Quantitative Analysis for Management Linear Programming Models:.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(20)
(21)
(22)
(23)
(24)
(25)
(26)
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
(35)
(36)
(37)
(38)
(39)
(40)
(41)
(42)
(43)
(44)
(45)
(46)
(47)
(1)
(2)
(3)
(4)
(5)
(6)
Quantitative Analysis
for Management
Linear Programming
Models:
(2)
Linear Programming
Problem
1. Tujuan adalah maximize or minimize
variabel dependen dari beberapa kuantitas
variabel independen (fungsi tujuan).
2. Batasan-batasan yang diperlukan guna
mencapai tujuan.
Tujuan dan Batasan dinyatakan dalam
persamaan linear.
(3)
Basic Assumptions of Linear
Programming
Certainty
Proportionality
Additivity
Divisibility
(4)
Flair Furniture Company
Data -
Table 7.1
Hours Required to Produce One Unit
Department
X
1
Tables
X
2
Chairs
Available
Hours This
Week
Carpentry
Painting/Varnishing
4
2
3
1
240
100
(5)
Flair Furniture Company
Data -
Table 7.1
Constraints:
Maximize:
Objective:
)
varnishing
&
(painting
100
1
2
X
1
+
X
2
2
1
5
7
X
+
X
)
(carpentry
240
3
4
X
1
+
X
2
STEP 1:
(6)
Flair Furniture Company
Feasible Region
120
100
80
60
40
20
0
Number
of
Chai
rs
20
40
60
80
100
Number of Tables
Painting/Varnishing
Carpentry
Feasible
Region
(7)
Flair Furniture Company
Isoprofit Lines
Numb
er
of
Chai
rs
120
100
80
60
40
20
0
20
40
60
80
100
Number of Tables
Painting/Varnishing
Carpentry
7
X
1
+ 5
X
2
= 210
7
X
1
+ 5
X
2
= 420
(8)
Flair Furniture Company
Optimal Solution
Number
of
Chairs
120
100
80
60
40
20
0
20
40
60
80
100
Number of Tables
Painting/Varnishing
Carpentry
Solution
(
X
1
= 30,
X
2
= 40)
Corner Points
1
2
3
(9)
Test Corner Point Solutions
Point 1) (0,0) => 7(0) + 5(0) = $0
Point 2) (0,100) => 7(0) + 5(80) = $400
Point 3) (30,40) => 7(30) + 5(40) = $410
(10)
Solve Equations Simultaneously
4X1 + 3X2 <= 240
2X1 + 1X2 <= 100
X1 = 60 - 3/4 X2 X1 = 50 - 1/2 X2
60 - 3/4 X2 = 50 - 1/2 X2
60 - 50 = 3/4 X2 - 1/2 X2
10 = 1/4 X2
40 = X2;
so, 4X1 + 3(40) = 240
4X1 = 240 - 120
X1 = 30
To get X1 & X2 values for Point 3:
(11)
Special Cases in LP
Infeasibility
Unbounded Solutions
Redundancy
Degeneracy
(12)
A Problem with No Feasible
Solution
X
2
X
1
8
6
4
2
0
2
4
6
8
Region Satisfying
3rd Constraint
(13)
A Solution Region That is
Unbounded to the Right
X
2
X
1
15
10
5
0
5 10 15
Feasible Region
X
1
>
5
X
2
< 10
(14)
A Problem with a
Redundant Constraint
X
2
X
1
30
25
20
15
10
5
0
5
10
15
20
25
30
Feasible
Region
2
X
1
+
X
2
< 30
X
1
< 25
X
1
+
X
2
< 20
Redundant
Constraint
(15)
An Example of Alternate
Optimal Solutions
8
7
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
Optimal Solution Consists of All
Combinations of
X
1
and
X
2
Along
the
AB
Segment
Isoprofit Line for $12
Overlays Line Segment
Isoprofit Line for $8
A
B
(16)
Marketing Applications
Media Selection - Win Big Gambling Club
Medium
Audience
Reached
Per Ad
Cost
Per
Ad($)
Maximum
Ads Per
Week
TV spot (1 minute)
5,000
800
12
Daily newspaper
(full-page ad)
8,500
925
5
Radio spot (30
seconds, prime time)
2,400
290
25
Radio spot (1 minute,
afternoon)
(17)
Win Big Gambling Club
:
to
Subject
)
spots/week
TV
(max
12
X
1
expense)
radio
(max
1800
380X
290X
3
+
4
)
spots/week
radio
(min
+
4
5
3
X
X
ad
budget)
weekly
+
+
+
925
290
380
8000
800
X
1
X
2
X
3
X
4
(
)
spots/week
radio
min.
