1

4.7 Alternate Optimal Solutions If an LP has more than one optimal solution, it has

multiple optimal solutions ( 多重最佳解 ) or alternative optimal solutions( 擇一最佳解 ).

If there is no NBV with a zero coefficient in row 0 of the optimal tableau, the LP has a unique optimal solution.

Even if there is a NBV with a zero coefficient in row 0 of the optimal tableau, it is possible that the LP may not have alternative optimal solutions.

p.152

2

Example： Dakota Furniture Company (p.140)

max z = 60x1 + 35x2 + 20x3

s.t. 8x1+ 6x2 + x3 ≤ 48 (lumber constraint) 4x1+ 2x2 + 1.5x3 ≤ 20 (finishing constraint) 2x1+1.5x2 + 0.5x3 ≤ 8 (carpentry constraint)

x2 ≤ 5 (table demand constraint) x1, x2, x3 ≥ 0

p.152

3

Table 13

BV z x1 x2 x3 s1 s2 s3 s4 RHS Ratio

z 1 -60 -35 -20 0 0 0 0 0s1 0 8 6 1 1 0 0 0 48s2 0 4 2 1.5 0 1 0 0 20s3 0 2 1.5 0.5 0 0 1 0 8s4 0 0 1 0 0 0 0 1 5

max z = 60x1 + 35x2 + 20x3

s.t. 8x1+ 6x2 + x3 ≤ 48 4x1+ 2x2 + 1.5x3 ≤ 20 2x1+1.5x2 + 0.5x3 ≤ 8

x2 ≤ 5 x1, x2, x3 ≥ 0

z - 60x1 - 35x2 - 20x3 = 0 8x1+ 6x2 + x3 + s1 = 48 4x1+ 2x2 + 1.5x3 + s2 = 20 2x1+1.5x2 + 0.5x3 + s3 = 8

x2 + s4 = 5 x1, x2, x3, s1, s2, s3, s4 ≥ 0

4

Table 13

BV z x1 x2 x3 s1 s2 s3 s4 RHS Ratio

Z 1 -60 -35 -20 0 0 0 0 0s1 0 8 6 1 1 0 0 0 48 8s2 0 4 2 1.5 0 1 0 0 20 5s3 0 2 1.5 0.5 0 0 1 0 8 4 LVs4 0 0 1 0 0 0 0 1 5

EVTable 14

BV z x1 x2 x3 s1 s2 s3 s4 RHS Ratio

Z 1 0 10 -5 0 0 30 0 240s1 0 0 0 -1 1 0 -4 0 16 Nones2 0 0 -1 0.5 0 1 -2 0 4 8 LVx1 0 1 0.75 0.25 0 0 0.5 0 4 16s4 0 0 1 0 0 0 0 1 5 None

EV

5

Table 15

BV z x1 x2 x3 s1 s2 s3 s4 RHS BV

Z 1 0 0 0 0 10 10 0 280 z=280s1 0 0 -2 0 1 2 -8 0 24 s1=24x3 0 0 -2 1 0 2 -4 0 8 x3=8x1 0 1 1.25 0 0 -0.5 1.5 0 2 x1=2s4 0 0 1 0 0 0 0 1 5 s4=5

Table 16

BV z x1 x2 x3 s1 s2 s3 s4 RHS BV

Z 1 0 0 0 0 10 10 0 280 z=280s1 0 1.6 0 0 1 1.2 -5.6 0 27.2 s1=27.2x3 0 1.6 0 1 0 1.2 -1.6 0 11.2 x3=11.2x2 0 0.8 1 0 0 -0.4 1.2 0 1.6 x1=1.6s4 0 -0.8 0 0 0 0.4 -1.2 1 3.4 s4=3.4

6

Any point on the line segment joining two optimal extreme points will also be optimal.

1 0 23211

61612

21161

0 1

802

21161

02

802

1

3

2

1

3

2

1

3

2

1

cforc..c..

c

..)c(c

xxx

..

xxx

intpoextremeOptimal

xxx

intpoextremeOptimal

7

Exercise, p.154, 5 min

Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5

8

4.8 Unbounded LPs For some LPs, there exist points in the feasible

region for which z assumes arbitrarily large (in max problems) or arbitrarily small (in min problems) values. The LP is unbounded solution( 無限解 ).

An unbounded LP occurs in a max problem if there is a NBV with a negative coefficient in row 0 and there is no constraint that limits how large we can make this NBV.

