Chapter 1
Example 1.1
The Apex Television Company has to decide on the number of 27- and 20-inch
sets to be produced at one of its factories. Market research indicates that at most
40 of the 27-inch sets and 10 of the 20-inch sets can be sold per month. The
maximum number of work-hours available is 500 per month. A 27-inch set
requires 20 work-hours and a 20-inch set requires 10 work-hours. Each 27-inch set
sold produces a profit of $120 and each 20-inch set produces a profit of $80. A
wholesaler has agreed to purchase all the television sets produced if the numbers
do not exceed the maxima indicated by the market research.
(a) Formulate a linear programming model for this problem.
The decisions that need to be made are the number of 27-inch and 20-inch TV sets
to be produced per month by the Apex Television Company. Therefore, the
decision variables for the model are
x1 = number of 27-inch TV sets to be produced per month,
x2 = number of 20-inch TV sets to be produced per month.
Also let
Z = total profit per month.
The model now can be formulated in terms of these variables as follows.
The total profit per month is Z = 120 x1 + 80 x2.
The resource constraints are:
(1) Number of 27-inch sets sold per month: x1 40
(2) Number of 20-inch sets sold per month: x2 10
(3) Work-hours availability: 20 x1 + 10 x2 500.
Nonnegativity constraints on TV sets produced: x1 0, x2 0
With the objective of maximizing the total profit per month, the LP model for this
problem is
Maximize Z = 120 x1 + 80 x2
subject to
x1
 40
x2  10
20 x1 + 10 x2 500
and x1 0, x2 0
(b) Use the graphical method to solve this model.
Example 1.2
Dwight is an elementary school teacher who also raises pigs for supplemental
income. He is trying to decide what to feed his pigs. He is considering using a
combination of pig feeds available from local suppliers. He would like to feed the
pigs at minimum cost while also making sure each pig receives an adequate supply
of calories and vitamins. The cost, calorie content, and vitamin content of each feed
are given in the table below.
Contents
Calories (per pound)
Vitamins (per pound)
Cost (per pound)
Feed Type A
800
140 units
$0.40
Feed Type B
1,000
70 units
$0.80
Each pig requires at least 8,000 calories per day and at least 700 units of vitamins.
A further constraint is that no more than one-third of the diet (by weight) can
consist of Feed Type A, since it contains an ingredient which is toxic if consumed in
too large a quantity.
(a) Formulate a linear programming model for this problem.
Let A and B be the quantity (pounds) of Feed Type A and Feed Type B,
respectively, used per day. Also let Z be the total daily cost of the feed per pig.
Then, the daily cost is
Z = $0.4 A + $0.8 B.
The constraints on the minimum daily requirements of calories and vitamins are
(1) Calories requirement:
800 A + 1000 B 8,000.
(2) Vitamins requirement: 140 A + 70 B  700.
Also, Dwight needs to avoid using too much of Feed Type A because of the
toxic ingredient in it. The toxic constraint is A  1/3 (A+B), which
reduces to 2/3 A - 1/
3B≤0.
Nonnegativity constraints: A 0, B 0.
The resulting linear programming model for this problem is
Minimize Z = 0.4 A + $0.8 B,
subject to
800 A + 1000 B  8000
140 A + 70 B  700
2/3 A - 1/3 B 
0
and A 0, B 0.
(b) Use the graphical method to solve this model. What is the resulting daily
cost per pig?
As shown below, the optimal solution is (A, B) = (20/7, 40/7). The resulting
daily cost per pig is Z = 40/7 = $5.71.
Example 1.3
A furniture store has set aside 800 square feet to display its sofas and chairs. Each
sofa utilizes 50 sq. ft. and each chair utilizes 30 sq. ft. At least five sofas and at least
five chairs are to be displayed.
a. Write a mathematical model representing the store's constraints.
b. Suppose the profit on sofas is $200 and on chairs is $100. On a given day, the
probability that a displayed sofa will be sold is .03 and that a displayed chair
will be sold is .05. Mathematically model each of the following objectives:
1. Maximize the total pieces of furniture displayed.
2. Maximize the total expected number of daily sales.
3. Maximize the total expected daily profit.
Solution:
Let s and c be the number of sofa and chair respectively.
a. 50s + 30c < 800
s> 5
c> 5
b. (1) Max s + c
(2) Max 0.03s + 0.05c
(3) Max 6s + 5c
Example 1.4
Maxwell Manufacturing makes two models of felt tip marking pens. Requirements
for each lot of pens are given below.
Plastic
Ink Assembly
Molding Time
Fliptop Model Tiptop Model
3
4
5
4
5
2
Available
36
40
30
The profit for either model is $1000 per lot.
a. What is the linear programming model for this problem?
b. Find the optimal solution.
c. Will there be excess capacity in any resource?
Solution:
a. Let F = the number of lots of Fliptip pens to produce
Let T = the number of lots of Tiptop pens to produce
Max 1000F + 1000T
s.t. 3F + 4T < 36
5F + 4T < 40
5F + 2T < 30
F,T > 0
b.
T
15
10
5
F
0
0
5
10
15
The complete optimal solution is：F = 2, T = 7.5, Z = 9500
c.There is an excess of 5 units of molding time available.
Chapter 2
Example 2.1
Consider the following linear programming model.
Maximize Z = 3x1 + 2 x2, subject to
x1
≤4
x1 + 3x2
≤15
2x1 + x2
≤10
and x1 ≥0,x2 ≥0.
(a) Use graphical analysis to identify all the corner-point solutions for this
model. Label each as either feasible or infeasible.
The graph showing all the constraint boundary lines and the corner-point
solutions at their intersections is shown below.
The exact value of (x1, x2) for each of these corner-point solutions (A, B, ..., I)
