1. Introduction to Network Design – Optimization Background
ECE 780
Introduction to Network Design – Optimization Background
Includes slide materials developed by Wayne D. Grover, John Doucette, Dave Morley
© Wayne D. Grover 2002, 2003, 2004

2. Outline Optimization Mesh-Restoration Concept Mathematical Programming
Linear Programming (LP)
Formulating LP Problems
Solving LP Problems
Integer Programming
Solving MIP Problems
Algebraic Expression of LP/IP Problems
Mesh-Restoration Concept
Terminology
Spare Capacity Sharing
Spare Capacity Placement (SCP)
SCP Integer Programming problem

3. Mathematical Programming
Definition
“A Mathematical Programming Model is a mathematical decision model for planning (programming) decisions that optimize an objective function and satisfy limitations imposed by mathematical constraints.” 1
General Symbolic Model
Maximize (or minimize):
Objective
Subject to:
Constraints …
… where
are the decision variables.
1 T.W. Knowles, Management Science: Building and Using Models, Irwin, 1989.

4. Mathematical Programming
Types of Mathematical Programs:
Linear Programs (LP): the objective and constraint functions are linear and the decision variables are continuous.
Integer (Linear) Programs (IP): one or more of the decision variables are restricted to integer values only and the functions are linear.
Pure IP: all decision variables are integer.
Mixed IP (MIP): some decision variables are integer, others are continuous.
1/0 MIP: some or all decision variables are further restricted to be valued either “1” or “0”.
Nonlinear Programs: one or more of the functions is not linear.

5. Linear Programming General Symbolic Form Maximize: Subject to:
Objective
Subject to:
Constraints …
Bounds
… where
are the model parameters.

6. Linear Programming General Restrictions
All decision variables must be nonnegative,
Constant terms cannot appear on the LHS of a constraint.
No variable can appear on the RHS of a constraint.
No variable can appear more than once in a function, i.e. objective or constraint.
Steps for Formulating LP Models
Construct a verbal model.
Define the decision variables.
Construct the symbolic model.

7. Formulating LP Problems
An example2
A steel company must decide how to allocate production time on a rolling mill. The mill takes unfinished slabs of steel as input and can produce either of two products: bands and coils. The products come off the mill at different rates and also have different profitabilities:
Tons/ Profit/
hour ton
Bands 200 $25
Coils 140 $30
The weekly production that can be justified based on current and forecast orders are:
Maximum tons: Bands 6,000
Coils 4,000
2 from, R. Fourer, D. Gay, B. Kernighan, AMPL, Boyd & Fraser, 1993, pp

8. Formulating LP Problems
An example (cont’d)
The question facing the company is: If 40 hours of production time are available, how many tons of bands and coils should be produced to bring in the greatest total profit?
Constructing the Verbal model
Put the objective and constraints into words.
For constraints, use the form
{a verbal description of the LHS} {a relationship} {an RHS constant}
Maximize: total profit
Subject to: total number of production hours  40
tons of bands produced  6,000
tons of coils produced  4,000

9. Formulating LP Problems
Defining the Decision Variables
XB number of tons of bands produced.
XC number of tons of coils produced.
Construct the Symbolic Model
Maximize:
Subject to:

10. Solving LP Problems Graphical Solution Method Constraints Profit 6000
Coils
Coils
Constraints
220K
Profit
6000
6000
Hours
192K
4000
4000
Optimal solution
Feasible region
2000
2000
120K
Bands
Bands
2000
4000
6000
8000
2000
4000
6000
8000

11. Solving LP Problems 4 Possible Outcomes Unique Optimal Solution
Alternate Optimal Solutions
Unbounded Optimal Solution
No Feasible Solution

12. Solving LP Problems Simplex method AMPL and CPLEX
Efficient algorithm to solve LP problems by performing matrix operations on the LP Tableau.
Developed by George Dantzig (1947).
Can be used to solve small LP problems by hand.
AMPL and CPLEX
AMPL: modeling language (and software) for designing large and complex LP/IP problems.
CPLEX: software package (“solver”) to solve large and complex LP/IP problems.
Sub-Optimal Algorithms (Heuristics)
Simulated annealing.
Genetic algorithms.
Tabu search.
Many others, often very specific to the type of problem.

