2. Introduction
• Our world is filled with limited resources.
• Oil
• Land
• Human
• Time
• Business
• Resource (Budget)
• Manufacturing
• m/c , number of worker
• Restaurant
• Space available for seating
3. Introduction
• How best to used the limited resources
available ?
• How to allocate the resource in such a way as
to maximize profits or minimize costs ?
4. Introduction
• Mathematical Programming(MP) is a field of
management science or operations research
that fines most efficient way of using limited
resources/ to achieve the objectives of a
business.
Optimization
5. Characteristics of Optimization
Problems
• One or more decisions
• Restrictions or constraints
e.g. Determining the number of products to
manufacture
a limited amount of raw materials
a limited amount of labor
• Objective
– The production manager will choose the mix of
products that maximizes profits
– Minimizing the total transportation cost
6. Expressing optimization problems
mathematically
• Decision variables
X1,X2 ,X3, … , Xn
e.g. the quantities of different products
Index n = the number of product types
• Constraints
– a less than or equal to constraint : f(X1,X2 ,X3, … , Xn) < b
– a greater than or equal to constraint : f(X1,X2 ,X3, … , Xn) > b
– an equal to constraint : f(X1,X2 ,X3, … , Xn) = b
• Objective
– MAX(or MIN) : f(X1,X2 ,X3, … , Xn)
7. Mathematical formulation of an
optimization problem
MAX(or MIN) : f(X1,X2 ,X3, … , Xn)
Subject to:
f(X1,X2 ,X3, … , Xn) < b1
….
f(X1,X2 ,X3, … , Xn) > bk
….
f(X1,X2 ,X3, … , Xn) = bm
note : n variables , m constraints
8. Mathematical programming techniques
The function in model : f(x) -Linear
-Nonlinear
Decision variable -Fractional value e.g. 2.33
-Integer value e.g. 1,2,3,4..
-Binary e.g. 0,1
Test : Linear functions?
1.
2.
3.
5032 321 ≥−+ xxx
602 21 ≥+ xx
75)3/1(4 21 =+ xx
9.0
323
321
321
≤
++
−+
xxx
xxx
4.
5.
4573 2
2
1 ≤+ xx
9. Problem
• Blue Ridge Hot Tubs manufactures and sells two models of hot
tubs : Aqua Spa and the Hydro-Lux.
– Howie Jones, the owner and manager of the company,
needs to decide how many of each type of hot tubs to
produce
– 200 pumps available
– Howie expects to have 1,566 production labor hours and
2,880 feet of tubing available.
– Aqua-Spa requires 9 hours of labor and 12 feet of tubing
– Hydro-Lux require s 6 hours of labor and 16 feet of tubing
Assuming that all hot tubs can be sold
To maximize profits, how many Aqua-Spas and Hydra-Luxs
should be produce?
10.
11. Formulating LP Models
1. Understand the problem
2. Indentify the decision variable
3. State the objective as a linear combination of decision variables
Max : 350x1 + 300x2
4. State the constraints as linear combinations of the decision variable
4.1 Pumps available
4.2 labor available
4.3 Tubing available
5. Identify any upper or lower bounds on the decision variable
x1 > 0
x2 > 0
12. Max : 350x1 + 300x2
Subjectto:
x1+x2 < 200 , Pumps available
9x1+ 6x2 < 1,566 , Labors available
12x1+ 16x2 < 2,880 , Tubing available
x1 > 0
x2 > 0
13. General Form of an LP model
MAX(or MIN) : C1X1+ C2X2+ , … , CnXn
Subject to:
a11X1+ a12X2+ , … , a1nXn < b1
ak1X1+ ak2X2+ , … , aknXn > bk
am1X1+ am2X2+ , … , amnXn = bm
14. Notations
a11X1+ a12X2+ , … , a1nXn < b1
X1+ X2+ , … ,+ Xn = b1
X11 + X21+ X31 = b1
X12 + X22+ X32 = b2
1
1
1 bxa i
n
i
i ≤∑=
1
1
bx
n
i
i =∑=
j
i
ij bx =∑=
3
1
j∀,
20. Using level curves
• MAX 350x1 + 300x2
1. Set a number of objective
e.g. Obj = 35,000
2. Finding points (x1,x2) which has obj = 35,000
x1 = 100 , x2 = 0
x1 = 0 , x2 = 116.67
23. Special Conditions in LP Models
9x1+ 6x2 ≤ 1,566 (2)
x1+x2 ≤ 200 (1)
12x1+ 16x2 ≤ 2,880 (3)
MAX 450x1 + 300x2
1. Alternate Optimal Solutions
- having more than one optimal solution
24. Special Conditions in LP Models
9x1+ 6x2 ≤ 1,566 (2)
x1+x2 ≤ 225 (1)
12x1+ 16x2 ≤ 2,880 (3)
MAX 350x1 + 300x2
2. Redundant Constraints
A constraint that plays no role in determining the feasible
region of the problem.