can be identified by obtaining the simultaneous solution of the corresponding
two constraint boundary equations. The results are summarized as...
Corner-point solutions
A
B
C
D
E
F
G
H
I
(x1, x2)
(0, 5)
(0,10)
(3, 4)
(4, 11/3)
(4, 2)
(4, 0)
(5, 0)
(15, 0)
(0, 0)
Feasibility
Feasible
Infeasible
Feasible
Infeasible
Feasible
Feasible
Infeasible
Infeasible
Feasible
(b) Calculate the value of the objective function for each of the CPF solutions.
Use this information to identify an optimal solution.
The objective value of each corner-point feasible solution is calculated in the
following table:
Corner-point
feasible solutions
A
C
E
F
I
(x1, x2)
Objective Value Z
(0, 5)
(3, 4)
(4, 2)
(4, 0)
(0, 0)
3*0+2*5 = 10
3*3+2*4 = 17
3*4+2*2 = 16
3*4+2*0 = 12
3*0+0*0 = 0
Since point C has the largest value of Z, (x1, x2) = (3, 4) must be an optimal
solution.
(c) Use the simplex method to get the solution and to identify which sequence of
CPF solutions would be examined by the simplex method to reach an
optimal solution.
Therefore, the sequence of CPF solutions examined by the simplex method would
be I → F→ E→ C.
Example 2.2
Consider the following linear programming model.
Maximize Z = x1 + 2x2,
subject to
6x1 － 2x2 ≤ 3
2x1 ＋ 3x2 ≤ 6
x1 + x2 ≤ 3
and
x1 ≥0,x2 ≥0.
(a) Use graphical analysis to identify all the corner-point solutions for this model.
Label each as either feasible or infeasible.
(b) Calculate the value of the objective function for each of the CPF solutions. Use
this information to identify an optimal solution.
(c) Use the simplex method to get the solution and to identify which sequence of
CPF solutions would be examined by the simplex method to reach an optimal
solution.
Solution:
(a)
x1 + x2 ＝ 3
3
6x1－ 2x2 ＝ 3
2
x1 + 2x2 =0
Feasible
Region
1
2x1 ＋ 3x2 = 6
0
x1
0
1
2
(a)
corner-point feasibility
(0, 0)
feasible
(0.5, 0)
feasible
(21/22, 15/11) feasible
(0, 2)
feasible
(c) 1st tableau
(b)
obj. value
0
0.5
3.68
4
1
x1
6
2
1
0
1
2
x2
-2
3
1
0
2
0 0 0
sl1 sl2 sl3
bi
i
－
1 0 0
3
0 1 0
6
2
0 0 1
3
3
0 0 0 ∑icBibi= 0
0 0 0
1
x1
2
x2
0 0 0
sl1 sl2 sl3
bi
1 cB1= 0 xB1= sl1 22/3
0
1 2/3 0
7
2 cB2= 2 xB2= x2
3 cB3= 0 xB3= sl3
fj
2/3
1/3
4/3
1
0
2
-1/3
0
0 1/3 0
2
0 -1/3 1
1
0 2/3 0 ∑icBibi= 4
0 -2/3 0
i
1
2
3
Max Cj
cB
xB
cB1= 0 xB1= sl1
cB2= 0 xB2= sl2
cB3= 0 xB3= sl3
fj
Cj-fj
3
2nd tableau (final)
i
Max
cB
Cj
xB
Cj-fj
Corner point (0, 0) → Corner point (0, 2)
i
Chapter 3
Example 3.1
The Fagersta Steelworks currently is working two mines to obtain its iron ore. This
iron ore is shipped to either of two storage facilities. When needed, it then is shipped
ont
ot
h
ec
ompa
ny
’
ss
t
e
e
lp
l
a
nt
.Thedi
a
gr
a
m be
l
owde
p
i
c
t
st
h
i
sdi
s
t
r
i
bu
t
i
o
nne
t
wor
k,
where M1 and M2 are the two mines, S1 and S2 are the two storage facilities, and P is
the steel plant. The diagram also shows the monthly amounts produced at the mines
and needed at the plant, as well as the shipping cost and the maximum amount that
can be shipped per month through each shipping lane.
$2,000/ton
30 tons max.
40 tons
produced M1
S1
$400/ton
70 tons max.
$1,700/ton
30 tons max.
P
$1,600/ton
50 tons max.
60 tons
M2
produced
$1,100/ton
50 tons max.
S2
100 tons
needed
$800/ton
70 tons max.
Management now wants to determine the most economic plan for shipping the iron
ore from the mines through the distribution network to the steel plant.
(a) Formulate a linear programming model for this problem.
The decision variables are defined as follows:
xm1-s1 : number of units (tons) shipped from Mine 1 to Storage Facility 1,
xm1-s2 : number of units (tons) shipped from Mine 1 to Storage Facility 2,
xm2-s1 : number of units (tons) shipped from Mine 2 to Storage Facility 1,
xm2-s2 : number of units (tons) shipped from Mine 2 to Storage Facility 2,
xs1 -p : number of units (tons) shipped from Storage Facility 1 to the Plant,
xs2 -p : number of units (tons) shipped from Storage Facility 2 to the Plant.
The total shipping cost is:
Z = 2000 xm1-s1 + 1700 xm1-s2 + 1600 xm2-s1 + 1100 xm2-s2 + 400 xs1-p + 800 xs2-p
The constraints we need to consider are:
(1) Supply constraint on M1 and M2:
xm1-s1 + xm1-s2 = 40
xm2-s1 + xm2-s2 = 60
(2) Conservation-of-flow constraint on S1 and S2:
xm1-s1 + xm2-s1 - xs1-p = 0
xm1-s2 + xm2-s2 - xs2-p = 0
(3) Demand constraint on P:
xs1-p + xs2-p = 100
(4) Capacity constraints:
xm1-s1 30, xm1-s2 30
xm2-s1 50, xm2-s2 50
xs1-p 70, xs2-p 70
(5) Nonnegativity constraints:
xm1-s1 0, xm1-s2 0
xm2-s1 0, xm2-s2 0
xs1-p 0, xs2-p 0
The resulting linear programming model for this problem is:
Maximize Z = 2000 xm1-s1 + 1700 xm1-s2 + 1600 xm2-s1 + 1100 xm2-s2
+ 400 xs1-p + 800 xs2-p,
subject to
xm1-s1 + xm1-s2 = 40
xm2-s1 + xm2-s2 = 60
xm1-s1 + xm2-s1 - xs1-p = 0
xm1-s2 + xm2-s2 - xs2-p = 0
xs1-p + xs2-p = 100
xm1-s1 30, xm1-s2 30
xm2-s1 50, xm2-s2 50
xs1-p 70, xs2-p 70
and xm1-s1 0, xm1-s2 0, xm2-s1 0, xm2-s2 0, xs1-p 0, xs2-p 0
Example 3.2
Minimize f = －1x1 + 2x2 － 3x3 subject to
x1 + x2 ＋ x3 = 6
－1x1 + x2 ＋ 2x3 = 4
2x2 ＋ 3x3 = 10
x3 2
and
x1, x2, x3 0
Find the optimal solution by big M and two phase methods.
Solution:
The first tableau of big M method:
Min
i
1
2
3
Cj
-1
cB
xB
x1
cB1= M xB1=a1 1
cB2= M xB2=a2 -1
cB3= M xB3=a3 0
4 cB4= 0 xB3=s1
fj
Cj-fj
2
x2
1
1
2
-3
x3
1
2
3
M
a1
1
0
0
M M
a2 a3
0 0
1 0
0 1
0
s1
0
0
0
bi
6
4
10
0
0
1
0
0
0
1
2
0
4M
6M
M
M M
0
0
0
0
-1 2-4M -3-6M
0
i
∑icBibi= 20M+2
The first tableau of Phase I in two phase method:
Min
C’
j
0
0
0
1
1
1
0
xB
x1
x2
x3
a1
a2 a3
s1
bi
i
1
1
1
1
0
0
0
6
6
2 cB2= 1 xB2=a2 -1
1
2
0
1
0