13. Integer Programming Convert Example to Integer Program
Assume that orders for bands and coils are placed (and filled) in 1,000s of pounds only.
Although feasible region is greatly reduced, problem becomes much more difficult.
New Symbolic Model
Let the new decision variables be the number of 1000 pound “units” or orders of bands and coils.
Maximize:
Subject to:
integer
integer

14. Integer Programming Graphical Solution Method 2 4 6 82 4 6 8
Feasible integer solutions
Bands
Coils
$185K
Optimal integer solution ($185K)

15. Solving MIP Problems Branch-and-Bound Procedure
The solution space consists of a tree of LP subproblems, in which each integer variable is either fixed or its integrality constraint is “relaxed.”
The root node of the tree is the LP relaxation of the problem, i.e. all integer variables are relaxed.
The relaxation can result in an all integer solution, or a fractional solution (some decision variables are non-integer).
If the solution of the relaxation has fractional-valued integer variables, a fractional variable is selected for branching and two new subproblems are generated, each with more restrictive bounds on the branching variable.
The subproblems can result in an all integer solution, an infeasible problem or another fractional solution.
If the solution is fractional, the process is repeated.
Branches are “fathomed” if the subproblem is infeasible, the objective value is worse than the current best integer solution or the subproblem gives an integer solution.

16. Solving MIP Problems
Branch-and-Bound Tree for the Bands and Coils IP Problem
Supplement: see “Appendix” slides at end for detailed working of this example 1 2 3 6 4 5
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
6<=XB<=6
2<=XC<=4
*Infeasible
0<=XB<=5
3<=XC<=4
Obj. = 183K
XB = 3.00
XC = 3.71
0<=XC<=4
Obj. = 192K
XC = 1.40
Obj. = 189K
XB = 5.14
XC = 2.00
XB = 5.00
XC = 2.10
2<=XC<=2
Obj. = 185K 8 7 9 10

17. Algebraic Expression of LP/IP Problems
Why use it?
Most LP/IP problems are quite large and it becomes very cumbersome to describe them by explicitly giving each linear function, equality, and inequality in full.
It is desirable to model problems in a more general fashion (e.g. give an IP for optimally designing a mesh-restorable network in general as opposed to doing so for a specific network).

18. Algebraic Expression of LP/IP Problems
Basic Production Model (Revisited)
Original Model
Algebraic Model
Maximize:
Subject to:
Given:
a set of products
hours available at the mill
tons per hour of product j, for each
profit per ton of product j, for each
maximum tons of product j, for each
Define variables:
tons of product j to be made, for each
\Problem name: prob.lp
Maximize
25 XB + 30 XC
Subject To
0.005 XB XC <= 40
Bounds
0 <= XB <= 6000
0 <= XC <= 4000
End

19. Solving Network Flow Problems with LP/IP Models

20. Network flow LP formulations
Example: LP to find max flow between nodes 1-5 1 5 2 4 3 20 10 8
Edge capacities:
Define directional flow variables:
x42 1 5 2 4 3
x12
x13
x32
x23
x45
x24
x35
“trans-shipment nodes”
source
sink

21. Network flow LP formulations (2)
“transportation problem” or “arc-flow” approach
To maximize (1  5) flow : /* using lp_solve syntax */
max: x12 + x13 ; /* (or x35 + x45) */
subject to constraints:
c1: x12 + x13 = x45 + x35 ; /* source = sink */
c2: x12 + x32 - x23 + x42 - x24 = 0 ; /* node 2 trans-shipment*/
c3: x13 + x23 - x32 - x35 = 0 ; /* node 3 trans-shipment*/
c4: x24 - x45 - x42 = 0 ; /* node 4 trans-shipment*/
x42 1 5 2 4 3
x12
x13
x32
x23
x45
x24
x35

22. Network flow LP formulations (2)
“transportation problem” or “arc-flow” approach
Continued....
Also subject to (span capacity constraints):
x12 <= 20 ;
x13 <= 8 ;
x24 <= 10 ;
x42 <= 10 ;
x23 <= 2 ;
x32 <= 2 ;
x35 <= 20 ;
x45 <= 5 ; 1 5 2 4 3 20 10 8
x42 1 5 2 4 3
x12
x13
x32
x23
x45
x24
x35

