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MS&E 211 Midterm Review 2008-2009 L&Y: Chapters 1-4, Chapter 6.1-6.4, 6.8PowerPoint Presentation

MS&E 211 Midterm Review 2008-2009 L&Y: Chapters 1-4, Chapter 6.1-6.4, 6.8

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MS&E 211 Midterm Review 2008-2009 L&Y: Chapters 1-4, Chapter 6.1-6.4, 6.8

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MS&E 211 Midterm Review 2008-2009 L&Y: Chapters 1-4, Chapter 6.1-6.4, 6.8. LP Models. Know how to recognize an LP in verbose or matrix form; standard or otherwise; max or min Know how to set up an LP

MS&E 211 Midterm Review 2008-2009 L&Y: Chapters 1-4, Chapter 6.1-6.4, 6.8

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MS&E 211Midterm Review2008-2009L&Y:Chapters 1-4, Chapter 6.1-6.4, 6.8

- Know how to recognize an LP in verbose or matrix form; standard or otherwise; max or min
- Know how to set up an LP
- Understand how to make various conversions between different types of LPs (max/min; <, >, =; >, free; |abs. value|; etc.)

- Modeling Process
- Understand the problem
- Collect information and data
- Identify and define the decision variables
- formulate the objective function
- Isolate and formulate the constraints

- Portfolio Management for World Cup Assets
- Five securities in a market for open trading at fixed prices and pay-offs, share limit on investment and short is not allowed
- We’d like to decide how many shares to purchase to maximize the pay-off when the game is realized.

x: shares purchased for each security; s: worst-case return

Return when it’s rainy

Return otherwise

What does it mean that a variable has a negative value?

Minimize the “purchase cost” under the condition that future has no loss under any circumstance

Return when it’s rainy

Return otherwise

It’s to sell for profit.

Maximize the profit under the condition that future has no loss under any circumstance

This is just one way to check if arbitrage exists or not …

Return when it’s rainy

Return otherwise

s: the new variable represents the worst case return under any circumstance.

- Understand the geometrical interpretation of LPs and associated intuition
- Plot a feasible region in 2D
- Plot the optimal iso-profit lines
- Relate how the Simplex Method moves from corner point to adjacent corner point in improving direction
- Understand the geographical interpretation of the various LP terms – (e.g. active constraints, basic variables, multiple optima, infeasibility, unbounded-ness,, etc.)

An LP problem falls in one of three cases:

- Problem is infeasible: Feasible region is empty.
- Problem is unbounded: Feasible region is unbounded towards the optimizing direction.
- Problem is feasible and bounded: then there exists an optimal point; an optimal point is on the boundary of the feasible region; and there is always at least one optimal corner point (if the feasible region has a corner point).
When the problem is feasible and bounded,

- There may be a unique optimal point or multiple optimizers (alternative optimizers, adjacent or not).
- If a corner point is not “worse” than all its neighbor corners, then it is optimal.

- Understand why we emphasize basic solutions and basic feasible solutions (corner points).
- Understand the equivalent (canonical) form of the LP and where it came from
- Understand what the different elements of this form mean: reduced cost coefficients, shadow prices, canonical constraint matrix, RHS
- Know when to terminate
- Know its relation with dual

- Recall the canonical form of the LP: all reduced cost coefficients for basic variables are zero, and reduced constraint matrix for basic variable form an identity matrix in certain order
- Why do we represent the problem this way?
- By looking at the objective function in terms of the non-basic variables only we can get insight into whether or not changing any of them might improve the objective function
- By looking at the constraints in this way, we can see
- on the RHS we see whether or not the BS is feasible or not, and
- in the constraint coefficients, how changing any non-basic variables will necessitate that we must change basic variables in response

- Initial Tableau:
x slack RHS

Basis - c 00

AI b (>0)

- Intermediate and final tableau:
Basis - c +y*Ay*cBAB-1b

AB-1AAB-1AB-1b (0,b-bar)

Where y*= cBAB-1, the dual solution to the production-max problem, AB is the current basis selected from [ A I ]

