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Integer Programming
Decision Variables take Integer
Values
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X1
X2
Area of Feasibility
1
2
3
4
5
1 2 3 4 5 6
Point at Intersectionare feasible points
as they
are integers
Acceptable valuesFor X1, X2 are
(1,1), (2,1), (1,2), (2,2) ..
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A cost minimization example
Minimize Cost = 0.01X1 + 0.07X2
Subject to :
6X1 + 2X2 >= 18 (Constraint 1) 8X1 + 10X2 >= 40 (Constraint 2)
X2 >= 1 (Constraint 3)
X1, X2 >= 0 (Non Negativity)
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Minimization Problem
X2
X1
X2 >= 1
Objective Function
Optimum X1 = 2.27, X2 = 2.19
Z = 0.10x2.27 + 0.07x2.19 = 0.38Z (Integer) = 0.10*2 + 0.07*3 = 0.41
A
B
C
4
2
8
10
2 4 6 8 10
3
Optimum (Integer) X1 = 2, X2 = 3
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Another Example
Maximize Z = 40000X1 + 30000X2
Subject to:
10X1 + 3X2
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Types of Integer ProgrammingProblems
Pure Integer Problems (Number of each type of plane toproduce, number of each type of house to construct, andso on..)
Mixed Integer Problems (have some variables requiring
integer values, and some may have continous (decimal)values, like a landscape design which requires thepercentage area to plant grass (continous variable) andnumber of trees to plant (integer variable)
0-1 Integer Problems where variables require values as0 or 1 (binary), or Yes/No type. Examples are a bankconsidering possible locations for a branch, schedulingof jobs to machines or workers to jobs.
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Formulating Integer programmingProblems with 0-1 constraints
Either - or Alternatives
k out of n alternatives
If Then Alternatives Either Or Constraints
Variables that have minimum level
requirements
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Either Or Alternatives
A manufacturer may need a machine toreplace one that recently has failed. Twoalternatives X1 and X2 are being
considered, but only one will be needed.
Constraint : X1 + X2 = 1
If neither machine will be acquired thenconstraint is X1 + X2
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K out of N Alternatives
A decision maker (DM) must choose a specifiednumber of alternatives. Say choose 2 machinesfrom a list of 5 alternatives.
X1 + X2 + X3 + X4 + X5 = 2 (exactly 2) X1 + X2 + X3 + X4 + X5 >= 2 (atleast 2)
X1 + X2 + X3 + X4 + X5 = 2 (lower bound)
X1 + X2 + X3 + X4 + X5
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If Then Alternatives
DM has to take some action which necessitatesanother action that supports the initial decision
Purchase of X2 machine may necessitate
purchase of another X1 machine Purchase X2 (X2=1) will lead to purchase of X1(X1=1). If X2 is not purchased, (X2=0) will leadto X1 not being purchased (X1=0)
X1 >= X2 or X1 X2 >= 0 (Reverse is not truePurchase of X1 does not lead to purchase of X2)
If purchase of either machine requires thepurchase of the other then X1 X2 = 0
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Either Or Constraints
Situations may arise in which a constraint will applyonly if a particular alternative is chosen A certain machine may necessitate special power
requirements. Hence it can be able to turn-on orturn-off a constraint.
A machine X3 requires the constraint 5X1 + 3X2 >=100. Formulate as 5X1 + 3X2 >= 100X3, X3 is 0-1 variable
5X1 + 3X2 100X3 >= 0 If X3 is not chosen, constraint 5X1 + 3X2 >= 50 is
required. Formulate as : 5X1 + 3X2 >= 100X3 (if X3 is chosen) 5X1 + 3X2 >= 50(1-X3) (if X3 is not chosen)
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Either Or ConstraintsChoice of X3
If X3 is ON (X3 = 1)
5X1 + 3X2 >= 100
Formulate as 5X1 +3X2 >= 100X3, X3 is0-1 variable 5X1 +
3X2 100X3 >= 0 X3 is 0-1 variable
If X3 is OFF (X3 = 0)
5X1 + 3X2 >= 50
Formulate 5X1 + 3X2>= 50(1-X3) (if X3 isnot chosen)
X3 is 0-1 variable
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Variables that have Minimum LevelRequirements
At times, a variable either will have a zerovalue or an amounts that exceeds aspecified value.