Specifically, an unbounded LP for a max problem occurs when a variable with a negative coefficient in row 0 has a non positive coefficient in each constraint.

p.154

9

Example 3. Breadco Bakeries x1=number of loaves of French bread baked

x2=number of loaves of sourdough bread baked x3=number of yeast packets purchased x4=number of ounces of flour purchased

Objective function ： Revenue = 36x1+30x2, Costs = 3x3 + 4x4

z = Revenue – Costs = 36x1 + 30x2 － 3x3 － 4x4 Constraints ： x1 +x2 ≤ 5 + x3 x1 +x2 － x3 ≤ 5 6x1 + 5x2 ≤ 10 + x4 6x1 + 5x2 － x4 ≤ 10 Max z = 36x1 + 30x2 － 3x3 － 4x4

s.t. x1 + x2 － x3 ≤ 5 6x1 + 5x2 － x4 ≤ 10

x1, x2 , x3 , x4 ≥0

p.154

10

max z = 36x1 + 30x2 – 3x3 – 4x4

s.t. x1 + x2 – x3 ≤ 5 6x1 + 5x2 –x4 ≤ 10

x1, x2, x3, x4 ≥ 0

max z = 36x1 + 30x2 – 3x3 – 4x4

s.t. x1 + x2 – x3 + s1 =5 6x1 + 5x2 –x4 + s2 =10

x1, x2, x3, x4, s1, s2 ≥ 0

Table 19BV z x1 x2 x3 x4 s1 s2 RHS BV Ratioz 1 -36 -30 3 4 0 0 0 z=0s1 0 1 1 -1 0 1 0 5 s1=5 5s2 0 6 5 0 -1 0 1 10 s2=10 5/3

11

Table 20

BV z x1 x2 x3 x4 s1 s2 RHS Ratio

Z 1 0 0 3 -2 0 6 60s1 0 0 1/6 -1 1/6 1 -1/6 10/3 20 LVx1 0 1 5/6 0 -1/6 0 1/6 5/3 none

EV

Table 19

BV z x1 x2 x3 x4 s1 s2 RHS Ratio

Z 1 -36 -30 3 4 0 0 0s1 0 1 1 -1 0 1 0 5 5s2 0 6 5 0 -1 0 1 10 10/6 LV

EV

Table 21

BV z x1 x2 x3 x4 s1 s2 RHS Ratio

Z 1 0 2 -9 0 12 4 100x4 0 0 1 -6 1 6 -1 20 nonex1 0 1 1 -3 0 1 0 5 none

12

Exercise, p.158 , 5 min

Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5 Exercise 6

13

4.11 Degeneracy and Convergence of the Simplex AlgorithmTheoretically, the simplex algorithm can fail to find an

optimal solution to an LP.The following are facts:

If (value of entering variable in new bfs) > 0, then (z-value for new bfs) > (z-value for current bfs)

If (value of entering variable in new bfs) = 0, then (z-value for new bfs) = (z-value for current bfs)

p.168

14

Assume that in each of the LP’s bfs all BV are

positive. An LP is a nondegenerate (非退化 ) LP. An LP is degenerate(退化 ) if it has at least one bfs

where a BV is equal 0. Any bfs that has at least one BV equal to zero is a

degenerate bfs( 退化基本可行解 ). When the same bfs is encountered twice it is called

cycling( 循環 ). If cycling occurs, then we will loop, or cycle,

forever among a set of basic feasible solutions and never get to an optimal solution.

p.169

15

Example ： Degenerate LP

max z = 5x1 + 2x2 s.t. x1 + x2 ≤ 6 x1 – x2 ≤ 0

x1, x2 ≥ 0

max z = 5x1 + 2x2 s.t. x1 + x2 + s1 = 6 x1 – x2 + s2 = 0

x1, x2, s1, s2 ≥ 0

Table 29

BV z x1 x2 s1 s2 RHS

Z 1 -5 -2 0 0 0s1 0 1 1 1 0 6s2 0 1 -1 0 1 0

p.169

16

Table 29

BV z x1 x2 s1 s2 RHS Ratio

z 1 -5 -2 0 0 0s1 0 1 1 1 0 6 6s2 0 1 -1 0 1 0 0 LV

EVTable 30

BV z x1 x2 s1 s2 RHS Ratio

z 1 0 -7 0 5 0s1 0 0 2 1 -1 6 6/2 LVx1 0 1 -1 0 1 0 None

EVTable 31

BV z x1 x2 s1 s2 RHS Ratio

z 1 0 0 3.5 1.5 21x2 0 0 1 0.5 -0.5 3x1 0 1 0 0.5 0.5 3

17

Figure 12 (p.171)

There are 3 sets of BVs that correspond to 1 extrem point.

Table 32BV BFS Extreme pointx1 , x2 x1 = 3, x2 = 3, s1 = s2 = 0 Dx1 , s1 x1 = 0, s1 = 6, x2 = s2 = 0 Cx1 , s2 x1 = 6, s2 =-6, x2 = s1 = 0 Infeasiblex2 , s1 x2 = 0, s1 = 6, x1 = s2 = 0 Cx2 , s2 x2 = 6, s2 = 6, x1 = s1 = 0 Bs1 , s2 s1 = 6, s2 = 0, x1 = x2 = 0 C

18

If an LP has many degenerate BFS (or many BV equal zero), then the simplex algorithm is often inefficient.

