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Lecture 3 Linear Programming: Tutorial Simplex Method

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

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)

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)

4 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)

5 Steps for simplex method
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 )

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)

7 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)

8 Standard format ≤ constraints ≥ constraints = constraints
x1 + x2 ≤ x1 + x2 + s1 = 3 ………(e1) ≥ constraints x1 + x2 ≥ x1 + x2 - s2 + A2= 3 …..(e2) = constraints x1 + x2 = x1 + x2 + A3 = 3 …….(e3) We will tell you why we need this format later! Where, S is slack, A is artificial variable Consider the LP problem of …… (to p9)

9 Sample of an LP problem maximize Z=$40x1 + 50x2 subject to
1x1 + 2x2  40 ………(e1) 4x2 + 3x2  120 ……..(e2) x1  ……...(e3) x2  ………(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)

10 Standard format Max Z=$40x1+50x2+0s1+0s2 1x1 + 2x2 + s1 = 40 ………(e1)
subject to 1x1 + 2x2 + s1 = 40 ………(e1) 4x2 + 3x s2 = 120 …..(e2) x1, x2  0 The different is … (to p11)

11 maximize Z=$40x1 + 50x2 subject to 1x1 + 2x2  40 4x2 + 3x2  120
Z=$40x1+50x2+0s1+0s2 subject to 1x1 + 2x2 + s = 40 4x2 + 3x s2 = 120 x1,x2  0 Original LP format Standard LP format (to p9) Extra!

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)

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

14 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)

15 Transfer all values Max Z=$40x1+50x2+0s1+0s2 subject to
These values read from s1 and s2 here (to p5) (to p16) Basic variables

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

17 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)

18 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

19 Compute cj-zj values All Cj are listed on this row (to p17)

20 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

21 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)

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)

23 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

24 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)

25 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)

26 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)

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

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

29 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

30 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)

31 Example maximize Z=$40x1 + 50x2 subject to 1x1 + 2x2  40
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 (to p30)

32 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)

33 Example, Degeneracy Degeneracy (to p32)

34 Based on Excel Then, we select Tools with “solver”
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

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

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)

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

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