Example is the minimum order size for apurchased part might be required by avendor. Say min qty for X1 is 200 units.
X1 >= 200Y1 (Y1 is 0-1 variable) X1 200Y1 >= 0, X1 = integer, Y1 = 0 or 1 If X1 = 0, the constraint would force Y1 to be zero
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Specialized Integer ProgrammingProblems
Fixed Charge problem
Set Covering problem
Knapsack problem Facility Location problem
Traveling Salesperson problem
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Fixed Charge Problem
A company makes 3 products A,B,C. Unit profit for A,B,C are $6,$10,$5respectively. The products can be manufactured using one of the twoprocesses. Demand for A is predicted to be between 50 and 100 units perweek. For product B, demand is between 150 to 200 units per week.Similarly for product C, it is predicted to be between 100 to 150 units per
week. Process 1 has a capacity of 2000 hours per week, with product A, B, C
taking 4 hours, 6 hours and 3 hours per unit. The setup cost for process 1 is$100 and setup time is 24 hours.
Process 2 has a capacity of 2400 hours per week, with product A, B, Ctaking 5 hours, 7 hours and 4 hours per unit. The setup cost for process 2 is
$80 and setup time is 18 hours.
Determine the production schedule that will maximize the profit and alsodetermine which of the two processes will be utilized.
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Fixed Charge Problem
Let X1, X2, X3 be units of Product A,B, C
Let Y1, Y2 be the 0-1 variables for Process 1 and 2
Maximize Z = 6X1 + 10X2 + 5X3 100Y1 80Y2
Subject to
4X1 + 6X2 + 3X3 + 24Y1
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Process 1
Process 2
Any one Process to beselected
Incur Set up CostAnd
Set up Time
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Set Covering Problem
A telecommunication company is considering expanding its cable and internetservices operations into a new area. The area is divided into 10 neighborhoods.The company is considering 7 location nodes to reach all 10 neighborhoods
The cost for opening the seven nodes are $125, $85, $70, $60, $90, $100 and$110.
Seven nodes can reach the following neighborhoods :
Node 1 : Neighborhoods 1,3,4,6,9, 10 Node 2 : Neighborhoods 2,4,6,8
Node 3 : Neighborhoods 1,2,5
Node 4 : Neighborhoods 3,6,7,10
Node 5 : Neighborhoods 2,3,7,9
Node 6 : Neighborhoods 4,5,8,10
Node 7 : Neighborhoods 1,5,7,8,9 Determine which nodes should be opened to provide coverage to all
neighborhoods at a minimum cost
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1 2 3 4 5 6 7 8 9 10
1 Y Y Y Y Y Y
2 Y Y Y Y3 Y Y Y
4 Y Y Y
5 Y Y Y Y6 Y Y Y Y Y
7 Y Y Y Y Y
Neighbourhood
Node
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Set Coverage Problem
Let Cjbe cost of operating or setting up node j (j = 1,2,3,7) Let Xj= 1 if node j provides service (j = 1,2,3,7) Let Xj= 0 if node j does not provide service (j = 1,2,3,7) Minimize Z = 125X1 + 85X2 + 70X3 + 60X4 + 90X5 + 100X6 + 110X7 Subject to : X1 + X3 + X7 >= 1 (coverage of neighborhood 1)
X2 + X3 + X5 >= 1 (coverage of neighborhood 2) X1 + X4 + X5 >= 1 (coverage of neighborhood 3) X1 + X2 + X6 >= 1 (coverage of neighborhood 4) X3 + X6 + X7 >= 1 (coverage of neighborhood 5) X1 + X2 + X6 >= 1 (coverage of neighborhood 6) X4 + X5 + X7 >= 1 (coverage of neighborhood 7) X2 + X6 + X7 >= 1 (coverage of neighborhood 8)
X1 + X5 + X7 >= 1 (coverage of neighborhood 9) X1 + X4 + X6 >= 1 (coverage of neighborhood 10) X1, X2, X3, X4, X5, X6, X7 = 0 or 1 Optimal Answer X2 = 1, X4 = 1, X7 = 1 Cost = $255
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KnapSack Problem
The Knapsack problem is defined as howmany units of different kinds of items orproducts to put in a knapsack with a given
capacity in order to maximize profit.