25. Special Conditions in LP Models
-x1+ 2x2 ≤ 400 (2)
x1+ x2 ≥ 400 (1)
x1 ≥ 0
MAX x1 + x2
3. Unbounded Solutions
The objective function can be made infinitely large.
x2 ≥ 0
26. Special Conditions in LP Models
4. Infeasibility
x1+ x2 ≤ 150 (1)
x1+ x2 ≥ 200 (2)
x1 ≥ 0
MAX x1 + x2
x2 ≥ 0
?
– No ways to satisfy all of the constraints
32. Modeling LP Problems
1. Make vs. Buy Decisions
- The Electro-Poly corporation received a $750,000 order for
various quantities of 3 types of slip rings.
- Each slip ring requires a certain amount of time to wire and
hardness.
Model1 Model2 Model3
Number ordered 3,000 2,000 900
Hours of wiring required per unit 2 1.5 3
Hours of harnessing required per unit 1 2 1
The company has only 10,000 hours of wiring capacity.
The company has only 5,000 hours of harnessing capacity.
33.
34. Make vs. Buy Decisions (Continue)
The company can sub contract to one of its competitors.
Determine the number of slip rings to make and the number to buy
in order to fill the customer order at the least possible cost
Cost($) Model1 Model2 Model3
Cost to make 50 83 130
Cost to buy 61 97 145
35. Make vs. Buy Decisions (Continue)
• Defining the decision variables
mi = number of model i slip rings to make in-house
bi = number of model i slip rings to buy from competitor
• Defining the objective function
– To minimize the total cost
Min : 50m1 + 83m2 + 130m3 + 61b1+ 97b2 + 145b3
Subjectto:
2m1 +1.5m2 + 3m3 < 10,000 , wiring constraint
1m1 +2m2 + m3 < 5,000 , harness constraint
m1 + b1 = 3,000 , Required order of model1
m2 + b2 = 2,000 , Required order of model2
m3 + b3 = 900 , Required order of model3
m1 ,m2 ,m3 , b1 ,b2 ,b3 , > 0
36. Make vs. Buy Decisions
Parameters Cost to buy, Cost to make, Number ordered, Hours of
wiring , Hours of harnessing
Decision Variable - number of model i slip rings to make in-house
- number of model i slip rings to buy from competitor
Objective Total Cost
Constraint wiring constraint
harness constraint
Required order of each model
39. A transportation problem
• How many product should be shipped from r/m area to the processing
plant, the trucking company charges a flat rate for every mile.
