1. Lecture 6 – Integer Programming Models
Topics
General model
Logic constraint
Defining decision variables
Continuous vs. integral solution
Applications: staff scheduling, fixed charge, TSP
Piecewise linear approximations to nonlinear functions
J. Bard and J. W. Barnes
Operations Research Models and Methods
Copyright All rights reserved
8/14/04

2. } Linear Integer Programming - IP
Maximize/Minimize z = c1x1 + c2x2 +    + cnxn { }
s.t. ai1x1 + ai2x2 +    + ainxn
0 £ xj £ uj, j = 1,…,n
xj integer for some or all j =1,…,n
bi, i = 1,…,m =

3. Decision Variables in IP Models
An IP is a mixed integer program (MIP)
if some but not all decision variables are integer.
If all decision variables are integer we have a pure IP.
A binary decision variable must be 0 or 1
(a yes-no decision variable).
If all decision variables are binary the IP is a
binary IP (BIP)
Decision variables not integer-constrained are
continuous decision variables

4. Why study IP?
LP divisibility assumption (fractional solutions are permissible) is not always valid.
(2) Binary variables allow powerful new techniques like logical constraints.

5. Call Center Employee Scheduling
Day is divided into 6 periods, 4 hours each
Demand/period = (15, 10, 40, 70, 40, 35)
Workforce consists of full-timers and part-timers
FT = 8-hour shift, $121.6/ shift
PT = 4-hr shift, $51.8/shift
One PT = 5/6 FT
In any period, at least 2/3 of the staff must be FT employees
Problem: Find minimum cost workforce

6. Call Center Employee IP Model
Decision variables:
xt = # of full-time employees that work shift t
yt = # of part-time employees that work shift t
Min z =
121.6(x1 +
+x6
) (y1 +
+y6) 5
s.t. x6
+ x1 + y1 15 6 5 2 x1
+ x2 + y2 10 x6
+ x1
(x6
+ x1
+ y1) 6 3 . . . . . . 5 2 x5
+ x6 + y6 35 x5
+ x6
(x5
+ x6
+ y6) 3 6
xt ³ 0, yt ³ 0, t = 1,2, …,6

7. Optimal LP solution Feasible – Yes, Optimal? We do not know!
x = [ 7.06, 0, 40, , , ]
y = [ 0, 3.53, 0, , 0, 0 ]
z = 12, Not feasible to IP model
A correction method: round continuous solution
x = [ 8, 0, 40, 13, 27, 8 ]
y = [ 0, 3, 0, 21, 0, 0 ]
z = 12,916.8
Feasible – Yes, Optimal? We do not know!

8. ( )  100% = 0.95% How good is this solution? from the optimum.z *
= is a lower bound on integer optimum
Hence, the rounded solution is no more than LP – (
)  100% = 0.95%
from the optimum.
Optimal solution is
x = ( 10, 0, 40, 20, 20, 5 )
z*IP = 12,795.2
y = ( 10, 0, 0, 12, 0, 12 )
Here the optimal LP and IP objective functions have the same value.
This is not commonly true.

9. Sometimes there is no “obvious” feasible solution that can be obtained by roundingX2     
 optimal LP solution         X1
iso-cost line
optimal
IP solution
The IP solution can be “far” from the LP rounded solution
even when the rounded solution is feasible.

10. The Days-Off Scheduling Problem
Each employee works 5 days per week and is given 2 consecutive days off [ (5,7)-cycle problem]
cj = weekly cost of pattern j per employee
ri = number of required on day i
xj = number of employees assigned to days-off pattern j
Note: There are 7 days-off patterns; i, j = 1,…,7

11. Days-Off Mathematical Model7
j =1
Minimize z =  cjxj
subject to (  xj ) – xi – xi-1  ri , i = 1,…7
xj  0 and integer, j = 1,…,7; x0 = x7 7
j =1
Solve problem to get x*j
Minimum cost workforce W =  x*j 7
j =1

12. Compact Mathematical Model
Minimize z = cx
subject to
x  0 and integer

13. Logic Constraints Either-Or Constraints Either f1(x1,…,xn) £ b1
or f2(x1,…,xn) £ b2
or both.
IP formulation
Let y Î {0,1}
f1(x1,…,xn) £ b1 + My
f2(x1,…,xn) £ b2 + M(1 – y)
y = 0  first constraint must hold
y = 1  second constraint must hold
Optimization process will choose the y value.

14. K out of N constraints must hold
f1(x1,…,xn) £ b1 
fN(x1,…,xn) £ bN
f1(x1,…,xn) £ b1 + M(1 – y1) 
fN(x1,…,xn) £ bN + M(1 – yN)
At least
K out of N
must be
satisfied N
S yi = K
i =1
yi Î {0, 1}, i = 1,…,N

15. Example of K out of N Constraints
A production system has N potential quality control inspection strategies
Management has decided that K of these strategies should be adopted

16. } Region 1 constraints } Region 2 constraints } Region 3 constraints
Compound Alternatives
f1(x1,…,xn) £ b1 + M (1 – y1)
f2(x1,…,xn) £ b2 + M (1 – y1)
} Region 1 constraints
f3(x1,…,xn) £ b3 + M (1 – y2)
f4(x1,…,xn) £ b4 + M (1 – y2)
} Region 2 constraints
f5(x1,…,xn) £ b5 + M (1 – y3)
f6(x1,…,xn) £ b6 + M (1 – y3)
} Region 3 constraints
y1 + y2 + y3 = 1, y1, y2, y3 Î {0,1}

17. Min å fj(xj) where fj(xj) = {
Fixed-Charge Problem n
Min å fj(xj) where fj(xj) = {
kj + cjxj if xj > 0
if xj = 0
j =1
kj = set-up cost, cj = per unit cost
IP formulation: n
Min
å (cjxj + kjyj )
j =1
s.t.
xj £ Myj , j = 1,…,n
yj Î {0,1} , j = 1,…,n
xj ³ 0 , j = 1,…,n