-1
(max
20
4
X
)
spots/week
radio
sec.
-30
max
25
3
(
X
ads/week)
newspaper
max
5
2
(
X
:
(18)
Manufacturing Applications
Variety
of Tie
Selling
Price per
Tie ($)
Monthly
Contract
Minimum
Monthly
Demand
Material
Required
per Tie
(Yds)
Material
Require
ments
All silk
6.70
6000
7000
0.125 100%
silk
All
polyester
3.55
10000
14000
0.08 100%
polyester
Poly-cotton
blend 1
4.31
13000
16000
0.10 50%
poly/50%
cotton
Polycotton
-blend 2
4.81
6000
8500
0.10 30%
poly/70%
cotton
(19)
Fifth Avenue
4.00X
3.56X
3.07X
4.08X
:
Maximize
1
+
2
+
3
+
4
1)
blend
max,
(contract
X
3
16000
2)
blend
max,
(contract
8500
4
X
2)
blend
min,
(contract
6000
4
X
1)
blend
min,
(contract
13000
3
X
polyester)
all
max,
(contract
14000
2
X
polyester)
all
min,
(contract
1000
2
X
silk)
max,
(contract
7000
1
X
silk)
min,
(contract
6000
1
X
:
to
Subject
cotton)
(yards
1600
07
0
05
0
.
X
3
+
.
X
4
polyester)
(yards
3000
03
0
05
0
08
0
.
X
2
+
.
X
3
+
.
X
4
silk)
of
(yards
800
125
(20)
Manufacturing Applications
Truck Loading - Goodman Shipping
Item
Value ($)
Weight
(lbs)
1
22,500
7,500
2
24,000
7,500
3
8,000
3,000
4
9,500
3,500
5
11,500
4,000
(21)
Goodman Shipping
1
1
1
1
1
1
10000
3500
4000
3500
3000
7500
7500
9750
11500
9500
8000
24000
22500
6
5
4
3
2
1
6
5
4
3
2
1
6
5
4
3
2
1
+
+
+
+
+
+
+
+
+
+
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
(Capacity)
:
to
Subject
:
value
load
Maximize
(22)
Flair Furniture Company
Hours Required to Produce One Unit
Department
X
1
Tables
X
2
Chairs
Available
Hours This
Week
Carpentry
Painting/Varnishing
4
2
3
1
240
100
Profit/unit
Constraints:
$7
$5
Maximize:
Objective:
)
varnishing
&
(painting
100
1
2
X
1
+
X
2
2
1
5
7
X
+
X
)
(carpentry
240
3
(23)
Flair Furniture Company's
Feasible Region & Corner Points
Numb
er
of
Chairs
100
80
60
40
20
0
20
40
60
80
100
X
X
2
Number of Tables
B = (0,80)
C = (30,40)
D = (50,0)
Feasible
Region
240
3
4
X
1
+
X
2
100
1
(24)
Flair Furniture - Adding
Slack Variables
)
varnishing
&
(painting
100
1
2
X
1
+
X
2
)
(carpentry
240
3
4
X
1
+
X
2
Constraints:
Constraints with Slack Variables
)
varnishing
&
(painting
)
(carpentry
100
1
2
240
3
4
2
2
1
1
2
1
=
+
+
=
+
+
S
X
X
S
X
X
2
1
5
7
X
+
X
Objective Function
Objective Function with Slack Variables
2
1
2
1
5
1
1
(25)
Flair Furniture’s Initial Simplex
Tableau
Profit
per
Unit
Column
Production
Mix
Column
Real
Variables
Columns
Slack
Variables
Columns
Constant
Column
C
j
Solution
Mix
X
1
X
2
S
1
S
2
Quantity
$7
$5
$0
$0
Profit per
unit row
2
1
1
0
