Post on 20-Oct-2015
transcript
Software Operation
• (1) Select “Linear Programming” and “OK”
• (2) Select “File” and Click “New”
• (3) Specify Number of Decision Variables
• (4) Specify Number of Constraints
• (5) Specify Objective Type and “OK”
• (6) Put “Coefficients”
• (7) Solve
ExampleEmbassy Motorcycle (EM) manufactures two lightweight motorcycles designed for easy handling and safety. The EZ-Rider model has a new engine and a low profile that make it easy to balance. The Lady-Sport model is slightly larger, uses a more traditional engine, and is specifically designed to appeal to women riders. Embassy produces the engines for both models at its Des Moines, Iowa, plant. Each EZ-Rider engine requires 6 hours of manufacturing time and each Lady-Sport engine requires 3 hours of manufacturing time. The Des Moines plant has 2100 hours of engine manufacturing time available for the next production period. Embassy’s motorcycle frame supplier can supply as many EZ-Rider frames as needed.
However, the Lady-Sport frame is more complex and the supplier can provide only up to 280 Lady-Sport frames for the next production period. Final assembly and testing requires 2 hours for each EZ-Rider model and 2.5 hours for each Lady-Sport model. A maximum of 1000 hours of assembly and testing time are available for the next production period. The company’s accounting department projects a profit contribution of $2400 for each EZ-Rider produced and $1800 for each Lady-Sport produced. Formulate a linear programming model that can be used to determine the number of units of each model that should be produced in order to maximize the total contribution to profit. Find the optimal solution using the graphical solution procedure. Which constraints are binding.
Example(a)
Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced
Max 2400E + 1800L
s.t.
6E + 3L 2100 Engine time
L 280 Lady-Sport maximum
2E + 2.5L 1000 Assembly and testing
E, L 0
Example(b)
0
L
Profit = $960,000
Optimal Solution
100
200
300
400
500
600
700
100 200 300 400 500E
Engine Manufacturing Time
Frames for Lady-Sport
Assembly and Testing
E = 250, L = 200
Number of Lady-Sport Produced
Num
ber
of E
Z-R
ider
Pro
duce
d
From a geometric viewpoint
: CPF solutions (Corner-Point Feasible): Corner-point infeasible solutions
1823 21 xx
1x0 2 4 6 8 10
2x
2
4
6
8
Feasibleregion
122 2 x
0x2
01 x
)6,4(
4x1
Optimality test:
There is at least one optimal solution.
If a CPF solution has no adjacent CPF solutions
that are better (as measured by Z) than itself,
then it must be an optimal solution.
The Key Solution Concepts
Solution concept 1:
The simplex method focuses solely on CPF
solutions.
For any problem with at least one optimal
solution, finding one requires only finding a best
CPF solution.
Solution concept 2:
The simplex method is an iterative algorithm ( a systematic solution procedure that keeps repeating a fixed series of steps, called an iteration).
Solution concept 3:
The initialization of the simplex method
chooses the origin to be the initial CPF
solution.
Solution concept 4:
Given a CPF solution, it is much quicker
computationally to gather information about its
adjacent CPF solutions than other CPF
solutions.
Therefore, each time the simplex method
performs an iteration to move from the current
CPF solution to a better one, it always chooses a
CPF solution that is adjacent to the current one.
Solution concept 5:
After the current CPF solution is identified, the simplex method identifies the rate of improvement in Z that would be obtained by moving along edge.
Solution concept 6:
The optimality test consists simply of checking whether any of the edges give a positive rate of improvement in Z. If no improvement is identified, then the current CPF solution is optimal.
Simplex Method
To convert the functional inequality
constraints to equivalent equality constraints,
we need to incorporate slack variables.
,x5x3Z 21
1x
1823 21 xx
122 2 x
4
0,0 21 xx
,x5x3Z 21
1x
21 23 xx
12
4
and
Maxs.t.
,0jx
3x
18
4x
5x22x
.5,4,3,2,1jfor
Original Form of Model
Augmented Form of the Model
Slack variables
and
Maxs.t.
A basic solution is an augmented corner-point solution.
A Basic Feasible (BF) solution is an augmented CPF solution.
Properties of BF Solution1. Each variable is designated as either a nonbasic
variable or a basic variable.
2. # of nonbasic variables = # of functional
constraints.
3. The nonbasic variables are set equal to zero.
4. The values of the basic variables are obtained
from the simultaneous equations.
5. If the basic variables satisfy the
nonnegativity constraints, the basic solution
is a BF solution.
Simplex in Tabular Form
1x 2x 3x 4x 5xZ
3x4x5x
(0)(1)(2)(3)
21 53 xxZ 1x 3x
4x21 23 xx 5x 18
12
4
0
(0)
(1)(2)(3)
(a) Algebraic Form
(b) Tabular FormCoefficient of: Right
SideBasic
Variable ZEq.1000
-3103
-5022
0100
0010
0001
04
1218
2x2
1x 2x 3x 4x 5xZ
3x4x5x
(0)(1)(2)(3)
(b) Tabular FormCoefficient of: Right
SideBV ZEq.1000
-3103
-5022
0100
0010
0001
04
1218
62
12
92
18
minimumThe most negative coefficient
1x 2x 3x 4x 5xZ
3x
4x
5x
(0)
(1)
(2)
(3)
(b) Tabular FormCoefficient of: Right
SideBV ZEq.1
0
0
0
-3
1
0
3
-5
0
2
2
0
1
0
0
0
0
1
0
0
0
0
1
0
4
12
18
minimum
The most negative coefficient
Iteration
0
12
18
62
12
92
18
1x 2x 3x 4x 5xZ
3x
2x
5x
(0)
(1)
(2)
(3)
(b) Tabular FormCoefficient of: Right
SideBV ZEq.1
0
0
0
-3
1
0
3
0
0
1
0
0
1
0
0
0
-1
0
0
0
1
30
4
6
6
minimum
The most negative coefficient
Iteration
1
4
6
41
4
23
6
25
21
1x 2x 3x 4x 5x
Z3x
2x
1x
(0)
(1)
(2)
(3)
(b) Tabular FormCoefficient of: Right
SideBV ZEq.
