Question #173496

3. Solve the following LPP by the two-phase simplex method. (10) 

 Max 1 2 3 Z = x + x − x

 Subject to 4x1 + x2 + x3 = 4

3 2 6 x1 + x2 − x4 =

 x1

, x2

, x3 ≥ 0



1
Expert's answer
2021-03-31T12:07:59-0400

Solution.

May be the condition of the task next:

z=x1+x2x3max,4x1+x2+x3=4,3x1+2x2x3=6,x1,x2,x3>=0.z=x_1+x_2-x_3 - max, \newline 4x_1+x_2+x_3=4,\newline 3x_1+2x_2-x_3=6, \newline x_1,x_2,x_3>=0.

Introduce new variables x4,x5:x_4, x_5:

4x1+x2+x3+x4=4,3x1+2x2x3+x5=6,4x_1+x_2+x_3+x_4=4, \newline 3x_1+2x_2-x_3+x_5=6,

then z=Mx4Mx5==M(44x1x2x3)M(63x12x2+x3)==7Mx1+Mx210Mmaxz = -Mx_4-Mx_5=\newline =-M(4-4x_1-x_2-x_3)-M(6-3x_1-2x_2+x_3)=\newline =7Mx_1+Mx_2-10M - max

Introduce new variable x_0:

x0=7x1+3x2.x_0=7x_1+3x_2.

We will have:

x0=10+7x1+3x2,x_0=-10+7x_1+3x_2,

x4=44x1x2x3,x5=63x12x2+x3.x_4=4-4x_1-x_2-x_3,\newline x_5=6-3x_1-2x_2+x_3.

This plan is not optimal because there are positive elements in the expression for x0x_0 . Select a new variable x1,x_1, then

x1=114x214x314x4.x_1=1-\frac{1}{4}x_2-\frac{1}{4}x_3-\frac{1}{4}x_4.

From here

x0=3+54x274x374x4,x_0=-3+\frac{5}{4}x_2-\frac{7}{4}x_3-\frac{7}{4}x_4,

x5=354x2+74x3+34x4.x_5=3-\frac{5}{4}x_2+\frac{7}{4}x_3+\frac{3}{4}x_4.

The expression for x0x_0 has positive elements, the plan is not optimal. The largest coefficient is near x2x_2 , then choose x2x_2 as a new variable.

x2=125+75x3+35x445x5,x0=0x4x5,x1=2535x325x4+15x5.x_2=\frac{12}{5}+\frac{7}{5}x_3+\frac{3}{5}x_4-\frac{4}{5}x_5, \newline x_0=0-x_4-x_5, \newline x_1=\frac{2}{5}-\frac{3}{5}x_3-\frac{2}{5}x_4+\frac{1}{5}x_5.

The expression for x0x_0

has not positive elements, the plan is optimal.

We will deduce variables x4,x5x_4, x_5 from basis:

x0=14515x3,x_0=\frac{14}{5}-\frac{1}{5}x_3,

x1=2535x3,x_1=\frac{2}{5}-\frac{3}{5}x_3,

x2=125+75x3.x_2=\frac{12}{5}+\frac{7}{5}x_3.

Optimal plan:

x1=25,x2=225,x3=0.x_1=\frac{2}{5}, x_2=2\frac{2}{5}, x_3=0.

So, max z(x)=225.z(x)=2\frac{2}{5}.

Answer. max z(x)=225.z(x)=2\frac{2}{5}.


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