0
4
2
3 cB3= 1 xB3=a3
0
2
3
0
0
1
0
10
10/3
4 cB4= 0 xB3=s1
0
0
1
0
0
0
1
2
2
f
’
j
0
4
6
1
1
1
0
∑icBibi= 20
C’
’
j-f
j
0
-4
-6
0
0
0
0
i
cB
1 cB1= 1 xB1=a1
The final tableau of Phase I in two phase method:
Min
C’
j
0
0
0
1
1
1
0
xB
x1
x2
x3
a1
a2
a3
s1
bi
1 cB1= 0 xB1=x1
1
0
0
1/2 -1/2 0 1/2
2
2 cB2= 0 xB2=x2
0
1
0
1/2 1/2
0 -3/2
2
3 cB3= 1 xB3=a3
0
0
0
-1
-1
1
0
0
4 cB4= 0 xB3=x3
0
0
1
0
0
0
1
2
f
’
j
0
0
0
-1
-1
1
0
∑icBibi= 0
C’
’
j-f
j
0
0
0
2
2
0
0
i
cB
**extra constraint
The first & also final tableau of Phase II in two phase method:
Min
Cj
-1
2
-3
0
xB
x1
x2
x3
s1
bi
1 cB1= -1 xB1=x1
1
0
0
1/2
2
2 cB2= 2 xB2=x2
0
1
0
-3/2
2
4 cB4= -3 xB3=x3
0
0
1
1
2
fj
-1
2
-3
-7/2
Cj-fj
0
0
0
7/2
i
cB
i
∑icBibi= -4
i
Example 3.3
A&C Distributors is a company that represents many outdoor products companies and
schedules deliveries to discount stores, garden centers, and hardware stores. Currently,
scheduling needs to be done for two lawn sprinklers, the Water Wave and Spring
Shower models. Requirements for shipment to a warehouse for a national chain of
garden centers are shown below.
Month
Shipping
Capacity
March
8000
April
7000
May
6000
Product
Water Wave
Spring Shower
Water Wave
Spring Shower
Water Wave
Spring Shower
Requirement
3000
1800
4000
4000
5000
2000
Unit
Cost
to Ship
.30
.25
.40
.30
.50
.35
Per Unit
Inventory
Cost
.06
.05
.09
.06
.12
.07
Let Sij be the number of units of sprinkler i shipped in month j, where i = 1 or 2, and j
= 1, 2, or 3. Let Wij be the number of sprinklers that are at the warehouse at the end of
a month, in excess of the requirement.
(a.) Write the portion of the objective function that minimizes shipping costs.
(b.) An inventory cost is assessed against this ending inventory. Give the portion of
the objective function that represents inventory cost.
(c.) There will be three constraints that guarantee, for each month, that the total
number of sprinklers shipped will not exceed the shipping capacity. Write these
three constraints.
(d.) There are six constraints that work with inventory and the number of units shipped,
making sure that enough sprinklers are shipped to meet the requirements. Write
these six constraints.
Solutions
(a.) Min 0.3S11 + 0.25S21 + 0.40S12 + 0.30S22 + 0.50S13 + 0.35S23
(b.) Min 0.06W11 +0.05W21 +0.09W12 +0.06W22 +0.12W13 +0.07W23
(c.) S11 + S21 8000
S12 + S22 7000
S13 + S23 6000
(d.) S11 - W11 = 3000
S21 - W21 = 1800
W11 + S12 - W12 = 4000
W21 + S22 - W22 = 4000
W12 + S13 - W13 = 5000
W22 + S23 - W23 = 2000
Example 3.4
(a.) Find the optimal solution of (a.) in Example 3.3 by the big-M method.
(b.) Find the optimal solution of (b.) in Example 3.3 by the two phases method. (寫到
Phase II 的 1st tableau 就可以。)
Solution for (a.)
Min 0.3S11 + 0.25S21 + 0.40S12 + 0.30S22 + 0.50S13 + 0.35S23
S11 + S21 8000
S12 + S22 7000
S13 + S23 6000
S11 - W11 = 3000
S21 - W21 = 1800
W11 + S12 - W12 = 4000
W21 + S22 - W22 = 4000
W12 + S13 - W13 = 5000
W22 + S23 - W23 = 2000
S11,~, S23, W11,~, W23 0
S11 + S21 +s1 = 8000
S12 + S22 + s2 = 7000
S13 + S23 +
s3 = 6000
S11 - W11 +
a1 = 3000
S21 - W21 +
a2= 1800
W11 + S12 - W12 +
a3= 4000
W21 + S22 - W22 +
a4 = 4000
W12 + S13 - W13 +
a5 = 5000
W22 + S23 - W23 +
a6 = 2000
S11,~, S23, W11,~, W23, s1,~, s3, a1,~, a6 0
(a) Simplex 1st tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0 xB1=s1
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0 8000 8000
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=M xB3=a1
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000
-
5 cB5=M xB3=a2
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800 1800
6 cB6=M xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000
-
7 cB7=M xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=M xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
M M M M M M
0
0 -M 0
0 -M 0
0
0
M M M M M M
0
0
0
0
0
fj
Cj-fj
0.3 0.4 0.5 0.25 0.3 0.35
0
-M -M -M -M -M -M
M
0
M
0
0
0
0
0
0
Simplex 2nd tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0 xB1=s1
1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
-1
0
0
0
0 6200 6200
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=M xB3=a1
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000
-
5 cB5=0.25 xB3=S21 0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=M xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000
-
7 cB7=M xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000 4000
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=M xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
M M M 0.25 M M
0
0 -M +M 0 -M 0
0
0
M 0.25 M M M M
0
0
0 +M 0
fj
Cj-fj
0.3 0.4 0.5
0.3 0.35
0
0
-M -M -M
-M -M
-0.25
0
0.25
M -M 0
M
0
-0.25
0
0
0
Simplex 3rd tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0 xB1=s1
1
0
0
0
-1
0
0
0
0
0
1
0
1
0
0
0
-1
0
-1
0
0 2200 2200
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=M xB3=a1
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000
-
5 cB5=0.25 xB3=S21 0
0
0
1
1
0
0
0
0
0
-1
0
0
0
0
0
1
0
1
0
0 5800
-
6 cB6=M xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000
-
7 cB7=0 xB3=W21 0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=M xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000 2000
M M M 0.25 0.25 M
0
0 -M 0 +M -M 0
0
0
M 0.25 M 0.25 M M
0
0
0 +M 0 +M 0
fj
Cj-fj
0.35
0.3 0.4 0.5
0 0.05
0
-M
-M -M -M
-0.25
0
M
0
0.25
-M
M
0
-0.25
-0.25
0
Simplex 4th tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0 xB1=s1
1
0
0
0
-1 -1
0
0
0
0
0
1
1
0
0
0
-1
0
-1
0
-1 200 200
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=M xB3=a1
