...} \cr {0,0,0,...,{X_{n - 1}},1} \cr {1,1,1,1,1,1} \cr}} \right| =...} \hfill \cr}} \cr {\matrix{{= \left({Xn + 1} \right)\left({{X_{n - 1}} + 1} \right)...\left({X3 + 1} \right)\left| {\matrix{{X1,1,1} \cr {0,X2,1} \cr {1,1,1} \cr}} \right|} \hfill \cr}} \cr {\matrix{{= 1 + X1 +... + Xn + X1X2 +... + X1Xn +... + X1X2X3 +... +} \hfill \cr}} \cr {\matrix{{{X_{n - 2}}{X_{n - 1}}Xn +... + X1X2X3...Xn} \hfill \cr}} \cr}

It can be seen that xk (k= 1,2... , n), then the determinant is exactly all of the combinations of n tree vectors. According to the definition of tree vector and theoretic analysis above, since its product is 0, it does not constitute the feedback basis module, so the result obtained by calculating this determinant is all the feedback basis modules of the subsystem. Combined with the above analysis of the theorem, it can be concluded that if the X-0-1 determinant is used to obtain the feedback fundamental mode of the system, it can not only obtain more calculation information, but also simplify the actual calculation steps [3].

R (R = 1,2... The fundamental conditions for multiplying tree vectors to obtain the feedback fundamental mode are shown in the above definition, and the simple calculation steps and results using the determinant X-0-1 are shown as follows: |(Ri(t),L2−6−7(t)),1,1,1,1,1,1,1,10,(R2(t),L1−3−8(t)),1,1,1,1,1,1,10,0,(R3(t),LA−7−3(t)),1,1,1,1,1,10,0,0,(R4(t),L3−5−3(t)),1,1,1,1,10,0,0,0,(R5(t),L7−3(t)),1,1,1,10,0,0,0,0,(R6(t),L7−3(t)),1,1,10,0,0,0,0,0,(R7(t),L2−4(t)),1,10,0,0,0,0,0,0,(R8(t),L2−4(t)),11,1,1,1,1,1,1,1,1|=(R2(t),L1−3−8(t)+1)×(R1(t),L2−6−7(t)+1)×|1,1,1,1,1,1,1(R(t),LA−7−3(t)),1,1,1,1,1,10,(R(t),L3−5−3(t)),1,1,1,1,10,0,(R(t),L7−3(t)),1,1,1,10,0,0,(R(t),L7−8(t)),1,1,10,0,0,0,(R(t),L2−4(t)),1,10,0,0,0,0,(R(t),L2−4(t)),1| \matrix{{\left| {\matrix{{\left({Ri\left(t \right),L2 - 6 - 7\left(t \right)} \right),1,1,1,1,1,1,1,1} \cr {0,\left({R2\left(t \right),L1 - 3 - 8\left(t \right)} \right),1,1,1,1,1,1,1} \cr {0,0,\left({R3\left(t \right),LA - 7 - 3\left(t \right)} \right),1,1,1,1,1,1} \cr {0,0,0,\left({R4\left(t \right),L3 - 5 - 3\left(t \right)} \right),1,1,1,1,1} \cr {0,0,0,0,\left({R5\left(t \right),L7 - 3\left(t \right)} \right),1,1,1,1} \cr {0,0,0,0,0,\left({R6\left(t \right),L7 - 3\left(t \right)} \right),1,1,1} \cr {0,0,0,0,0,0,\left({R7\left(t \right),L2 - 4\left(t \right)} \right),1,1} \cr {0,0,0,0,0,0,0,\left({R8\left(t \right),L2 - 4\left(t \right)} \right),1} \cr {1,1,1,1,1,1,1,1,1} \cr}} \right|} \cr {= \left({R2\left(t \right),L1 - 3 - 8\left(t \right) + 1} \right) \times \left({R1\left(t \right),L2 - 6 - 7\left(t \right) + 1} \right)} \cr {\times \left| {\matrix{{1,1,1,1,1,1,1} \cr {\left({R\left(t \right),LA - 7 - 3\left(t \right)} \right),1,1,1,1,1,1} \cr {0,\left({R\left(t \right),L3 - 5 - 3\left(t \right)} \right),1,1,1,1,1} \cr {0,0,\left({R\left(t \right),L7 - 3\left(t \right)} \right),1,1,1,1} \cr {0,0,0,\left({R\left(t \right),L7 - 8\left(t \right)} \right),1,1,1} \cr {0,0,0,0,\left({R\left(t \right),L2 - 4\left(t \right)} \right),1,1} \cr {0,0,0,0,0,\left({R\left(t \right),L2 - 4\left(t \right)} \right),1} \cr}} \right|} \cr}

Combined with the above definition analysis, we can finally get the method of calculating all the feedback fundamental modes, which can be divided into the following points: First, we should build the basic flow rate into the tree model; Second, we have to construct the X minus 0 minus 1 determinant. Since the flow rate obtained basically corresponds to a tree vector in all the trees in the tree model, tree vectors (Ri (t), Lj1-J2 -...) can be used in calculation. Jm (t)) (I = 1,2...... , n) instead of Xk (k= 1,2... , n); Thirdly, the determinant is calculated and the feedback schema of all subsystems is obtained.