4
3
0
1
$0
$0
$0
$0
$7
$5
$0
$0
$0
$0
S
1
S
2
Z
j
C
j
-
Z
j
100
240
$0
$0
Constraint
equation
rows
Gross
profit row
Net
profit row
(26)
Pivot Row, Pivot Number Identified
in the Initial Simplex Tableau
C
j
Solution
Mix
X
1
X
2
S
1
S
2
Quantity
$7
$5
$0
$0
2
1
1
0
4
3
0
1
$0
$0
$0
$0
$7
$5
$0
$0
$0
$0
S
1
S
2
Z
j
C
j
-
Z
j
100
240
$0
$0
Pivot row
Pivot number
(27)
Completed Second Simplex Tableau
for Flair Furniture
C
j
Solution
Mix
X
1
X
2
S
1
S
2
Quantity
$7
$5
$0
$0
1
1/2
1/2
0
0
1
-2
1
$7
$7/2
$7/2
$0
$0
$3/2
-$7/2
$0
$7
$0
X
1
S
2
Z
j
C
j
-
Z
j
50
40
$350
(28)
Pivot Row, Column, and Number
Identified in Second Simplex Tableau
C
j
Solution
Mix
X
1
X
2
S
1
S
2
Quantity
$7
$5
$0
$0
1
1/2
1/2
0
0
1
-2
1
$7
$7/2
$7/2
$0
$0
$3/2
-$7/2
$0
$7
$0
X
1
S
2
Z
j
C
j
-
Z
j
50
40
$350
(Total
Profit)
Pivot row
Pivot number
(29)
Calculating the New
X
1
Row
for Flair’s Third Tableau
(
Number
in new
X
1
row
)
Number
in old
X
1
row
(
)
above pivot
Number
number
(
)
Corresponding
number in
new
X
2
row
(
)
=
-
x
=
-
x
1
0
3/2
-1/2
30
1
1/2
1/2
0
50
(1/2)
(1/2)
(1/2)
(1/2)
(1/2)
(0)
(1)
(-2)
(1)
(40)
=
-
x
=
-
x
=
-
x
(30)
Final Simplex Tableau for
the Flair Furniture Problem
C
j
Solution
Mix
X
1
X
2
S
1
S
2
Quantity
$7
$5
$0
$0
1
0
3/2
-1/2
0
1
-2
1
$7
5
$1/2
$3/2
$0
$0
-$1/2 -$3/2
$7
$5
X
1
X
2
Z
j
C
j
-
Z
j
30
40
$410
(31)
Simplex Steps for
Maximization
1. Choose the variable with the greatest positive
C
j
- Z
j
to enter the solution.
2. Determine the row to be replaced by selecting that
one with the smallest (non-negative)
quantity-to-pivot-column ratio.
3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).
5. Calculate the C
j
and C
j
- Z
j
values for this tableau.
If there are any C
j
- Z
j
values greater than zero,
(32)
Surplus & Artificial Variables
Constraints
900
30
25
210
8
10
5
2
1
3
2
1
=
+
+
+
X
X
X
X
X
900
30
25
210
8
10
5
2
2
1
1
1
3
2
1
=
+
+
=
+
-+
+
A
X
X
A
S
X
X
X
Constraints-Surplus & Artificial Variables
Objective Function
3
2
1
9
7
5
X
+
X
+
X
:
Min
2
1
1
3
2
1
9
7
0
5
X
+
X
+
X
+
S
+
MA
+
MA
:
Min
(33)
Simplex Steps for
Minimization
1. Choose the variable with the greatest negative
C
j
- Z
j
to enter the solution.
2. Determine the row to be replaced by selecting that
one with the smallest (non-negative)
quantity-to-pivot-column ratio.
3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).
5. Calculate the C
j
and C
j
- Z
j
values for this tableau.
If there are any C
j
- Z
j
values less than zero, return
to Step 1.
