chapter 4 simplex method for linear programming shi-shang jang chemical engineering department...
TRANSCRIPT
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Chapter 4 Simplex Method for Linear Programming
Shi-Shang Jang
Chemical Engineering Department
National Tsing-Hua University
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Example 1: Inspector Problem
Assume that it is desired to hire some inspectors for monitoring a production line. A total amount of 1800 species of products are manufactured every day (8 working hours), while two grades of inspectors can be found. Maximum, 8 grade A inspector and 10 grade B inspector are available from the job market. Grade A inspectors can check 25 species/hour, with an accuracy of 98 percent. Grade B inspectors can check 15 species/hour, with an accuracy of 95 percent. Note that each error costs $2.00/piece. The wage of a grade A inspector is $4.00/hour, and the wage of a grade B inspector is $3.00/hour. What is the optimum policy for hiring the inspectors?
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Problem Formulation
Assume that the x1 grade A inspectors x2 grade B inspects are hired, then
total cost to be minimized 48 x1 +38 x2 +2580.022 x1 +1580.052 x2
=40 x1 +36 x2 manufacturing constraint 258 x1 +158 x2 1,800 200 x1 +120 x2 1,800 no. of inspectors available: 0 x1 8 0 x2 10
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The Graphical Solution
X2=10
X1 =8
B(8,10)
(8,5/3)
C(3,10)
15
10
5
5 10
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Theorem
Property: If there exists an optimal solution to a LP, then at least one of the corner point of the feasible region will always qualify to be an optimal solution.
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Special Cases
Alternate Solutions (non-unique solutions) Max x1+2x2
s.t. x1+2x210
x1+ x21
x24
x10, x20
X1+2X
2=10
X1+2X
2=6X1+2X
2=2X1+X
2=1
5
4
106
X2=4
X2
X1
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Special Cases - continued
Unbounded Optima : A system has a feasible region with open boundaries such that the optima may appear at the infinity.
Example: For the previous example, in case the constraint x1+2x210 is not given, then moving far away from the origin increases the objective function x1+2x2, and the maxim Z would be +
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Unbounded Optima
0 1 2 3 4 5 6 7 8 9 100
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
x1
x2
x1+x2=1
x2=4
x1+2x2=2
x1+2x2=6
x1+2x2=10
x1+2x2=15
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Example 2: Student Fab
The RIT student-run microelectronic fabrication facility is taking orders for four indigenously developed ASIC chips that can be used in (1) touch sensors ($6, s4hr, m1hr, v30), (2)LCD($10,s9hr, m1hr, v40), (3) pressure sensors($9, s7hr, m3hr, v20), and (4) controllers($20, s10hr, m8hr, v10).
Constraints : Student hr 600, machine ≦hr 420, space 800≦ ≦
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The LP Problem
80010204030
42083
60010794..
209106max
4321
4321
4321
4321
xxxx
xxxx
xxxxts
xxxxx
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4-2 The Basic Approach
Standard Form of Linear Programming
mnbbbxxx
bxaxaxa
bxaxaxa
bxaxaxats
xcxcxcy
mn
mnmnmm
nn
nn
nnxxx n
,0,,, ,0,,,
..
min
2121
2211
22222121
11212111
2211,,, 21
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Handling of in-equality Constraints
Case 1: Slack Variable:
x1+2x2+3 x3 +4x425 – Modified to
x1+2x2+3x3+4x4+ x5=25 x50 is a slack variable.
Case 2: Surplus Variable:
2 x1+ x2-3 x312– Modified to
2 x1+ x2-3 x3- x4=12
x40 is a surplus variable.
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Handling of Equality Constraints
If s is unrestricted, i.e., s can be positive or negative, then we set
s=s+-s-
such that s+0, s- 0.
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Example
0,
533
2
7 ..
32max
21
321
321
321
321
xx
xxx
xxx
xxxts
xxxz
Modify to
0,,,,,
5333
2
7 ..
