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Fikalo mislim da je to element skroz gore desno u tablici, tj. prvi element vektora b

sheriffHorsey aha ok fala, samo kolko vidim obrnuti predznak tog broja

Fikalo mislim da se obrnuti predznak dodaje zbog maksimizacije, na 30. slajdu su preformulirali problem iz maksimizacijskog u minimizacijski

ako radim right backbone pa uravnotezavnje jel bi stablo trebalo ispast isto kao da sam radio left backbone pa uravnotezavanje?

Gledam ovoga sta radi DSW na auditornima i on racuna visinu sa ceil, a u prezi je sa floor. I kome sad vjerovat. Onomo tko je pisao skriptu ili skripti.

Marko racuna se sa ceil jer jedino tako dobijes tocan broj rotacija koje treba obaviti.
U onoj njihovoj AADS skripti pise isto ceil

Marko

anon00
razlika je u tome što na slajdovima imamo floor i k = 2^h - 1, dok oni u skripti imaju ceil i k = 2^h-1-1

trebalo bi biti oboje ispravno dok god je konzistentno

ima li negdje rjesenja starih meduispita?

Abcdef potraži malo git https://github.com/studosi-fer/NAISP

jel itko rijesio ovaj s AVL-om iz prošlogodišnjeg MI?

[obrisani korisnik] ja sam rješavao, ali mi nije jasno njihovo rješenje za brisanje.
Ako se koristi metoda Deletion by Copying dobije se ovakav rezultat:

što je drugačije nego njihovo…

MrDREDD a cek di su njihova rjesenja? nisam ih vidio

MrDREDD oni su na mjesto roditelja stavljali neposrednog sljedbenika, kod brisanja 11 je to 15, a ovaj avl tree visualization stavlja neposrednog prethodnika, brise 11 i stavlja 3 na njegovo mjesto

kasnije u b stablima kod brisanja (slajd 29) se stavlja neposredni prethodnik pa mislim da je dobro držat se toga, ali mislim da je zapravo svejedno u tom avl-u oces uzet prethodnika il sljedbenika, bitno da stablo ostane uravnotezeno

Ovako sam rješavala simplex metodu za jednu knjigu pa ako će kome pomoć i bit malo jasnije 🙂

To turn the i-system into an e-system, add slack variables to the left side of the problem constraint inequalities.

The slack variable $s$ is the difference between the left and right sides of inequality so if we add it to the left side, the inequality sign becomes a sign of equality.

There are two problem constraint inequalities, so we will add $s_1$ to the first one and $s_2$ to the second one.

\begin{aligned} 2x_1+x_2 +s_1&=10&&\text{}\\ x_1+3x_2 +s_2&=10&&\text{}\\ \end{aligned}

To complete the initial system, move terms on the left of the objective function equation to the right.
-15x_1-10x_2+P=0

(B)

Fill the matrix with coefficients from Step 2 to get the initial simplex tableau.

Label rows in tableau with basic variables, $s_1,s_2, P$ in such a way that the intersection element of the same label row and column is 1.

\begin{array}{rcl} &\\ &\begin{array}{c}&x_1&x_2&s_1&s_2&P&&\end{array}\hspace{1em} \\ \begin{matrix}s_1\\ s_2\\ P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}2& \phantom{-}1&1 & 0 &0 &\big| & 10&\\ &\phantom{-}1& \phantom{-}3&0& 1 & 0 &\big| & 10 &\\\hdashline &-15& -10&0 &0& 1 &\big| &0& \\\end{bmatrix} &\hspace{-0.5em} \\ &\end{array}

The column with the most negative indicator is called the pivot column. The label of that column is called entering variable.

Divide each element in the last column with corresponding positive ones in the pivot column. The row in which the smallest quotient is obtained is called the pivot row. The label of that row is called the exiting variable.

\begin{array}{rcl} &\begin{array}{c} \scriptstyle Entering\;variable&&&&&&&&\end{array}\\ &\begin{array}{c}\downarrow&&&&&&&&&&\end{array}\\ &\begin{array}{c}&x_1&x_2&s_1&s_2&P&&\end{array}\hspace{1em} \\ \begin{matrix}\scriptstyle Exiting\;variable\displaystyle \rightarrow&s_1\\ &s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}2& \phantom{-}1&1 & 0 &0 &\big| & 10&\\ &\phantom{-}1& \phantom{-}3&0& 1 & 0 &\big| & 10 &\\\hdashline &-15& -10&0 &0& 1 &\big| &0& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \frac{10}{2}=5\phantom{..} \\\\\frac{10}{1}=10\\\ \end{matrix}\hspace{-0.5em} \\ &\end{array}

The first column is the pivot column. The first row is the pivot row.

