INTEGER LINEAR PROGRAMMING

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INTEGER LINEAR PROGRAMMING
There are many LP problems in which the decision
variables will take only integer values. If all
the decision variables will only take integer
values it is called a pure integer LPP otherwise
the problem is called a mixed integer LPP. We
discuss Land-Doig Branch and Bound algorithm to
solve an integer LPP.
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Branch-and-Bound Algorithm (Problem 2a problem
set 9.2A page 378) Maximize
Subject to
and integers
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The set of feasible region consists of the
Lattice points (0,0),(1,0),(2,0),(0,1),(1,1). The
associated LPP, LP0 is defined by removing the
integer restrictions. Solving we get x11.69,
x21.13, z7.31. Because the optimum LP0 solution
does not satisfy the integer requirements, the
BB algorithm modifies the solution space in a
manner that eventually identifies the integer
optimum solution. First we select one of the
variables whose optimum value at LP0 is not an
integer. We select x1.
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We replace the original LP0 with two new LPPs,
LP1 and LP2 defined as LP1 space LP0 space
LP2 space LP0 space
We set lower bound -?
We solve the LP1 problem (which is given by
adding to LP0, the constraint
The solution is x1 1, x2 1.4,
z 5.80
Now x1 is integer but x2 is not.
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So we replace the LP1 problem with two new LPs,
LP3,LP4, defined as
LP3 space LP1 space LP4 space LP1 space
We now solve the LP3 problem. The solution is
x1 1, x2 1, z 5
Thus we have now got an all integer solution. So
we need not consider this node any further as we
will not get a better optimum since we will be
looking at smaller subsets. We say this node is
fathomed. We also update the lower bound to 5.
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Now we solve the LP4 problem. We find it is
infeasible. We say this node is also
fathomed. Now we look at LP2 problem. Solving we
get x1 2, x2 0.5, z 7
We now replace the LP2 problem with two new LPs,
LP5, LP6 defined as
LP5 LP2 LP6 LP2
Solving LP5 we get x1 2.25, x2 0, z 6.75
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Thus we replace the LP5 problem with two new LPs
LP7, LP8 defined as
LP7 LP5 LP8 LP5
Solving LP7 we get the all-integer solution
x1 2, x2 0, z 6 . And this node is
fathomed.
We update the lower bound to 6.
We find LP8 is infeasible. Thus this is also
fathomed. We also find LP6 is infeasible. Thus
this is also fathomed. Thus ultimately we got the
all-integer optimum solution as
x1 2, x2 0, z 6.
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All the above are depicted graphically as
follows
LP0 x11.69,x21.13,z7.31
LP1 x11,x21.4,z5.8
LP2 x12,x20.5,z7.0
LP3 x11,x21,z5
LP4 Infeasible
LP6 Infeasible
(fathomed)
(fathomed)
(fathomed)
LP5 x12.25,x20, z6.75
LP8 Infeasible
LP7 x12,x20,z6
(fathomed)
(fathomed)
Optimum Solution
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Summary of the BB Algorithm
Assume the problem is a maximization one. Set an
initial lower bound z - ? on the optimal value
of the ILPP. Set i 0.
Step 1. (Fathoming/bounding) Select LPi, the
next problem to be examined. Solve LPi and
attempt to fathom it using one of the three
conditions
(a) The optimal z-value of LPi cannot yield a
better objective value than the current lower
bound.
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(b) LPi yields a better feasible integer solution
than the current lower bound. (c) LPi has no
feasible solution.
Two cases will arise.
(i) If LPi is fathomed and a better solution is
found, then update the lower bound. If all
subproblems have been fathomed, stop. The optimum
ILP is associated with the current lower bound,
if any. Otherwise set i i 1 and repeat Step 1.
(ii) If LPi is not fathomed, go to step 2 for
branching.
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Step 2. (Branching) Select one of the integer
variables xj, whose optimum value in the
LPi solution is not integer. Eliminate the
region
by creating two LP subproblems that correspond to
and
Set i i 1. Go to step 1.
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Problem 4 Problem 9.1C Page 369 The
treasures of King Tut are on display in a museum
in New Orleans. The layout of the museum is shown
in the figure below, with different rooms joined
by open doors. A guard standing at a door can
watch two adjoining rooms. The museum wants to
ensure guard presence in every room, using the
minimum number possible. Formulate the problem
as an ILPP, and find the optimum solution.
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Figure for the problem
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x12
2
1
x13
x25
4
5
3
x34
x45
x58
x67
x78
7
8
6
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We define xij 1 if guard is posted between
rooms i and j and 0 otherwise. We have one
constraint per room. Thus we have the LPP
Minimize
Subject to
xij 1 or 0 for all j.
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Using TORA, we get the optimal solution (as
expected) as x121, x341, x671, x781 z 4.