Defeningen Operations Research
exercises
ad
Formulating LP
2
Step Define
-
1 D
M1 ML M3 M4
↓ 11 ,
i
↳
22
s
C4
84
24
Cj : amount invested in construction in month j-
amount invested in bands in I
month
Gj government
:
-
1j : amount invested
locally in month
I
Step 2
Define objective function
Max intrest
Max z =
0 , 06 ·
[j (j + 002 ·
[jgj + 00075 ·
Zjlj
Step 3 Defini functional constraints
Balancing budget M1 Cr + g1 + he 20 000 000
·
: : =
. .
be
M2 (2
+ 2 + 12 1 + 0 0075
: =
,
M3 : (3 +
93 + 13
= 1 + 0007512 + 1 + 0,
0291
M4 (y
gu + 14 1+ 0 , 087513 + 1 + 01
8292 + 1 + 0,86C
: + =
At least 10 000 000 at 00751410
M5 1 , 06 C2 +
110293 + 1 000 Onc
·
:
. .
, . .
·
At most 8 000.
. 000 in C Or G :
= 8 000 000 8 000 000
91
M1 : Ch .
M1 : = .
. .
: C + Ch 8 000 000 8 000 000
M2
g2
= M2 : =
.
.
g1 + .
.
M3 : (1 + C2 +3 = 8 000 000 M3 : = 8 000 000
.
.
g1 + g2 + 93 .
.
M4 C (2 + (3 + CH = 8 000 000 M4 - 8 000 000
g1 gz g3 94
: + .
.
: + + + .
.
Step 4
Sign restrictions :
(j . gjibj 0 Xje [1 , 2 ,
3 , 43
,3
Step 1 Define DV
x1 :
amount of Speedhawk
2 :
amount of silverbial
x3 : amount of Catman
Xu : amount of
classy
Step 2
Define objective function
Max
profit
Max z
Ej daily profit of type j Xi
= .
=
7011 + 8012 + 5013 + MOX4
Step 3
Define functional constraints
Maximum
budget 6000X1 700042 + 5000x3 900044 - 420 000
·
: + + .
Purchasing constant :1 + 12 + x3 + x4L 50
·
,
Balancing constraint : 11 + 12 x3 + x4(EX1 + 12 x4 0
·
= -
x3 - =
Seating capacity 311 : + 5/2 + 2x3 + 6x4 200
·
Step 4 :
Sign restrictions
1
xj48VgE [1 ,
2, 3, 43
,2) Graphical solution method exercises
a MAX Ca11 + X2
:
S t .
.
11 + 12 - 6
x1 + 2x2 18
Step 1 Determine functional region
a Draw constraints
·
x1 + x2 = 6
18
x1 + 2x2 18
·
=
b Determine
region that satisfies all construits
=> optimal solution is always a CD ·
if :· c =
12 E optimal solution his on
C1
· =
1 = Optimal solution his on
XB
c not in these slides 1 480- .
&
Min z =
100XA + 80 X 1 .
280 -
S t 2XA- XB0 W
it
. .
1 000
. -
XA + XB31009
Step 1 Determine
functional region
a Draw constraits 400--
H
·
LYA- XB = 0 200 -
·
XA + XB =
1000 XA
2no >
200 48 soo ado"1
b Determine
000
1
of points that satisfies
. .
set constrants
Step 2 Determine optimal solution
a) Draw objective function line
=> ex : P1200 , 0) = z(12000) =
100 1200 + 80 0 . - =
120 000
.
=> 100XA + 80XB =
120 000
.
b line
E E
Move objective function 2xa -
X =
0 Xa =
333 , 33
> Optimal solution
C stop moving= :A + XB =
1 . 000 XB =
666167
, 3 Simplex method theory lecture
A slide 4 Max z =
5x1 + 7x2
S t . .
X1 16
211 + 3X2419
x1 + x2x8
*1 , X238
a with
geoph
Iteration 0 : Initialization
Select origin : X1 ,
x2 =
0, 0
Iteration 1 : Current <1 , x2 =
0, 0
Step 1
Optimality test
00 = 200 =
0
adj . CPFS :
0 ZS On t
Step 2 Determine entering variable
* = rate
of improvement
Along x 0 X2
·
= :
mmx2 + =
1 = 20 ,
1 =
7 = x =
z(01) -
z(0 , 8) =
7
#
Along +2 0 Xi
·
= :
-x1 + =
1 = 210 =
5 = x =
2( , 8) -
z(0 , 8) =
5
Xx2) variable
= X2 is
entering
=>
111
Step 3 Determine leaving variable
constraints move along 1 =
0
X2 =
0 , 0
0 current
&
x1 =
6
211 + 342 =
19 , 65
0 min65;8 =
65
X1 + xz = 2 , 8
0
exercises
ad
Formulating LP
2
Step Define
-
1 D
M1 ML M3 M4
↓ 11 ,
i
↳
22
s
C4
84
24
Cj : amount invested in construction in month j-
amount invested in bands in I
month
Gj government
:
-
1j : amount invested
locally in month
I
Step 2
Define objective function
Max intrest
Max z =
0 , 06 ·
[j (j + 002 ·
[jgj + 00075 ·
Zjlj
Step 3 Defini functional constraints
Balancing budget M1 Cr + g1 + he 20 000 000
·
: : =
. .