(34)
Special Cases
Infeasibility
C
j
5
8
0
0
M
M
Solution
Mix
X
1
X
2
S
1
S
2
A
1
A
2
Qty
5
X
1
1
0 -2
3
-1
0
200
8
X
2
0
1
1
2
-2
0
100
M
A
2
0
0
0
-1
-1
1
20
Z
j
5
8 -2 31-M -21-M
M 1800+20M
(35)
Special Cases
Unboundedness
C
j
6
9
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
9
X
2
-1
1
2
0
30
0
S
2
-2
0
-1
1
10
Z
j
-9
9
18
0
270
C
j
-Z
j
15
0
-18
0
(36)
Special Cases
Degeneracy
C
j
5
8
2
0
0
0
Solution
Mix
X
1
X
2
X
3
S
1
S
2
S
3
Qty
8
X
2
1/4
1
1
-2
0
0
10
0
S
2
4
0
1/3
-1
1
0
20
0
S
3
2
0
2
2/5
0
1
10
Z
j
2
8
8
16
0
0
80
C
j
-Z
j
3
0
6
16
0
0
(37)
Special Cases
Multiple Optima
C
j
3
2
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
2
X
2
3/2
1
1
0
6
0
S
2
1
0
1/2
1
3
Z
j
3
2
2
0
12
(38)
Sensitivity Analysis
High Note Sound Company
60
1
3
80
4
2
120
50
2
1
2
1
2
1
+
+
+
X
X
X
X
X
X
:
to
Subject
:
Max
(39)
Sensitivity Analysis
(40)
Simplex Solution
High Note Sound Company
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60
120
30
0
2400
C
j
-Z
j
0
0
-30
0
(41)
Simplex Solution
High Note Sound Company
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60
120
30
0
2400
(42)
Nonbasic Objective Function
Coefficients
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60
120
30
0
2400
(43)
Basic Objective Function
Coefficients
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120+
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60+1/2
120+
30+1/4
0
2400+20
(44)
Simplex Solution
High Note Sound Company
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60
120
30
0
2400
C
j
-Z
j
0
0
-30
0
Objective increases by 30 if 1 additional
hour of electricians time is available.
(45)
Steps to Form the Dual
To form the Dual:
If the primal is max., the dual is min., and vice
versa.
The right-hand-side values of the primal constraints
become the objective coefficients of the dual.
The primal objective function coefficients become
the right-hand-side of the dual constraints.
The transpose of the primal constraint coefficients
become the dual constraint coefficients.
(46)
Primal & Dual
60
1
3
80
4
2
120
50
2
1
2
1
2
1
+
+
+
X
X
X
X
X
X
:
to
Subject
:
Max
Primal:
Dual
120
1
4
50
3
2
60
80
2
1
2
1
2
1
+
+
+
U
U
U
U
U
U
:
to
Subject
:
Min
(47)
Comparison of the Primal and
Dual Optimal Tableaus
Du
al’
s Op
tima
l So
lut
ion
C
j
Solution
Mix
Quantity
$7
$5
X
2
S
2
Z
j
C
j
-
Z
j
20
40
$2,400
X
1
X
2
S
1
S
2
$50
$120
$0
$0
1/2
1
1/4
0
5/2
0
-1/4
1
60
120
30
0
-10
0
-30
0
Primal’
s Op
tima
l So
lut
ion
C
j
Solution
Mix
Quantity
$7
$5
U
1
S
1
Z
j
C
-
Z
30
10
$2,400
X
1
X
2
S
1
S
2
80
60
$0
$0
1
1/4
0
-1/4
0
-5/2
1
-1/2
80
20
0
-20
$0
40
0
20
A
1
A
2
M
M
0
1/2
-1
1/2
0
40
M
M
-40
(1)
Nonbasic Objective Function
Coefficients
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60
120
30
0
2400
(2)
Basic Objective Function
Coefficients
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120+
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60+1/2
120+
30+1/4
0
2400+20
(3)
Simplex Solution
High Note Sound Company
C
j
50
120
0
0
Solution
Mix
X
1
X
2
S
1
S
2
Qty
120
X
2
1/2
1
¼
0
20
0
S
2
5/2
0
-1/4
1
40
Z
j
60
120
30
0
2400
C
j
-Z
j
0
0
-30
0
(4)
Steps to Form the Dual
To form the Dual:
If the primal is max., the dual is min., and vice
versa.
The right-hand-side values of the primal constraints
become the objective coefficients of the dual.
The primal objective function coefficients become
the right-hand-side of the dual constraints.
The transpose of the primal constraint coefficients
become the dual constraint coefficients.
(5)
Primal & Dual
60
1
3
80
4
2
120
50
2
1
2
1
2
1
+
+
+
X
X
X
X
X
X
:
to
Subject
:
Max
Primal:
Dual
120
1
4
50
3
2
60
80
2
1
2
1
2
1
+
+
+
U
U
U
U
U
U
:
to
Subject
:
Min
(6)