Chapter 4 Simplex Method for Linear Programming

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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
• 4?8? x1 3?8 ? x2 25?8?0.02?2? x1 15?8?0.05?2 ?
  x2
• 40 x1 36 x2
• manufacturing constraint
• 25?8? x1 15?8? 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
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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 x12x2
• s.t. x12x2?10
• x1 x2?1
• x2?4
• x1?0, x2?0

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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 x12x2?10 is not given, then moving
  far away from the origin increases the objective
  function x12x2, and the maxim Z would be ?

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Unbounded Optima
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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
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4-2 The Basic Approach

• Standard Form of Linear Programming

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Handling of in-equality Constraints

• Case 1 Slack Variable
• x12x23 x3 4x4?25
• Modified to
• x12x23x34x4 x525
• x5?0 is a slack variable.
• Case 2 Surplus Variable
• 2 x1 x2-3 x3?12
• Modified to
• 2 x1 x2-3 x3- x412
• x4?0 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
• ss-s-
• such that s?0, s- ?0.

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Example
Modify to
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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. z0cTx is maximum or minimum in where Z is
  termed by the optimal value.
• Definition Alternate optimal solution is a set
  X?S, s.t. all x?X has the same objective value z0
  and for all x?S, and zcTx, 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.

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4-3 The Simplex Method
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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, xm1, ?,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
where
? Definition
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
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
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 i1,?,m,
and, for all jm1, ?,n, j?s. xj0
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Theorem 1 (Inner Product Rule)
Relative cost,
(inner product rule)
More (1) In a minimization problem, a basic
feasible solution is optimal if 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 ?z0.
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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

The above minimum happens at ir.
Corollary The above rule fails if there exist
unbounded optima.
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Example The LP Problem
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The Standard Form
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Table 1(s4,r6,rc2)Basic5 6 7Nonbasic1 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 (600/1060)
0 x6 1 1 3 8 0 1 0 420(420/852.5)
0 x7 30 40 20 10 0 0 1 800(800/1080)
?Z 6 10 9 20 - - - Z0
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Table 2(s2,r7,rc3)Basic5 4 7Nonbasic1 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(52.5/0.125420)
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 - Z1050
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Table 3Basic5 4 2Nonbasic1 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 Z1103.226
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2. The Two Phase Simplex Method-Example
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The Standard Form
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Two Phase Approach-Phase I
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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(11/111)
1 x6 -4 1 2 0 -1 1 0 3(3/21.5)
1 x7 -2 0 1 0 0 0 1 1(1/11)
?Z   6 -1 -3 - 1 - -  Z4
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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   - - - - - - -  Z0
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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(12/34)
1 x2 0 1 0 0 -1 1(1/0?)
1 x3 -2 0 1 0 0 1(1/-2lt0)
?Z   -1 - - - 5  Z2
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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
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
Formulate a linear program to determine the most
profitable product mix.
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Problem Formulation
Lets denote the variables as shown in the table,
then we have the following
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MATLAB PROGRAM

• f-10 -9 -8 -7 -6 -4 -3 -5 -4'
• A1 1 1 1 2 2 2 1 1 10 10 10 10 4 4 4 5 53 3 3
  3 2 2 2 1 1
• b100700400
• Aeqbeq
• LB0 0 0 0 0 0 0 0 0
• UB40 60 50 Inf 50 50 Inf 100 Inf
• X,FVAL,EXITFLAG,OUTPUT,LAMBDALINPROG(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
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Optimal Solution and Sensitivity Analysis

• x133.33, x266.67,x30,Z733.33
• Shadow prices for row 13.33, row 20.67, row 30
• Opportunity Costs for x32.67
• Ranges on the objective function coefficients
  6?c1(10)?15, 4?c2(6)?10,
• -8?c3(4)?6.67

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Optimal Solution and Sensitivity Analysis-
Continued

• 60?b1(100)?150, 400?b2(600)?1000,
• 200?b3(300)?8

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100 Rules

• 100 rule for objective function coefficients
• 100 rule for RHS constants

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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?(dc1-1, ?c1-4, dc21, ?c24,
  dc31, ?c32.67)
• Simultaneous variation of 10 hr decrease on labor
  100 lb increase in material and 50hr decrease on
  administration