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Use Implicit Enumeration to solve the following 0-1 IP.MAX Z=3X1+X2+2X3-X4+X5
ST 2X1+X2-3X4<=1 X1+2X2-3X3-X4+2X5=>2
Xi=0 or 1Please show enumeration tree. If a node is fathomed, indicate why it is fathomed. Provide the sequence of problems solved, eg. P1->P3->P4. Clearly indicate what the optimal solution is. Branch on x1 first.I need to understand the process, so no LINGO solution or any other program please. Steps have to be done by hand.MAX Z=3X1+X2+2X3-X4+X5
ST 2X1+X2-3X4<=1 X1+2X2-3X3-X4+2X5=>2
Xi=0 or 1
Branch on x1
P1: x1=1
P2:x1=0At P1 x1=1,
The subproblem is
MAX Z=3+x2+2×3-x4+x5
s.t.
2+x2-3×4<=1 1+2x2-3x3-x4+2x5>=2Or
MAX Z=3+x2+2×3-x4+x5
s.t.
x2-3×4<= -1 2x2-3x3-x4+2x5>=1
By looking at the first constraint, x4 has to be 1, otherwise x2<=-1 is invalid, since x2=1 or 0. So x4=1 Therefore X2-3<=-1 2x2-3x3-1+2x5>=1
Or
X2<=2 2x2-3x3+2x5>=2
The first constraint is redundant, since x2= 1 or 0, both satisfying x2<=2. By looking at the second constraint, we see x3 has to be 0, otherwise 2x2-3+2x5>=2
2×2+2×5>=5 is infeasible, since 2×2+2×5<=2+2=4. So x4=1, x3=0, x1=1. To maximize 3X1+X2+2X3-X4+X5, we choose x2=x5=1 Therefore the optimal solution at P1 is (1,1,0,1,1), with objective value=3+1+0-1+1=4 This is the most current lower bound of the original problem. Any other sub-problem with objective value<=4 will be fathomed.At P2: x1=0 The subproblem is MAX Z=x2+2x3-x4+x5 s.t. x2-3x4<=1 2 x2-3 x3-x4+2 x5>=2
We then branch on x2.
At P3: x2=1
The subproblem is
MAX Z=1+2×3-x4+x5
s.t.
1-3×4<=1 2 -3 x3-x4+2 x5>=2
Or
3×4>=0 (redundant)
3×3+x4-2×5<=0Branch on x3 At P4: x3=1 The subproblem is MAX Z=1+2-x4+x5 s.t. 3x4>=0 (redundant)
3+x4-2×5<=0 Or -x4+2x5>=3
This is infeasible, since 2×5-x4<=2*1-x4<=2. So P4 is fathomed.At P5: x3=0 The subproblem is MAX Z=1+0-x4+x5 s.t. 3x4>=0 (redundant)
x4-2×5<=0 Branch on x4At P6: x4=1 The subproblem is MAX Z=1+0-1+x5 s.t. 1-2x5<=0 Or 2x5>=1, so x5=1
The solution is x5=1, with objective value=1-1+1=1, which less than the current lower bound=4
So this branch is fathomedAt P7: x4=0
The subproblem is
MAX Z=1+0-0+x5
s.t.
0-2×5<=0 Or 2x5>=0, so x5=1 is the optimal solution, with objective value=1+1=2, , which less than the current lower bound=4
So this branch is fathomedNow we start from branching at x2=0
At P8: x2=0
The subproblem is
MAX Z=0+2×3-x4+x5
s.t.
-3×4<=1 -3 x3-x4+2 x5>=2
Or
3×4>= – 1
3×3+x4-2×5<=2 The first constraint is invalid, since x4>=0
So this branch is fathomed.Therefore the only unfathomed branch is
(x1,x2,x3,x4,x5)=(1,1,0,1,1), with objective value=4, which is optimal.Branch tree:

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