332max
765421
5421
75421
65421
5421
xxxxxx
xxxx
xxxxx
xxxxxts
xxxxz
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Definitions
Definition: A feasible region, denoted by S is the set of all feasible solution. Mathematically, .
Definition: An optimal solution is a vector x*S, s.t. z0=cTx* is maximum or minimum in where Z is termed by the optimal value.
Definition: Alternate optimal solution is a set XS, s.t. all xX has the same objective value z0 and for all xS, and z=cTx, z z0.
Definition: If the solution set of LP contains only one element, it is termed the unique optimum.
Definition: If the optimum value z approaches to infinity, then the LP is said to have unbounded optimum.
0, xbAxxS
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4-3 The Simplex Method
sconstraint for theApproach Variable Artificial
andn EliminatioJordan Gauss
0,,,
..
min
21
2211
22222121
11212111
2211,,, 21
n
mnmnmm
nn
nn
nnxxx
xxx
bxaxaxa
bxaxaxa
bxaxaxats
xcxcxcyn
mnmnsmsmmmm
nnssmm
nnssmm
bxaxaxax
bxaxaxax
bxaxaxax
11,
22211,22
11111,11
Form Canonical
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Definitions
Definition: A pivot operation is sequence of elementary row operations that reduce the coefficients of a specified variable to unity in one of the equation and zero elsewhere.
Definition: In the above canonical form, x1,,xm are termed the basic variables or dependent variables, xm+1, ,xn are called nonbasic variables or the independent variables.
Definition: The solution obtained from a canonical form is by setting the nonbasic variable or independent variable to zero is called a basic solution.
Definition: A basic feasible solution is a basic solution in which the basic or dependent variables are non-negative.
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Property
Remark: )!(!
!solutions basic of no.
mnm
n
m
n
where nmjxnonbasicmibxbasic iii ,,1,0 ,,,1,
Definition: mBmB ccxx ,,,,, 11 cx
Property: A feasible basic solution is a simplex of the feasible region.
Note: Given a canonical form and feasible basic solution, then the objective function:
m
iiiB
TB bcz
1
bc
where xB is a basic variable.
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Approach (Simplex Method):
Start with an initial basic feasible solution in canonical form.
Improve the solution by finding another basic feasible solution if possible.
When a particular basic feasible solution is found, and cannot be improved by finding new basic feasible solution, the optimality is reached.
Definition: An adjacent basic solution differs from a basic solution is exactly one basic variable.
Question: If one wants to find an adjacent feasible basic solution from one feasible basic solution (i.e., switch to another simplex), which adjacent basic solution gives lowest objective function?
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Derivation of Inner Product Rule
msmsm
ss
ss
bxax
bxax
bxax
222
111
Supposing, one wants to replace one of the original basic variable with nonbasic variable xs, we firstly, increase xs from zero to one,then for all i=1,,m,
1
s
isii
x
abx
and, for all j=m+1, ,n, js. xj=0
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Theorem 1 (Inner Product Rule)
s
m
iisiinew cabcz
1
Relative cost,
m
iisisnew acczzz
1(inner product rule)
More: (1) In a minimization problem, a basic feasible solution is optimalif the relative costs of its all nonbasic variable are all positive or zero.(2) One should choose an adjacent basic solution from which the relative cost is the minimum.
Corollary: The alternate optima exists if z=0.
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Theorem 2: (The Minimum Ratio Rule )
Given a nonbasic variable xs is change into the basic variable set, then one of the basic variable xr should leave from the basic variable set, such that:
ia
bx
is
i
as
is
,minmax
0
The above minimum happens at i=r.
Corollary: The above rule fails if there exist unbounded optima.
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Example: The LP Problem
80010204030
42083
60010794..
209106max
4321
4321
4321
4321
xxxx
xxxx
xxxxts
xxxxx
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The Standard Form
80010204030
42083
60010794..