Some degenerate LPs have a special structure that enables us to solve them by methods other than the simplex. (chapter 7)

19

Exercise ： Degenerate LP

max z = 7x1 + 8x2 s.t. x1 + 2x2 ≤ 10 3x1 + 2 x2 ≤ 18 x2 ≤ 5

x1, x2 ≥ 0

max z = 7x1 + 8x2 s.t. x1 + 2x2 + s1 = 10 3x1 + 2x2 + s2 = 18 2x2 + s3 = 5

x1, x2 , s1, s2 ≥ 0

BV z x1 x2 s1 s2 s3 RHS Ratio

z 1 -7 -8 0 0 0 0s1 0 1 2 1 0 0 10 5s2 0 3 2 0 1 0 18 9s3 0 0 1 0 0 1 5 5

20

BV z x1 x2 s1 s2 s3 RHS Ratio

z 1 -7 0 0 0 8 40s1 0 1 0 1 0 -2 0 0s2 0 3 0 0 1 -2 8 8/3x2 0 0 1 0 0 1 5 -

BV z x1 x2 s1 s2 s3 RHS Ratio

z 1 -7 -8 0 0 0 0s1 0 1 2 1 0 0 10 5s2 0 3 2 0 1 0 18 9s3 0 0 1 0 0 1 5 5

BV z x1 x2 s1 s2 s3 RHS Ratio

z 1 0 0 7 0 -6 40x1 0 1 0 1 0 -2 0 0s2 0 0 0 -3 1 4 8 2x2 0 0 1 0 0 1 5 5

21

BV z x1 x2 s1 s2 s3 RHS Ratio

z 1 0 0 7 0 -6 40x1 0 1 0 1 0 -2 0 -

s2 0 0 0 -3 1 4 8 2

x2 0 0 1 0 0 1 5 5

BV z x1 x2 s1 s2 s3 RHS Ratio

z 1 0 0 5/2 3/2 0 52x1 0 1 0 -1/2 1/2 0 4s3 0 0 0 -3/4 1/4 1 8x2 0 0 1 3/4 -1/4 0 3

22

Exercise, p.170, 5 min

Exercise 1 Exercise 2

23

Summery (1)

1. 進入變數平手指 2 個 NBV 具有相同且最負的z 列係數。任意選一個為進入變數( 無有效方法 ) 。

2. 離開變數平手 --- 退化解有 2 個以上相同之最小比率。未被選擇離開之 BV，其值為 0 ，該 BV 即degenerate BV ，含 degenerate BV 之 bfs 即是degenerate bfs。

理論上，退化 bfs 易產生 cycle，使得 z值不變，即 bfs1bfs2bfs3 bfs1而無法求得最佳解。 (有書專門討論此類問題 )

24

3. 無離開變數 --- 無窮解 進入變數之欄無正值，使得無離開變數。 實務上，遇到此情況，代表該 LP 之模式有誤。( 若 z 代表利潤，則 z 不可能∞ )

4. 最佳單形表含 z 列係數為 0 之 NBV --- 多重最佳解 有任何 NBV 之 z列係數為 0 若讓為 0 之 NBV 為進入變數，則下一個 z值不會改變，但 BFS 將由不同之 BV 組成。亦即不同之解，卻有相同之目標函數值。

25

LP 的各種變形與解法 Standard Form

0..

ixBAXts

CXZMax

Min:(1)Method 1 (Z=-Z) (2)Method 2

變形 2: “≧”

Unrestricted-in-sign (urs) variable:Let "'

iii xxx

變形 1: Min

變形 3: “urs”

(1)Big-M Method(2)Two-Phase Method

26

Artificial Variables ( 人工變數 )

為使 simplex method 得以繼續沿用，以人工變數對「≥」之「 = 」之限制式進行轉換。

處裡人工變數之方法有二： Big M method (大M法 ) Two-phase Simplex method (雙階法 )

目的在盡量使人工變數為 0, 使得到之人工問題最佳解即為原問題之最佳解。

27

4.12 The Big M Method( 大 M 法 ) The simplex method requires a starting bfs. Previous problems have found starting bfs by

using the slack variables as BV. If an LP have ≥ or = constraints, however,

a starting bfs may not be readily apparent. In such a case, the Big M method may be used to

solve the problem.

p.172

28

Description of the Big M Method1.Modify the constraints so that the rhs of each constraint

is nonnegative. Identify each constraint that is now an = or ≥ constraint.

2.Convert each inequality constraint to standard form (add a slack variable for ≤ constraints, add an excess variable for ≥ constraints).

3.For each ≥ or = constraint, add artificial variables. Add sign restriction ai ≥ 0.

