Lecture 3
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Lecture 3. Linear Programming: Simplex Method Computer Solutions Based on Excel Based on QM software Tutorial. (to p2). (to p34). (to p36). (to p37). Simplex Method. Last week, we covered on using the graphical approach in deriving solutions for LP problem Question:

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Lecture 3

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Lecture 3

Lecture 3

  • Linear Programming:

    • Simplex Method

    • Computer Solutions

      • Based on Excel

      • Based on QM software

  • Tutorial

(to p2)

(to p34)

(to p36)

(to p37)


Simplex method

Simplex Method

  • Last week, we covered on using the graphical approach in deriving solutions for LP problem

  • Question:

    • Can we solve all LP problems using graphical approach ?

(to p3)


Answer is no

Answer is “NO”

  • Why?

    Consider the following equation:

    x1+x2+x3+x4 = 4

  • It is extremely difficult for us to use graph to represent this equation

  • Thus, we need another systematic approach to solve an LP problem – known as Simplex Method

(to p4)


Simplex method1

Simplex Method

  • It is a method which applies Linear Algebra technique to determine values of decision variables of a set of equations

  • Steps for simplex method

  • Special/irregular cases

  • Other cases

(to p5)

(to p24)

(to p28)

(to p1)


Steps for simplex method

Steps for simplex method

  • Step 1

    • Convert all LP resource constraints into a standard format

  • Step 2

    • Form a simplex tableau

    • Transfer all values of step 1 into the simple tableau

    • Determine the optimal solution for the above tableau by following the simplex method algorithm

(to p6)

(to p13)

(to p15)

(to p17)

(to p4 )


Step 1

Step 1

  • Convert LP problem into standard format!

  • We refer standard format here as.

    • All constraints are in a form of “equation”

    • i.e. not equation of ≥ or ≤

    • Procedural steps

(to p7)


Standard format

Standard format

  • Consider the following 3 types of possible equation in a LP problem:

  • ≤ constraints

    • Such as x1 + x2 ≤ 3 ….(e1)

  • ≥ constraints

    • Such as x1 + x2 ≥ 3 ….(e2)

  • = constraints

    • Such as x1 + x2 = 3 ….(e3)

  • We need to change all these into a standard format as such:

  • (to p8)


    Standard format1

    Standard format

    • ≤ constraints

      • x1 + x2 ≤ 3 x1 + x2 + s1 = 3 ………(e1)

  • ≥ constraints

    • x1 + x2 ≥ 3 x1 + x2 - s2 + A2= 3 …..(e2)

  • = constraints

    • x1 + x2 = 3 x1 + x2 + A3 = 3 …….(e3)

  • We will tell you why we need this format later!

    Where, S is slack,

    A is artificial variable

    (to p9)

    Consider the LP problem of ……


    Sample of an lp problem

    Sample of an LP problem

    • maximize Z=$40x1 + 50x2

      subject to

      1x1 + 2x2  40 ………(e1)

      4x2 + 3x2  120 ……..(e2)

      x1  0 ……...(e3)

      x2  0 ………(e4)

      Convert them into a standard format will be like …

      (We can leave e3 and e4 alone as it is an necessary constraints for LP solution!)

      More example….

    (to p10)

    (to p12)


    Standard format2

    Standard format

    Max Z=$40x1+50x2+0s1+0s2

    subject to

    1x1 + 2x2 + s1 = 40 ………(e1)

    4x2 + 3x2 +s2 = 120 …..(e2)

    x1, x2 0

    • The different is …

    (to p11)


    Lecture 3

    maximize

    Z=$40x1 + 50x2

    subject to

    1x1 + 2x2  40

    4x2 + 3x2  120

    x1,x2  0

    Max

    Z=$40x1+50x2+0s1+0s2

    subject to

    1x1 + 2x2 + s1 = 40

    4x2 + 3x2 + s2 = 120

    x1,x2  0

    Original LP format

    Standard LP format

    (to p9)

    Extra!


    More example of standard format

    Consider the following:

    More example of standard format

    M refer to very big value, -ve value here means

    that we don’t wish to retain it in the final solution

    (to p5)


    Forming a simplex tableau

    Forming a simplex tableau

    • A simple tableau is outline as follows:

    (to p14)

    What are these?


    Forming a simplex tableau1

    Forming a simplex tableau

    LP Decision variables

    • A simple tableau is outline as follows:

    Cost in the obj. func.

    We will compute this value later

    It is known as marginal value

    we will discuss onhow to use them very soon!

    (to p5)


    Transfer all values

    Transfer all values

    Max

    Z=$40x1+50x2+0s1+0s2

    subject to

    1x1 + 2x2 + s1 = 40

    4x2 + 3x2 + s2 = 120

    x1,x2  0

    These values read from s1 and s2 here

    (to p5)

    (to p16)

    Basic variables


    Basic variables

    Basic variables

    • A decision variable is a basic variable in a tableau when

      • it is the only variable that has a coefficient of value “1” in that column and that others have values “0”

    (to p15)

    S1 has value “1” in this column only!


