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Lecture 3 – Classic LP Examples
Topics• Employee scheduling problem• Energy distribution problem• Feed mix problem• Cutting stock problem• Regression analysis• Model Transformations
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 FTEs1 0-4 152 4-8 103 8-12 404 12-16 705 16-20 406 20-0 35
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 $12.95. 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.
Constraints for Employee Scheduling
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. 56 y1 1556 y2 1056 y3 4056 y4 7056 y5 4056 y6 35
(8 15.20) (4 12.95)
All shifts must be covered
PT employee is 5/6 FT employee
x1 + x6 +
x1 + x2 +
x2 + x3 +
x3 + x4 +
x4 + x5 +
x5 + x6 +
x1 + x6 23
23
... 2
3
xt yt t =1,2,…,6
At least 2/3 workers must be full time
More constraints:
Nonnegativity
(x6 + x1 + y1)
x1 + x2(x1 + x2 + y2)
(x5 + x6 + y6)x5 + x6
• 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:
Ingredient, iVitamins Protein Calcium Crude Fat
Corn 8 10 6 8Limestone 6 5 10 6Soybeans 10 12 6 6Fish Meal 4 18 6 9
Let aik = quantity of nutrient k per kg of ingredient i
Nutrient, k
Feed Mix Problem
• The mill has (firm) contracts for the following demands.
Demand (kg) Cattle Sheep Chicken10,000 6,000 8,000
• There are limited availabilities of the raw ingredients.
Supply (kg) Corn Limestone Soybeans Fish Meal6,000 10,000 4,000 5,000
• The different feeds have “quality” bounds per kilogram.
Vitamins Protein Calcium Crude fatmin max min max min max min max
Cattle 6 -- 6 -- 7 -- 4 8Sheep 6 -- 6 -- 6 -- 4 8Chicken 4 6 6 -- 6 -- 4 8
si
dj
The above values represent bounds: ljk and ujk
Constraints
• 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 }
Costs and Notation
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 jupper bound on number of nutrients of type k per kg of product jcost 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
ujk
ci
min cixij
s.t.
xij
xij = dj
xij si
j J
iI jJ
iI
i IjJ
aikxij ujkdjiI
j J, kK
aikxij ljk djiI
j J, kK
i I, j J
LP Formulation of Feed Mix Problem
Raw Materials QualitiesBlended
commoditiescorn, limestone,
soybeans, fish mealprotein, vitamins,calcium, crude fat
feed
butane, catalyticreformate,
heavy naphtha
octane, volatility,vapor pressure
gasoline
pig iron,ferro-silicon,
carbide, variousalloys
carbon,manganese,
chrome content
metals
2 raw ingredients 1 quality 1 commodity
Generalization of feed Mix Problem GivesBlending Problems
• 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 Length1 5 10,0002 7 30,0003 9 20,000
• Assumption: Lengthwise strips can be taped together
• Goal: Throw away as little as possible
Trim-Loss or Cutting Stock problem
Problem: What is trim-loss?
Decision variables: xj = length of roll cut using
pattern, j = 1, 2, … ?
7
10
9'5'
5
20
5000'
10 roll 20 roll
x1
5 2 0 0 4 2 2 1 0 07 0 1 0 0 1 0 2 1 09 0 0 1 0 0 1 0 1 2
Trim loss 0 3 1 0 3 1 1 4 2
x2 x3 x4 x5 x6 x7 x8 x9
min z = 10(x1+x2+x3) + 20(x4+x5+x6+x7+x8+x9)
s.t. 2x1 + 4x4 + 2x5 + 2x6 + x7 10,000
x2 +x5 + 2x7 + x8 30,000
x3 + x6 + x8 + 2x9 20,000
xj0, j = 1, 2,…,9
Patterns Possible
Minimize Trim Loss + Overproduction
min z = 3x2 + x3 + 3x5 + x6 + x7+ 4x8
+ 5y1 + 7y2 + 9y3
s.t. 2x1 + 4x4 + 2x5 + 2x6 + x7 – y1 = 10,000
x2 + x5 + 2x7 + x8 – y2 = 30,000
x3 + x6 + x8 + 2x9 – y3 = 20,000
xj0, j = 1,…,9; yi 0, i = 1, 2, 3
where yi is overproduction of width i
+ 2x9
Alternative Formulation
Minimizing Piecewise Linear Convex Functions
• Definition of convexity
• Examples of objective functions
1. f (x) = maxk=1,…,p (ckx + dk)
2. f (x) = j=1,n cj|xj|, cj > 0 for all j
3. f (x) = separable, piecewise linear, convex
Definition of a Convex/Concave Function
• A function f : n is called convex if for every x and y n, and every [0,1], we have
f (x + (1 – )y) ≤ f (x) + (1 – )f (y)
• A function f : n is called concave if for every x and y n, and every [0,1], we have
f (x + (1 – )y) ≥ f (x) + (1 – )f (y)
• If f (x) is convex, then –f (x) is concave
Minimizing the Maximum of Several Affine Functions
Problem: min maxk=1,…,p (ckx + dk)
s.t. Ax ≥ b
Transformed problem:min z
s.t. z ≥ ckx + dk, k = 1,…,p
Ax ≥ b
x
f(x) = max
Problems Involving Absolute Values: Minimizing the L1-Norm
Problem: min j=1,n cj|xj|, cj > 0 for all j
s.t. Ax ≥ b
Transformation 1:
min j=1,n cjzj
s.t. Ax ≥ b zj ≥ xj, j = 1,…,n
zj ≥ –xj, j = 1,…,n
Transformation 2:
min j=1,n cj(xj+ + xj
-)
s.t. Ax + – Ax - ≥ b
x + ≥ 0, x - ≥ 0where xj = xj
+ – xj- for all j
Data Fitting Example
• Problem: We are given p data points of the form (ak, bk), k = 1,…,p, where ak n and bk , and wish to build a model that predicts the value of the variable b from knowledge of the vector a.