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000 3000
5 cB5=0.25 xB3=S21 0
0
0
1
1
1
0
0
0
0
0
-1
0
0
0
0
1
0
1
0
1 7800
-
6 cB6=M xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000
-
7 cB7=0 xB3=W21 0
0
0
0
1
1
0
0
0
1
0
-1
0
0
0
0
0
0
1
0
1 6000
-
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=0 xB3=W22 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-0.25
fj
M M M 0.25 0.25 0.25 0
0 -M 0
0
0
0
0
M 0.25 M 0.25 M 0.25
Cj-fj
0.3 0.4 0.5
0 0.05 0.05 0
-M -M -M
0
0 0.25 0
0
0
0 +M 0 +M 0 +M
M
0
-0.25
-0.25
-0.25
Simplex 5th tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0.3 xB1=S11 1
0
0
0
-1 -1
0
0
0
0
0
1
1
0
0
0
-1
0
-1
0
-1 200
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000 7000
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
4 cB4=M xB3=a1
0
0
0
0
1
1
-1
0
0
0
0
-1 -1
0
0
1
1
0
1
0
1 2800 2800
5 cB5=0.25 xB3=S21 0
0
0
1
1
1
0
0
0
0
0
-1
0
0
0
0
1
0
1
0
1 7800 7800
6 cB6=M xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000
7 cB7=0 xB3=W21 0
0
0
0
1
1
0
0
0
1
0
-1
0
0
0
0
0
0
1
0
1 6000 6000
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=0 xB3=W22 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
0
M +M M +M M +M
0
0 0.05 0 0.05 0 0.05
fj
Cj-fj
-0.05 -0.05
0.3 M M 0.25 +M +M 0
0.35 0.4
0.4 0.5
0
0
0
-M -M
-M -M
0 -M 0
0
M
0
-0.05 0.3
0 -M
0
-M
0.05 -0.3
0
0
+M +M
-0.05
-0.05
-0.05
-
-
-
Simplex 6th tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0.3 xB1=S11 1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000
2 cB2=0 xB2=s2
0
1
0
0
0
-1
1
0
0
0
0
1
1
1
0
-1 -1
0
0
0
-1 4200 4200
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=0.3 xB3=S22 0
0
0
0
1
1
-1
0
0
0
0
-1 -1
0
0
1
1
0
1
0
1 2800
-
5 cB5=0.25 xB3=S21 0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
-1
0
0
0
0
0 5000 5000
6 cB6=M xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000 4000
7 cB7=0 xB3=W21 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200 3200
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=0 xB3=W22 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
fj
Cj-fj
-0.35
0.3 M M 0.25 0.3 0.3 +M 0 -M 0
0.35
0.4 0.5
0
0 0 0.05
0 M 0
-M
-M -M
0 -0.3 -0.05 0
0 0.35 0.3 M 0.3 M 0.3
0 0.3 0.05 0
0
0
-0.3
+M
0
-0.3
+M
0
-0.3
+M
-
Simplex 7th tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0.3 xB1=S11 1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
-1
0
0
0
0 6200
2 cB2=0 xB2=s2
0
1
0
0
0
-1
0
0
0
-1
0
1
0
1
0
0
0
0
0
0
-1 1000 1000
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=0.3 xB3=S22 0
0
0
0
1
1
0
0
0
1
0
-1
0
0
0
0
0
0
1
0
1 6000
-
5 cB5=0.25 xB3=S21 0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=M xB3=a3
0
1
0
0
0
0
0
-1
0
-1
0
0
-1
0
0
1
1
1
0
0
0
7 cB7=0 xB3=W11 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200
-
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=0 xB3=W22 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
0 -M -M 0 -0.3
0
-M
0
M +M M 0.3 M 0.3
0
0 0.05 0
fj
Cj-fj
0.3 M M 0.25 0.3 0.3 0
0
0.4 0.5
0
-M -M
0 0.05 0
0
0.35
0.3
-0.35
-0.3
M +M 0 0.3
0
+M
-0.05
-0.3
+M
0
-0.3
+M
-
800 800
Simplex 8th tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
i
1 cB1=0.3 xB1=S11 1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0 6200
2 cB2=0 xB2=s2
0
0
0
0
0
-1
0
1
0
0
0
1
1
1
0
-1 -1 -1
0
0
-1 200 200
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=0.3 xB3=S22 0
0
0
0
1
1
0
0
0
1
0
-1
0
0
0
0
0
0
1
0
1 6000
-
5 cB5=0.25 xB3=S21 0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=0.4 xB3=S12 0
1
0
0
0
0
0
-1
0
-1
0
0
-1
0
0
1
1
1
0
0
0
800
-
7 cB7=0 xB3=W11 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200
-
8 cB8=M xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000 5000
9 cB9=0 xB3=W22 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
fj
Cj-fj
0.3 0.4 M 0.25 0.3 0.3 0
0
0
0.5
0
-M
-0.4
-M -0.05 0 -0.3 -0.1 0
+M
0.4
0 0.05 0
M 0.05 0 0.3 0.1 0
-M
-1
bi
0 0.4 0.35 0.4 0.3 M 0.3
0
-0.4 -0.35 -0.4 -0.3
-0.3
0
+M +M +M +M
+M
-
-
Simplex 9th tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
i
1 cB1=0.3 xB1=S11 1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0 6200
-
2 cB2=0 xB2=W12 0
0
0
0
0
-1
0
1
0
0
0
1
1
1
0
-1 -1 -1
0
0
-1 200
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000 6000
4 cB4=0.3 xB3=S22 0
0
0
0
1
1
0
0
0
1
0
-1
0
0
0
0
0
0
1
0
1 6000 6000
5 cB5=0.25 xB3=S21 0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=0.4 xB3=S12 0
1
0
0
0
-1
0
0
0
-1
0
1
0
1
0
0
0
0
0
0
-1 1000
-
7 cB7=0 xB3=W11 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200
-
8 cB8=M xB3=a5
0
0
1
0
0
1
0
0
-1
0
0
-1 -1 -1
0
1
1
1
0
1
1 4800 4800
9 cB9=0 xB3=W22 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
1 2000 2000
0
0 -M
-0.05
0
0.1 0.3 0.4
0
-M -M -M
0
0
fj
Cj-fj
0.3 0.4 M 0.25 0.3
0
0
0.5
0
-M
0
-0.1
+M
0.45
-M
M 0.05 0
0
0
-0.1 -0.3 -0.4
+M +M +M
0
-1
bi
-0.1
-0.05
M +M M 0.3 M
+M
0
0.05
0
-0.3
+M
0 0.1
Simplex 10th tableau
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
i
1 cB1=0.3 xB1=S11 1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0 6200
-
2 cB2=0 xB2=W12 0
0
0
0
0
0
0
1
0
0
1
0
1
1
0
-1 -1 -1
0
0
0 2200
-
3 cB3=0 xB3=s3
0
0
1
0
0
0
0
0
0
0
-1
1
0
0
1
0
0
0
0
0
-1 4000 4000
4 cB4=0.3 xB3=S22 0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
5 cB5=0.25 xB3=S21 0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=0.4 xB3=S12 0
1
0
0
0
0
0
0
0
-1
1
0
0
1
0
0
0
0
0
0
0 3000
-
7 cB7=0 xB3=W11 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200
-
8 cB8=M xB3=a5
0
0
1
0
0
0
0
0
-1
0
-1
0
-1 -1
0
1
1
1
0
1
0 2800 2800
