1. 数理計画法： Mathematical Programming- ( 最適化問題： Optimization)

目的関数　 Objective function: Z(x1,x2,…,xn)制約条件式Constraints function(s): b1(x1,x2,…,xn)=0

…..bm1(x1,x2,…,xn)=0bm1+1(x1,x2,…,xn) 0≦ …..bm1+m2 (x1,x2,…,xn) 0≦

独立変数： Dependent variable: Z(x1,x2,…,xn)従属変数： Independent variable: x1,x2,…,xn

3. Combinational Optimization * Dynamic Programming * Relaxation Method * Branch and Bound Method

4. Probabilistic Methods * Queuing Theory * Game Theory * Markov Chains

Discrete Variable Continuous Variable (differentiable)

Constrained UnconstrainedLinear Function Non Linear FunctionDeterministic Probabilistic ･･････････　　　　　　　･････････

数理計画法

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1. Linear Programming

* Linear Programming

* Integer Linear Programming

2. Non Linear Programming

* Unconstrained Optimization

* Constrained Optimization

2.1 線形計画法： Linear Programming

Min. Z(xi) = Σ ai ・ xi

s.t. Σb1i ・ xi 　≦ b01

……Σbmi ・ xi 　≦ b0m

x1

x2

輸送問題： Transport Problem

輸送問題： Transport Problem

輸送問題： Transport Problem

xij [ton] ： i 倉庫 から j 百貨店への輸送荷物量Cij [B/ton] : i 倉庫 から j 百貨店への単位重量あたり輸送コストSi [ton] : i 倉庫における在庫量 Cj [ton] : j 百貨店における消費量

目的関数　 Min. Σ Σ Cij ・ xij

制約条件　 s.t. Σ xij 　≦ Si [ton] 　　　　　　　　　 Σ xij ＞＝ Cj [ton]

線形計画問題の条件

1. 目的関数とすべての制約が線形関数2. 制約条件はすべて ≦ , , or = ≧3. 変数は連続変数

c.f. 整数計画法

線形計画法の解法

1. 図形解法 2. 代数解法3. シンプレックス法

Max. Z=60x1+80x2

s.t. x1 ＋ x2 10≦

4x1 ＋ 6x2 48≦

　 0 x≦ 1, 0 x≦ 2

[ 例 ]

x1 ＋ x2 10≦

4x1 ＋ 6x2 48≦

0 ≦ x1

0 ≦ x2

2.1.1 図形解法 (1)

0

8

10

0 10

x1 ＋x2=10

4x1 ＋6x2=48

x1 = 0

x2 = 012

x2

x1

Feasible Region

図形解法 (2)

0

8

10

0

x2

Feasible Region

Z=60x1+80x2

x2=-3/4x1+Z/80

x2=-3/4x1+Z4/80

x2=-3/4x1+Z2/80x2=-3/4x1+Z3/80

x2=-3/4x1+Z1/80

),(y

Z

x

ZZ

=(3,4)

x1 ＋ x2=10

4x1 ＋ 6x2=48

x1 =6x2 =4Z =680

端点： Corner Point(1)

0

8

10

0 10 12

x2

x1

(6,4)

Feasible Region

Corner Point

実行可能領域Feasible Region制約条件を全て満たす点の集合

端点 (2)(2)

x1

x2

xx3=tx=tx1+(1-t)x+(1-t)x2 S : Convex Set

For all t [0,1] and all xx1,xx2 S : Convex Set

1-t1-t

tt

x2x1 1-t t

端点 (3)

端点定理

線形計画問題の最適解は，端点で得られる

もし，端点以外で，最適解が得られるとすると，　　　 x0 =αx1 ＋ βx2 　　　 α ＋ β = １　　 α≧ ０， β≧ ０　　　　Ｚ = f(x0)

　　　　 = ｃ x0

　　　　 = ｃ αx1 ＋ｃ βx2

　　　　 = αf(x1) ＋ βf(x2)

　　　　 = α （ f(x1)−f(x2) ）＋ f(x2)

　　　　 = β （ f(x2)−(x1) ）＋ f(x1)

　　 If f(x1) f(x≦ 2) then f(x1) f(x≦ 0) f(x≦ 2)

If f(x2) f(x≦ 1) then f(x2) f(x≦ 0) f(x≦ 1)

　　　 In either case, f(x0) is not a optimal solution.