23. Network flow LP formulations (3)
“transportation problem” or “arc-flow” approach
Symbolically....
s.t.
Where:
E = set of edges that exist
N = set of nodes
sij = spare capacity on edge ij (= sji)

24. Network flow LP formulations (4)
Alternate approach: “flow assignment to routes” or “arc-path” approach
Symbolically....
s.t.
Where:
E = set of edges that exist (indexed by k)
P 15= set of “eligible” distinct routes between nodes 1 and 5 (source-sink)
sk = spare capacity on edge k
if the i th distinct route crosses span k. Zero otherwise

25. Network flow LP formulations (5)
“flow assignment to routes” or “arc-path” approach - example
Identify all distinct routes between source- sink (set P15) .... f1 f4 2 4 2 4 1 1 f2 3 5 3 5 f3
Route associated flow variable f1 f2
Route associated flow variable f3 f2

26. Network flow LP formulations (6)
“flow assignment to routes” or “arc-path” approach - example (2)
To maximize (1 -> 5) flow :
max: f1 + f2 + f3 + f4 ;
subject to constraints:
c1: f1 + f2 <= 20 ; /*span 12 capacity */
c2: f4 + f3 <= 8 ; /* span 13 capacity */
c3: f4 + f2 <= 2 ; /* span 23 (discuss this ) */
c4: f1 + f4 <= 10 ; /* span 24 */ f1 f4 20 10 5 8 2 4 2 4 20 10 5 8 1 1 2 2 f2 3 5 3 5 f3

27. Network flow LP formulations (6)
“flow assignment to routes” or “arc-path” approach - example (2)
What are the remaining constraints ? :
- for span ?:
- for span ?
c5: f3 + f2 <= 20; /*span 35 capacity */
c6: f4 + f1 <= 5; /*span 45 capacity */
(note this makes prior constraint f4 + f1<=10 redundant ) f1 f4 20 2 10 4 2 20 10 4 1 1 2 2 5 5 f2 8 3 5 8 3 5 20 20 f3

28. Network flow LP formulations (7)
“flow assignment to routes” or “arc-path” approach - example (3)
Note that the “indicator” parameters do not appear explicitly in the executable model.
Really they just represent our knowledge of the topology and the routes being considered.
Implicitly above, we only wrote the flow variables that had non-zero coefficients.
Examples:
Hence f1 is in the first constraint
Hence f3 is in the fifth constraint, etc.

29. LP/IP for Mesh-Restoration Design
Networks are Inherently Mesh-Like
Level(3)
North American Network
Distributed mesh-restoration exploits network connectivity to allow sharing of redundancy.

30. Spare Capacity Sharing
Consider 2 different failure scenarios X X
Restoration is allowed to follow multiple distinct routes.
Restoration route for both failure scenarios have several spans in common.
Spare capacity on each span contributes to restorability of many spans.

31. Spare Capacity Placement (SCP)
Optimal Design
Objective is to find least costly way to place sufficient spare capacity on a network such that all spans are fully restorable.
Can we use LP/IP?
Reference:
M. Herzberg and S. J. Bye, “An Optimal Spare-Capacity Assignment Model for Survivable Networks with Hop Limits”, IEEE Globecom’94, 1994
Integer Programming Approach
Objective Function:
Minimize Cost of Spare Capacity Placement
Constraints:
Each possible span failure has enough restoration flow for full restoration.
Enough spare capacity exists on each span to accommodate restoration flows.