209106max
743213
643212
543211
4321
xxxxxg
xxxxxg
xxxxxgts
xxxxf
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Table 1(s=4,r=6,rc=2)Basic={5 6 7};Nonbasic={1 2 3 4}
6 10 9 20 0 0 0
CB Basis x1 x2 x3 x4 x5 x6 x7 constraints
0 x5 4 9 7 10 1 0 0 600 (600/10=60)
0 x6 1 1 3 8 0 1 0 420(420/8=52.5)
0 x7 30 40 20 10 0 0 1 800(800/10=80)
Z 6 10 9 20 - - - Z=0
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Table 2(s=2,r=7,rc=3)Basic={5 4 7};Nonbasic={1 2 3 6}
6 10 9 20 0 0 0
CB Basis x1 x2 x3 x4 x5 x6 x7 constraints
0 x5 2.75 7.75 3.25 0 1 -1.25 0 75(75/7.75=9.6774)
20 x4 0.125 0.125 0.375 1 0 0.125 0 52.5(52.5/0.125=420)
0 x7 28.75 38.75 16.25 0 0 -1.25 1 275(275/38.75=7.0798)
Z 3.5 7.5 1.5 - - -2.5 - Z=1050
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Table 3Basic={5 4 2};Nonbasic={1 7 3 6}
6 10 9 20 0 0 0
CB Basis x1 x2 x3 x4 x5 x6 x7 constraints
0 x5 -3 0 0 0 1 -1 -0.2 20
20 x4 0.0323 0 0.3226 1 0 0.129 -0.0032 51.6129
10 x2 0.7419 1 0.4194 0 0 -0.0323 0.0258 7.0968
Z -2.0645 - -0.1935 - - -1.6452 -2.2581 Z=1103.226
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2. The Two Phase Simplex Method-Example
0,,
12
334
112..
3min
321
31
321
321
321
xxx
xx
xxx
xxxts
xxx
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The Standard Form
0,,,,
12
324
112 ..
3min
54321
31
5321
4321
321
xxxxx
xx
xxxx
xxxxts
xxx
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Two Phase Approach-Phase I
12
324
112..
min
731
65321
4321
76
xxx
xxxxx
xxxxts
xx
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Two Phase Approach-Phase ITable 1
0 0 0 0 0 1 1
CB Basis x1 x2 x3 x4 x5 x6 x7 constraints
0 x4 1 -2 1 1 0 0 0 11(11/1=11)
1 x6 -4 1 2 0 -1 1 0 3(3/2=1.5)
1 x7 -2 0 1 0 0 0 1 1(1/1=1)
Z 6 -1 -3 - 1 - - Z=4
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Two Phase Approach-Phase IFinal
0 0 0 0 0 1 1
CB Basis x1 x2 x3 x4 x5 x6 x7 constraints
0 x4 3 0 0 1 -2 2 -5 12
0 x2 0 1 0 0 -1 1 -2 1
0 x3 -2 0 1 0 0 0 -1 1
Z - - - - - - - Z=0
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Two Phase Approach-Phase IITable 1
-3 1 1 0 0
CB Basis x1 x2 x3 x4 x5 constraints
0 x4 3 0 0 1 -2 12(12/3=4)
1 x2 0 1 0 0 -1 1(1/0=)
1 x3 -2 0 1 0 0 1(1/-2<0)
Z -1 - - - 5 Z=2
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Two Phase Approach-Phase IIFinal
-3 1 1 0 0
CB Basis x1 x2 x3 x4 x5 constraints
-3 x1 1 0 0 0.3333 -0.667 4
1 x2 0 1 0 0 -1 1
1 x3 0 0 1 0.6667 -1.333 9
Z - - - - - Z=-2
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Example: Multi-Products Manufacturing
A company manufactures three products: A, B, and C. Each unit of product A requires 1 hr of engineering service, 10 hr of direct labor, and 3lb of material. To produce one unit of product B requires 2hr of engineering, 4hr of direct labor, and 2lb of material. In case of product C, it requires 1hr of engineering, 5hr of direct labor, and 1lb of material. There are 100 hr of engineering, 700 hr of labor, and 400 lb of material available. Since the company offers discounts for bulk purchases, the profit figures are as shown in the next slide:
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Example- Continued
Formulate a linear program to determine the most profitable product mix.
