1. Lecture 3 – Classic LP Examples
Topics
Employee scheduling problem
Energy distribution problem
Feed mix problem
Cutting stock problem
Regression analysis
Model Transformations

2. Examples of LP Formulations
1. Employee Scheduling
Macrosoft has a 24-hour-a-day, 7-days-a-week toll free hotline that is being set up to answer questions regarding a new product. The following table summarizes the number of full-time equivalent employees (FTEs) that must be on duty in each time block.
Interval
Time
FTEs 1
0-4 15 2
4-8 10 3
8-12 40 4
12-16 70 5
16-20 40 6
20-0 35

3. Formulate an LP to determine how to staff the hotline at minimum cost.
Constraints for Employee Scheduling
Macrosoft may hire both full-time and part-time employees. The former work 8-hour shifts and the latter work 4-hour shifts; their respective hourly wages are $15.20 and $ Employees may start work only at the beginning of 1 of the 6 intervals.
Part-time employees can only answer 5 calls in the time a full-time employee can answer 6 calls. (i.e., a part-time employee is only 5/6 of a full-time employee.)
At least two-thirds of the employees working at any one time must be full-time employees.
Formulate an LP to determine how to staff the hotline at minimum cost.

4. All shifts must be covered
Decision Variables
xt = # of full-time employees that begin the day at the start of interval t and work for 8 hours
yt = # of part-time employees that are assigned interval t
min
121.6(x1 + • • • + x6) +
51.8(y1 + • • • + y6)
s.t. 5 6 y1 15 y2 10 y3 40 y4 70 y5 y6 35
(8  15.20)
(4  12.95)
All shifts must be covered
PT employee is 5/6 FT employee
x x6 +
x x2 +
x x3 +
x x4 +
x x5 +
x x6 +

5. At least 2/3 workers must be full time (x1 + x2 + y2) x1 + x2 . . .
More constraints:
x1 + x6 2
(x6 + x1 + y1) 3 2
At least 2/3 workers must be full time
(x1 + x2 + y2)
x1 + x2 3 . . .
x5 + x6 2
(x5 + x6 + y6) 3
xt ³ 0, yt ³ 0 t =1,2,…,6
Nonnegativity

6. 2. Energy Generation Problem (with piecewise linear objective)
Austin Municipal Power and Light (AMPL) would like to determine optimal operating levels for their electric generators and associated distribution patterns that will satisfy customer demand. Consider the following prototype system
Demand requirements 1
4 MW 1
Demand
sectors
Plants
7 MW 2 2 3
6 MW
The two plants (generators) have the following (nonlinear) efficiencies:
Plant 1
[ 0, 6 MW]
[ 6MW, 10MW]
Unit cost ($/MW)
$10
$25
Plant 2
[ 0, 5 MW]
[5MW, 11MW]
Unit cost ($/MW) $8
$28
For plant #1, e.g., if you generate at a rate of 8MW (per sec), then the cost ($) is = ($10/MW)(6MW) + ($25/MW)(2MW) = $110.

7. Problem Statement and Notation
Formulate an LP that, when solved, will yield optimal power generation and distribution levels.
Decision Variables x
= power generated at plant
1 at operating level 1 11 x 1 12 2 x 2 1 21 x 2 2 22 y
= power sent from plant
1 to demand sector 1 11 y
1 ² ² ² 2 12 y 3 13 1 y 2 21 1 y 22 2 2 y 2 3 23

8. Formulation min 10x11 + 25x12 + 8x21 + 28x22 s.t. y y y = x11 + x1213
Flow balance y + y + y
= x x22 21 22 23 y y 11 + 21
= 4 y y 12 +
Demand 22
= 7 y y 13 + 23
= 6
0 £ x11 £ 6, 0 £ x12 £ 4
0 £ x21 £ 5, 0 £ x22 £ 6
y11, y12, y13, y21, y22, y32  0
Note that we can model the nonlinear operating costs as an LP only because the efficiencies have the right kind of structure. In particular, the plant is less efficient (more costly) at higher operating levels. Thus the LP solution will automatically select level 1 first.