Maximize Z = C1X1 + C2X2+ . + CnXn Subject to :
A1X1 + A2X2+ .. + AnXn = 0 and integer
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Knapsack Example
TriState must purchase three possible major food items : poultry,ice-cream, yoghurt in batches to realize quantity discounts for fillingup vacant 5 tons capacity in its super freezer.
One batch of ice-cream weighs 2 tons and profit per batch is $90.One batch of poultry weights 3 tons and profit per batch is $150,while one batch of yoghurt weighs 1 ton and profit per batch is $30.
Let X1,X2,X3 = batch of ice-cream, poultry, yoghurt
Maximize Z = 90X1 + 150X2 + 30X3
Subject to :
2X1 + 3X2 + X3 = 0 and integer
Optimal Solution : X1 = 1, X2 = 1, Max Profit = $240
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Facility Location Problem
Here we consider many new locations andalso take capacity considerations
0-1 mixed integer programming is utilized.
Used for location of plants, hospitals,healthcare facilities, fast food restaurants,schools, police and fire stations.
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Facility Location Problem
M&R Manufacturing Company is considering a major expansion. It has 2plants A & B with capacity of 26000 units and 30000 units per annumrespectively. The company sells through its retail outlets located at P, Q, Rand S with annual demands of 27000, 32000, 23000 and 30000 units perannum.
The company is in the process of considering four new manufacturing plantlocations at I, J, K and L. The capacities and cost of purchasing or buildingthe facility is as : I 30000 units / $220000, J 33000 units / $260000, K26000 units / $ 200000, L 37000 units / $ 280000.
The shipping costs from the existing plants and the 4 new plant locations tothe existing retail outlets is given.
Determine the optimal shipping schedule that minimizes the total cost whichincludes the shipping cost and cost of building/purchasing, such that thetotal demand at all retail outlets is satisfied.
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Plant RO P RO Q RO R RO S Supply
A 7 5 4.5 5.5 26000
B 5 7 12 11 30000
I (New) 9 6.5 2 3.5 30000
J (New) 6 3.5 5 3 33000
K (New) 8 6 2.5 4 26000
L (New) 6.5 4.5 5 3 37000
Demand 27000 32000 23000 30000
Shipping Cost per Unit
Total Demand = 112000, while Existing Capacity = 56000.New Capacity to be added should be 56000 or more
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Facility Location Problem
m = number of sources of supply (m = 1,2) k = number of sources of new supply (k = 1,2,3,4) j = number of destinations (j = 1,2,3,4) Cij = Shipping cost of one unit from source i to destination j Xij = Units shipped from source I to destination j
Pk = Purchase cost of plant k Min Z = CijXij + PkYk for all i, j and k Subject to : Xij
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Plant RO P RO Q RO R RO S Supply
A 26000 26000
B 27000 3000 30000
I (New) 30000 30000
J (New) 33000
K (New) 3000 23000 26000
L (New) 37000
Demand 27000 32000 23000 30000 112000
Shipping Schedule
Total Demand = 112000, while Existing Capacity = 56000.New Capacity to be added should be 56000 and added with plant I and K
The cost of building or Purchasing plant I and K is 220000 + 200000 = $420000
The total Minimum Cost is $ 886500. Shipping Cost = $466500
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Traveling Salesperson Problem
It attempts to minimize total cost, distanceor time of departing location I andreturning to same location I in a tour,
visiting all locations once.
It is like the transportation problem, butrelatively complicated.
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Traveling Salesperson Example
The City Garbage collection team wants todetermine the best route for their garbagetrucks. There are 4 sections in the city under
consideration. The time of travel (in minutes)between the various sections of the city is given.
If location 1 is the garage from which garbagetrucks leave and to which they return, formulate
this problem as a traveling salesperson problem
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From/To
1 2 3 4
1 0 8 6 5
2 8 0 3 7
3 6 3 0 9
4 5 7 9 0
Travel Time (in minutes) between locations
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Traveling Salesperson Example
Let Xijk = 0 or 1 for travel from start location i toend location j in journey leg k.