Reduced Cost Vector

A-bar

- Understand the Simplex tableau and the canonical form and what it all means relative to sensitivity
- Changing the cost vector and RHS: Know how far each entry can be changed before the optimal basis is changed
- Non basic variable objective coefficient (check reduced cost)
- Basic variable objective coefficient (interval analysis on the basic row of A-bar and reduced cost for non-basic columns)
- RHS entry (interval analysis on column of AB-1 and b-bar)

- Don’t just memorize formulas!
- Understand where they come from and why they work
- Understand the importance of what reduced costs and shadow prices mean

- The Concrete Products Corporation has the capability of producing four types of concrete blocks. Each block must be subjected to three processes: batch mixing, mold vibrating and inspection. The plant manager desires to maximize the profit during the next month. During the upcoming thirty days, he has 800 machine hours available on the batch mixer, 1000 hours on the mold vibrator and 340 man-hours of inspection time. The problem is formulated as follows:max 8x1 + 14x2 + 30x3 + 50x4 S.t. x1 + 2 x2 + 10x3 + 16x4 ≤ 800 1.5x1 + 2 x2 + 4 x3 + 5 x4 ≤ 1000 0.5x1 + 0.6x2 + x3 + 2 x4 ≤ 340where xi is the production level of the ith product; i = 1, 2, 3, 4.

Shadow prices (for maximization)

- c +y*A

Basicx1x2x3x4x5x6x7RHS

B0028405206000

20111191.5-10200

110-12-22-220400

700-41.6.1-.4120

AB-1

AB-1b

- By how much must the unit profit on block 3 be increased before it would be profitable to manufacture it?
: This can be answered by simply checking the reduced cost of the final tableau for x3, which is 28. The same question for x4 is 40

2. What must the minimum unit profit on block 2 be so that the optimal basis remains the same?

: Take the reduced cost row and the row with x2 as the basic variable for non-basic columns

(28405 2)

-λ(11191.5-1) 0,

(2.542.13.3-2) , -2 ≤ λ≤2.1

-14+2.1=-11.9; thus, the answer is 11.9

3. If the 800 machine-hours on the batch mixer is uncertain, for what range of machine hours will the optimal basis remains the same ?

: We need to find λ such that AB-1(b+λe1) 0, where e1 is the vector with 1 at the first position and 0 everywhere else. This is equivalent to AB-1b+λAB-1e1 0.

But AB-1b is the RHS of the final tableau (200;400;20), and AB-1e1 is the first column of AB-1, which in this problem is (1.5;-2;0.1).

Thus we have

200 + 1.5λ 0, 400 - 2 λ 0, 20 + 0.1 λ 0, which imply that -133 ≤ λ≤ 200. Then the answer is [800-133=667, 800+200=1000]

4. A competitor located next door has offered the manager additional batch-mixing time at a rate of $4.50 per hour. Should he accept his offer?

: Easy, take it since the shadow price for this resource is $5

5. Suppose instead that the competitor offers the manager 250 hours of batch-mixing time for a total of $1,100, Should he accept his offer? (The manager can only accept or reject the extra 250 hours.)

: We know that the first 200 hours worth 5*200=1000 but don’t know how much the next 50 worth for. Sorry, we need to resolve the LP problem to answer this question.

6.The owner has approached the manager with a thought about producing a new type of concrete block that would require 4 hours of batch mixing, 4 hours of molding and 1 hour of inspection per pallet. What should be the profit per pallet if block 5 is to be manufactured?

: The current shadow prices are 5, 2, and 0, respectively. So the answer is that it should sell by at least 5*4+2*4+0*1=28 per pallet

7. If the next door competitor would like to buy the resources from Concrete Products next month, what would be the fair prices? Formulate it as a Linear Program.

: This is the liquidation price problem which is the dual of this production problem.