4.Let M denote a very large positive number. Add (for each artificial variable) Mai to min problem objective functions or -Mai to max problem objective functions

5.Since each artificial variable will be in the starting basis, all artificial variables must be eliminated from row 0 before beginning the simplex. Remembering M represents a very large number, solve the transformed problem by the simplex.

29

Example 4: BevcoBevco manufactures an orange-flavored soft drink

called Oranj by combining orange soda and orange juice. Each orange soda contains 0.5 oz of sugar and 1 mg of vitamin C.

Each ounce of orange juice contains 0.25 oz of sugar and 3 mg of vitamin C.

It costs Bevco 2¢ to produce an ounce of orange soda and 3¢ to produce an ounce of orange juice.

Bevco’s marketing department has decided that each 10-oz bottle of Oranj must contain at least 20 mg of vitamin C and at most 4 oz of sugar.

Use LP to determine how Bevco can meet the marketing department’s requirements at min cost.

p.172

30

Example 4: Solution

x1 = number of ounces of orange soda in a bottle of Oranj

x2 = number of ounces of orange juice in a bottle of Oranj

The LP is:min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 ≤ 4 (sugar constraint)

x1 + 3x2 ≥ 20 (Vitamin C constraint)

x1 + x2 = 10 (10 oz in 1 bottle of Oranj)

x1, x2 ≥ 0

Step 1

31

The LP in standard form has z and s1 which could be used for BVs but row 2 would violate sign restrictions and row 3 no readily apparent BV.

In order to use the simplex method, a bfs is needed. Artificial variables ( 人工變數 ) are created.

Row 0: z - 2x1 - 3x2 = 0Row 1: 0.5x1 + 0.25x2 + s1 = 4Row 2: x1 + 3x2 - e2 = 20Row 3: x1 + x2 = 10

Step 2

32

In the optimal solution, all artificial variables must be set equal to zero. For a min LP, a term Mai is added to the objective

function for each artificial variable ai.For a max LP, the term –Mai is added to the objective

function for each ai. M represents some very large number.

Row 0: z - 2x1 - 3x2 = 0

Row 1: 0.5x1 + 0.25x2 + s1 = 4

Row 2: x1 + 3x2 - e2 + a2 = 20

Row 3: x1 + x2 + a3 = 10

Step 3

33

The modified Bevco LP in standard form then becomes:

Modifying the objective function this way makes it extremely costly for an artificial variable to be positive. The optimal solution should force a2 = a3 =0.

Row 0: z - 2x1 - 3x2 -Ma2 - Ma3 = 0

Row 1: 0.5x1 + 0.25x2 + s1 = 4

Row 2: x1 + 3x2 - e2 + a2 = 20

Row 3: x1 + x2 + a3 = 10

Step 4

34

Because a2 and a3 are in starting BFS, they must be eliminated from row 0.

Replace row0 by row0 +M(row2)+M(row3)

Row 0: z - 2x1 - 3x2 -Ma2 - Ma3 = 0

Row 1: 0.5x1 + 0.25x2 + s1 = 4

Row 2: x1 + 3x2 - e2 + a2 = 20

Row 3: x1 + x2 + a3 = 10

Row 0’: z + (2M - 2)x1 + (4M - 3)x2 - Me2 = 30M

Step 5

35

Table 33BV z x1 x2 s1 e2 a2 a3 RHS RatioZ 1 2M-2 4M-3 0 -M 0 0 30Ms1 0 1/2 1/4 1 0 0 0 4 16a2 0 1 3 0 -1 1 0 20 20/3 LVa3 0 1 1 0 0 0 1 10 10

EVTable 34BV z x1 x2 s1 e2 a2 a3 RHS RatioZ 1 0 0 0s1 0 5/12

0 1 1/12 -1/12 0 7/3 28/5

x2 0 1/3 1 0 -1/3 1/3 0 20/3 20a3 0 2/3 0 0 1/3 -1/3 1 10/3 5 LV

EV

332 M

33M

645 M

31060 M

36

Table 35BV z x1 x2 s1 e2 a2 a3 RHS Ratioz 1 0 0 0 -1/2 25s1 0 0 0 1 -1/8 1/8 -5/8 1/4x2 0 0 1 0 -1/2 1/2 -1/2 5x1 0 1 0 0 1/2 -1/2 3/2 5

221 M

223 M

The optimal solution for Bevco is z=25, x1 = x2 = 5, s1 = 1/4, e2 = 0.

This means that Bevco can hold the cost of producing a 10-oz. bottle of Oranj to $.25 by mixing 5 oz of orange soda and 5 oz of orange juice.