There are 4 locations : 1, 2, 3, 4 In leg 1, start from node 1 and go to node 2,
node 3, node 4 In leg 4, come to node 1 from node 2, node 3,
node 4. In leg 2, go from node 2 to node 3, node 4 or go
from node 3 to node 4 or node 2 In leg 3, go from node 2 to node 3, node 4 or go
from node 3 to node 4 or node 2
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Start Leg 1 Leg 2 Leg 3 Leg 4
1 2 3 4 1
1 2 4 3 1
1 3 2 4 1
1 3 4 2 1
1 4 3 2 1
1 4 2 3 1
Various Routes when started from Node 1
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Leg DV1
DV2
DV3
DV4
DV5
DV6
1 X121 X131 X141
2 X232 X242 X342 X422 X432 X322
3 X233 X243 X343 X423 X433 X323
4 X214 X314 X414
Decision Variables (DV) for each leg in a 4 node problem
Xijk = travel from start location i to end location j in journey leg k.
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Constraints
For every leg of journey (1, 2, 3,and 4), no ofjourneys = 1
No of Visits from Node 2, 3 and 4 = 1
No of Visits to Node 2, 3 and 4 = 1 Leg 2 starts where Leg 1 ends for Node 2, 3 and
4
Leg 3 starts where leg 2 ends for Nodes 2, 3 and
4 Leg 4 starts where leg 3 ends for Nodes 2, 3 and
4
P bl F l ti
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Problem Formulation
Minimize Z1 = 8X121 + 6X131 + 5X141Minimize Z2 = 3X232 + 7X242 + 9X342 + 7X422 + 9X432 + 3X322Minimize Z3 = 3X233 + 7X243 + 9X343 + 7X423 + 9X433 + 3X323Minimize Z4 = 8X214 + 6X314 + 5X414
Minimize Z = Z1 + Z2 + Z3 + Z4Subject to :X121 + X131 + X141 = 1 (Starting node is 1 or Leg 1 visits limited to 1)X214 + X314 + X414 = 1 (Ending Node is 1 or Leg 4 visits limited to 1)X232 + X242 + X342 + X422 + X322 + X432 = 1 (leg 2 visits limited to 1)X233 + X243 + X343 + X423 + X323 + X433 = 1 (leg 3 visits limited to 1)X214 + X232 + X242 + X233 + X243 = 1 (visits from node 2 are limited to 1)X314 + X322 + X323 + X342 + X343 = 1 (visits from node 3 are limited to 1)X414 + X422 + X432 + X423 + X433 = 1 (visits from node 4 are limited to 1)X121 + X322 + X422 + X323 + X423 = 1 (visits to node 2 are limited to 1)X131 + X232 + X233 + X432 + X433 = 1 (visits to node 3 are limited to 1)X141 + X242 + X243 + X342 + X343 = 1 (visits to node 4 are limited to 1)X121 = X232 + X242 (leg 1 ends where leg 2 starts)X131 = X322 + X342 (leg 1 ends where leg 2 starts)X141 = X422 + X432 (leg 1 ends where leg 2 starts)
If leg 2 of trip ends at location 2,3 or 4, leg 3 of trip must start at node 2, 3 or 4X322 + X422 = X233 + X243 (Node 2)X232 + X432 = X323 + X343 (Node 3)X242 + X342 = X423 + X433 (Node 4)X323 + X423 = X214 (last leg of trip must end at location 1)X233 + X433 = X314 ( --- do ---)X243 + X343 = X414 (---- do ---)Xijk = 0 or 1 for all I,j,k
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From/To
1 2 3 4
1 0 8 6 5
2 8 0 3 7
3 6 3 0 9
4 5 7 9 0
Travel Time (in minutes) between locations
Total Time = 6 + 3 + 7 + 5 = 21Route is location 1 to 3 to 2 to 4 to 1
Optimal Solution
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Difficulties in Solving IntegerProgramming Problems
In Integer Programming, a finite number ofinteger points are considered which are morethan the feasible points in linear programming.
As the number of variables (n) increase, the no
of solutions are 2n in 0-1 programming. In pure integer programming, the number of
solutions taking integer values are still more ascompared to 0-1 variables in 0-1 programming.
The computational procedures for integerprogramming are not as simple as that in linearprogramming.