端点 (4)

a11x1+a21x2=0a12x1+a22x2=0

a11x1+a21x2 + a31x3=0a12x1+a22x2 + a32x3 =0a13x1+a23x2 + a33x3 =0

凸領域凸領域

x1

x2

xx3=tx=tx1+(1-t)x+(1-t)x2 S

For all t [0,1] and all xx1,xx2 S

1-t1-t

tt

x1

x2

For all xx1,xx2 S

xx3= tx= tx1+ (1- t)x+ (1- t)x2 S

2.2.2 代数解法

スラック変数Slack Variable : x3, x4

Z=60x1+80x2

s.t. x1 ＋ x2 10≦

4x1 ＋ 6x2 48≦

　 0 x≦ 1, 0 x≦ 2

Z=60x1+80x2

s.t. x1 ＋ x2 ＋ x3 = 10

4x1 ＋ 6x2 ＋ x4 =48

　 0 x≦ 1, 0 x≦ 2 , 0 x≦ 3 , 0 x≦ 4

0

8

10

0 10

x1 ＋x2=10

4x1 ＋6x2=48

x1 = 0

x2 = 0

12

x2

x1(0, 0,10,48)

(x1 , x2 , x3 , x4 )

(10, 0, 0,8)

(12, 0,-2, 0)

(6, 4, 0, 0)

(0,10,0,-12)

(0,8,2,0)

x1 ＋ x2 ＋ x3 = 10

4x1 ＋ 6x2 ＋ x4 =48

標準系

xi ＝ 0xj ＝ 0

‥xk ＝ 0

n

非基底解　： Non-Basic Solutions非基底変数： Non-Basic Variables

xi’ ＝ c1

xj’ ＝ c2

‥xk’ ＝ cm

m

基底解　： Basic Solutions 　基底変数　： Basic Variables

ａ 11x1 + ａ 12x2 + ‥‥ + ａ 1nxn ＋ xn+1 ＝ ｂ 1

ａ 21x1 + ａ 22x2 + ‥‥ + ａ 2nxn ＋ xn+2 ＝ ｂ2

‥‥ ‥‥

ａ m1x1 + ａ m2x2 + ‥‥ + ａ mnxn ＋ xn+m ＝ ｂ m

0

8

10

0 10 12

x2

x1(0, 0,10,48) (10, 0, 0,8)

(12, 0,-2, 0)

(6, 4, 0, 0)

(0,10,0,-12)

(0,8,2,0)

Ｘ1 Ｘ2 Ｘ3 Ｘ4 Ｚ① 0 0 10 48 0② 0 10 0 -12 800③ 0 8 2 0 640④ 10 0 0 8 600⑤ 12 0 -2 0 720⑥ 6 4 0 0 680

①

②

③

④ ⑤

⑥

非基底変数 xi =0基底変数 xi =c=0

x1 ＋ x2 ＋ x3 = 10

4x1 ＋ 6x2 ＋ x4 =48

2.2.3 シンプレックス法Simplex Method

シンプレックス法の概念

Z

Z(x11,x21,x31)

Z(x12,x22,x32)

≦

Z(x15,x25,x35)

隣接端点 (1)

a41x1+a42x2+a43x3-b1 =0

a31x1+a32x2+a33x3-b3=0a21x1+a22x2+a23x3-b2=0

a11x1+a12x2+a13x3-b1=0

a41x1+a42x2+a43x3-b1 =0

a21x1+a22x2+a23x3-b2=0a31x1+a32x2+a33x3-b3=0 a21x1+a22x2+a23x3-b2=0

a31x1+a32x2+a33x3-b3=0

a11x1+a12x2+a13x3-b1=0

(2)

(1)

(3)

(4)

隣接端点 (2)

