M328 lec2 2015

Math 328: Linear Optimization, Jan.13-May.6, 2015 MSU, Lecture Note #02 1

Math 328: Linear Optimization

Hongxia Yin

Department of Mathematics and Statistics

Minnesota State University, Mankato

271 Wissink Hall 389-2216

Email: [email protected]


1 General Linear Programming Model

max(min) z = c1x1 + c2x2 + · · ·+ cnxn (1)

subject to (s.t.) a11x1 + a12x2 + · · ·+ a1nxn{≤,=,≥}b1, (2)

a21x1 + a22x2 + · · ·+ a1nxn{≤,=,≥}b2, (3)

· · · , (4)

am1x1 + am2x2 + · · ·+ amnxn{≤,=,≥}bm, (5)

xj {≥,≤}0 j = 1, 2, · · · , n. (6)


A linear programming problem (LP) is an optimization problem for which we do

the following:

1. We attempt to maximize (or minimize) a linear function of the decision

variables. The function that to be maximized or minimized is called the

objective function.

2. The value of the decision variables must satisfy a set of constraints. Each

constraint must be a linear equation or linear inequality.

3. A sign restriction is associated with each variable. For any variable xi, the

sign restriction specifies either that xi must be nonnegative (xi ≥ 0) or that

xi may be unrestricted in sign (urs).


2 Characteristics of LP

• Proportionality and additivity: The contribution of the objective function

from each decision variable is proportional to the value of the decision

variable.

The contribution to the objective function for any variable is independent of

the value of the other variables. same for the left hand sides of constraints

• Divisibility: variables should be allowed to assume fractional values.

• Certainty: each parameter is known with certainty.


? Feasible region of LP: The set in which all the points satisfy the constraints

and signal restriction

Example 1:

max z = 3x1 + 2x2

subject to (s.t.) 2x1 + x2 ≤ 100

x1 + x2 ≤ 80

x1 ≤ 40

x1 ≥ 0

x2 ≥ 0

(7)

(x1 = 40, x2 = 20) is a point in the feasible set, we call it a feasible point.

(x1 = 15, x2 = 70), (x1 = 40, x2 = −20) is not in the feasible set, we

call it an infeasible point.

– bounded set; unbounded set; empty set.


? Optimal Solution of LP: For Maximize problem, the optimal solution of LP is

the point at which the objective function have its maximum value.

For Minimize problem, the optimal solution of LP is the point at which the

objective function have its minimum value.

An LP problem may

– have unique optimal solution;

– have no optimal solution; the problem is infeasible;

– have more that one, even infinite number of optimal solutions;

– be unbounded.

Example 1: has unique optimal solution (x∗1 = 20, x∗

2 = 60), Optimal

objective value is:

z∗ = 3x∗1 + 2x∗

2 = 3× 20 + 2× 60 = 180.


? binding constraints/active constraints and nonbinding constraints/inactive

constraints

Example 1: 2x∗1 + x∗

2 = 2× 20 + 60 = 100 =⇒ Finishing constraint is

a binding constraint.

x∗1 = 20 < 40 =⇒ Constraint on demand for soldiers is an nonbinding

constraint.


3 Other forms of LP

? Brief Form:

max(ormin)∑n

j=1 cjxj

s.t.∑n

j=1 aijxj ≤ (=,≥)bi, (i = 1, · · · ,m)

xj ≥ 0 (j = 1, · · · , n).

(8)

Let

c = (c1, · · · , cn), x =

x1

...

xn

, pj =

a1j

...

amj

, b =

b1...

bn

, (9)


A =

a11 a12 . . . a1n

a21 a22 . . . a2n...

......

am1 am2 . . . amn

(10)

? Vector Form:

max(ormin) cx

s.t.∑n

j=1 pjxj ≤ (=,≥)b,x ≥ 0.

(11)

? Matrix-vector Form(A is called the coefficient matrix of the constraints)

max(ormin) cx

s.t. Ax ≤ (=,≥)b,x ≥ 0.

