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# 数学代写|运筹学代写Operations Research代考|OPR561 ONTINUOUS VARIABLES—HIGHER DEGREE

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## 数学代写|运筹学代写Operations Research代考|ONTINUOUS VARIABLES—HIGHER DEGREE

Minimize $X_1^3-5 X_1^2+8 X_1+X_2^3-2 X_2^2-10 X_2+10$
Subject to
$$\begin{array}{r} X_1+X_2 \leq 4 \ X_1, X_2 \geq 0 \end{array}$$
Stage: Each variable
State: Resource available for allocation
Decision variable: Values of $X_1$ and $X_2$
Criterion of effectiveness: Minimize $Z$
In this problem we first solve for variable $X_1$ and then for variable $X_2$. The objective function is rewritten as:
Minimize $X_2^3-2 X_2^2-10 X_2+X_1^3-5 X_1^2+8 X_1+10$
One more stage to go $n=1$
\begin{aligned} f_1\left(s_1, X_1\right) & =X_1^3-5 X_1^2+8 X_1+10 \ f_1^\left(s_1\right) & =\text { Minimize } X_1^3-5 X_1^2+8 X_1+10 \ 0 & \leq X_1 \leq s_1 \end{aligned} Differentiating with respect to $X_1$ and equating to zero, we get or $$\begin{array}{r} 3 X_1^2-10 X_1+8=0 \ X_1=2 \text { or } X_1=\frac{4}{3} \end{array}$$ Second derivative is $6 X_1-10$ and takes positive value for $X_1=2$ indicating minimum. Therefore, $X_1{ }^=2$ if $s_1 \geq 2$ and $X_1^=s_1$ if $s_1<2$. \begin{aligned} f_1^\left(s_1\right) & =8-20+16+10=14 \text { if } s_1 \geq 2 \ & =s_1^3-5 s_1^2+8 s_1+10 \text { if } 0=s_1<2 \end{aligned}
Since the function is cubic, we also verify the value of the function at $X_1=0$. At $X_1=0$, the value of $f_1\left(s_1\right)=10$ which is less than 14 and since $s_1$ can take only non-negative values we have $X_1^=0$ and $f_1^\left(s_1\right)=10$.

## 数学代写|运筹学代写Operations Research代考|ACTORIZING THE TERMS

Maximize $2 X_1+3 X_2+X_1 X_2$
Subject to
$$\begin{array}{r} X_1+X_2 \leq 2 \ X_1, X_2 \geq 0 \end{array}$$
Stage: Each variable
State: Resource available for allocation
Decision variable: Values of $X_1$ and $X_2$
Criterion of effectiveness: Maximize $Z$

In this example the term $X_1 X_2$ makes it difficult to separate the objective function in terms of separable functions of the variables. We factorize the objective function as:
$$\text { Maximize }\left(X_1+3\right)\left(X_2+2\right)-6$$
We can leave out the constant from the objective function and write the problem as:
Maximize $\left(X_1+3\right)\left(X_2+2\right)$
Subject to
$$\begin{array}{r} X_1+X_2 \leq 2 \ X_1, X_2 \geq 0 \end{array}$$
One more stage to go $n=1$
\begin{aligned} f_1\left(s_1, X_2\right) & =X_2+2 \ f_1^\left(s_1\right) & =\text { Maximize } X_2+2 \end{aligned} Subject to $0 \leq X_2 \leq s_1$ Here, the maximum value is at $X_2^=s_1$ and $f_1^\left(s_1\right)=s_1+2$ Two more stages to go $n=2$ \begin{aligned} f_2\left(2, X_1\right)= & \left(X_1+3\right) f_1^\left(2-X_1\right) \ f_2^\left(s_1\right)= & \text { Maximize }\left(X_1+3\right)\left(2-X_1+2\right) \ & \text { Subject to } 0 \leq X_1 \leq 2 \end{aligned} Maximize $\left(X_1+3\right)\left(4-X_1\right)$ Subject to $0 \leq X_1 \leq 2$ Maximize $-X_1^2+X_1+12$ Differentiating with respect to $X_1$ and equating to zero, we get $X_1=1 / 2$. The second derivative is negative indicating maximum. We have $X_1^=1 / 2, s_1=3 / 2$ and $X_2^*=3 / 2$ with $Z=49 / 4-$ $6=25 / 4$ (for the original problem)

## 数学代写|运筹学代写Operations Research代考|ONTINUOUSVARIABLESHIGHER DEGREE

$$X_1+X_2 \leq 4 X_1, X_2 \geq 0$$

$$f_1\left(s_1, X_1\right)=X_1^3-5 X_1^2+8 X_1+10 f_1^{\left(s_1\right)} \quad=\text { Minimize } X_1^3-5 X_1^2+8 X_1+100 \leq X_1 \leq s_1$$

$$3 X_1^2-10 X_1+8=0 X_1=2 \text { or } X_1=\frac{4}{3}$$

$$f_1^{\left(s_1\right)}=8-20+16+10=14 \text { if } s_1 \geq 2 \quad=s_1^3-5 s_1^2+8 s_1+10 \text { if } 0=s_1<2$$

## 数学代写|运筹学代写Operations Research代考|ACTORIZING THE TERMS

$$X_1+X_2 \leq 2 X_1, X_2 \geq 0$$

$$\text { Maximize }\left(X_1+3\right)\left(X_2+2\right)-6$$

$$X_1+X_2 \leq 2 X_1, X_2 \geq 0$$

$$f_1\left(s_1, X_2\right)=X_2+2 f_1^{\left(s_1\right)} \quad=\text { Maximize } X_2+2$$

$$f_2\left(2, X_1\right)=\left(X_1+3\right) f_1^{\left(2-X_1\right)} f_2^{\left(s_1\right)}=\quad \text { Maximize }\left(X_1+3\right)\left(2-X_1+2\right) \text { Subject to } 0 \leq X_1 \leq 2$$

## MATLAB代写

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