(1) a21x1+a22x2+a23x3+x4 =b2

(2) a21x1+a22x2+a23x3 + x5 =b2

(3) a31x1+a32x2+a33x3 +x6 =b3

(4) a41x1+a42x2+a43x3 +x7 =b4

x1=a1

x2=a2

x3=a3

x4=a4

x5=0x6=0x7=0

x1=c1

x2=c2

x3=c3

x4=0x5=0x6=0

x7= c7

Basic Non-Basic Variables Variables

Basic Non-Basic Variables Variables

(2)

(1)

(3)

(4)

a21x1+a22x2+a23x3-b2=0

隣接端点 (3)

n dimensions * the number of non-basic variables at the corner points is n. * n-1 non-basic variables are common between the adjacent corner point

0

8

10

0 10 12

x2

x1(0, 0,10,48) (10, 0, 0,8)

(6, 4, 0, 0)

① ④

⑥

Non-Basic Variables xi =0Basic Variables xi =c=0

x1 ＋ x2 ＋ x3 = 10

4x1 ＋ 6x2 ＋ x4 =48

隣接端点 (4)

Basic VariablesNon-Basic Variables

xn+2 → Non-Basic Variables, x1 → Basic Variables

x1 + a12’x2 + + a‥‥ 1n’xn + xn+1 + a1n’xn+2 = b1 ’ 　

(1) ’

a22’x2 + + a‥‥ 2n’xn + a1n’xn+2 = b2 ’ (2) ’

‥‥ ‥‥ am2’x2 + +a‥‥ mn’xn +a1n’xn+2 +xn+m= bm ’ (m) ’

a11x1 + a12x2 + + a‥‥ 1nxn ＋ xn+1 ＝ b1 　 (1)

a21x1 + a22x2 + + a‥‥ 2nxn ＋ xn+2 ＝ b2 (2)

‥‥ ‥‥am1x1 + am2x2 + + a‥‥ mnxn 　　　＋ xn+m ＝ bm (m)

隣接端点 (5)

a11x1 + a12x2 + + a‥‥ 1nxn ＋ xn+1 ＝ b1 　 (1)

a21x1 + a22x2 + + a‥‥ 2nxn ＋ xn+2 ＝ b2 (2)

‥‥ ‥‥am1x1 + am2x2 + + a‥‥ mnxn ＋ xn+m ＝ bm (m)

(2)/a21

x1 + a12/a21x2 + + a‥‥ 1n /a21xn ＋ 1 /a21xn+1 ＝ b1 /a21 (2)’

(1) － a11×(2)’

　　 0+(a22-a11a12/a21)x2 + + (a‥ 2n-a11a1n /a21)xn － a11 /a21xn+1 ＋ xn+2 ＝ b2 － a11b1 /a21 (1)’

(m) － am1×(1)’

‥

0+(am2-am1a12/a21)x2 + + (a‥ mn-am1a1n /a21)xn － am1 /a21xn+1 ＋ xn+m ＝ bm － am1b1 /a21 (m)’

a12’x2 + + a‥‥ 1n’xn + xn+1 + a1n’xn+2 = b1 ’ 　 (1) ’

x1 + a22’x2 + + a‥‥ 2n’xn + a1n’xn+2 = b2 ’ (2) ’

‥‥ ‥‥

am2’x2 + +a‥‥ mn’xn +a1n’xn+2 +xn+m = bm ’ (m) ’

隣接端点 (6)

a12’x2 + + a‥‥ 1n’xn + xn+1 + a1n’xn+2 = b1 ’ 　 (1) ’

x1 + a22’x2 + + a‥‥ 2n’xn + a1n’xn+2 = b2 ’ (2) ’

‥‥ ‥‥

am2’x2 + +a‥‥ mn’xn +a1n’xn+2 +xn+m = bm ’ (m) ’