32. SCP Integer Programming Problem
Parameters (Inputs)
Cj: Cost of each unit of capacity on span j
Li: Target Restoration level for span i (Li = 1 assumed)
S: Set of spans in the network
Pi: Set of eligible routes for restoration of span i
wj: Number of working links (capacity units) on span j
i,jp: Equal to 1 if pth eligible route for span i uses span j
Variables (Outputs)
fip: Restoration flow assigned to pth route for span i
sj: Number of spare capacity units placed on span j

33. SCP Integer Programming Problem
Objective Function:
Subject To: 1. 2.

34. Appendix Added October 5, 2004 to better explain the branch-and-bound procedure

35. Solving MIP Problems
Initial completely relaxed LP problem and solution
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40 1 3 2
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 1 2 4 10
Down-forcing branch direction
Up-forcing branch direction 9
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10
Bounds
6<=XB<=6
2<=XC<=4
Solution
*Infeasible 3 6 5 7 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 8
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4 5

36. XC is fractional so “branch” on it by adding bounds…
Coils 6
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40 4 2
XC is fractional so “branch” on it by adding bounds…
Bands 2 4 6 8

37. This bound is changed to force XC “down”
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40
This bound is changed to force XC “down” 1 3 2
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 1 2 4 10 9
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10
Bounds
6<=XB<=6
2<=XC<=4
Solution
*Infeasible 3 6 5 7 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 8
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4 5

38. Found an integer solution – best so far too
Coils 6 4
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00 2
Found an integer solution – best so far too
Bands 2 4 6 8

39. Now test forcing XC “up” to at least two
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40
Now test forcing XC “up” to at least two 1 3 2
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 1 2 4 10 9
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10
Bounds
6<=XB<=6
2<=XC<=4
Solution
*Infeasible 3 6 5 7 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 8
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4 5

40. Coils 6
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 4
Dang ! , now the solution has gone fractional on XB..ok…So now we will branch on XB 2
Bands 2 4 6 8

41. Xb became 5.14, so now test forcing XB “down” to at least five
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40 1 3 2
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 1 2
Xb became 5.14, so now test forcing XB “down” to at least five 4 10 9
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10
Bounds
6<=XB<=6
2<=XC<=4
Solution
*Infeasible 3 6 5 7 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 8
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4 5

42. Coils 6
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10 4
Ok now XC is fractional again on us. so now we will branch on XC again…force it down first… 2
Bands 2 4 6 8

43. XC became 2.10, so now test forcing XC “down” to at least two
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40 1 3 2
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 1 2 4 10 9
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10
Bounds
6<=XB<=6
2<=XC<=4
Solution
*Infeasible
XC became 2.10, so now test forcing XC “down” to at least two 3 6 5 7 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 8
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4 5

44. Coils 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 4
Found another fully integer solution..This one takes over as best so far (185k$)
In fact this is the optimum, but we don’t know it yet…have to go on to prove this.. 2
Bands 2 4 6 8

45. To continue, now force XC from last node “up” to at least three
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40 1 3 2
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 1 2 4 10 9
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10
Bounds
6<=XB<=6
2<=XC<=4
Solution
*Infeasible 3 6
To continue, now force XC from last node “up” to at least three 5 7 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 8
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4 5

46. Fractional solution with lower objective value. Go back to last node…
Coils 6
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4
Fractional solution with lower objective value. Go back to last node… 2
Bands 2 4 6 8

47. Result is no feasible solution.
Bounds
0<=XB<=6
0<=XC<=4
Solution
Obj. = 192K
XB = 6.00
XC = 1.40 1 3
This brings us back to this node, where XB was fractional and we now to force it “up” to at least 6.
Result is no feasible solution.
Tree is exhausted, so the integer solution at node 4 is now “proven” optimal. 2
Bounds
0<=XB<=6
0<=XC<=1
Solution
Obj. = 180K
XB = 6.00
XC = 1.00
Bounds
0<=XB<=6
2<=XC<=4
Solution
Obj. = 189K
XB = 5.14
XC = 2.00 1 2 4 10 9
Bounds
0<=XB<=5
2<=XC<=4
Solution
Obj. = 189K
XB = 5.00
XC = 2.10
Bounds
6<=XB<=6
2<=XC<=4
Solution
*Infeasible 3 6 5 7 6
Bounds
0<=XB<=5
2<=XC<=2
Solution
Obj. = 185K
XB = 5.00
XC = 2.00 8
Bounds
0<=XB<=5
3<=XC<=4
Solution
Obj. = 183K
XB = 3.00
XC = 3.71 4 5

48. The Overall Search trajectory
Coils 6 4 2
Bands 2 4 6 8