The pivot element is placed at the intersection of the pivot column and pivot row and it is circled in the tableau below.

\begin{array}{rcl} &\begin{array}{c} \scriptstyle Entering\;variable&&&&&&&&\end{array}\\ &\begin{array}{c}\downarrow&&&&&&&&&&\end{array}\\ &\begin{array}{c}&x_1&x_2&s_1&s_2&P&&\end{array}\hspace{1em} \\ \begin{matrix}\scriptstyle Exiting\;variable\displaystyle \rightarrow&s_1\\ &s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}\textcircled{2}& \phantom{-}1&1 & 0 &0 &\big| & 10&\\ &\phantom{-}1& \phantom{-}3&0& 1 & 0 &\big| & 10 &\\\hdashline &-15& -10&0 &0& 1 &\big| &0& \\\end{bmatrix} &\hspace{-0.5em} \\ &\end{array}

Index of the pivot element is $(1,1)$.
Entering variable is $x_1$.
Exiting variable is $s_1$.

©

Perform pivot operation. $R$ represents a row.

\begin{array}{rcl} &\\ &\begin{array}{c}&x_1&x_2&s_1&s_2&P&&\end{array}\hspace{1em} \\ \begin{matrix}s_1\\ s_2\\ P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}\textcircled{2}& \phantom{-}1&1 & 0 &0 &\big| & 10&\\ &\phantom{-}1& \phantom{-}3&0& 1 & 0 &\big| & 10 &\\\hdashline &-15& -10&0 &0& 1 &\big| &0& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \frac{R_1}{2}\scriptstyle\to R_1\phantom{...} \\ \\ \\\end{matrix}\hspace{-0.5em} \\ &\end{array}

\begin{array}{rcl} &\\ &\begin{array}{c}&x_1&x_2&s_1&s_2&P&&\end{array}\hspace{1em} \\ \begin{matrix}&s_1\\ \sim& s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}\textcircled{1}& 0.5&0.5 & 0 &0 &\big| & 10&\\ &\phantom{-}1& \phantom{-}3&0& 1 & 0 &\big| & 10 &\\\hdashline &-15& -10&0 &0& 1 &\big| &0& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \\ \scriptstyle -R_1+R_2\to R_2 \\ \scriptstyle 15R_1+R_3\to R_3 \end{matrix}\hspace{-0.5em} \\ &\end{array}

\begin{array}{rcl} &\\ &\begin{array}{c}&&x_1&&x_2&&s_1&s_2&P&&&&\end{array}\hspace{1em} \\ \begin{matrix}&s_1\\ \sim& s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}\textcircled{1}& \phantom{-}0.5&\phantom{-}0.5& 0 &0 &\big| & 10&\\ &\phantom{-}0& \phantom{-}2.5&-0.5& 1 & 0 &\big| & 0 &\\\hdashline &\phantom{-}0& -2.5&\phantom{-}7.5 &0& 1 &\big| &150& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \\ \phantom{\scriptstyle -1R_1+R_2\to R_2} \\ \end{matrix}\hspace{-0.5em} \\ &\end{array}
—————–

Since there are negative indicators in the bottom row, repeat the whole procedure explained in (B).

New basic variables are $x_1,s_2,P.$ Relabel the rows.

\begin{array}{rcl} &\begin{array}{c} \scriptstyle Entering\;variable&&&&&&&&\end{array}\\ &\begin{array}{c}\downarrow&&&&&&&\end{array}\\ &\begin{array}{c}&&x_1&&x_2&&s_1&s_2&P&&&&\end{array}\hspace{1em} \\ \begin{matrix}&x_1\\\scriptstyle Exiting\;variable\displaystyle \rightarrow& s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}1& \phantom{-}0.5&\phantom{-}0.5& 0 &0 &\big| & 10&\\ &\phantom{-}0& \phantom{-}\boxed{2.5}&-0.5& 1 & 0 &\big| & 0 &\\\hdashline &\phantom{-}0& -2.5&\phantom{-}7.5 &0& 1 &\big| &150& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \frac{10}{0.5}=20\phantom{..} \\\\\frac{0}{2.5}=0\phantom{....} \\\\ \end{matrix}\hspace{-0.5em} \\ &\end{array}

Index of the pivot element is $(2,2)$.
Entering variable is $x_2$.

Exiting variable is $s_2$.