- All inactive constraints have zero dual price
- In general, the dual price on a given active constraint is the rate of increase in the optimal value (OV) as the RHS of the constraint increases with all other data held fixed. If the RHS is decreased, it is the rate at which the OV is decreased.
- The constraint RHS ranges give the ranges of the constraint RHS over which no change in the dual price will occur.
- One of the “allowable increase and decrease” for an inactive constraint is infinite and the other equals the slack or surplus.
- In general, when the RHS of an active constraint changes, both the OV and OS will change.

- All basic variables have zero reduced cost
- In general, the reduced cost of any non-basic variable is the amount the objective coefficient of that variable would have to change, with all other data held fixed, in order for it to have a positive value in the optimum.
- The objective coefficient ranges give the ranges of the objective function over which no change in the OS will occur.
- One of the “allowable increase and decrease” for a non-basic variable is infinite and the other is the reduced cost. If a non-basic variable has zero reduced cost, then there exist alternative optimums.

- Know how to construct the dual
- Understand the economic interpretation of y (=cBAB-1) as shadow prices
- Using this y, understand why the dual feasibility condition is the primal optimality condition
- Know the duality theorems
- Know the primal-dual optimality conditions and complementary slackness

1. The number of variables in the dual problem is equal to the number of constraints in the original (primal) problem. The number of constraints in the dual problem is equal to the number of variables in the original problem.

2. Coefficient of the objective function in the dual problem come from the right-hand side of the original problem.

3. If the original problem is amax model, the dual is aminmodel; if the original problem is amin model, the dual problem is themaxproblem.

4.The coefficient of the first constraint function for the dual problem are the coefficients of the first variable in the constraints for the original problem, and the similarly for other constraints.

5. The right-hand sides of the dual constraints come from the objective function coefficients in the original problem.

6. The sense of the ith constraint in the dual is = if and only if the ith variable in the original problem is unrestricted in sign.

7. If the original problem is man (min ) model, then after applying Rule 6, assign to the remaining constraints in the dual a sense the same as (opposite to ) the corresponding variables in the original problem.

8. The ith variable in the dual is unrestricted in sigh if and only if the ith constraint in the original problem is an equality.

9. If the original problem is max (min) model, then after applying Rule 8, assign to the remaining variables in the dual a sense opposite to (the same as) the corresponding constraints in the original problem.

Max model Min modelxj 0 jth constraint xj≤ 0 jth constraint ≤ xj free jth constraint =

ith const ≤ yi 0 ith const yi ≤ 0 ith const = yi free

Use common sense to remember them!

Standard Primal Production Form:

Dual Form:

Standard Primal Production Form:

Dual Form:

Standard (Equality) Primal Form:

Dual Form:

1. The dual of the dual problem is again the primal problem.

2. Either of the two problems has an optimal solution if and only if the other does; if one problem is feasible but unbounded, then the other is infeasible; if one is infeasible, then the other is either infeasible or feasible/unbounded.

3. Weak Duality Theorem: The objective function value of the primal (dual) to be minimized evaluated at any primal (dual) feasible solution cannot be less than the dual (primal) objective function value evaluated at a dual (primal) feasible solution.

cTx >= bTy (in the standard equality form)

4. Strong Duality Theorem: When there is an optimal solution, the optimal objective value of the primal is the same as the optimal objective value of the dual.

cTx* = bTy*

5. Complementary Slackness Theorem:Consider an inequality constraint (nonnegative variable) in any LP problem. If that constraint is inactive (nonnegative variable is positive) for an optimal solution to the problem, the corresponding dual variable (inequality constraint) will be zero (active) in any optimal solution for the dual of that problem.

y*j (b-Ax*)j = 0, j=1,…,n;

x*i (c-ATy*)i = 0, i=1,…,m;

Phase I Problem:

The dual of Phase I

If the dual objective value is positive, y is a certificate of primal in feasibility.

- Portfolio Management for World Cup Assets
- Five securities in a market for open trading at fixed prices and pay-offs, and short is allowed
- We’d like to decide how many shares to purchase or sell to maximize the pay-off when the game is realized.

No share limit, and short is allowed:

Is the primal feasible?

P: the state prices

If the dual is infeasible, then the primal is unbounded or it has an arbitrage