Z=c1 x1+ c2 x2+ + cn xn

= c1(b2 ’ - a22’x2 - - a‥‥ 2n’xn - a1n’xn+2 ) + c2 x2+ + cn xn

= c1 b2 ’ +(c2 - a22’ ) x2 +(‥ c2 - a22’ ) xn - a1n’xn+2

例Z=60x1+80x2

s.t. x1 ＋ x2 10≦4x1 ＋ 6x2 48≦

　 0 x≦ 1, 0 x2≦

Z=60x1+80x2

s.t. x1 ＋ x2 ＋ x3 = 10 (1)

4x1 ＋ 6x2 ＋ x4 =48 (2)

　 0 x≦ 1, 0 x≦ 2 , 0 x≦ 3 , 0 x≦ 4

0

8

0①

(0, 0,10,48)

① x1 ,x2 x3 , x4

③ x1 , x4 x2 , x3

Non-Basic Variables Basic Variables

1/3x1 ＋ x3 ｰ 1/6x4 = 2 (1’)

2/3x1 ＋ x2 ＋ 1/6x4 = 8 (2’)

(0,8,2,0)

③

1/3x1 ＋ x3 ｰ 1/6x4 =2 (1’)(1’)=(1) - (2’)

Z=60x+80x2=60x1+80 (8-2/3x1-1/6x4)

(2’)=(2)/a22

2/3x1 ＋ x2 ＋ 1/6x4 =8 (2’)

Z = 20/3 x1 – 40/3x4 + 640

Simplex Method Phase I

n 次元 * ひとつの端点には，　　個の隣接端点がある． どの端点を選ぶか？ （どの変数が次の基底解になるか？）

Z

Z

),,,,()(21 ni x

ZxZ

xZ

xZ

xZ

),,,,(21 ni x

ZxZ

xZ

xZ

Max

Simplex Method Phase I (cont.)

0

8

0 10

x2

(0, 0,10,48) (10, 0, 0,8)

(0,8,2,0)

①

③

④

Z=60x1+80x2

)80,60(),()(21

xZ

xZ

xZ

80)80,60( Max

x2 is selected as a new basic variable

),(21 x

ZxZ

Z

Simplex Method Phase II

ｍ個の制約条件式 * m 個の基底変数 . m 個の基底変数の内，どの基底変数が非基底変数になるか？

0

8

0 10

x2

(0, 0,10,48)

(0,8,2,0)

①

③

(0,10,0,-12)②x3, x4

x28 10

48

-12

0

10 x3

x4

x1 ＋ x2 ＋ x3 = 10

4x1 ＋ 6x2 ＋ x4 =48

x1 = 0

x2 ＋ x3 = 10

6x2 ＋ x4 =48

x4 is selected as a new non-basic variable ③ is selected as a corner point

シンプレックス法の解法手順

Step.1 初期実行可能解を求める .

Step.2 目的関数を改善する隣接端点を調べる無い場合：その点が最適解ある場合： Step 3 へ

Step.3 目的関数を最も改善する方向を決定する．　　　　（どの変数が次の基底解になるかを決める）

Step.4 制約条件を満たす中で，目的関数を最も改善する端点を決定する．　　　　（どの変数が次の非基底解になるかを決める）　 Step 2 へ

解なし［実行可能領域無し］

x2

x1 ＋ x2 10≧2x1 ＋ x2 8≦0 x≦ 1

0 x2≦

4 10

10

8

x1

How to solve LP with Excel?

Spread Sheet

Example

A company produces two products P1 and P2 with three materials M1, M2, and M3. 1kg of M1, 2kg of M2, and 3 kg of M3 are used to produce 1 kg of P1. Similarly, 7kg of M1, 4kg of M2, and 2 kg of M3 are used to produce 1 kg of P2. However, they have only 140kg of M1, 100kg of M2, and 120 kg of M3. Obtain the best combination of product P1 and P2, when the benefits of 1kg of P1

and that of P2 are 3US$ and 5US$ respectively?