37

Example 4’:

min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 ≤ 4 (sugar constraint) x1 + 3x2 ≥ 36 (Vitamin C constraint) x1 + x2 = 10 (10 oz in 1 bottle of Oranj) x1, x2 ≥ 0

Row 0: z - 2x1 - 3x2 -Ma2 - Ma3 = 0Row 1: 0.5x1 + 0.25x2 + s1 = 4Row 2: x1 + 3x2 - e2 + a2 = 36Row 3: x1 + x2 + a3 = 10

Row 0’: z + (2M - 2)x1 + (4M - 3)x2 - Me2 = 46M

p.177

38

Table 36BV z x1 x2 s1 e2 a2 a3 RHS Ratioz 1 2M-2 4M-3 0 -M 0 0 46Ms1 0 1/2 1/4 1 0 0 0 4 16a2 0 1 3 0 -1 1 0 36 12a3 0 1 1 0 0 0 1 10 10 LV

EV

Table 37BV z x1 x2 s1 e2 a2 a3 RHSz 1 1-2M 0 0 -M 0 3-4M 30+6Ms1 0 1/4 0 1 0 0 -1/4 3/2a2 0 -2 0 0 -1 1 -3 6x2 0 1 1 0 0 -0 1 10

39

If all artificial variables equal zero in the optimal

solution, the solution is optimal. (exercise 4) If any artificial variables are positive in the optimal

solution, the solution is infeasible. Then the original LP has no feasible solution.

(exercise 5) 由於M為一相當大的值，對電腦而言，易產生數值上的問

題 ( 極大數與極小數相加，後果？ ) ，因此商業軟體少用此法，而多使用 Two-Phase method.

40

Exercise ： p.178 #1~#6

min z = 4x1 + 4x2 + x3

s.t. x1 + x2 + x3 ≤ 2 2x1 + x2 ≤ 3 2x1 + x2 + 3x3 ≥ 3 x1 , x2 , x3 ≥ 0

min z = 4x1 + 4x2 + x3 + Ma3

s.t. x1 + x2 + x3 + s1 = 2 2x1 + x2 + s2 = 3 2x1 + x2 + 3x3 ‑ e3 + a3 = 3

x1 , x2 , x3 , s1 , s2 , e3 , a3 ≥ 0

41

Eliminating the BV a3 from z ‑ 4x1 ‑ 4x2 ‑ x3 ‑ Ma3 = 0 z + (2M ‑ 4)x1 + (M ‑ 4)x2 + (3M ‑ 1)x3 ‑ Me3 = 3M.

z x1 x2 x3 s1 s2 e3 a3 rhs1 2M‑4 M‑4 3M‑ 1 0 0 ‑M 0 3M0 1 1 1 1 0 0 0 20 2 1 0 0 1 0 0 30 2 1 3 0 0 -1 1 3

min z = 4x1 + 4x2 + x3 + Ma3

s.t. x1 + x2 + x3 + s1 = 2 2x1 + x2 + s2 = 3 2x1 + x2 + 3x3 ‑ e3 + a3 = 3 ×(-M)

x1 , x2 , x3 , s1 , s2 , e3 , a3 ≥ 0

42

z x1 x2 x3 s1 s2 e3 a3 rhs r1 2M‑4 M‑4 3M‑ 1 0 0 ‑M 0 3M0 1 1 1 1 0 0 0 2 20 2 1 0 0 1 0 0 3 none0 2 1 3 0 0 -1 1 3 1z x1 x2 x3 s1 s2 e3 a3 rhs1 -10/3 -11/3 0 0 0 ‑1/3 1/3-M 10 1/3 2/3 0 1 0 1/3 ‑1/3 10 2 1 0 0 1 0 0 30 2/3 1/3 1 0 0 -1/3 1/3 1The optimal solution is z = 1, s1 = 1, s2 = 3, x3 = 1, x2 = x1 = e3 = 0.

43

Exercise, p.178, 15min

Exercise 1 (Z=1, x1=0, x2=0, x3=1) Exercise 2 Exercise 3 (Z=5, x1=1, x2=2,s1=0) Exercise 4 (Infeasible LP) Exercise 5 (Z=2, x1=2, x2=0, x3=0) Exercise 6 (Z=2, x1=2, x2=0)

44

4.13 Two-Phase Simplex MethodPhase I 之目標函數僅考慮人工變數。 (p.178)

無論 Min or Max 問題，最好之 z值為 0 ，故不可能為無窮解讓所有原始變數為 0 即為一可行解，故不可能為無可行解所以必有最佳解。

當得到最佳解時，若所有人工變數均為 0 ，則至 phase II ；若有任何人工變數不為 0 ，則原問題無可行解。Phase II 中：

刪除全部人工變數，若人工變數仍為 BV ，則保留。以 phase I 之最佳解做為 phase II 之起始 bfs ，並回復原問題之目標函數係數

p.178

45

Phase I LP In this method, artificial variables are added to the

same constraints, then a bfs to the original LP is found by solving Phase I LP.

The objective function is to minimize the sum of all artificial variables.

At completion, reintroduce the original LP’s objective function and determine the optimal solution to the original LP.

46

Phase II LPBecause each ai ≥ 0, solving the Phase I LP will result in

one of the following three cases:Case 1.