9 cB9=0.35 xB3=S23 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1 2000
fj
Cj-fj
0.3 0.4 M 0.25 0.3 0.35 0
0
0
0.5
0
-M
0
0
0
0 -M
0
-0.05
0.45
-M
-0.45
-0.35
0
0.3 0.4
0
-M -M
-0.3 -0.4
M 0.05 +M 0.35
0
+M +M
-1
bi
-0.05
M +M M 0.3 M 0.35
0
0.05
0
-0.3
+M
-0.35
0 +M
Simplex 11th tableau - Final
i
Min
Cj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0
M M M M M M
cB
xB
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2 a3
a4 a5
a6
i
1 cB1=0.3 xB1=S11 1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0 6200
-
2 cB2=0 xB2=W12 0
0
0
0
0
0
0
1
0
0
1
0
1
1
0
-1 -1 -1
0
0
0 2200
-
3 cB3=0 xB3=s3
0
0
0
0
0
0
0
0
1
0
0
1
1
1
1
-1 -1 -1
0
-1 -1 1200
4 cB4=0.3 xB3=S22 0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
5 cB5=0.25 xB3=S21 0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=0.4 xB3=S12 0
1
0
0
0
0
0
0
0
-1
1
0
0
1
0
0
0
0
0
0
0 3000
-
7 cB7=0 xB3=W11 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200
-
8 cB8=0.5 xB3=S13 0
0
1
0
0
0
0
0
-1
0
-1
0
-1 -1
0
1
1
1
0
1
0 2800
9 cB9=0.35 xB3=S23 0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1 2000
fj
Cj-fj
0.3 0.4 0.5 0.25 0.3 0.35 0
0
0
0
0
0
0
0
0 -M
0
-0.05
0.45
-M
-0.45
-0.35
0
0.3 0.4
0
-M -M
-0.3 -0.4
M 0.05 +M 0.35
0
+M +M
-1
bi
-0.05
M +M M 0.3 M 0.35
0
0.05
0
-0.3
+M
-0.35
0 +M
Optimal cost: 0.3*6200 + 0.25*1800 + 0.40*3000 + 0.30*4000 + 0.50*2800 + 0.35*2000 = 6810
Solution for (b) Phase I : 1st tableau (由於考慮 Wij 所以 pivoting 以 Wij 優先)
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
0
0
0
0
0
0
0
1
1
1
1
1
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0 xB1=s1
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0 8000 8000
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=1 xB3=a1
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000 3000
5 cB5=1 xB3=a2
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=1 xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000
-
7 cB7=1 xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
8 cB8=1 xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=1 xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
1
1
1
1
1
1
0
0
-1
0
0
-1
0
0
0
1
1
1
1
1
1
-1 -1 -1 -1 -1 -1
0
0
1
0
0
1
0
0
0
0
0
0
0
0
0
fj
Cj-fj
Phase I : 2nd tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
0
0
0
0
0
0
0
1
1
1
1
1
a1 a2 a3
a4 a5
a6
bi
i
1 cB1=0 xB1=s1
0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
-1
0
0
0
0
0 5000 5000
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=0 xB3=S11 1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000
-
5 cB5=1 xB3=a2
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=1 xB3=a3
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000 4000
7 cB7=1 xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
8 cB8=1 xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
-
9 cB9=1 xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
fj
0
1
1
1
1
1
1
0
-1
0
0
-1
0
0
0
0
1
1
1
1
1
Cj-fj
0
-1 -1 -1 -1 -1 -1
0
1
0
0
1
0
0
0
1
0
0
0
0
0
Phase I : 3rd tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
a3 a4
a5 a6
0
0
0
0
0
0
0
1
1
1
1
1
bi
i
1 cB1=0 xB1=s1
0
-1
0
1
0
0
0
1
0
0
0
0
1
0
0
-1
0
-1
0
0
0 1000 1000
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=0 xB3=S11
1
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
1
0
1
0
0
0 7000
-
5 cB5=1 xB3=a2
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=0 xB3=W11 0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0
0
0 4000
-
7 cB7=1 xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
8 cB8=1 xB3=a5
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000 5000
9 cB9=1 xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
fj
0
0
1
1
1
1
0
1
-1
0
0
-1
0
0
0
0
1
0
1
1
1
Cj-fj
0
0
-1 -1 -1 -1
0
-1
1
0
0
1
0
0
0
1
0
1
0
0
0
-
Phase I : 4th tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
a3 a4
a5 a6
0
0
0
0
0
0
0
1
1
1
1
1
bi
i
1 cB1=0 xB1=W12 0
-1
0
1
0
0
0
1
0
0
0
0
1
0
0
-1
0
-1
0
0
0 1000
2 cB2=0 xB2=s2
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000 7000
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
-
4 cB4=0 xB3=S11
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0 8000
-
5 cB5=1 xB3=a2
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
-
6 cB6=0 xB3=W11 0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
-1
0
0
0
0
0 5000
-
7 cB7=1 xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
8 cB8=1 xB3=a5
0
1
1
-1
0
0
0
0
-1
0
0
0
-1
0
0
1
0
1
0
1
0 4000 4000
9 cB9=1 xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
fj
0
1
1
0
1
1
0
0
-1
0
0
-1 -1
0
0
1
1
1
1
1
1
Cj-fj
0
-1 -1
0
-1 -1
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
-
-
Phase I : 5th tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
a3 a4
a5 a6
0
0
0
0
0
0
0
1
1
1
1
1
bi
i
1 cB1=0 xB1=W12 0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
0
1
0 5000
2 cB2=0 xB2=s2
0
0
-1
1
1
0
0
0
1
0
0
0
1
1
0
-1
0
-1
0
-1
0 3000 3000
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
4 cB4=0 xB3=S11
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0 8000 8000
5 cB5=1 xB3=a2
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800 1800
6 cB6=0 xB3=W11 0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
-1
0
0
0
0
0 5000 5000
7 cB7=1 xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000
-
8 cB8=0 xB3=S12
0
1
1
-1
0
0
0
0
-1
0
0
0
-1
0
0
1
0
1
0
1
0 4000
-
9 cB9=1 xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
fj
0
0
0
1
1
1
0
0
0
0
0
-1
0
0
0
0
1
0
1
0
1
Cj-fj
0
0
0
-1
-1 -1
0
0
0
0
0
1
0
0
0
1
0
1
0
1
0
-
-
Phase I : 6th tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
a3 a4
a5 a6
0
0
0