9. General Formulation of Power Distribution Problem
The above formulation can be generalized for any number of plants, demand sectors, and generation levels.
Indices/Sets
i Î I
plants
j Î J
demand sectors
k Î K
generation levels
Data
Cik = unit generation cost ($/MW) for plant i at level k
uik = upper bound (MW) for plant i at level k
dj = demand (MW) in sector j
Decision Variables
xik = power (MW) generated at plant i at level k
yij = ower (MW) sent from plant i to sector j

10. General Network Formulationå å
cikxik
min
iÎI
kÎK
s.t. å å
xik
yij =
" i Î I
jÎJ
kÎK å
yij = dj
" j Î J
iÎI
£ xik £ uik " i Î I, k Î K
0 £ yij " i Î I, j Î J

11. Let aik = quantity of nutrient k per kg of ingredient i
3. Feed Mix Problem
An agricultural mill produces a different feed for cattle, sheep, and chickens by mixing the following raw ingredients: corn, limestone, soybeans, and fish meal.
These ingredients contain the following nutrients: vitamins, protein, calcium, and crude fat in the following quantities:
Nutrient, k
Vitamins
Protein
Calcium
Crude Fat
Ingredient, i
Corn 8 10 6 8
Limestone 6 5 10 6
Soybeans 10 12 6 6
Fish Meal 4 18 6 9
Let aik = quantity of nutrient k per kg of ingredient i

12. Constraints
The mill has (firm) contracts for the following demands.
Demand (kg)
Cattle
Sheep
Chicken dj
10,000
6,000
8,000
There are limited availabilities of the raw ingredients.
Supply (kg)
Corn
Limestone
Soybeans
Fish Meal si
6,000
10,000
4,0 00
5,000
The different feeds have “quality” bounds per kilogram.
Vitamins
Protein
Calcium
Crude fat
min max
min max
min max
min max
Cattle
Sheep
Chicken
The above values represent bounds: Ljk & Ujk

13. Costs and Notation Cost per kg of the raw ingredients is as follows:
Corn
Limestone
Soybeans
Fish meal
cost/kg, ci
20¢
12¢
24¢
12¢
Formulate problem as a linear program whose solution yields desired feed production levels at minimum cost.
Indices/sets
i  I
ingredients { corn, limestone, soybeans, fish meal }
j  J
products { cattle, sheep, chicken feeds }
k  K
nutrients { vitamins, protein, calcium, crude fat }

14. Data dj
demand for product j (kg) si
supply of ingredient i (kg)
Ljk
lower bound on number of nutrients of type k per kg of product j
Ujk
upper bound on number of nutrients of type k per kg of product j ci
cost per kg of ingredient i
aik
number of nutrients k per kg of ingredient i
Decision Variables
xij
amount (kg) of ingredient i used in producing product j

15. å å å å å å " j  J " i  I " j  J, kK " j  J, kK " i  I, j  J
LP Formulation å å
cixij
min
iÎI
jÎJ å
xij = dj
s.t.
" j  J
iÎI å
xij £ si
" i  I
jÎJ å
aikxij ³ Ljk dj
iÎI
" j  J, kK å
aikxij £ Ujkdj
iÎI
" j  J, kK
xij ³ 0
" i  I, j  J

16. Generalization of feed Mix Problem Gives
Blending Problems
Blended
commodities
Raw Materials
Qualities
corn, limestone,
protein, vitamins,
feed
soybeans, fish meal
calcium, crude fat
butane, catalytic
octane, volatility,
gasoline
reformate,
vapor pressure
heavy naphtha
pig iron,
carbon,
metals
ferro-silicon,
manganese,
carbide, various
chrome content
alloys
2 raw ingredients
1 quality
1 commodity

17. 4. Trim-Loss or Cutting Stock problem
Three special orders for rolls of paper have been placed at a paper mill. The orders are to be cut from standard rolls of 10¢ and 20¢ widths.
Order
Width
Length 1 5
10,000¢ 2 7
30,000¢ 3 9
20,000¢
Assumption: Lengthwise strips can be taped together
Goal: Throw away as little as possible

18. Problem: What is trim-loss?20 10
5000' 5' 5 9' 7
Decision variables: xj = length of roll cut using pattern, j = 1, 2, … ?