18. Example: Warehouse Location Problem
A company has m potential warehouse sites and n customers
Data:
dj : demand for customer j
si : capacity (supply) of warehouse i
Decision variables:
yi : build a warehouse at site i? (1 = yes, 0 = no)
xij : shipment from warehouse i to customer j

19. Warehouse Location IP Modelå å
Min
cijxij + å
kiyi
i =1
j =1
i =1 m
å xij = dj j = 1,…,n
s.t.
Satisfy each customer’s
demand
i =1 n
å xij £ siyi i = 1,…,m
each warehouse can
ship no more than its
supply if it is built
j =1
xij ³ 0, i = 1,…,m, j = 1,…,n
yi Î {0,1}, j = 1,…,m

20. Example: Facility Location Problem (with forcing constraints)
Problem: open set of facilities & assign each customer to one facility at minimum cost.
No shipping here:
Cost of assigning customers to facilities could be based on a response time (e.g., locating fire stations).

21. Indices/Sets Data Decision Variables
potential facility location, j Î J with |J | = n
customers to be served, i Î I with |I | = m
Data
kj : cost of opening a facility at location j
cij : cost of serving customer i from facility j
Decision Variables
yj : open facility at location j (1 = yes, 0 = no)
xij : assign customer i to location j (1 = yes, 0 = no)

22. Strong Formulation å å cijxij + å kjyj å xij = 1, " i Î I Min
iÎI jÎJ
jÎJ
Each customer assigned to exactly one facility
Forcing constraint
å xij = 1, " i Î I
s.t.
jÎJ
xij £ yj , " i Î I, j Î J
xij , yj Î {0,1} , " i Î I, j Î J
Can assign customer i to facility j only if we open facility j.

23. { Mathematically Equivalent Weak formulation å å cijxij + å kjyj
Min
iÎI jÎJ
jÎJ
å xij = 1, " i Î I
s.t.
jÎJ {
xij must be 0 for all iÎI
if yj = 0;
up to |I |xij can be 1 if
yj = 1.
å xij £ |I |yj , j Î J
iÎI
xij, yj Î {0,1} " i Î I, j Î J
 Less computational efficiency

24. Traveling Salesman Problem (TSP)1 2 4 3
Problem: Find minimum distance tour that starts at a city, visits every other city exactly once, and returns to the initial city

25. Decision Variables Initial Formulation
xij = 1 if tour includes arc (i, j )
= 0 otherwise
Initial Formulation
Min
50x12 +
44x13 +
25x14
20x41 +
45x42 +
40x43
s.t. x + x + x
= 1 12 13 14
xij Î {0,1}, i ¹ j x + x + x
= 1 21 31 41 x + x + x
= 1 21 23 24 1 x + x + x
= 1 12 . . 32 42 . 2 4 x + x + x
= 1 41 42 43 x + x + x
= 1 3 14 24 34
This is incomplete because subtours are allowed.

26. Add Subtour Elimination Constraints
Let S1 = {1,3} & S2 = {2,4}
and require at least 1 arc from S1 to S2
x12 + x14 + x32 + x34 ³ 1
Disallows the “2 loop” solution.
Alternative formulation: x13 + x31 £ 1 and x24 + x42 £ 1
To generalize, let N = {1,…,n}, and let S  N
SEC: S xij £ |S| – 1, 2 £ |S| £ n/2
(ij)S

27. Example of SEC Let n = 10 and S = { 2, 5, 6, 9 }. Then |S| = 4.
SEC: x25 + x26 + x29 + x52 + x56 + x59 +
x62 + x65 + x69 + x92 + x95 + x96 £ 4 – 1 = 3 6 2 5 9
In general, there exponentially many subtour elimination constraints.

28. Sequencing problems with sequence-dependent
set-up times can be modeled as a TSP
Testing integrated circuits (ICs): A machine is used to test several batches of ICs of differing types.
After each batch the machine must be reset.
The change-over time depends on what type of IC we just tested and what we’ll test next.
Change-Over Times
IC Type 1 2 3 4 -- 10 17 15 20 19 18 50 44 25 45 40

29. If the order of testing is
IC1  IC2  IC4  IC3  IC
The total change-over time for one cycle is
= 98
The goal is to sequence the testing order so that the throughput (i.e., minimize cycle time) is maximized for fixed batch sizes
(Note that in this example the “travel times” are not symmetric)

30. Graph for IC Testing Example1
Dummy node 2 44 17 3 18 4
Sample path: 0  1  3  2  4  0
Cost of path: = 79

31. General Piecewise Linear Approximations
Given: fj(xj), 0  xj  uj
Let r = number of grid points
Let (dij, fij) be ith grid point, i = 1,…,r

32. Linear Transformation for jth Variable
Let xj = S i dij
and fj(xj) = S i fij
where S i = 1, i  0, i = 1,…,r
Not sufficient to guarantee solution is on one of the line segments. r
i=1 r
i=1

33. Additional Constraints for Piecewise Linear Approximation
Requirement: No more than two i can be positive; also i’s must be adjacent; i.e., i and i+1
a1 ≤ y1
ai ≤ yi-1 + yi , i = 2,…,r–1
ar ≤ yr-1
y1 + y2 + · · · + yr-1 = 1
yi = 0 or 1, i = 1,...,r–1

34. What you Should Know about Integer Programming
How to convert a problem statement in an IP model
How to define the decision variables
How to convert logic statements into constraints
How to formulation fixed charge problems, scheduling problems, covering problems,TSP, piece-wise linear approximation to nonlinear functions