0
0
0
0
1
0
1
1
1
1
bi
i
1 cB1=0 xB1=W12 0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0 5000
2 cB2=0 xB2=s2
0
0
-1
0
1
0
0
0
1
1
0
0
1
1
0
-1 -1 -1
0
-1
0 1200 1200
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000
4 cB4=0 xB3=S11
1
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
-1
0
0
0
0 6200 6200
5 cB5=0 xB3=S21
0
0
0
1
0
0
0
0
0
-1
0
0
0
0
0
0
1
0
0
0
0 1800
6 cB6=0 xB3=W11 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200 3200
7 cB7=1 xB3=a4
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
1
0
0 4000 4000
8 cB8=0 xB3=S12
0
1
1
0
0
0
0
0
-1 -1
0
0
-1
0
0
1
1
1
0
1
0 5800
-
9 cB9=1 xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
fj
0
0
0
0
1
1
0
0
0
1
0
-1
0
0
0
0
0
0
1
0
1
Cj-fj
0
0
0
0
-1 -1
0
0
0
-1
0
1
0
0
0
1
1
1
0
1
0
-
-
-
Phase I : 7th tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
a3 a4
a5 a6
0
0
0
0
0
0
0
1
1
1
1
bi
i
1 cB1=0 xB1=W12 0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0 5000
-
2 cB2=0 xB2=W21 0
0
-1
0
1
0
0
0
1
1
0
0
1
1
0
-1 -1 -1
0
-1
0 1200
-
3 cB3=0 xB3=s3
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0 6000 6000
4 cB4=0 xB3=S11
1
0
1
0
-1
0
0
0
-1
0
0
0
0
-1
0
1
0
1
0
1
0 5000
-
5 cB5=0 xB3=S21
0
0
-1
1
1
0
0
0
1
0
0
0
1
1
0
-1
0
-1
0
-1
0 3000
-
6 cB6=0 xB3=W11 0
0
1
0
-1
0
1
0
-1
0
0
0
0
-1
0
0
0
1
0
1
0 2000
-
7 cB7=1 xB3=a4
0
0
1
0
0
0
0
0
-1
0
-1
0
-1 -1
0
1
1
1
1
1
0 2800
-
8 cB8=0 xB3=S12
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
9 cB9=1 xB3=a6
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000 2000
fj
0
0
1
0
0
1
0
0
-1
0
0
-1 -1 -1
0
1
1
1
1
1
1
Cj-fj
0
0
-1
0
0
-1
0
0
1
0
0
1
0
0
0
0
0
0
0
1
1
0
1
Phase I : 8th tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
a3 a4
a5 a6
0
0
0
0
0
0
0
1
1
1
1
bi
i
1 cB1=0 xB1=W12 0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
0
1
0 5000 5000
2 cB2=0 xB2=W21 0
0
-1
0
1
0
0
0
1
1
0
0
1
1
0
-1 -1 -1
0
-1
0 1200
3 cB3=0 xB3=s3
0
0
1
0
0
0
0
0
0
0
-1
1
0
0
1
0
0
0
0
0
-1 4000 4000
4 cB4=0 xB3=S11
1
0
1
0
-1
0
0
0
-1
0
0
0
0
-1
0
1
0
1
0
1
0 5000 5000
5 cB5=0 xB3=S21
0
0
-1
1
1
0
0
0
1
0
0
0
1
1
0
-1
0
-1
0
-1
0 3000
6 cB6=0 xB3=W11 0
0
1
0
-1
0
1
0
-1
0
0
0
0
-1
0
0
0
1
0
1
0 2000 2000
7 cB7=1 xB3=a4
0
0
1
0
0
0
0
0
-1
0
-1
0
-1 -1
0
1
1
1
1
1
0 2800 2800
8 cB8=0 xB3=S12
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000
-
9 cB9=0 xB3=S23
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
fj
0
0
1
0
0
0
0
0
-1
0
-1 -1 -1 -1
0
1
1
1
1
1
0
Cj-fj
0
0
-1
0
0
0
0
0
1
0
1
0
0
0
0
0
0
1
1
1
1
0
1
-
-
Phase I : 9th tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
a3 a4
a5 a6
0
0
0
0
0
0
0
1
1
1
1
1
bi
i
1 cB1=0 xB1=W12 0
0
0
0
1
0
-1
1
0
0
0
0
0
1
0
0
0
-1
0
0
0 3000 3000
2 cB2=0 xB2=W21 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200
3 cB3=0 xB3=s3
0
0
0
0
1
0
-1
0
1
0
-1
1
0
1
1
0
0
-1
0
-1 -1 2000 2000
4 cB4=0 xB3=S11
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000
-
5 cB5=0 xB3=S21
0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
-1
0
0
0
0
0 5000
-
6 cB6=0 xB3=S13
0
0
1
0
-1
0
1
0
-1
0
0
0
0
-1
0
0
0
1
0
1
0 2000
-
7 cB7=1 xB3=a4
0
0
0
0
1
0
-1
0
0
0
-1
0
-1
0
0
1
1
0
1
0
0
8 cB8=0 xB3=S12
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 7000 7000
9 cB9=0 xB3=S23
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
fj
0
0
0
0
1
0
-1
0
0
0
-1
0
-1
0
0
1
1
0
1
0
0
Cj-fj
0
0
0
0
-1
0
1
0
0
0
1
0
1
0
0
0
0
1
0
1
1
-
800 800
-
Phase I : 10th tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0
0
0
1
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23 s1
s2
s3
a1 a2
0
0
0
0
0
0
0
1
1
1
a3 a4
1
1
a5 a6
bi
i
1 cB1=0 xB1=W12 0
0
0
0
0
0
0
1
0
0
1
0
1
1
0
-1 -1 -1 -1
0
0 2200
-
2 cB2=0 xB2=W21 0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
-1 -1
0
0
0
0 3200
-
3 cB3=0 xB3=s3
0
0
0
0
0
0
0
0
1
0
0
1
-1
1
1
1
1
-1
1
-1 -1 2800
-
4 cB4=0 xB3=S11
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
1
0
0
0
0
0 3000
-
5 cB5=0 xB3=S21
0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
-1
0
0
0
0
0 5000
-
6 cB6=0 xB3=S13
0
0
1
0
0
0
0
0
-1
0
-1
0
-1 -1
0
1
1
1
1
1
0 2800
-
7 cB7=0 xB3=S22
0
0
0
0
1
0
-1
0
0
0
-1
0
-1
0
0
1
1
0
1
0
0
800
-
8 cB8=0 xB3=S12
0
1
0
0
0
0
1
0
0
0
1
0
1
1
0
-1 -1
0
-1
0
0 6200
-
9 cB9=0 xB3=S23
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
0
0
0
0
0
1 2000
-
fj
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Cj-fj
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
(b.) Min 0.06W11 +0.05W21 +0.09W12 +0.06W22 +0.12W13 +0.07W23
Phase II : 1st tableau
i
Min
Cj
cB
xB
0
0
0
0
0
0.06 0.09 0.12 0.05 0.06 0.07
0
0
0
S11 S12 S13 S21 S22 S23 W11 W12 W13 W21 W22 W23
s1
s2
s3
bi
0
1 cB1=0.09 xB1=W12
0
0
0
0
0
0
0
1
0
0
1
0
1
1
0
2200
2 cB2=0.05 xB2=W21
0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
3200
3
cB3=0
xB3=s3
0
0
0
0
0
0
0
0
1
0
0
1
-1
1
1
2800
4
cB4=0
xB3=S11
1
0
0
0
0
0
-1
0
0
0
0
0
0
0
0
3000
5
cB5=0
xB3=S21
0
0
0
1
0
0
1
0
0
0
0
0
1
0
0
5000
6
cB6=0
xB3=S13
0
0
1
0
0
0
0
0
-1
0
-1
0
-1
-1
0
2800
7
cB7=0
xB3=S22
0
0
0
0
1
0
-1
0
0
0
-1
0
-1
0
0
800
8
cB8=0
xB3=S12
0
1
0
0
0
0
1
0
0
0
1
0
1
1
0
6200
9
cB9=0
xB3=S23
0
0
0
0
0
1
0
0
0
0
1
-1
0
0
0
2000
fj
0
0
0
0
0
0
0.05 0.09
Cj-fj
0
0
0
0
0
0
0.01
0
0
0.12
0.05 0.09
0
0
0.14 0.09
0
-0.03 0.07 -0.14 -0.09
0
i
Chapter 4
Example 4.1
Consider the following primal linear programming model.
Minimize f = 6x1 + 8x2
subject to
3x1 + x2 ≤ 4
5x1 + 2x2 ≤10
x1 + 2x2 ＝ 3
and x1, x2 ≥0
Find its dual linear programming model.
Solution:
Change into canonical form:
－3x1 － x2 －4
－5x1 － 2x2 －10
x1 ＋ 2x2 3
－x1 － 2x2 －3
and x1, x2 ≥0
Thus,
4 

