Mathematical Programming Introduction

© 2013 OptiRisk India (P) Ltd, All rights reserved
Bala. Padmakumar
Director & CEO
OptiRisk India
Ph: +91 98406 18472/ +91 44 4501 8472
Web: http://www.optiriskindia.com
Email : optimize@optiriskindia.com

1. Operations Research
2. Optimization
3. Linear Programming
4. Case Study (Linear Programming )
5. Summary
2© 2010-13 OptiRisk India (P) Ltd, All rights reserved

 Analysing Decision Problems,
 Formulating Mathematical Models
 Coming-up the best possible (optimal) solutions
 Testing the solution for sensitivity, “what-if”, etc
OR deals with Optimization

2. Optimization
5. Summary

Optimization is a
decision making process
and a set of related tools that
employ mathematics,
Algorithms, and computer software not
only to sort and organize data, but to
use that data to make
Recommendations faster and better
than humans can.
- The Optimization Edge, Steve Sashihara

Route
#
Possible
Routes
Distance(KM)
1 A – B – C 245
2 A – C – B 315
3 B – A – C 370
4 B – C – A 315
5 C – A – B 370
6 C – B – A 245
Example : Find the shortest path for delivery
Deliveries Possible Routes Time to process
manually
3 3! = 6
10 10! = 3,628,800
13 13! = 6,227,020,800 3 centuries
20 20! = 11 billion
centuries
10 seconds to calculate each route manually
A
B
C
Warehouse
80
35
115
60
70
105
Best Routes

Wherever decisions are made !Wherever decisions are made !
Where, Decisions require resources for implementation !!Where, Decisions require resources for implementation !!
Capital AllocateCapital Allocate
People Acquire, schedule, assign, TrainPeople Acquire, schedule, assign, Train
Equipment Acquire, schedule, LocateEquipment Acquire, schedule, Locate
Facilities Locate, scheduleFacilities Locate, schedule
Vehicles Acquire, route, scheduleVehicles Acquire, route, schedule
Raw Material Acquire, assignRaw Material Acquire, assign
Time Allocate, ScheduleTime Allocate, Schedule

FacilityLayoutPlanning
OR Key
Areas
Production
Planning
Allocation
Job,
Machines
Inventory
Optimization
Transportation
(Routing,
Scheduling)
Network
Optimization

SC Network optimization
Locate facilities
Match Supply & Demand
Managing seasonality, and
Reduce carbon footprint.
Benefits:
5-15% reduction in SC costs
Better service level
Sourcing Selection
Select Suppliers
Determine Order Quantities
Benefits :
Enhanced S&OP capability,
2-5% reduction in mfg costs
Transportation Scheduling
Fleet Selection
Route Plan / Schedule
Benefits :
10-30% reduction in Transport
cost Increased on-time deliveries
Production planning
Optimal production schedule
Meet plant floor constraints.
Benefits :
Improved throughput
Reduced costs,
Reduced inventory
Leaner plants
Operational Inventory Management
Order quantity
Order time
Supplier identification
Benefits :
10-30% reduction in inventory
Improved service level
Demand Planning
Choose best price
Indentify Optimal Commission
Benefits :
Higher sales volume
Better Margins

What-If Analysis
Collaboration
Courtesy: IBM Limited

2 Chilean Forestry firms* Timber Harvesting $20M/yr + 30% fewer trucks
UPS* Air Network Design $40M/yr + 10% fewer planes
South African Defense* Force/Equip Planning $1.1B/yr
Motorola* Procurement Mgmt $100M-150M/yr
Samsung Electronics* Semiconductor Mfg 50% reduction in cycle times
SNCF (French RR)* Scheduling & Pricing $16M/yr rev + 2% lower op ex
Continental Airlines* Crew Re-scheduling $40M/yr
AT&T* Network Recovery 35% reduction spare capacity
Grant Mayo van Otterloo* Portfolio Optimization $4M/yr
Indeval* Securities Trade Settlement $150M/yr financing costs
Midwest ISO* Energy Grid Management $2.1B to $3B savings over 4 years
*Franz Edelman Competition Finalists, Science of Better, http://www.scienceofbetter.org , Published Case
Studies
12© 2010-13 OptiRisk India (P) Ltd, All rights reserved 12© 2010-13 OptiRisk India (P) Ltd, All rights reserved

