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1 Lecture 3 Linear Programming: –Simplex Method –Computer Solutions Based on Excel Based on QM software Tutorial (to p2) (to p34) (to p36) (to p37)

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2 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)

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3 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)

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4 Simplex Method determine values of decision variablesIt 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 p1) (to p5) (to p24) (to p28)

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5 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 simplex method algorithm –Determine the optimal solution for the above tableau by following the simplex method algorithm (to p4 ) (to p6) (to p13) (to p15) (to p17)

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6 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)

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7 Standard format Consider the following 3 types of possible equation in a LP problem: 1.≤ constraints Such as x1 + x2 ≤ 3 ….(e1) 2.≥ constraints Such as x1 + x2 ≥ 3 ….(e2) 3.= constraints Such as x1 + x2 = 3 ….(e3) We need to change all these into a standard format as such: (to p8)

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8 Standard format 1.≤ constraints x 1 + x 2 ≤ 3 x 1 + x 2 + s 1 = 3 ………(e1) 2.≥ constraints x 1 + x 2 ≥ 3 x 1 + x 2 - s 2 + A 2 = 3 …..(e2) 3.= constraints x 1 + x 2 = 3 x 1 + x 2 + A 3 = 3 …….(e3) We will tell you why we need this format later! Consider the LP problem of …… Where, S is slack, A is artificial variable (to p9)

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9 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)

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10 Standard format Max Z=$40x 1 +50x 2 +0s 1 +0s 2 subject to 1x 1 + 2x 2 + s 1 = 40 ………(e1) 4x 2 + 3x 2 +s 2 = 120 …..(e2) x 1, x 2 0 The different is … (to p11)

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11 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 Extra! (to p9)

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12 More example of standard format Consider the following: M refer to very big value, -ve value here means that we don’t wish to retain it in the final solution (to p5)

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13 Forming a simplex tableau A simple tableau is outline as follows: What are these? (to p14)

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14 Forming a simplex tableau A simple tableau is outline as follows: we will discuss onhow to use them very soon! Cost in the obj. func. LP Decision variables We will compute this value later It is known as marginal value (to p5)

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15 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 Basic variables (to p5) (to p16)

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16 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” S1 has value “1” in this column only! (to p15)

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17 simplex method algorithm Compute z j values Compute c j -z j values enteringDetermine the entering variable leavingDetermine 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 c j -z i are all negative values –example (to p18) (to p19) (to p20) (to p22) (to p21) (to p23) (to p5)

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18 Compute z j values Compute by multiplying cj column values by the variable column values and summing: Z 1 is sum of multiple of these two columns (to p17)

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19 Compute c j -z j values All C j are listed on this row (to p17)

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20 entering Determine the entering variable It is referred to largest positive c j -z j value –The variable (i.e. the column) with the largest positive c j -z j value –Also known as “Pivot column” Max value, ie higher marginal cost contribute to the obj fuc This mean, we will next introduce x 2 as a basic variable In next Tableau (to p17)

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21 leaving Determine the leaving variable Min value of ratio of quantity values by the pivot column of entering variable Also known as “Pivot row” Min value = min (40/2, 120/3) = mins (20,40), thus pick the first value This mean, s1 will leave as basic variable in next Tableau (to p17)

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22 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)

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23 until all c j -z i are all negative values The following is the optimal tableau, and the solution is: All negative values, STOP And s1 =0 and s2= 0 (to p17)

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24 Irregular cases How to realize the following cases from the simplex tableau: 1.Multiple/alternative solutions 2.Infeasible LP problem 3.Unbound LP problem (to p4) (to p25) (to p26) (to p27)

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25 Multiple/alternative solutions Alternative solution is to consider the non-basic variable that has c j -z j = 0 as the next pivot column and repeat the simplex steps Note: S 1 is not a basic variable but has value “0” for c j- z j (to p24)

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26 Infeasible LP problem Infeasible means the LP problem is not properly formulated and that a feasible region cannot be identified. M value appear in final solution representing infeasible solution (to p24)

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27 Unbound LP problem Cannot identifying the Pivot row (i.e. leaving basic variable) (for s1 as pivot column) (to p24)

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28 Other cases 1.Minimizing Z 2.When a decision variable is –either ≤ or ≥ 3.Degeneracy (to p29) (to p32) (to p30) (to p4)

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29 Minimizing Z The tableau of Max Z still applied but we change the last row c j -z j into z j -c j Select this as Pivot column (to p28)

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30 either ≤ or ≥ If x1 is either ≤ or ≥, then We adopt a transformation as such: let x 1 = x’ 1 – x’’ 1 And then substitute it into the LP problem, and then follow the normal procedure Example … (to p28) (to p31)

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31 Example maximize Z=$40x x 2 subject to 1x 1 + 2x 2 40 4x 2 + 3x 2 120 x 1 0 maximize Z=$40x (x’ 2 -x’’ 2 ) subject to 1x 1 + 2(x’ 2 -x’’ 2 ) 40 4x 2 + 3(x’ 2 -x’’ 2 ) 120 x 1, x’ 2, x’’ 2 0 (to p30)

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32 Degeneracy It refers to the n th 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 n th tableau and select the other one tie-value variable as pivot column/row (to p28) (to p33)

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33 Example, Degeneracy Degeneracy (to p32)

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34 Based on Excel Xi >= 0 We type them in an Excel file Then, enter formulate here Then, we select Tools with “solver” An solution is obtained (to p35)

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35 Solution using Excel How to read them? (to p1)

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36 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)

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37 Tutorial Chapter 4, Edition 8 th : #12, #17, #19 Edition 9 th : #7, #11, #12 And from appendix A/B P3, p20, p25, p32

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