Z=3x1+5x2

s.t. x1 ＋ 7x2 140≦

2x1 ＋ 4x2 100≦

3x1 ＋ 2x2 120≦

　 0 x≦ 1, 0 x≦ 2

Z=3x1+5x2

s.t. x1 ＋ 7x2 ＋ x3 =140

2x1 ＋ 4x2 ＋ x4 =100

3x1 ＋ 2x2 ＋ x5 =120

0 x≦ 1, 0 x≦ 2 ,0 x≦ 3, 0 x≦ 4 , 0 x≦ 5

Simplex Table(1)

Z=3x1+5x2

s.t. x1 ＋ 7x2 ＋ x3 =140

2x1 ＋ 4x2 ＋ x4 =100

3x1 ＋ 2x2 ＋ x5 =120

0 x≦ 1, 0 x≦ 2 ,0 x≦ 3, 0 x≦ 4 , 0 x≦ 5

Marginal

B.V. Z x1 x2 x3 x4 x5

(1) Z 0 1 -3 -5 0 0 0

I (2) x3 140 0 1 7 1 0 0 140/7 = 20

(3) x4 100 0 2 4 0 1 0 100/4 = 25

(4) x5 120 0 3 2 0 0 1 120/2=60

Variable

Z - 3x1 - 5x2 =0

Simplex Table(2)

Marginal

Z x1 x2 x3 x4 x5

(1) Z 0 1 -3 -5 0 0 0

I (2) x3 140 0 1 7 1 0 0 140/7 = 20

(3) x4 100 0 2 4 0 1 0 100/4 = 25

(4) x5 120 0 3 2 0 0 1 120/2=60

(5) Z 100 1 -16/7 0 5/7 0 0 (1) - (6)× (-5)

II (6) x2 20 0 1/7 1 1/7 0 0 20/ (1/7) = 140 (2)÷7

(7) x4 20 0 10/7 0 -4/7 1 0 20/ (10/7) =14 (3)-(6)×4

(8) x5 80 0 19/7 0 -2/7 0 1 80/ (19/7)≒ 30 (4)-(6)×2

Variable

Variable

Basic

Simplex Table(3)

Marginal

Z x1 x2 x3 x4 x5

(5) Z 100 1 -16/7 0 5/7 0 0 (1) - (6)× (-5)

II (6) x2 20 0 1/7 1 1/7 0 0 20/ (1/7) = 140 (2)÷7

(7) x4 20 0 10/7 0 -4/7 1 0 20/ (10/7) =14 (3)-(6)×4

(8) x5 80 0 19/7 0 -2/7 0 1 80/ (19/7)≒ 30 (4)-(6)×2

(9) Z 132 1 0 0 -1/5 8/5 0 (5) - (11)×(-16/7)

III (10) x2 18 0 0 1 1/5 -1/10 0 18/ (1/5) =90 (6) - (11)×1/7

(11) x3 14 0 1 0 -2/5 7/10 0 (7) ÷10/7

(12) x5 42 0 0 0 4/5 -19/10 1 42/ (4/5) = 52.5 (8)-(11)×19/7

(13) Z 142.5 1 0 0 0 9/8 1/4 (9)-(16)×(-1/5)

IV (14) x2 7.5 0 0 1 0 3/8 -1/4 (10)-(16)×1/5

(15) x1 35 0 1 0 0 -1/4 1/2 (11) - (16)×(-2/5)

(16) x3 52.5 0 0 0 1 -19/8 5/4 (12)÷ 4/5

Variable

VariableBasic

1

Shadow Price (Marginal Value)

Z=3x1+5x2

s.t. x1 ＋ 7x2 ＋ x3 =140

2x1 ＋ 4x2 ＋ x4 =100

3x1 ＋ 2x2 ＋ x5 =120

0 x≦ 1, 0 x≦ 2 ,0 x≦ 3, 0 x≦ 4 , 0 x≦ 5

(x*, x2*, x3

*, x4 *, x5

*)=(35,7.5, 52.5, 0, 0)

Transport Problem

Transport Problem

Transport Problem 　 (3 depots(i=1..3) & 4 customers(j=1..4))

xij [ton] ： Amount of Transported Cargo from i warehouse to j department

Cij [B/ton] : Transport Cost per ton from i warehouse to j department

Si [ton] : Amount of Stock at i warehouse Cj [ton] : Amount of Consumption at j department