be
M2 (2
+ 2 + 12 1 + 0 0075
: =
,
M3 : (3 +
93 + 13
= 1 + 0007512 + 1 + 0,
0291
M4 (y
gu + 14 1+ 0 , 087513 + 1 + 01
8292 + 1 + 0,86C
: + =
At least 10 000 000 at 00751410
M5 1 , 06 C2 +
110293 + 1 000 Onc
·
:
. .
, . .
·
At most 8 000.
. 000 in C Or G :
= 8 000 000 8 000 000
91
M1 : Ch .
M1 : = .
. .
: C + Ch 8 000 000 8 000 000
M2
g2
= M2 : =
.
.
g1 + .
.
M3 : (1 + C2 +3 = 8 000 000 M3 : = 8 000 000
.
.
g1 + g2 + 93 .
.
M4 C (2 + (3 + CH = 8 000 000 M4 - 8 000 000
g1 gz g3 94
: + .
.
: + + + .
.
Step 4
Sign restrictions :
(j . gjibj 0 Xje [1 , 2 ,
3 , 43
,3
Step 1 Define DV
x1 :
amount of Speedhawk
2 :
amount of silverbial
x3 : amount of Catman
Xu : amount of
classy
Step 2
Define objective function
Max
profit
Max z
Ej daily profit of type j Xi
= .
=
7011 + 8012 + 5013 + MOX4
Step 3
Define functional constraints
Maximum
budget 6000X1 700042 + 5000x3 900044 - 420 000
·
: + + .
Purchasing constant :1 + 12 + x3 + x4L 50
·
,
Balancing constraint : 11 + 12 x3 + x4(EX1 + 12 x4 0
·
= -
x3 - =
Seating capacity 311 : + 5/2 + 2x3 + 6x4 200
·
Step 4 :
Sign restrictions
1
xj48VgE [1 ,
2, 3, 43
,2) Graphical solution method exercises
a MAX Ca11 + X2
:
S t .
.
11 + 12 - 6
x1 + 2x2 18
Step 1 Determine functional region
a Draw constraints
·
x1 + x2 = 6
18
x1 + 2x2 18
·
=
b Determine
region that satisfies all construits
=> optimal solution is always a CD ·
if :· c =
12 E optimal solution his on
C1
· =
1 = Optimal solution his on
XB
c not in these slides 1 480- .
&
Min z =
100XA + 80 X 1 .
280 -
S t 2XA- XB0 W
it
. .
1 000
. -
XA + XB31009
Step 1 Determine
functional region
a Draw constraits 400--
H
·
LYA- XB = 0 200 -
·
XA + XB =
1000 XA
2no >
200 48 soo ado"1
b Determine
000
1
of points that satisfies
. .
set constrants
Step 2 Determine optimal solution
a) Draw objective function line
=> ex : P1200 , 0) = z(12000) =
100 1200 + 80 0 . - =
120 000
.
=> 100XA + 80XB =
120 000
.
b line
E E
Move objective function 2xa -
X =
0 Xa =
333 , 33
> Optimal solution
C stop moving= :A + XB =
1 . 000 XB =
666167
, 3 Simplex method theory lecture
A slide 4 Max z =
5x1 + 7x2
S t . .
X1 16
211 + 3X2419
x1 + x2x8
*1 , X238
a with
geoph
Iteration 0 : Initialization
Select origin : X1 ,
x2 =
0, 0
Iteration 1 : Current <1 , x2 =
0, 0
Step 1
Optimality test
00 = 200 =
0
adj . CPFS :
0 ZS On t
Step 2 Determine entering variable
* = rate
of improvement
Along x 0 X2
·
= :
mmx2 + =
1 = 20 ,
1 =
7 = x =
z(01) -
z(0 , 8) =
7
#
Along +2 0 Xi
·
= :
-x1 + =
1 = 210 =
5 = x =
2( , 8) -
z(0 , 8) =
5
Xx2) variable
= X2 is
entering
=>
111
Step 3 Determine leaving variable
constraints move along 1 =
0
X2 =
0 , 0
0 current
&
x1 =
6
211 + 342 =
19 , 65
0 min65;8 =
65
X1 + xz = 2 , 8
0