Product A Product B Product C
Sales units Unit profit variables
Sales units Unit profit variables Sales units Unit profit variables
0-40 10 X1 0-50 6 X5 0-100 5 X8
40-100 9 X2 50-100 4 X6 Over 100 4 X9
100-150 8 X3 Over 100 3 X7
Over 150 7 X4
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Problem Formulation
Let’s denote the variables as shown in the table, then we have the following:
0
600
400
400)(1)(2)(3
700)(5)(4)(10
100)(1)(2)(1..
4534678910max
9
2
1
987654321
987654321
987654321
987654321,, 101
X
X
X
XXXXXXXXX
XXXXXXXXX
XXXXXXXXXts
XXXXXXXXXXX
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MATLAB PROGRAM
f=[-10 -9 -8 -7 -6 -4 -3 -5 -4]'; A=[1 1 1 1 2 2 2 1 1; 10 10 10 10 4 4 4 5 5;3 3 3 3 2
2 2 1 1]; b=[100;700;400]; Aeq=[];beq=[]; LB=[0 0 0 0 0 0 0 0 0]; UB=[40 60 50 Inf 50 50 Inf 100 Inf]; [X,FVAL,EXITFLAG,OUTPUT,LAMBDA]=LINPROG(
f,A,b,Aeq,beq,LB,UB)
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Solution
X’= 40.0000 22.5000 0.0000 0.0000 18.7500 0.0000 0.0000 0.0000 0.0000
FVAL =-715.0000 EXITFLAG =1 OUTPUT = iterations: 7 cgiterations: 0 algorithm: 'lipsol' LAMBDA = ineqlin: [3x1 double] eqlin: [0x1 double] upper: [9x1 double] lower: [9x1 double]
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3. Sensitivity Analysis
Shadow Prices: To evaluate net impact in the maximum profit if additional units of certain resources can be obtained.
Opportunity Costs: To measure the negative impact of producing some products that are zero at the optimum.
The range on the objective function coefficients and the range on the RHS row.
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Example
A factory manufactures three products, which require three resources – labor, materials and administration. The unit profits on these products are $10, $6 and $4 respectively. There are 100 hr of labor, 600 lb of material, and 300hr of administration available per day. In order to determine the optimal product mix, the following LP model is formulated and solve:
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Basic LP Problem
0,,
ation)(administr 300622
(material) 6005410
(labor) 100..
4610max
321
321
321
321
321
xxx
xxx
xxx
xxxts
xxxZ
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Optimal Solution and Sensitivity Analysis
x1=33.33, x2=66.67,x3=0,Z=733.33 Shadow prices for row 1=3.33, row 2=0.67,
row 3=0 Opportunity Costs for x3=2.67 Ranges on the objective function coefficients:
6 c≦ 1(10) 15, 4 c≦ ≦ 2(6) 10, ≦
-∞ c≦ 3(4) 6.67≦
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Optimal Solution and Sensitivity Analysis- Continued
60 b≦ 1(100) 150, 400 b≦ ≦ 2(600) 1000, ≦
200 b≦ 3(300) ∞≦
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100% Rules
100% rule for objective function coefficients
100% rule for RHS constants
%100
j j
j
c
c
%100
j j
j
b
b
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Examples
Unit profit on product 1 decrease by $1, but increases by $1 for products 2 and 3, will the optimum change?(δc1=-1, Δc1=-4, δc2=1, Δc2=4, δc3=1, Δc3=2.67)
Simultaneous variation of 10 hr decrease on labor 100 lb increase in material and 50hr decrease on administration
1875.067.2
1
4
1
4
1
1100
50
400
100
40
10