Transportation Cost Cij

1 2 3 41 2 1 1 32 1 1 2 33 1 2 3 1

j(S1 , S2 , S3 )=(100,200,150)(C1 , C2 , C3 , C4 )=(100,60,50,80)

Transport Problem 　 (3 depots(i=1..3) & 4 customers(j=1..4))

Total Cost ΣCij xij= 2x11 +x12 +x13 +3x14

+x21 +x22 +2x23 +3x24

+x31 +2x32 +3x33 +x34

x11 +x21 +x31 C≧ 1 = 100x12 +x22 +x32 ≧ C2 = 60x13 +x23 +x33 ≧ C3 = 50x14 +x24 +x34 ≧ C4 = 80

x11 +x12 +x13 +x14 S≦ 1 =100

x21 +x22 +x23 +x24 S≦ 2 = 200

x31 +x32 +x33 +x34 S≦ 3 = 150

0 50 50 030 10 0 070 0 0 80

xij

Transport Problem

xij [ton] ： Amount of Transported Cargo from i warehouse to j department

Cij [B/ton] : Transport Cost per ton from i warehouse to j department

Si [ton] : Amount of Stock at i warehouse Cj [ton] : Amount of Consumption at j department

Min. Σ Σ Cij ・ xij

s.t. Σ xij 　≦ Si [ton] Σ xij C≦ j [ton]

DP is a mathematical technique used for the optimization of multistage decision process. In this technique, the decisions that affect the process are optimized in stage rather than simultaneously. This is done by dividing the original decision problem into small sub-problems that can be handled much more efficiently from a computational standpoint. DP is a systematic procedure for determining the combination of decisions that maximizes overall effectiveness or minimizes overall disutility. It is based on the principle of optimality enunciated by Bellman(1957).

Dynamic Programming(1)

Bellman's Principle of Optimality: An optimal policy has the property that whatever the

initial state and the initial decisions are, the remaining decisions must constitute an optimal policy with regard to the state resulting from the first decision.

Dynamic Programming(2)

○D.P. can be applied to problems that are ･ are non-linear. ･ have non-convex feasible region. ･ have discontinuous variables.

×D.P. is limited to dealing with problems with relatively few constraints

Example: Shortest Path

OD

DP complements LP

Queuing Theory(1)

Arrival ServiceAverage Arrival Rate 　 λ Average Arrival Rate μProbability of k arrivals Probability of k services in t hours in t hours

Demand vs. Supply

Pn: Probability that there are n customers in the system Pn= (1- λ/μ)(λ/μ) n

Lq: Average number of customers in the queue Lq= λ 2/(μ(μ- λ))

Wq: Average time a customer spends in the queueWq= λ/(μ(μ- λ))

Pw: Probability that an arriving customer must for services

Pw= λ /μ

!

et)( λtk

k

!

et)( tk

k

Queuing Theory(2)

Arrival ServiceAverage Arrival Rate 　 λ Average Arrival Rate μProbability of k arrivals Probability of k services in t hours in t hours

Demand vs. Supply

Pn: Probability that there are n customers in the system Pn= (1- λ/μ)(λ/μ) n

Lq: Average number of customers in the queue Lq= λ 2/(μ(μ- λ))

Wq: Average time a customer spends in the queueWq= λ/(μ(μ- λ))

Pw: Probability that an arriving customer must for services

Pw= λ /μ

!

et)( λtk

k

!

et)( tk

k

0

10

20

30

40

50

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Lq

λ /μ

Queuing Theory(3)

0

1

2

3

4

5

6

7

8

9

10

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

λ /μ

Lq

Convex Function(Strictly Convex Function)

0 dx

Z(x)d 3.

)z(x )x-(xdx

)dZ(x) Z(x2.

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Hessian Matrix

a. One variable

b. Multi variables

Report

1. Make a transportation problem on the condition that each number of shippers and customers is more than 3.

2. Solve it with Excel.

3. Make “what-if ” analysis.

By the dead line, 25th July.

Assignment