The optimal value of w’ is greater than zero.The original LP has no feasible solution.

Case 2.The optimal value of w’ is equal to zero, and no AVs are in the optimal Phase I basis.

Drop all columns in the optimal Phase I tableau that correspond to the artificial variables.

Now combine the original objective function with the constraints from the optimal Phase I tableau.

Case 3. The optimal value of w’ is equal to zero and at least one AV is in the optimal Phase I basis.

47

Example 5 ： case 2min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 ≤ 4 x1 + 3x2 ≥ 20 x1 + x2 = 10 x1, x2 > 0

min w’ = a2 + a3 (w’ - a2- a3=0) ---- (1) s.t. 0.5x1 + 0.25x2 + s1 = 4

x1 + 3x2 – e2 + a2 = 20 ---- (2) x1 + x2 + a3 = 10 ---- (3) x1, x2, s1, e2, a2, a3 > 0

(1) + (2) + (3) min w’ + 2x1 + 4x2 - e2 = 30

p.179

48

Table 38BV w’ x1 x2 s1 e2 a2 a3 RHS Ratiow’ 1 2 4 0 -1 0 0 30s1 0 1/2 1/4 1 0 0 0 4 16a2 0 1 3 0 -1 1 0 20 20/3 LVa3 0 1 1 0 0 0 1 10 10

EV

Table 39BV w’ x1 x2 s1 e2 a2 a3 RHS Ratiow’ 1 2/3 0 0 1/3 -4/3 0 3/10s1 0 5/12

0 1 1/12 - /12 0 7/3 28/5

x2 0 1/3 1 0 -1/3 1/3 0 20/3 20a3 0 2/3 0 0 1/3 -1/3 1 10/3 5 LV

EV

49

Table 40BV w’ x1 x2 s1 e2 a2 a3 RHSw’ 1 0 0 0 0 -1 -1 0s1 0 0 0 1 -1/8 1/8 -5/8 1/4x2 0 0 1 0 -1/2 1/2 -1/2 5x1 0 1 0 0 1/2 -1/2 3/2 5

Drop all AVs.

Phase II: reintroduce Z, reduce the BV’s of Row 0,The optimal solution is z=25, x1=5, x2=5, s1=1/4, e2=0

50

Example 6 ： case 1

min z = 2x1 + 3x2

s.t. 0.5x1 + 0.25x2 ≤ 4 x1 + 3x2 ≥ 36 x1 + x2 = 10 x1, x2 > 0

min w’ = a2 + a3 (w’ - a2- a3=0) ---- (1) s.t. 0.5x1 + 0.25x2 + s1 = 4

x1 + 3x2 – e2 + a2 = 36 ---- (2) x1 + x2 + a3 = 10 ---- (3) x1, x2, s1, e2, a2, a3 > 0

(1) + (2) + (3) min w’ + 2x1 + 4x2 - e2 = 46

p.181

51

Table 41BV w’ x1 x2 s1 e2 a2 a3 RHS Ratiow’ 1 2 4 0 -1 0 0 46s1 0 1/2 1/4 1 0 0 0 4 16a2 0 1 3 0 -1 1 0 36 12a3 0 1 1 0 0 0 1 10 10 LV

EVTable 42BV w’ x1 x2 s1 e2 a2 a3 RHS Ratiow’ 1 -2 0 0 -1 0 -4 6s1 0 1/4 0 1 0 0 -1/4 3/2x2 0 -2 0 0 -1 1 -3 6a3 0 1 1 0 0 -0 1 10

w’ =6 > 0, so the original LP has no feasible solution.

52

Example 7 ： case 3max z = 40x1 + 10x2 + 7x5 + 14x6s.t. x1 - x2 +2x5 = 0 -2x1 + x2 - 2x5 = 0

x1 + x3 + x5 - x6 = 3 2x2 + x3 + x4 + 2x5 + x6 = 4

x1, x2, x3, x4, x5, x6 > 0

(1) + (2) + (3) + (4) min w’ –x3 – x5 + x6 = 0

min w’-a1-a2-a3 = 0 ----(1)s.t. x1 - x2 + 2x5 + a1 = 0 ----(2) -2x1 + x2 - 2x5 + a2 = 0 ----(3)

x1 + x3 + x5 - x6 + a3 = 3 ----(4) 2x2 + x3 + x4 + 2x5 + x6 = 4

x1, x2, x3, x4, x5, x6, a1, a2, a3 > 0

p.183

53

Table 43BV w’ x1 x2 x3 x4 x5 x6 a1 a2 a3 RHS Rw’ 1 0 0 1 0 1 -1 0 0 0 3a1 0 1 -1 0 0 2 0 1 0 0 0a2 0 -2 1 0 0 -2 0 0 1 0 0a3 0 1 0 1 0 1 -1 0 0 1 3 3 LVx4 0 0 2 1 1 2 1 0 0 0 4 4