10 
6

, c  , and A =
b
3  
8


 
3 
3 1 


5 2

, i.e., AT =
1
2


1 2

So, the dual form is:
Maximize g = －4y1－10y2＋3y3－3y4
－3y1－5y2＋ y3－ y4 ≤6
－ y1－2y2＋2y3－2y4 ≤8
y1, y2, y3, y4 ≥0
3 5 1 1 


1 2 2 2


Example 4.2
Consider the the following linear programming model:
Min f = 2x1 ＋ 4x2
Subject to x1 ＋ 5x2 80
4x1 ＋ 2x2 6
x1 ＋ x2 = 3
x1, x2 0
(a) Rewrite the linear programming model into its canonical form.
(b) Change into its dual form.
Solution for (a)
Min f = 2x1 ＋ 4x2
Subject to － x1 － 5x2 －80
4x1 ＋ 2x2 6
－x1 － x2 －3
x1 ＋ x2 3
x1, x2 0
Solution for (b)
Max f
’= 80y1 －6y2＋ 3y3 －3y4
y1 －4y2 ＋ y3 －y4 2
5y1 －2y2 ＋ y3 －y4 4
y1, y2, y3, y4 0
Chapter 5
Example 5.1
Consider the following primal linear programming model.
Maximize f = 4x1 － x2 + x3
subject to 2x1 ＋ x2 ＋ x3 ≤6
－2x1 ＋ 2x2
≤4
and
x1, x2, x3 ≥0
The following is its simplex final tableau.
Cj
4
i cB
xB
x1
1 cB1=4 xB1=x1 1
2 cB2=0 xB2=s2 0
fj
4
Cj-fj
0
-1
1
0 0
x2 x3 s1 s2
bi
0.5 0.5 0.5 0
3
3
1
1 1
10
2
2
2 0 ∑icBibi=12
-3 -1 -2 0
Find the adjustable range of the follows such that the basic variables in the optimal
solution can keep invariant.
(1) The coefficient’
s range of non-basic variable x2 in objective function.
(2) The coefficient’
s range of basic variable x1 in objective function.
(3) The range of b1.
Solutions:
1
0 0
(1)
Cj
4
c2
i cB
xB x1 x2
x3 s1 s2
bi
1 cB1=4 xB1=x1 1 0.5 0.5 0.5 0
3
2 cB2=0 xB2=s2 0
3
1
1 1
10
fj
4
2
2
2 0 ∑icBibi=12
Cj-fj
0 c2 - 2 -1 -2 0
Let c2 be the coefficient of x2 in the objective function, then the following
inequality c2 –2 ≤0  c2 ≤2 must be hold to keep x2 staying outside of the
basic variables. This also implies that if the unit profit of x2 is greater than $2, the
optimality should be changed.
(2)
Cj
i cB
xB
1 cB1=c1 xB1=x1
2 cB2=0 xB2=s2
fj
Cj-fj
-1
1
0
x1
x2
x3
s1
1
0.5
0.5
0.5
0
3
1
1
c1 0.5c1 0.5c1 0.5c1
0 -1-0.5c1 1-0.5c1 -0.5c1
c1
0
s2
bi
0
3
1
10
0 ∑icBibi=12
0
Let c1 be the coefficient of x1 in the objective function, then the following
inequalities must be hold to keep the basic variables unchanged.
-1-0.5c1 ≤0  1+0.5c1 0  c1 -2
1-0.5c1 ≤0  c1 2
-0.5c1 ≤0  c1 0 The concluding result is c1 2.
If the unit profit of x1 is smaller than $2, x1 should leave from basic variables, and
x3 will get into as a basic variable.
(3)
i cB
Cj
xB
4
x1
-1
x2
1
x3
0
s1
0
s2
bi
1 cB1=4 xB1=x1
1
0.5 0.5 0.5 0
3
2 cB2=0 xB2=s2
0
3
1
1
1
fj
4
2
2
2
0 ∑icBibi=12
Cj-fj
0
-3
-1
-2
0
10
Observe that NxN + BxB = b  B-1NxN + B-1BxB = B-1b  xB = B-1b － B-1NxN
-1
Since xN = 0 and xB 0, so we conclude that xB = B b 0, thus the range of b1 is,
0.5 0
b1  
0

1 14 
 0.5b10 and b1 + 4 0  b1 -4, the result is: b10.
0

  
Example 5.2
Maximize f = 9 x1＋12 x2
subject to
3x1＋2x2 40
2x1＋3x2 30
2x1＋ x2 250
and x1, x2 0
The following table is the final table of the above model.
Max Cj 9 12
i cB x B x 1 x 2
0
s1
0
s2
0
s3
bi
1 9 x1
1
0
3/5
-2/5
0
12
2 12 x2
0
1
-2/5
3/5
0
2
3 0
s3
0
0
-4/5
5/9
1
224
fj
9 12
3/5
18/5
0
0
-3/5 -18/5 0
Cj-fj
0
132
i
Under the concern of keeping the basic variables invariant, do the following
exercises
(a) Find the allowable range of coefficient’
s values (say, c1 and c2) for x1 and x2.
(b) Find the allowable range of resource amount (say, b1 and b2) for 1st and 2nd
resources.
(c) Find the marginal values of 1st and 2nd resources.
Solution:
(a) －0.6c1＋4.8 0；0.4c1－7.2 0  c1 8；c1 18
 8 c1 18
－5.4＋0.4c2 0；0.4－0.6c2 0  c2 1.35；c2 ⅔
 ⅔ c2 1.35
(b)
0.6 0.4 0
b1   0.6b1 12 


30  0.4b 18 0
0.4 0.6 0
1

  

5



0.8
0.8b1 266.67 
1
250

 


9

 0.6b1－12 0；－0.4b1＋18 0；－0.8b1＋266.67 0
 b1 20；b1 45；b1 333.34  20 b1 45
0.6 0.4 0
24 0.4b2 
40  



b  16 0.6b 0

0
.
4
0
.
6
0
2


2  


b
5


0.8 5 9 1
250
218  2 

 