Min the total distance ~ Min total cost of transportation
• Defining the decision variable
xij = number of R/M to ship from node i to node j
• Defining the objective function
Min : 21x11+50x12+40x13+35x21+30x22+22x23 +55x31+20x32+25x33
• Defining the constraints
x11+x21 + x31 < 200,000 capacity restriction for j1
x12+x22 + x32 < 600,000 capacity restriction for j2
x13+x23 + x33 < 225,000 capacity restriction for j3
x11+x12 + x13 = 275,000 supply restriction for i1
x21+x22 + x23 = 400,000 supply restriction for i2
x31+x32 + x33 = 300,000 supply restriction for i3
x > 0 , for all i and j
40. A transportation problem
Parameters Distance from I to j, no supply , no capacity
Decision Variable number of R/M to ship from node i to node j
Objective Min the total distance
Constraint capacity restriction
supply restriction
42. A Blending problem
• Agri-Pro stocks bulks amounts of four types of
feeds that can mix to meet a given customer’s
specifications.
Nutrient(%) feed1 feed2 feed3 feed4
Corn (i=1) 30% 5% 20% 10%
Grain(i=2) 10% 30% 15% 10%
Minerals(i=3) 20% 20% 20% 30%
Cost per Pound($) 0.25 0.30 0.32 0.15
43. A Blending problem
• Agri-Pro has just received an order from a local chicken farmer for 8,000
pounds of feed.
• The farmers wants this feed to contain at least 20% corn, 15% grain , and
15% minerals.
• What should Agri-Pro do to fill this order at minimum cost?
• Defining the decision variable
xi = pounds of feed (j) to use in the mix
• Defining the objective function
Min : 0.25x1+0.30x2+0.35x3+0.15x4 , total cost
• Defining the constraints
x1+x2 + x3 + x4 = 8,000
0.3x1+0.05x2 + 0.20x3 + 0.10x4 > 0.20 × 8,000 --- Corn
0.1x1+0.30x2 + 0.15x3 + 0.10x4 > 0.15× 8,000 --- Grain
0.2x1+0.20x2 + 0.20x3 + 0.30x4 > 0.15 × 8,000 --- Mineral
x1 , x2, x3 , x4 > 0
44. A Blending problem
Parameters Cost per Pound($),
Nutrient(%) of each feed
Order quantity ,
Required Nutrient(%)
Decision Variable pounds of feed (j) to use in the mix
Objective Minimize Cost
Constraint Satisfy the Required Nutrient(%)
46. A production and inventory planning
problem
• Upton Corporation is trying to plan its production and inventory levels for
the next 6 months
• Given the size of Upton’s warehouse, a maximum of 6,000 units
• The owner like to keep at least 1,500 units in inventory any months as
safety stock
• The company wants to produce at no less than half of its maximum
production capacity each month.
• Estimate that the cost of carrying a unit in any given month is equal to
1.5% of the unit production cost in the same month.
m1 m2 m3 m4 m5 m6
Unit production cost (Cm) 240 250 265 280 280 260
Unit Demanded(Dm) 1000 4500 6000 5500 3500 4000
Maximum Production(Maxp m) 4000 3500 4000 4500 4000 3500
47. A production and inventory planning
problem
• The number of units carried in inventory using the
averaging the beginning and ending inventory for
each month
• Current inventory 2,750 units
• To minimize production and inventory costs, the
company wants to identify the production and
inventory plan.
• Defining the decision variable
pm = the number of units to product in month m
bm = the beginning inventory for month m
48. A production and inventory planning
problem
• Defining the objective function
Min : ,
Subject to
2/))(015.0( 1
6
1
6
1
+
==
+×+ ∑∑ mmm
mm
mm bbcpC
mm Maxpp ≤ m∀,
m∀,
m∀,
mmmm Dpbb −+=+1
m∀,
m∀,
mm Maxpp )2/1(≥
000,6≤−+ mmm Dpb
500,1≥−+ mmm Dpb
49. A production and inventory
planning problem
Parameters Maximum Production(Maxp),
Unit production cost(Cm) ,
Unit Demanded(Dm)
Decision Variable the number of units to product in month m , pm
the beginning inventory for month m, bm
Objective Minimize Cost
Constraint Upper bound and Lower bound of the production and
inventory , Balance constraint
