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ENMG 6150 Development And Management

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ENMG 6150 Development And Management

Course Code: ENMG6150
University: The University Of New Orleans

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Country: United States

Question:
Solve or prove that there is no solution to the following LP problems by verifying any graphical results using the Simplex method.1) Minimize: z = x1 – x2 Subject to: x1 + x2 ≤ –1 With: x1 & x2 nonnegative.2) Minimize: z = x1 – x2 Subject to: x1 + x2 ≥ –1 With: x1 & x2 nonnegative.3) Maximize: z = x1 – x2 Subject to: x1 + x2 ≥ –1 With: x1 & x2 nonnegative.
Redo problems 1-3 imposing the extra constraint that all basic variables are also integer.Consider the following integer programs. Using Chvatal cuts formulate associated LP’s and solve them graphically to obtain an optimal solution if it exists.4) Maximize: z = 10×1 + x2 Subject to: 2×1 + 5×2 ≤ 11 With: x1 & x2 nonnegative & integer.5) Maximize: z = 3×1 + 4×2 Subject to: 2×1 + x2 ≤ 62×1 + 3×2 ≤ 9 With: x1 & x2 nonnegative & integer.6) Maximize: z = x (5π – x) on [0, 20].7) Maximize: z = |x2 – 8| on [–4, 4].8) Maximize: z = x1 (x2 – 1) + x3 (x32 – 3) over the entire real line.
Derive algebraically the Kuhn-tucker conditions of the problem. Then, plot these conditions graphically as well as level contours of the objective function. Finally, using the plot obtain a solution to the optimization problem considered.10) In general, there are two sizes of oil tankers: those that can go through the Suez canal and those that cannot and in effect have to travel around Africa to get from the Persian Gulf region to a NW European oil terminal. In this problem we are interested in calculating the minimum capacity (in weight/mass of cargo carried) of Suez incompatible tankers so that they are competitive to Suez compatible ones. The following values are of interest.
L, B, T: Principal vessel dimensions?: Vessel mass displacement F: Fare charged to customer per unit mass of cargo for single transit C: Cargo single transit cost per unit mass of cargo R: Empty ship seagoing cost per nm traveled WS: Average waiting time per Suez Canal transit XS: Fee per single Suez Canal transit P: Port time (for loading or unloading) per unit mass of cargo V: Vessel’s service speed in open seas Z: Distance in nm between Gulf and European port of call
For simplicity assume that cargo capacity is equal to mass displacement ? and that ? = ρgLBT.Denote variables related to the Suez compatible vessel by an index of S and those related to the Suez incompatible one by I.Suez Canal constraints: LS <= Lmax, BS <= Bmax, TS <= Tmax. Also: ZI = 5ZS.Determine ?I for economic feasibility if the following are known:ρ,g, Lmax, Bmax, Tmax, ZS, WS, XS, FS, CS/I, RS/I, PS/I, VS/I Answer: Z= x1-x2 Our Constraint is             x1+x2 ≤ - 1 It can be written as x1+x2 = - 1 x1 & x2 are non negative values. Simplex Method: Z1= x1-x2 + OS, x1 + x2 + s1 = -1 s1 is the slack variable. Initial Table: CBi Ci 1 -1 0 Solution Rate Basic Variable x1 x2 s1 0 Si 1 1 1 -1 -1/1 = -1   Zi 0 0 0 0   Ci - Zi                                         1 -1 0 Zi =  For Minimization: All Ci - Zi         ≥ 0 Iteration 1: Here we check the Ci - Zi value as maximum that is 1 first column.  The corresponding column value is Si = 1. This is called key column. CBi Ci 1 -1 0 Solution Rate Basic Variable x1 x2 s1 0 Si 1 1 1 -1     Zi 1 1 1 0   Ci - Zi                                          0 -2 -1 The above Ci – Zi values does not satisfy minimization constraint. All Ci - Zi         ≥ 0 So, we cannot find the solution for this problem. Question – 2  Minimize:             Z= x1-x2 Our Constraint is             x1+x2 ≥ - 1 It can be written as x1+x2 = - 1 x1 & x2 are non negative values. Simplex Method: Z1= x1-x2 + OS, x1 + x2 + s1 = -1 s1 is the slack variable. Initial Table: CBi Ci 1 -1 0 Solution Rate Basic Variable x1 x2 s1 0 Si 1 1 1 -1     Zi 0 0 0 0   Ci - Zi                                          1 -1 0  For Minimization: All Ci - Zi         ≥ 0 Iteration 1: Here we check the Ci - Zi value as maximum that is 1 first column.  The corresponding column value is Si = 1. This is called key column. CBi Ci 1 -1 0 Solution Rate Basic Variable x1 x2 s1 0 Si 1 1 1 -1     Zi 1 1 1 0   Ci - Zi                                             0 -2 -1 The above Ci – Zi values does not satisfy minimization constraint. All Ci - Zi         ≥ 0 So, we cannot find the solution for this problem. Question – 3  Minimize:             Z= x1-x2 Our Constraint is             x1+x2 ≥ - 1 x1 & x2 are non negative values. Simplex Method: Z1= x1-x2 + 0 x1 + x2 + s1 = -1 s1 is the slack variable. The Constraint can be written as, x1+x2 = - 1 x1 + x2 + s1 = -1 The constraint confliction is >=. So, the independent variables should be in negative.
-x1 – x2 – s1 = 1
Initial Table:

CBi

Ci

1 -1 0

Solution

Rate

Basic Variable

x1 x2 s1

0

Si

-1 -1 -1

-1

-1

Zi

1 -1 0

0

Ci – Zi                                           0 0 0
The solution for this problem is unbounded.
Question – 4
Z= 10×1+x2
Our Constraint is
2×1+5×2 ≤  11
It can be written as
2×1+5×2 =  11
Take x1=0,
2(0)+5×2 =  11
x2 = 11/5
x2 = 2.2
take x2 = 0
2×1+5(0) =  11
x1 = 11/2
x1 = 5.5

x1

0

2.2

X2

5.5

0

The shaded parts are the feasible solution.
(0, 2.2)
Z=10 x1 + x2
Z=10(0) +2.2
Z=2.2
(5.5, 0)
Z=10(5.5) + 0
Z=55
For maximization, Z=55 is the solution.
Question – 5
Maximize:
Z= 3×1+4×2
The constraints are,
2×1+x2 ≤ 6
2×1+3×2 ≤ 9
We can written as,
2×1+x2 = 6
2×1+3×2 = 9
If we take x1 = 0
2 (0) +x2 = 6
x2 = 6
If we take x2 = 0
2×1+0 = 6
x1 = 6/3
x1 =2

x1

0

3

X2

6

0

x1 = 0,
2×1+3×2 = 9
2(0) + 3×2 = 9
3×2 = 9
x2 = 9/3
x2 = 3
x2 = 0,
2×1+3×2 = 9
2 x1+ 3(0) = 9
2 x1 = 9
x1= 9/2
x1 = 4.5

x1

0

4.5

X2

3

0

point is (2.25, 1.5)
For maximization,
(0, 6)
Z = 3×1+4×2
Z = 3(0) + 4(6)
Z = 24
(4.5, 0)
Z = 3×1+4×2
Z = 3(4.5) + 4(0)
Z = 13.5
(2.25, 1.5)
Z = 3×1+4×2
Z = 3(2.25) + 4(1.5)
Z = 12.75
The solution is (0,6).
The shaded parts of the graph is feasible solution.
Question 6
Maximize: z = x (5π – x) on [0, 20]
The value of π is 3.14.
Z = x (5*3.14 – x)
Z = x (15.7 – x)
Z = 15.7x – x2
In [0, 20], the value of x is 0,
Z = 15.7 (0) – 0
Z = 0
The equation is,
Z = 15.7x – x2
The solution is x = 16 and x = 16
X=0                                                         X=0.064
V=0                                                         V=0
Question 7
Maximize: z = |x2 – 8| on [-4, 4]
In [-4, 4], the value of x is -4,
Z = |(-42)-8|
Z = |16-8|
Z = 8
The equation is,
Z = x2 – 8  X = -2.825             X = 0                   X = 2.825
Z = 0                   Z = 8
Question 8
Maximize: z = x1(x2-1) + x3(x32 – 3)
If we take x1 = 0, x2 = 1, x3 = -1
Z = 0(1-1) + (-1)((-12)-3)
Z = 2
If we take x1 = 1, x2 = -1, x3 = 0
Z = 1(-1-1) + 0(0-3)
Z = -2
If we take x1 = -1, x2 = 0, x3 = 1
Z = -1(0-1) + 1(1-3)
Z = 1 – 2
Z = -1

X1

X2

X3

Z

0

1

-1

2

1

-1

0

-2

-1

0

1

-1

Question 9
Minimize: f(x1, x2) = (x1 – 4)2 +(x2 – 4)2
The constraints are,
x1+x2 ≤ 4
x1+3×2 ≤ 9
We can written as,
x1+x2 = 4
x1+3×2 = 9
If we take x1 = 0
(0) +x2 = 4
x2 = 4
If we take x2 = 0
x1+0 = 4
x1 = 4
x1 =4

X1

0

4

X2

4

0

If we take x1 = 0
(0) +3×2 = 9
x2 = 9/3
x2 = 3
If we take x2 = 0
x1+0 = 9
x1 = 9
x1 =9

X1

0

3

X2

9

0

(0,0)                                      (4,0)                          (9,0)
We have,

L(x1, x2)

(x1 − 4)2 + (x2 − 4)2 − λ1(x1 + x2 − 4) − λ2(x1 + 3×2 − 9)

The Kuhn-Tucker conditions are,
−2(x1 − 4) − λ1 − λ2 = 0
−2(x2 − 4) − λ1 − 3λ2 = 0
x1 + x2 ≤ 4, λ1 ≥ 0, and λ1(x1 + x2 − 4) = 0
x1 + 3×2 ≤ 9, λ2 ≥ 0, and λ2(x1 + 3×2 − 9) = 0

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