Islamic University of Gaza
Faculty of Engineering
School of Industrial Engineering
EIND 2302 Operations Research 1
Midterm exam solutions
1.
X1
0
1
0
Z
X2
a
c
d
X3
0
0
1
X4
1
3
1
RHS
3
B
4
a) For what values of "a" is the solution optimal ? (2.5)
The solution is optimal if a >= 0
b) For what values of "b" is the solution feasible? (2.5)
The solution is feasible if b>= 0
C) For what values of "c" and "d" is the solution unbounded? (2.5)
The solution is unbounded if c and d <=0
d) For what values of "b" is the solution degenerate? (2.5)
The solution is degenerate if b=0
2. Given the following LP
Max 3X1+2X2+5X3
St
X1+2X2+X3<=430
3X1+2X3<=460
X1+4X2<=420
Let X4, X5, X6 represent the slacks in the constraints
The optimal basic variables and the associated inverse are given
X2, X3, x6
1/2 -1/4
0
1/2
-2
1
0
0
1
a) Compute the entire simplex tableau (10)
Z
X2
X3
X6
1
X1
C1=4
A1=-1/4
A2=3/2
A3=2
X2
0
1
0
0
X3
0
0
1
0
X4
Y1=1
1/2
0
-2
X5
Y2=2
-1/4
1/2
1
X6
0
0
0
1
RHS
Z=1350
B1=100
B2=230
B3=20
Y1, y2, y3 =
2
=
5 0
1 2
0
Coefficient of X1= C1= y1+3y2+y3-3= 1+3*2+0-3= 4
New RHS =
=
430
460
420
100
230
20
A1
A2
A3
1
3
1
=
-1/4
=
3/2
2
Z value = 3*0+ 2*100+5*230=1350
b) Determine the worth of increasing the time of each operation by 1 minute?
(3)
The worth of increasing the time of operation 1, 2, and 3 are y1=1, y2=2
y3=0 respectively.
and
c) Rank the three operations in order of priority of increase in time allocation
? (3)
The higher the dual value, the higher should be the priority of increase in time
allocation. So operation 2 give high priority, then operation 1 and finally
operation 3.
d) Check the optimality of the problem if the objective function coefficient are
changed to (5, 1, 7) and (3, 2, 1) ? (6)
Since the changes involve both x2, and x3 which happen to be basic in the
current solution, we must determine the new dual values.
2
(y1 y2 y3 )= (1 7 0)
= ( .5 3.25 0)
The next step is to recompute the z-equation coefficient by taking the difference
between the left and right side of the dual constraints.
X1 coefficient= y1+3y2+y3-5 = .5+3*3.25+0-5= 5.25
Since all the z equations coefficient are >=0(maximization), the change indicated in
the objective function will not change the optimum variables or their values .
The same calculations should be conducted if we change the objective function
coefficients to (3, 2, 1)
1. new dual values will be (1 0 0)
2. x1 coefficient = 1+3*0+0-3=-1
x2 coefficient = 2*1+4*0-2=0 (basic )
x4 coefficient= y1=1
x5 coefficient = y2= 0
Since the X1 coefficient is -2 <0(maximization), the change indicated in the
objective function will change the optimum variables and their values
e) Check the feasibility if the RHS are changed to (400 490 380)? (3)
400
490
380
New RHS =
=
77.5
245
70
Since the RHS elements remain nonnegative, the change indicated in the RHS
will not affect the feasibility and the current basic variables remain unchanged .
3
3) Given the problem
Max 2X2-5X3
St
X1+X3>=2
2X1+X2+6X3<=6
X1-X2+3X3=0
A) Write the problem in standard format? (4)
Max Z= 2X2-5X3-MR1-MR2 = 2X2-5X3-M(2-X1-X3+S1)-M(-X1+X23X3)= 2MX1 +(2-M)X2+(-5+4M)X3-MS1-2M
St
X1+X3-S1+R1=2
2X1+X2+6X3+S2=6
X1-X2+3X3+R2=0
R1= 2-X1-X3+S1
R2=-X1+X2-3X3
B) Write the dual (5)
MIN 2y1+6y2+oy3
St
Y1+2y2+y3>=0
Y2-y3>=2
Y1+6y2+3y3>=-5
-y1>=0
y1<=0
Y2>=0
Y3>=-M y3 is unrestricted
c) Solve the problem (dual or primal)? ( 6)
4