• Assume a linear model: b = ax + x0, where (x,
x0) is a parameter vector to be determined.
• Error: Given a particular values of (x, x0), the residual (predictive error) at the k th data point is defined by |akx + x0 – bk|.
• Objective: Find values of (x, x0) that best explain the available data; i.e., minimize the error.
Data Fitting Example (cont’d)
• Model 1: Minimize the largest residual
min maxk |akx + x0 – bk|
Transformed model 1: min z
s.t. z ≥ akx + x0 – bk , k = 1,…,p
z ≥ – akx – x0 + bk , k = 1,…,p
• Model 2: Minimize the sum of residuals
min k=1,p |akx + x0 – bk|
Transformed model 2:
min k=1,p zk
s.t. zk ≥ akx + x0 – bk , k = 1,…,p
zk ≥ – akx – x0 + bk , k = 1,…,p
Data (a,b) = { (1,2) , (3,4) , (4,7) }
We want to “fit” a linear function b = ax + x0 to these data points; i.e., we have to choose optimal values for x and x0.
7654321
1 2 3 4 5
b
a
Constrained Regression
Objective: Find parameters x and x0 that minimize the
maximum absolute deviation between the data ak and
the fitted line bk = akx + x0. bk andbk
In addition, we’re going to impose a priori knowledge that theslope of the line must be positive. (We don’t know about the intercept.)
Decision variables x = slope of line known to be positivex0 = b-intercept positive or negative
observedvalue
Predictedvalue
Let z = max { |bk bk| : k = 1, 2, 3 }
Optimization model:
min z
where bk = akx + x0
Objective function:
s.t. z |bk bk|, k = 1, 2, 3
min max { |bk bk| : k = 1, 2, 3 }
Note: 2 |x| iff 2 x and 2 x
Thus z |bk bk| is equivalent to
z akx + x0 bk and z akx – x0 + bk
Convert absolute value terms to linear terms:
Nonlinear constraints:
b1 b1 = 1x + x0 – 2
b2 b2 = 3x + x0 – 4
b3 b3 = 4x + x0 – 7
z
z
z
Letting x0 = x0+ x0
-, x0+ 0, x0
- 0, we finally get …
min z
s.t. x + x0+ x0
- z
x, x0+, x0
-, z 0
x x0+ x0
- z3x + x0
+ x0- z
3x x0+ x0
- z4x + x0
+ x0- z
4x x0+ x0
- z
Separable Piecewise Linear Functions
• Model: min f (x) = f1(x1) + f2(x2) + . . . + fp(xp)
• For each xj we are given r break points:
0 < aj1 < aj2 < . . . < ajr < ∞
• Let cjt be the slope in the interval aj,t-1 ≤ xj ≤ ajt for t =1,…,r+1, where aj0= 0 and aj,r+1 = ∞
• Let yjt be the portion of xj lying in the tth interval, t = 1,…,r+1
xjaj1 aj2 ajr
fj(xj)
aj,r-10
Transformation for fj(xj)
• Let xj = yj1 + yj2 + . . . + yj,r+1
• Model:
min cj1yj1 + cj2yj2 + . . . + cj,r+1 yj,r+1 + f1(x1) + . . .
s.t. 0 ≤ yj1 ≤ aj1
0 ≤ yj2 ≤ aj2 – aj1
. . .
0 ≤ yjr ≤ ajr – aj,r-1
0 ≤ yj,r+1
and for every t, if yjt > 0, then each yjk is equal to
its upper bound ajk – aj,k-1, for all k < t.
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
Plants
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) = $110.
2
1
3
2
1
Demandsectors
Demand requirements
4 MW
7 MW
6 MW
Energy Generation Problem (with piecewise linear objective)
Formulate an LP that, when solved, will yield optimal power generation and distribution levels.
Decision Variables
= power generated at plant 1 at operating level 1
1 2x21 2 1
x22 2 2
= power sent from plant 1 to demand sector 1
1 2
1 3
2 1
2 2
2 3
Problem Statement and Notation
w11
w12
w13
w21
w22
w23
x11
x12
Formulation
min 10x11 + 25x12 + 8x21 + 28x22
s.t. w11 + w12 + w13
w21 + w 22 + w23
w11 + w 21 = 4w12 + w22 = 7w13 + w23 = 60 x11 6, 0 x12 4
0 x21 5, 0 x22 6
w11, w12, w13, w21, w22, w32 0
Note that we can model the nonlinear operating costs as an LP only because the efficiencies have the right kind of structure. In particular, the plant is less efficient (more costly) at higher operating levels. Thus the LP solution will automatically select level 1 first.
= x11 + x12
= x21 + x22
Flow balance
Demand
The above formulation can be generalized for any number of plants, demand sectors, and generation levels.
Indices/Setsi I plants
demand sectorsgeneration levels
Data
Cik = unit generation cost ($/MW) for plant i at level kuik = 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 wij = power (MW) sent from plant i to sector j
j J
k K
General Formulation of Power Distribution Problem
min
s.t. wij
cikxik
xik
xik uik i I, k K
wij i I, j J
wij = dj
kKiI
kKjJ i I
j JiI
General Network Formulation
Model Transformations• Direction of optimization:
Minimize {c1x1 + c2x2 + … + cnxn}
Maximize {–c1x1 – c2x2 – … – cnxn}
• Unrestricted variables:
xj = y1j – y2j where y1j 0, y2j 0
• 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
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.