EVTable 44BV w’ x1 x2 x3 x4 x5 x6 a1 a2 a3 RHS Rw’ 1 -1 0 0 0 0 0 0 0 -1 0a1 0 1 -1 0 0 2 0 1 0 0 0a2 0 -2 1 0 0 -2 0 0 1 0 0x3 0 1 0 1 0 1 -1 0 0 1 3x4 0 -1 2 0 1 1 2 0 0 -1 1

54

Table 44BV w’ x1 x2 x3 x4 x5 x6 a1 a2 a3 RHS Ratio

w’ 1 -1 0 0 0 0 0 0 0 -1 0a1 0 1 -1 0 0 2 0 1 0 0 0a2 0 -2 1 0 0 -2 0 0 1 0 0x3 0 1 0 1 0 1 -1 0 0 1 3x4 0 -1 2 0 1 1 2 0 0 -1 1

Table 44’ ---- drop x1 and a3

BV w’ x1 x2 x3 x4 x5 x6 a1 a2 a3 RHS Ratio

w’ 1 0 0 0 0 0 0 0 0a1 0 -1 0 0 2 0 1 0 0a2 0 1 0 0 -2 0 0 1 0x3 0 0 1 0 1 -1 0 0 3x4 0 2 0 1 1 2 0 0 1

55

Table 45BV z x2 x3 x4 x5 x6 a1 a2 RHS Ratio

z 1 -10 0 0 -7 -14 0 0 0a1 0 -1 0 0 2 0 1 0 0a2 0 1 0 0 -2 0 0 1 0x3 0 0 1 0 1 -1 0 0 3x4 0 2 0 1 1 2 0 0 1

Table 44’BV w’ x2 x3 x4 x5 x6 a1 a2 RHSw’ 1 0 0 0 0 0 0 0 0a1 0 -1 0 0 2 0 1 0 0a2 0 1 0 0 -2 0 0 1 0x3 0 0 1 0 1 -1 0 0 3x4 0 2 0 1 1 2 0 0 1

56

Table 45BV z x2 x3 x4 x5 x6 a1 a2 RHS Ratio

z 1 -10 0 0 -7 -14 0 0 0a1 0 -1 0 0 2 0 1 0 0a2 0 1 0 0 -2 0 0 1 0x3 0 0 1 0 1 -1 0 0 3x4 0 2 0 1 1 2 0 0 1 1/2 LV

EVTable 46BV z x2 x3 x4 x5 x6 a1 a2 RHS Ratio

z 1 4 0 7 0 0 0 0 7a1 0 0 0 0 2 0 1 0 0a2 0 1 0 0 0 0 0 1 0x3 0 1 1 1/2 3/2 0 0 0 7/2x6 0 0 0 1/2 1/2 1 0 0 1/2

The optimal solution is z=7, x3 =7/2, x6 =1/2, x2 = x4 = x5 =0

57

Exercise, p.184(p.178), 15min

Exercise 1 Exercise 2 Exercise 3 Exercise 4 Exercise 5 Exercise 6

58

LP 的各種變形與解法 Standard Form

0..

ixBAXts

CXZMax

Min:(1)Method 1 (Z=-Z) (2)Method 2

變形 2: “≧”

Unrestricted-in-sign (urs) variable:Let "'

iii xxx

變形 1: Min

變形 3: “urs”

(1)Big-M Method(2)Two-Phase Method

59

4.14 Unrestricted-in-Sign VariablesAn LP with an unrestricted-in-sign (urs) variable

can be transformed to non-negative.For each urs variable, define two new variables

x’i and xn

i.Then substitute x’

i - xni for xi in each constraint

and in the objective function. Also add the sign restrictions.

The effect of this substitution is to express xi as the difference of the two nonnegative variables x’

i and xni.

No basic feasible solution can have both x’i ≥ 0

and xni ≥ 0.

p.184

60

For any basic feasible solution, each URS variable

xi must fall into one of the following three cases.1. x’

i > 0 and xni = 0

2. x’i = 0 and xn

i > 03. x’

i = xni = 0

61

Example 8 ： Using urs vars x1 = number of loaves of bread baked

x2 = number of ounces by which flour supply is increased by

Let x2 = x’2 – x”2

Max z = 30x1 - 4 x’2 – 4x”2

s.t. 5x1 ≤ 30 + x’2 – x”2 (Flour constraint) x1 ≤ 5 (Yeast constraint)

x1 ≥ 0 , x’2 , x”2 ≥ 0

Max z = 30x1 - 4x2

s.t. 5x1 ≤ 30 + x2 (Flour constraint) x1 ≤ 5 (Yeast constraint)

x1 ≥ 0 , x2 urs

p.185

62

Table 47BV z x1 x ’2 x”2 s1 s2 RHS Ratioz 1 -30 4 -4 0 0 0s1 0 5 -1 1 1 0 30 6s2 0 1 0 0 0 1 5 5 LV

