TRAN 650 Urban System Engineering
Professor. Lazar Spasovic
HW #02 Assignment
By: Ali M. Kadhim
Due Date: September 24th, 2025
Problem #01:
Using Simplex Tableau Method
Standard form
Maximize Z,
Z- x1 - x2 - x3
=0
Subject to
Row
(0)
(1)
(2)
(3)
x1 + 2x2 + 2x3 + s1
2x1 + x2 + 2x3 +
s2
2x1 + 2x2 + x3 +
x1,x2,x3,s1,s2,s3 ≥ 0
Basic
RHS
x1
Variable
Z
0
-1
s1
20
1
s2
20
2
s3
20
2
+s3
= 20
= 20
= 20
x2
x3
s1
s2
s3
-1
2
1
2
-1
2
2
1
0
1
0
0
0
0
1
0
0
0
0
1
Ratio
(Test)
0/-1 = 0
20/1 = 20
20/2 = 10
20/2 = 10
Pivoting:
(0’) = (2’) +(0)
(1’) = -(2’) +(1)
(2’) = (2) *{1/2}
(3’) = -(2’)*{2} +(3)
Row
(0’)
(1’)
(2’)
(3’)
Basic
Variable
Z
s1
x1
s3
RHS
x1
x2
x3
s1
s2
s3
10
10
10
0
0
0
1
0
-0.5
1.5
0.5
1
0
1
1
-1
0
1
0
0
0.5
-0.5
0.5
-1
0
0
0
1
Pivoting:
(0”) = (3’’)*{1/2}+(0’)
(1”) = -(3’’)*{1.5}+(1’)
(2”) = -(3’’)*{1/2}+(2’)
(3”) = (3’)
Ratio
(Test)
10/1.5=6.7
10/0.5=20
0/1=0
Row
Basic
Variable
RHS
x1
x2
x3
s1
s2
s3
Ratio
(Test)
(0’’)
(1’’)
(2’’)
(3’’)
Z
s1
x1
x2
10
10
10
0
0
0
1
0
0
0
0
1
-0.5
2.5
1.5
-1
0
1
0
0
0
1
1
-1
0.5
-1.5
-0.5
1
-20
4
6.67
0
RHS
x1
x2
x3
s1
s2
s3
Ratio
(Test)
12
4
4
4
0
0
1
0
0
0
0
1
0
1
0
0
0.2
0.4
-0.6
0.4
0.2
0.4
0.4
-0.6
0.2
-0.6
0.4
0.4
All ≥ 0
Pivoting:
(0’’’) = (1’’’)*{1/2}+(0’’)
(1’’’) = -(1’’)/ {2.5}
(2’’’) = -(1’’’)*{1.5}+(2’’)
(3’’’) = (1’’’)+ (3’’’)
Row
(0’’’)
(1’’’)
(2’’’)
(3’’’)
Basic
Variable
Z
x3
x1
x2
Reached Optimal!
x = (4,4,4,0,0,0)
Z * = 12
LP OPTIMUM FOUND AT STEP
3
OBJECTIVE FUNCTION VALUE
1)
12.00000
VARIABLE
VALUE
X1
4.000000
X2
4.000000
X3
4.000000
REDUCED COST
0.000000
0.000000
0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2)
0.000000
0.200000
3)
0.000000
0.200000
4)
0.000000
0.200000
NO. ITERATIONS=
3
RANGES IN WHICH THE BASIS IS UNCHANGED:
VARIABLE
X1
X2
X3
ROW
2
3
4
OBJ COEFFICIENT RANGES
CURRENT
ALLOWABLE
ALLOWABLE
COEF
INCREASE
DECREASE
1.000000
0.333333
0.500000
1.000000
0.333333
0.500000
1.000000
0.333333
0.500000
RIGHTHAND SIDE RANGES
CURRENT
ALLOWABLE
ALLOWABLE
RHS
INCREASE
DECREASE
20.000000
6.666667
10.000000
20.000000
6.666667
10.000000
20.000000
6.666667
10.000000
Problem #02:
Standard form
Maximize Z,
Z- 2x1 - x2 =0
Subject to
3x1 + 2x2 + s1
x1 + x2
+ s2
x1,x2,s1,s2, ≥ 0
=6
=4
x2 unrestricted in sign → x2 = x2’-x2’’
Row
Basic
Variable
RHS
x1
x2’
x2’’
s1
s2
Ratio
(Test)
(0)
(1)
(2)
Z
s1
s2
0
6
4
-2
3
1
-1
1
1
1
-1
-1
0
1
0
0
0
1
6/3 = 2
4/1 = 4
Pivoting:
(0’) = (1’)*2+(0)
(1’) = (1)/{3}
(2’) = -(1’)*(2)
Row
Basic
Variable
RHS
x1
x2’
x2’’
s1
s2
Ratio (Test)
(0’)
(1’)
(2’)
Z
4
2
2
0
1
0
-1/3
1/3
2/3
1/3
-1/3
-2/3
2/3
1/3
-1/3
0
0
1
2/(1/3)=6.67
2/(2/3)=3
Ratio
(Test)
All ≥ 0
x1
s2
Pivoting:
(0’’) = (2’’)*{1/3}+(0’)
(1’’) = -(2’’)/{3}+(1’)
(2’’) = (2’)*(3/2)
Row
Basic
Variable
RHS
x1
x2’
x2’’
s1
s2
(0’’)
(1’’)
(2’’)
Z
5
1
3
0
1
0
0
0
1
0
0
-1
0.5
0.5
-0.5
0.5
-0.5
3/2
x1
x2’
Reached Optimal!
x = (1,3,0,0) Z * = 5
LP OPTIMUM FOUND AT STEP
2
OBJECTIVE FUNCTION VALUE
1)
5.000000
VARIABLE
VALUE
X1
1.000000
X2
3.000000
REDUCED COST
0.000000
0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2)
0.000000
0.500000
3)
0.000000
0.500000
NO. ITERATIONS=
2
RANGES IN WHICH THE BASIS IS UNCHANGED:
VARIABLE
X1
X2
ROW
2
3
OBJ COEFFICIENT RANGES
CURRENT
ALLOWABLE
ALLOWABLE
COEF
INCREASE
DECREASE
2.000000
1.000000
1.000000
1.000000
1.000000
0.333333
RIGHTHAND SIDE RANGES
CURRENT
ALLOWABLE
ALLOWABLE
RHS
INCREASE
DECREASE
6.000000
6.000000
2.000000
4.000000
2.000000
2.000000