(12)


? Important concepts in LP

– decision variable/activity, data/parameter

– objective/goal/target

– constraint/limitation/requirement

– equality/inequality constraint

– constraint function/the right-hand side

– direction of inequality

– coefficient vector/coefficient matrix

– nonnegativity constraint

– satisfied/violated


– extreme point

A set of points S is a convex set if the line segment joining any pair of

points in S is wholly contained in S.

For any convex set S, a point is an extreme point if each line segment

that lied completely in S and contains the point P has P as an endpoint of

the line segment.

– slack/surplus (introduce later)


4 A Diet problem


Example 2 Suppose my diet requires that all the food I eat come from one of the

four ”basic food groups” (chocolate cake, ice cream, soda, and cheesecake). At

present, the following four foods are available for consumption: brownies,

chocolate ice cream, cola, and pineapple cheesecake. Each brownie costs 50c,

each scoop of chocolate ice cream costs 20c, each bottle of cola costs 30c, and

each piece of pineapple cheesecake costs 80c. Each day, I must ingest at least

500 calorie, 6 oz of chocolate, 10 oz of sugar, and 8 oz of fat. The nutritional

content per unit of each food is shown in the table below,

Calories Chocolate oz sugar (oz) fatoz

Brownie 400 3 2 2

Chocolate ice cream 200 2 2 4

Cola 150 0 4 1

Pineapple cheesecake 500 0 4 5

Formulate a mathematical model that can be used to satisfy my daily nutritional

requirements at minimum cost.


∗ Define decision variables

x1 =number of brownies eaten daily;

x2 =number of scoops of chocolate ice cream eaten daily;

x3 =bottles of cola drunk daily;

x4 = pieces of pineapple cheesecake eaten daily.

∗ Objective function; The total cost of the foot for each day:

z = 50x1 + 20x2 + 30x3 + 80x4.

∗ The decision variables must satisfy the following constraints

1: 400x1 + 200x2 + 150x3 + 500x4 ≥ 500, (Calorie constraint)

2: 3x1 + 2x2 ≥ 6, (Chocolate constraint)

3: 2x1 + 2x2 + 4x3 + 4x4 ≥ 10, (Sugar constraint)

4: 2x1 + 4x2 + x3 + 5x4 ≥ 8, (Fat constraint)

∗ Sign limitation xi ≥ 0, i = 1, 2, 3, 4.


Thus, we obtain the LP model for the problem

min z = 50x1 + 20x2 + 30x3 + 80x4

subject to (s.t.) 400x1 + 200x2 + 150x3 + 500x4 ≥ 500

3x1 + 2x2 ≥ 6

2x1 + 2x2 + 4x3 + 4x4 ≥ 10

2x1 + 4x2 + x3 + 5x4 ≥ 8

xi ≥ 0, i = 1, 2, 3, 4.

(13)


5 A work-scheduling problem


Example 3: A post office requires different number of full-time employees on

different days of the week. The number of full-time employees required on each

day is given in the following table:

Number of Employees Required

Day 1=Mon. 17

Day 2=Tue. 13

Day 3=Wed. 15

Day 4=Thur. 19

Day 5=Fri. 14

Day 6=Sat. 16

Day 7=Sun. 11

Each full-time employee must work 5 consecutive days and then receive 2 days

off. Formulate an LP to minimize the number of full-time employees must be hired.


Solution:

? Decision Variables:

– xi = the number of employees beginning work on day i i = 1, 2, · · · , 7.For example, x1 is the number of people beginning work on Monday, these

people work Monday to Friday.

? Objective Function:

Since each employee begins work on exactly one day of the week, thus we

can obtain the objective function:

– z = x1 + x2 + x3 + x4 + x5 + x6 + x7

? Constraints:

The post office must ensure that enough employees are working on each day

of the week. For example, at least 17 employees must be working on Monday.

Who is working on Monday?