    Simplex method algorithm

    simplex method algorithm

    • Compute zj values

    • Compute cj-zj values

    • Determine the entering variable

    • Determine the leaving variable

    • Revise a new tableau

      • Introducing cell that crossed by “pivot row” and “pivot column” that has only value “1” and the rest of values on that column has value “0”

    • Repeat above steps until all cj-zi are all negative values

      • example

    (to p18)

    (to p19)

    (to p20)

    (to p21)

    (to p22)

    (to p23)

    (to p5)


    Compute z j values

    Compute zj values

    • Compute by multiplying cj column values by the variable column values and summing:

    (to p17)

    Z1 is sum of multiple of these two columns


    Compute c j z j values

    Compute cj-zj values

    All Cj are listed on this row

    (to p17)


    Determine the entering variable

    Determine the entering variable

    • It is referred to

      • The variable (i.e. the column) with the largest positive cj-zj value

      • Also known as “Pivot column”

    This mean, we will next introduce x2 as a basic variable

    In next Tableau

    (to p17)

    Max value, ie higher marginal cost contribute to the obj fuc


    Determine the leaving variable

    Determine the leaving variable

    • Min value of ratio of quantity values by the pivot column of entering variable

    • Also known as “Pivot row”

    This mean, s1 will leave as basic variable in next Tableau

    Min value = min (40/2, 120/3) = mins (20,40), thus pick the first value

    (to p17)


    Revise a new tableau

    Revise a new tableau

    Note, this value is copied

    This row values

    divided by 2

    New row x 3 – old row (2), note quantity must > 0

    Resume z and c-j computation!

    (to p17)


    Until all c j z i are all negative values

    until all cj-zi are all negative values

    • The following is the optimal tableau, and the solution is:

    And s1 =0 and s2= 0

    (to p17)

    All negative values, STOP


    Irregular cases

    Irregular cases

    • How to realize the following cases from the simplex tableau:

      • Multiple/alternative solutions

      • Infeasible LP problem

      • Unbound LP problem

    (to p25)

    (to p26)

    (to p27)

    (to p4)


    Multiple alternative solutions

    Multiple/alternative solutions

    Note: S1 is not a basic variable but has value “0” for cj-zj

    • Alternative solution is to consider the non-basic variable that has cj-zj = 0 as the next pivot column and repeat the simplex steps

    (to p24)


    Infeasible lp problem

    Infeasible LP problem

    M value appear in final solution representing infeasible solution

    • Infeasible means the LP problem is not properly formulated and that a feasible region cannot be identified.

    (to p24)


    Unbound lp problem

    Unbound LP problem

    (for s1 as pivot column)

    • Cannot identifying the Pivot row (i.e. leaving basic variable)

    (to p24)


    Other cases

    Other cases

    • Minimizing Z

    • When a decision variable is

      • either ≤ or ≥

    • Degeneracy

    (to p29)

    (to p30)

    (to p32)

    (to p4)


    Minimizing z

    Minimizing Z

    • The tableau of Max Z still applied but we change the last row cj-zj into zj-cj

    (to p28)

    Select this as Pivot column


    Either or

    either ≤ or ≥

    If x1 is either ≤ or ≥, then

    We adopt a transformation as such:

    let x1 = x’1 – x’’1

    And then substitute it into the LP problem, and then follow the normal procedure

    Example …

    (to p31)

    (to p28)


    Example

    maximize

    Z=$40x1 + 50x2

    subject to

    1x1 + 2x2 40

    4x2 + 3x2 120

    x1 0

    maximize

    Z=$40x1 + 50(x’2-x’’2)

    subject to

    1x1 + 2(x’2-x’’2)  40

    4x2 + 3(x’2-x’’2)  120

    x1, x’2, x’’2 0

    Example

    (to p30)


    Degeneracy

    Degeneracy

    • It refers to the nth tableau and (n+1)th tableau is the same (repeated)

    • Two ways

      • A tie value when selecting the pivot column

      • A tie value when selecting the pivot row

        • Example, Degeneracy

    • Solution:

      • Go back to nth tableau and select the other one tie-value variable as pivot column/row

    (to p33)

    (to p28)


    Example degeneracy

    Example, Degeneracy

    Degeneracy

    (to p32)


    Based on excel

    Based on Excel

    Xi >= 0

    We type them in an Excel file

    Then, we select

    Tools with “solver”

    An solution is obtained

    (to p35)

    Then, enter formulate

    here


    Solution using excel

    Solution using Excel

    (to p1)

    How to read them?


    Qm software

    QM software

    • Install the QM software

    • Loan QM software

    • Select option of “Linear Programming” from the “Module”

    • Then select “open” from option “file” to type in a new LP problem

    • Following instructions of the software accordingly

    • See software illustration!

    (to p1)


    Tutorial

    Tutorial

    • Chapter 4,

    • Edition 8th: #12, #17, #19

    • Edition 9th: #7, #11, #12

      And from appendix A/B

    • P3, p20, p25, p32


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