EV

Max z = 30x1 - 4 x’2 – 4x”2s.t. 5x1 –x’2 + x”2 + s1 = 30 x1 + s1 = 5

x1 ≥ 0 , x’2 , x”2 ≥ 0 ,s1, s2 ≥ 0

Table 48BV z x1 x ’2 x”2 s1 s2 RHS Ratioz 1 0 4 -4 0 30 150s1 0 0 -1 1 1 -5 5 5 LVx1 0 1 0 0 0 1 5 none

EV

63

Table 49BV z x1 x ’2 x”2 s1 s2 RHS Ratioz 1 0 0 0 4 10 170

x”2 0 0 -1 1 1 -5 5x1 0 1 0 0 0 0 5

The optimal solution is z= 170, x1= 5, x”2= 5, x2’= 0, s1= s2= 0.

x2= x’- x”2 = -5 Because x2 = -5, the baker should sell 5 oz of flour.

64

Exercise ： p.189 #4

Note that in any BFS, z’and z” cannot both be positive.If 2x1-3x2>0,then z”=0 and the objective function will equal z=z’+z”=z’=2x1-3x2=|2x1-3x2|.If 2x1-3x2<0,z’=0 and z”=|2x1-3x2| so the objective function will equal z=z’+z”=z”=|2x1-3x2|.

max z = |2x1-3x2|s.t. 4x1+x2 ≤ 4 2x1–x2 ≤ 0.5 x1, x2 ≥ 0

max z = z’ + z''s.t. 4x1+x2 ≤ 4 2x1-x2 ≤ 0.5 2x1-3x2 = z’–z” x1, x2 , z’ , z” ≥ 0

65

其它形式x’ = x – 3

x’ = x – 2

x ≧ 3

2 ≦ x ≦ 5

x’ ≧0

x’, x” ≧0x” = 5 – x

x’ = 6 – xx ≦ 6 x’ ≧0

66

Using the simplex algorithm to solve different type of problems

67

(1) Max. z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn ≤ b1

a21x1 + a22x2 +···+ a2nxn ≤ b2

： ： am1x1+ am2x2 +···+ amnxn ≤ bm

xi ≥ 0 (i=1,2, ···,n)

Max. z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn + s1 = b1

a21x1 + a22x2 +···+ a2nxn + s2 = b2

： ： am1x1+ am2x2 +···+ amnxn + sn = bm

xi , si ≥ 0 (i=1,2, ···,n)

68

(2) Min. z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn ≤ b1

a21x1 + a22x2 +···+ a2nxn ≤ b2

： ： am1x1+ am2x2 +···+ amnxn ≤ bm

xi ≥ 0 (i=1,2, ···,n)

Max. -z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn ≤ b1

a21x1 + a22x2 +···+ a2nxn ≤ b2

： ： am1x1+ am2x2 +···+ amnxn ≤ bm

xi ai ≥ 0 (i=1,2, ···,n)

69

(3) Max. z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn ≤ b1 a21x1 + a22x2 +···+ a2nxn ≤ b2

： ： aj1x1 + aj2x2 +···+ ajnxn ≥ bj ： ： ak1x1 + ak2x2 +···+ aknxn = bk ： ： am1x1+ am2x2 +···+ amnxn ≤ bm xi ≥ 0 (i =1, 2, ···, n), 1 ≤ j, k ≤ m

Max. z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn + s1 = b1 a21x1 + a22x2 +···+ a2nxn + s2 = b2

： ： aj1x1 + aj2x2 +···+ ajnxn - ej + aj = bj ： ： ak1x1 + ak2x2 +···+ aknxn + ak = bk ： ： am1x1+ am2x2 +···+ amnxn + sm = bm

xi , si , ai ≥ 0 (i= 1, 2, ···, n), 1 ≤ j, k ≤ m

70

(4) Max. z = c1x1+ c2x2 +···+ cnxn

s.t. a11x1 + a12x2 +···+ a1nxn ≤ b1

a21x1 + a22x2 +···+ a2nxn ≤ b2

： ： am1x1+ am2x2 +···+ amnxn ≤ bm

xi ≥ 0, xj urs (0 ≤ i , j ≤ n, i ≠ j)

Max. z = c1x1+ c2x2 +···+ cj (x’j - x”j )+···+ cnxn

s.t. a11x1 + a12x2 +···+ cj (x’j - x”j )+···+ a1nxn ≤ b1 a21x1 + a22x2 +···+ cj (x’j - x”j )+···+ a2nxn ≤ b2

： ： am1x1+ am2x2 +···+cj (x’j - x”j )+···+ amnxn ≤ bm

xi ≥ 0, x’j ,x”j ≥ 0 (0 ≤ i , j ≤ n, i ≠ j)

xj = x’j - x”j