Everybody except the employee who begin work on Tuesday or on

Wednesday (they get, respectively, Sunday and Monday, and Monday and

Tuesday off). This means that

– Constraint 1: x1 + x4 + x5 + x6 + x7 ≥ 17.

– Constraint 2: at least 13 employees must be working on Tuesday⇒x1 + x2 + x5 + x6 + x7 ≥ 13.

– Constraint 3: at least 15 employees must be working on Wednesday⇒x1 + x2 + x3 + x6 + x7 ≥ 15.

– Constraint 4: at least 19 employees must be working on Thursday⇒x1 + x2 + x3 + x4 + x7 ≥ 19.

– Constraint 5: at least 14 employees must be working on Friday⇒x1 + x2 + x3 + x4 + x5 ≥ 14.

– Constraint 6: at least 16 employees must be working on Saturday⇒x2 + x3 + x4 + x5 + x6 ≥ 16.

– Constraint 7: at least 11 employees must be working on Sunday⇒


x3 + x4 + x5 + x6 + x7 ≥ 11.

? Sign Restrictions:

xi ≥ 0, i = 1, · · · , 7.


? LP model for post office work-scheduling problem:

min z = x1 + x2 + x3 + x4 + x5 + x6 + x7

s.t. x1 + x4 + x5 + x6 + x7 ≥ 17

x1 + x2 + x5 + x6 + x7 ≥ 13

x1 + x2 + x3 + x6 + x7 ≥ 15

x1 + x2 + x3 + x4 + x7 ≥ 19

x1 + x2 + x3 + x4 + x5 ≥ 14

x2 + x3 + x4 + x5 + x6 ≥ 15

x3 + x4 + x5 + x6 + x7 ≥ 11

xi ≥ 0, i = 1, · · · , 7.

(14)


The optimal solution for (14) is

x1 = 4/3, x2 = 10/3, x3 = 2, x4 = 22/3, x5 = 0, x6 = 10/3, x7 =

5, z = 67/3

Since we are only allowing full-time employees, the variables must be integers,

and the Divisibility Assumption is not satisfied.

In am attempt to find a reasonable answer in which all variable are integer, we

could try to round the fractional variables up, yielding the feasible solution:

x1 = 2, x2 = 3, x3 = 2, x4 = 8, x5 = 0, x6 = 4, x7 = 5. and z = 25.


Integer programming model ( in which the variables are integer) can be used

the solve this problem. An optimal solution to the post office problem is

x1 = 4, x2 = 4, x3 = 2, x4 = 6, x5 = 0, x6 = 4, x7 = 3, and z = 23.

Notice: There is no way that the optimal linear programming solution could have

been rounded to obtain the optimal all-integer solution.


6 Homework

Graph the feasible set and solve the problems

max z = x1 + x2

s.t. x1 + x2 ≤ 4

x1 − x2 ≥ 5

x1, x2 ≥ 0.

(15)

max z = 4x1 + x2

s.t. 8x1 + 2x2 ≤ 16

5x1 + 2x2 ≤ 12

x1, x2 ≥ 0.

(16)


max z = −x1 + 3x2

s.t. x1 − x2 ≤ 4

x1 + 2x2 ≥ 4

x1, x2 ≥ 0.

(17)

max z = 3x1 + x2

s.t. 2x1 + x2 ≤ 6

x1 + 3x2 ≥ 9

x1, x2 ≥ 0.

(18)


Homework:

In the post office example (Example 2 of Lecture 2), suppose that each full-time

employee works 8 hours per day. Thus, Mondays requirement of 17 workers may

be viewed as a requirement of 8(17)=136 hours. The post office may meet its

daily labor requirements by using both full-time and part-time employees. during

each week, a full-time employee works 8 hours a day for five consecutive days,

and a part-time employee works 4 hours a day for five consecutive days. A

full-time employee costs the post office $15per hour, whereas a part-time

employee (with reduced fringe benefits) costs the post office only 10 per hour.

Union requirements limit part-time labor to25% of weekly labor requirements.

Formulate an LP to minimize the post offices weekly labor costs.

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