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Previously in Chapter 4PowerPoint Presentation

Previously in Chapter 4

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Previously in Chapter 4

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Previously in Chapter 4

- Assignment Problems
- Network Flow Problems
- Sequential Decision Problems
- Vehicle Routing Problems
- Transportation Problems
- Staffing Problems
- Production Problems

Agenda

- Quiz
- Hardness
- Modeling with Binary Variables
- Issues with binary/integer variables
- Rounding may fail

Quiz

- 24 hour take-home
- Posted noon Monday
- Due by noon on Tuesday
- Coverage: through today’s lecture

Hardness

- LP with n variables
- can be solved in √n matrix operations

- 2n possibilities for n binary variables
- No really faster way knownfor some cases (NP hard problems)
- fame + $1m Clay prize for proving it

Binary Variables

- Piecewise linear functions
- If statements
- Discontinuous functions
- Set Covering
- Versions of the assignment problem

Knapsack Problem

- n items
- item i has weight wi, value vi
- maximize the value in the knapsack
- s.t. weight limit B is not exceeded

Knapsack Problem

max x1v1+…+xnvn

s.t.x1w1+…+xnwn ≤ B

xi binary

xi = 1 if item i in the knapsack

NP hard problem

Penalty

- Operating coal plant
- $3000 penalty (per day) if emissions > b
(emissions always < 88kg/day)

Penalty

- $3000 penalty (per day) if emissions > b
(emissions always < 88kg/day)

- emissions p
- f binary
- p ≤ 88 + (b-88)f
- penalty: (1-f)3000
- unintended option?

Fixed Cost

- Transportation Problem
- Fixed cost of $1000 for any shipment
(quantity shipped always less than 100)

Fixed Cost

- xij quantity shipped from i to j
- fij binary (1 if xij > 0)
- xij ≥ 0, xij ≤ 100 fij
- fixed cost of 1000 fij

Facility Location

Solution to Maximal Covering Problem w/ 10 facilities

Dc=300

Set covering – Find min. # needed to cover all demands

Max covering – Cover max # DEMANDS w/ fixed # facilities

P-center – Cover all demand nodes w/ fixed # facilities in smallest possible distance

Slide courtesy of Prof. Daskin

If statements (Part 2)

0 ≤x and

If x≤b, then y=c, else y=d

- create binary 0/1 variable z
- add the constraints
(b-x)/b ≤ z(if x≤b, then z=1)

z≤1+(b-x)/b(if x>b, then z=0)

y=cz+d(1-z)(if z=1, then y=c else y=d)

Other Integer/Binary issues

- Sensitivity Analysis
- Relaxation
- Rounding

Rounding may fail

Example courtesy of Prof. Daskin

Note that none of the points you would get to by rounding(9,9) (10,9), (9,8), (10,8)

is feasible!

Solution