Perform pivot operation.

\begin{array}{rcl} &\\ &\begin{array}{c}&&x_1&&x_2&&s_1&s_2&P&&&&\end{array}\hspace{1em} \\ \begin{matrix}\phantom{...}&x_1\\& s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}1& \phantom{-}0.5&\phantom{-}0.5& 0 &0 &\big| & 10&\\ &\phantom{-}0& \phantom{-}\boxed{2.5}&-0.5& 1 & 0 &\big| & 0 &\\\hdashline &\phantom{-}0& -2.5&\phantom{-}7.5 &0& 1 &\big| &150& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \\ \frac{R_2}{2.5}\scriptstyle\to R_2\phantom{...} \\ \\\end{matrix}\hspace{-0.5em} \\ &\end{array}

\begin{array}{rcl} &\\ &\begin{array}{c}&&x_1&&x_2&&s_1&s_2&P&&&&\end{array}\hspace{1em} \\ \begin{matrix}&x_1\\\sim & s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}1& \phantom{-}0.5&\phantom{-}0.5& 0 &0 &\big| & 10&\\ &\phantom{-}0& \phantom{-}\boxed{1}&-0.2&0.4 & 0 &\big| & 0 &\\\hdashline &\phantom{-}0& -2.5&\phantom{-}7.5 &0& 1 &\big| &150& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \scriptstyle -0.5R_2+R_1\to R_1 \\ \\ \scriptstyle 2.5R_2+R_3\to R_3 \end{matrix}\hspace{-0.5em} \\ &\end{array}

\begin{array}{rcl} &\\ &\begin{array}{c}&\;x_1&&x_2&&s_1&&s_2&&P&&&&\end{array}\hspace{1em} \\ \begin{matrix}&x_1\\\sim & s_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}1& \phantom{-}0&\phantom{-}0.1& -0.2 &0 &\big| &\phantom{-}10&\\ &\phantom{-}0& \phantom{-}\boxed{1}&-0.2&\phantom{-} 0.4 & 0 &\big| & \phantom{-}0 &\\\hdashline &\phantom{-}0& \phantom{-}0&\phantom{-}7&\phantom{-}1& 1 &\big| &\phantom{-}75& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \\ \phantom{\scriptstyle -2.5R_1+R_2\to R_2} .\\ \end{matrix}\hspace{-0.5em} \\ &\end{array}
——————–
New basic variables are $x_1,x_2,P.$ Relabel the rows.

\begin{array}{rcl} &\\ &\begin{array}{c}&\;x_1&&x_2&&s_1&&s_2&&P&&&&\end{array}\hspace{1em} \\ \begin{matrix}&x_1\\& x_2\\ &P\end{matrix}\hspace{-0.5em} &\begin{bmatrix} &\phantom{-}1& \phantom{-}0&\phantom{-}0.1& -0.2 &0 &\big| &\phantom{-}10&\\ &\phantom{-}0& \phantom{-}1&-0.2&\phantom{-} 0.4 & 0 &\big| & \phantom{-}0 &\\\hdashline &\phantom{-}0& \phantom{-}0&\phantom{-}7&\phantom{-}1& 1 &\big| &\phantom{-}75& \\\end{bmatrix} &\hspace{-0.5em} \begin{matrix} \\ \phantom{\scriptstyle -2.5R_1+R_2\to R_2} .\\ \end{matrix}\hspace{-0.5em} \\ &\end{array}
————————

There are no negative indicators in the last row, therefore we can read optimal solution from the tableau.

Read row labels with corresponding elements to the right of the vertical line to obtain a solution.

Max $P=75$ at $x_1=10$ and $x_2=0$

FERella
Ali x1 = 10 i x2 = 0 ne zadovoljava ogranicenje 2×1 + x2 <= 0 ili to ja nesto buncam?

Ja sam ovako rijesio i cini mi se da je ok:

Zašto se ovdje sjedinjuju lijevo i desno dijete od 13, a ne lijevo i desno dijete od 3?

[obrisani korisnik]
Sry, kasnije sam shvatio da sam ja gledao one primjere zadatka iz edgara, a ne MI.
Mislim da nema službenih rješenja za MI

Filemon ekipa šta downvota ovaj post, odite na link, razno, tutoriali i imate pdf sa valjda 100 rješenih zad sa starih MI , sortirano po gradivu

lp

Zna netko ovaj? Zadatak s 2020-2021 Primjeri Edgar, nikako ne mogu dobiti rješenje kao kod njih 🙁

Stubla nisam rješavala, ali vidim da je to dvofazni simpleks pa možda griješiš u tome, no mislim da to ni nije gradivo za MI

Stubla

*** rješavao sam ga bio kao dvofazni simplex, ali svejedno negdje je zapelo

*** da na obavijesti piše samo naivni, ali na auditornima smo prošli i dvofazni pa sam išao riješit