2. Optimization
5. Summary

 A Carpenter can produce 2 products:
Chairs and Tables.
 The cost of production of a chair is Rs
50 and it sells for Rs 60
 The cost of production of a table is Rs
60 and it sells for Rs 75
 The Carpenter want to know how much
he has to produced so that he can
maximize his profit.
A classic decision DilemmaA classic decision Dilemma

 Chair needs 1 log of wood and 3
steel rods
 Table needs 2 logs of wood and 2
steel rods
 And there are only 6 logs of wood
and 14 steel rods available
... Cont

Brute force
Graphical Method
Mathematical Programming
• Using Excel
• Using ILOG CPLEX
Brute force
Graphical Method
Mathematical Programming
• Using Excel
• Using ILOG CPLEX

Chair Table
Number to be
produced
x y
Revenue 60 75
Cost 50 60
Profit 10 15
Chair Table
Max
Available
Number to be
produced
x y
Wood
Required
1 2 6
Steel Required 3 2 14
Total Profit = yx 1510 + 62 ≤+ yx
1423 ≤+ yx
Objective FunctionObjective Function ConstraintsConstraints
Decision VariablesDecision Variables
Number of chairs x
Number of Tables y

Subject to:
Maximize z =
Objective function
Maximize total contribution(profit)
Constraint
Resource Availability
Constraints
Non negativity
Constraint
Production quantity
must be non negative
62 ≤+ yx
1423 ≤+ yx
0,0 ≥≥ yx
yx 1510 +

Chair Table
Costing
Revenue 60 75
Cost 50 60
Resource Requirement
Wood Req 1 2
Steel Reg 3 2
S. No # Chairs # Tables Wood Req Steel Reg Wood Avail Steel Avail Revenue Cost Profit Solution Optimal
1 0 0 0 0 6 14 0 0 0 Feasible
2 0 1 2 2 6 14 75 60 15 Feasible
3 0 2 4 4 6 14 150 120 30 Feasible
4 0 3 6 6 6 14 225 180 45 Feasible
5 0 4 8 8 6 14 300 240 60 Infeasible
6 1 0 1 3 6 14 60 50 10 Feasible
7 1 1 3 5 6 14 135 110 25 Feasible
8 1 2 5 7 6 14 210 170 40 Feasible
9 1 3 7 9 6 14 285 230 55 Infeasible
10 2 0 2 6 6 14 120 100 20 Feasible
11 2 1 4 8 6 14 195 160 35 Feasible
12 2 2 6 10 6 14 270 220 50 Feasible
13 2 3 8 12 6 14 345 280 65 Infeasible
14 3 0 3 9 6 14 180 150 30 Feasible
15 3 1 5 11 6 14 255 210 45 Feasible
16 3 2 7 13 6 14 330 270 60 Infeasible
17 4 0 4 12 6 14 240 200 40 Feasible
18 4 1 6 14 6 14 315 260 55 Feasible YES
19 4 2 8 16 6 14 390 320 70 Infeasible
20 5 0 5 15 6 14 300 250 50 Infeasible

3
Steel Constraints
Infeasible
Wood Constraints
62x ≤+ y
142x3 ≤+ y
Y
X
7
4.6
0
6
y15x10max +
Objective Function
Feasible
X – No of Chairs
Y – No of Tables
X – No of Chairs
Y – No of Tables
Optimal Point

Excel Demo

Model Code
ResultsDebug
Project Explorer

ILOG CPLEX Demo

modeling language/systemmodeling language/system
min f (x, y)
g(x, y) = 0
...model building...
Data
Know How
Experience
Variables
Constraints
Objective Function
Solver: CPLEX, FortMP, Gurobi, etc
Feasibility
Optimality
High Quality
Low Costs
h(x, y) 0>_
Model type: LP, IP, MILP,
Modeling Language: OPL, AMPL, GAMS

2. Optimization
5. Summary

XYZ Chemical Ltd manufactures three chemicals, namely CH-A, CH-B, and CH-C. All
three chemicals have to go through three processes namely MIXING, CONVERSION,
and BAGGING. Each process has finite daily capacity, which are given in the below
table. Each chemical requires different processing time, which are also given in the below
table.
Process
Production minutes per Kg of
product
Capacity of
production in
minutes per dayCH-A CH-B CH-C
MIXING
CONVERSION
BAGGING
2
3
1
2
1
4
1
1
0
400
450
390
Contribution /Kg Rs. 30 Rs. 55 Rs. 23
The company can sell whatever it produces (there is no constraint on demand). The
company’s management wants to find out how much of each chemical it should produce
to maximize the contribution (profit).