1
0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
1
0
36
2
6
2
None of the coefficient is negative.
Iteration
22
3
21
31
31 3
1
31
The optimal solution
6,2 21 xx
(a) Optimality Test:
The current BF solution is optimal
if and only if every coefficient in row 0 is
nonnegative .
Pivot Column:
A column with the most negative coefficient
)0(
(a) Optimality Test
(b) Minimum Ratio Test:
(b) Minimum Ratio Test:1. Pick out each coefficient in the pivot column
that is strictly positive (>0).
2. Divide each of these coefficients into
the right side entry for the same row.
3. Identify the row that has the smallest of
these ratios.
4. The basic variable for that row is the leaving
basic variable, so replace that variable by the
entering basic variable in the basic variable
column of the next simplex tableau.
Breaking in Simplex Method
(a) Tie for Entering Basic Variable
Several nonbasic variable have largest and
same negative coefficients.
(b) Degeneracy
Multiple Optimal Solution occur if a non
BF solution has zero or at its coefficient
at row 0.
1x 2x 3x 4x 5x
Z3x
4x
5x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
-3
1
0
3
-2
0
2
2
0
1
0
0
0
0
0
1
0
4
12
18
0
0
0
1
0
SolutionOptimal?
No
21 23 xxZ 1x 3x
22x 4x21 23 xx 5x 18
12
4
0
(0)
(1)(2)(3)
Algebraic Form
1x 2x 3x 4x 5x
Z1x
4x
5x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
0
1
0
0
-2
0
2
2
3
1
0
-3
0
0
0
1
12
4
12
6
1
0
0
1
0
SolutionOptimal?
No
1x 2x 3x 4x 5x
Z1x
4x
2x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
0
1
0
0
0
0
0
1
0
1
3
1
0
-1
18
4
6
3
2
0
0
1
0
SolutionOptimal?
Yes
23 2
1
1x 2x 3x 4x 5x
Z1x
3x
2x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
0
1
0
0
0
0
0
1
0
0
1
0
1
0
18
2
2
6
Extra
0
SolutionOptimal?
Yes
21
31
31
31
31
(c) Unbounded Solution
If An entering variable has zero in these coefficients
in its pivoting column, then its solution can be
increased indefinitely.
Z3x
3x2x1x(0)
(1)
Coefficient of: Rightside Ratio
BasicVariable ZEq.
None
1
0
-3
1
-5
0
0
1
0
4
(M: a large positive number.)
1823 21 xx
(a) Big M Method
21 53 xxZ
1x122 2 x4
and
Maxs.t.
0,0 21 xx
521 53 xMxxZ
1x
21 23 xx 124
and
Maxs.t.
,0jx
3x
184x
5x22x
.5,4,3,2,1jfor
Original Problem Artificial Problem
: Artificial Variable43 , xx
5x: Slack Variables
521 53 xMxx
21 23 xx 185x (3)
(0)
)2318(53 2121 xxMxx
(4)215 2318 xxx
MxMxM 18)52()33( 21
MxMxMZ 18)52()33( 21
Max:
or Max:
or Max:
Max:
s.t.
Eq (3) can be changed to
Put (4) into (0), then
1x 2x 3x 4x 5x
Z3x
4x
5x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
-3M-3
1
3
3
-2M-5
0
2
2
0
1
0
0
0
0
0
1
-18M
4
12
18
0
0
0
1
0
MxMxMZ 18)52()33( 21 Max
1x
21 23 xx 124s.t.
3x
184x
5x22x
(1)
(2)
(3)
1x 2x 3x 4x 5x
Z1x
4x
5x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
-2M-5
1
0
3
3M+3
0
2
2
0
1
0
0
0
0
0
1
-6M+12
4
12
18
1
0
0
1
0
1x 2x 3x 4x 5x
Z1x
4x
2x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
0
1
0
0
0
0
0
1
1
3
0
-1
27
4
6
3
2
0
0
1
0 21
25M2
9
23
1x 2x 3x 4x 5x
Z1x
3x
2x
(0)
(1)
(2)
(3)
Coefficient of: RightSideBV ZEq.
1
0
0
0
0
1
0
0
0
0
0
1
0
0
1
0 0
27
4
6
3
Extra 31
1M
31
31
31
21
23
(b) Variables with a Negative Value jjj xxx 0,0,
jj xx
21 53 xx Mins.t.
0,: 21 xURSx
4122 2 x
1823 21 xx
1x
211 533 xxx Mins.t.
00,,0 211 xxx
4122 2 x
18233 211 xxx
11 xx