19. Patterns possible ³ ³ ³ 10 ¢ roll 20 ¢ roll x1 x2 x3 x4 x5 x6 x7 x8 x95¢ 2 4 2 2 1 7¢ 1 1 2 1 9¢ 1 1 1 2
Trim loss 3 1 3 1 1 4 2
min
z =
10(x +x +x
) + 20(x +x +x +x +x +x ) 1 2 3 4 5 6 7 8 9
s.t. 2x + 4x + 2x
+ 2x
+ x
10,000 1 4 5 6 7 x + x + 2x
+ x
30,000 2 5 7 8 x + x
+ x
+ 2x
20,000 3 6 8 9
xj ³ 0, j = 1, 2,…,9

20. Alternative Formulation
min
z =
3x2 + x3 +
3x5
+ x6
+ x7 +
4x8
+ 2x9
+ 5y1 + 7y2 + 9y3
s.t.
2x1 + 4 x4 +
2x5
+ 2x6
+ x7 – y1 =
10,000 x2 + x5 +
2x7
+ x8
– y2 =
30,000 x3 + x6 + x8
+ 2x9
– y3 =
20,000
xj ³ 0, j = 1,…,9; yi ³ 0, i = 1, 2, 3
where yi is overproduction of width i

21. 5. Constrained Regression
Data (x,y) = { (1,2) , (3,4) , (4,7) } y 7 6 5 4 3 2 1 x
We want to “fit” a linear function y = ax + b to these data points; i.e., we have to choose optimal values for a and b.

22. yi Objective: Find parameters a and b that minimize the
maximum absolute deviation between the data yi and the fitted line yi = axi + b. Ù Ù yi
and yi Ù
observed
value
Predicted
value
In addition, we’re going to impose a priori knowledge that the
slope of the line must be positive. (We don’t know about the intercept.)
Decision variables
a = slope of line
known to be positive
b = y-intercept
positive or nega
tive
b = b+ - b-, b+  0, b-  0

23. Optimization model: Objective function:Ù
min max { |yi - yi| : i = 1, 2, 3 } Ù
where yi = axi + b Ù
Let w = max { |yi - yi| : i = 1, 2, 3 }
Optimization model:
min w Ù
s.t. w ³ |yi - yi|, i = 1, 2, 3

24. Nonlinear constraints:Ù
w ³ y1 - y1 =
 1a + b – 2  Ù
w ³ y2 - y2 =
 3a + b – 4  Ù
w ³ y3 - y3 =
 4a + b – 7 
Convert absolute value terms to linear terms:
Note: 2 ³|x| iff 2 ³ x and 2 ³ -x Ù
Thus w ³ y - y
is equivalent to i i
w ³ axi + b - yi and w ³ - axi – b + yi

25. so finally … min w s.t. a + b - b - w £ 2 - a - b + b - w £ - 2 3a + b4 - - 3a - b +
+ b - w - 4 -
4a + b + - b - w 7 - - 4a - b + + b - w - 7
a, b+, b-, w ³ 0

26. Model Transformations
Direction of optimization:
Minimize {c1x1 + c2x2 + … + cnxn}
Û Maximize {–c1x1 – c2x2 – … – cnxn}
Constant term in objective function  ignore
Nonzero lower bounds on variables:
xj > lj  replace with xj = yj + lj where yj  0
Nonpositive variable:
xj ≤ 0  replace with xj = –yj where yj  0
Unrestricted variables:
xj = y1j – y2j where y1j  0, y2j  0

27. What You Should Know About LP Problems
How to formulate various types of problems.
Difference between continuous and integer variables.
How to find solutions.
How to transform variables and functions into the standard form.