9 

 24－0.4b2 0；16＋0.6b2 0；218＋0.8b2 0
 b2 60；b2 －26.67；b2 －272.5  -26.67 b2 60
(c) Let y1, y2 and y3 be the marginal values of 1st, 2nd and 3rd resources, then the
dual form of the original model is as follows:
Min
Min
g = 40 y1＋30 y2＋250 y3 g = 40 y1＋30 y2＋250 y3＋Ma1＋Ma2
subject to
subject to
3y1+2y2+2y3－s1+a1 = 9
3y1+2y2+2y3 9
2y1+3y2+ y3 12
2y1+3y2+ y3－s2+a2 = 12
y1, y2, y2 0
y1, y2, y2 s1, s2, a1, a2 0
Min Cj
i cB x B
40
y1
30
y2
250
y3
0
s1
0
s2
M
a1
M
a2
bi
i
1 M a1
3
2
2
-1
0
1
0
9
3
2 M a2
2
3
1
0
-1
0
1
12
4
3M
-M
-M
M
M
M
M
0
0
fj
Cj-fj
5M 5M
40-5M 30-5M 250-3M
Min Cj
i cB x B
40
y1
30
y2
250
y3
0
s1
0
s2
M
a1
M
a2
bi
i
1 40 y1
1
2/3
2/3
-1/3
0
1/3
0
3
4.5
2 M a2
0
5/3
-1/3
2/3
-1
-2/3
1
6
3.6
fj
40
-40/3
+2M/3
40/32M/3
M
40/3 2M/3
-40/3
+2M/3
M
0
80/3 M/3
670/3
+M/3
-M
Cj-fj
80/3+
5M/3
10/35M/3
Min Cj
i cB x B
40
y1
30
y2
250
y3
0
s1
0
s2
M
a1
M
a2
bi
i
1 40 y1
1
0
0.4
-0.6 0.4 7/15 -0.4 0.6
2 30 y2
0
1
-0.2
0.4 -0.6 -0.2 0.6 3.6
fj
40
30
10
-12 -18
略
略
Cj-fj
0
0
240
12
略
略
Marginal value of Resource 1 is 0.6.
Marginal value of Resource 2 is 3.6.
18
0
Chapter 6
Example 6.1
某電子公司，主要訂單來自 A、B、C 三市，而該公司在這三市都設有倉庫。公司
的三座工廠與倉庫不在同一地方，分別設於 I、J、K 三地。由於市場不景氣，業務
嚴重萎縮，如下表所示：
工廠
每年最大產能(千個)
倉庫
每倉庫預估來年之需求量(千個)
I
J
K
210
140
290
A
B
C
80
200
200
各廠與倉庫間的運輸成本及其生產成本如下表：
倉庫
工廠
I
J
K
A
B
C
生產成本
2
4
3
4
3
6
4
4
4
11
14
12
公司目前的流通策略如下：所有 A 市倉庫的需求由 I 廠供應，所有 J 廠的產品都運
到 B 市倉庫，而 B 市倉庫不足的需求量都由 I 廠供應，C 市的需求量則完全由 K 廠
供應。試評估針對來年需求，採用現行策略的成本。以 VAM 法改善公司此一策
略。
Solution：
(1)、試評估針對來年需求，採用現行策略的成本。解釋公司為何採用此一策略？
令成本＝運輸成本＋生產成本
倉庫
工廠
I
J
K
Request
A
B
13
80
18
15
80
15
60
17
140
18
200
C
D(slack) Supply
15
0
70
18
0
16
200
0
200
90
160
Total cost＝13×80＋15×60＋17×140＋16×200＝7420(千元)。
210
140
290
640
Apply company strategy:
penalty
1 4 15
233
1 165
0
penalty
倉庫
工廠
A
B
C
D(slack)
13 2
I
17 1
J
15 2 1
K
Supply
13
10
18
15
200
17
15
0
18
15
70
18
16
200
0
140
0
20
290 270
70
200
140 20
640
Request 80 70
200
210 10
140
Total cost＝13×10＋15×200＋15×70＋16×200＝7380(千元)。
Example 6.2
下表所示為三處倉庫與四位顧客間的供需關係表。陰影部份是各相關之運輸成本。
顧客
倉庫
戒
定
慧
需求量
阿
彌
陀
佛
7
10
6
8
12
10
11
5
11
10
4
9
20
28
17
33
庫存量
98
30
45
35
110
以 VAM 法求出其可行解，並計算該解之總運算成本。
Solution:
penalty
顧客
penalty 倉庫
7
戒
4
定
6
慧
Request
1
2
6
5
阿
彌
陀
佛 D(slack) 庫存量
7
8
11
10
0
10
12
5
4
0
6
10
11
9
0
20
28
17
33
0
12
30
45
35
110
penalty
顧客
penalty 倉庫
17
戒
14
定
36
慧
Request
penalty
顧客
penalty 倉庫
1
戒
61
定
3
慧
Request
penalty
顧客
penalty 倉庫
1
戒
6
定
3
慧
Request
0
1
2
6
5
阿
彌
陀
佛 D(slack) 庫存量
12
7
8
11
10
0
10
12
5
4
0
6
10
11
9
0
30 18
45
35
20
28
17
33
12 0
1
2
6
5
0
阿
彌
陀
佛 D(slack) 庫存量
12
7
8
10
6
10
0
12
11
17
5
4
0
10
11
9
0
110
18
45 28
35
20
28
17 0
33
0
1
2
6
51
0
阿
彌
陀
佛 D(slack) 庫存量
110
12
7
8
10
6
20
10
28
4
0
12
11
17
5
10
11
9
0
28
0
33 5
0
0
18
28 0
35
110
1
2
6
1
阿
彌
陀
佛 D(slack) 庫存量
penalty
顧客
penalty 倉庫
12
戒
6
定
31
慧
12
7
8
10
20
6
Request 20 0
penalty
顧客
penalty 倉庫
2
戒
6
定
1
慧
Request
penalty
顧客
penalty 倉庫
2
戒
6
定
1
慧
Request
0
10
28
4
0
12
11
17
5
10
11
9
0
28
0
5
18
0
0
35 15
0
110
1
2
6
1
0
阿
彌
陀
佛 D(slack) 庫存量
12
18
7
8
10
28
4
0
12
11
17
5
10
20
6
10
11
9
0
0
0
28 10
0
5
0
1
-
6
-
0
阿
彌
陀
10
20
6
12
10
10
0
10 0
11
17
5
11
0
15
110
12
8
0
佛 D(slack) 庫存量
18
7
18 0
10
28
4
5
9
50
0
0
0
0
0
0
15 0
110
Final
penalty
顧客
penalty 倉庫
戒
定
慧
Request
阿
彌
陀
佛 D(slack) 庫存量
12
18
7
8
10
20
6
12
10
10
20
28
11
17
5
11
17
10
28
4
5
9
33
30
0
45
0
35
0
12
110
總運送成本： 20×6＋18×8＋10×10＋17×5＋28×4＋5×9 = 606。