Decision Variables:
= Number of Kgs of chemical p produced each day. p = 1, …, P
= Number of minutes per day available at process s.
ConstraintsConstraints
px
Constraints:
sC
= Number of minutes required by product p at production stage s. s = 1, …, S

= Profit per unit of product p
Parameters:
psa ,
pv
Objective Function:
z = Maximize total daily profit

∑=
P
p
pp xv
1
∑=
≤
P
p
spps Cxa
1
,
0≥px for p = 1, …, P=Max num of products
for s = 1, …, S=Max num of processes
Subject to:
Maximize z =
Objective function
Maximize total contribution (profit)
Constraint
Process Capacity
Constraints
Non negativity
Constraint
Production quantity

Subject to:
Maximize z =
Objective function
Maximize total contribution(profit)
Constraint
Process Capacity
Constraints
Non negativity
Constraint
Production quantity
321 22 xxx ++
3213 xxx ++
21 4xx +
0,0,0 321 ≥≥≥ xxx
321 235530 xxx ++
≤ 400
≤ 450
≤ 390

Objective function
Maximize Contribution
/*********************************************
* OPL 12.5 Model
* Author: OptiRisk
* Creation Date: 27-Aug-2013 at 10:02:32 AM
*********************************************/
// Input - indices
{string} Chemicals = ...;
{string} Processes = ...;
// Input parameters
float Capacity[Processes] = ...;
float ProcessTime[Chemicals][Processes] = ...;
float Contribution[Chemicals] = ...;
// Decision Variable
// Production Quantity of each Chemical
dvar float+ ChemicalQty[Chemicals];
// Objective Function
// Maximize total contribution
maximize
sum( p in Chemicals )
Contribution[p] * ChemicalQty[p];
Input Parameters
Decision Variables

Post Processing
Print Results
Constraint
Capacity Constraint
// Constraints
// 1. Capacity Constraints
subject to {
forall( s in Processes )
ctCapacity:
sum( p in Chemicals )
ProcessTime[p][s] * ChemicalQty[p] <= Capacity[s];
}
// Post processing to print results.
execute {
writeln("Maximum Contribution Possible = Rs. ",
cplex.getObjValue());
writeln("n----- RESULT: Optimal Product Mix---------");
// Print the optimal product quantities.
for( var p in Chemicals){
writeln(p, " ---> ", ChemicalQty[p], " Kgs");
}
writeln("n---- END: Optimal Product Mix-------------");
}

Input Data
Parameters
Index
/*********************************************
* OPL 12.5 Data
* Author: OptiRisk
* Creation Date: 27-Aug-2013 at 10:02:32 AM
*********************************************/
Chemicals = {"CH-A","CH-B","CH-C"};
Processes = {"MIXING", "CONVERSION", "BAGGING"};
// Capacity in minutes of each process
Capacity = [400, 450, 390];
// Processing time for each chemical, for every process
ProcessTime = [[2,3,1],[2,1,4],[1,1,0]];
// Contribution (profit) for each chemical
Contribution = [30, 55, 23];

// solution (optimal) with objective 10077.5
Maximum Contribution Possible = Rs. 10077.5
----------- RESULT: Optimal Product Mix---------
CH-A ---> 0 Kgs
CH-B ---> 97.5 Kgs
CH-C ---> 205 Kgs
----------- END: Optimal Product Mix-------------
// solution (optimal) with objective 10077.5
Maximum Contribution Possible = Rs. 10077.5
----------- RESULT: Optimal Product Mix---------
CH-A ---> 0 Kgs
CH-B ---> 97.5 Kgs
CH-C ---> 205 Kgs
----------- END: Optimal Product Mix-------------

2. Optimization
5. Summary

___________________________________________________________________________________________________________________________

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Bala. Padmakumar
Ph: +91 98406 18472 / +91 44 4501 8472
Email: optimize@optiriskindia.com
Web: http://www.optiriskindia.com/
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Mathematical Programming Introduction

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