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
x: shares purchased for each security; s: worst-case return
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 circumstance2. Security Cost Minimization
Return otherwise= Security Profit Maximization
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 otherwiseMore Reasonable Profit Maximization
s: the new variable represents the worst case return under any circumstance.
An LP problem falls in one of three cases:
When the problem is feasible and bounded,
x slack RHS
Basis - c 0 0
A I b (>0)
Basis - c +y*A y* cBAB-1b
AB-1A AB-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
Shadow prices (for maximization)
- c +y*A
Basic x1 x2 x3 x4 x5 x6 x7 RHS
B 0 0 28 40 5 2 0 6000
2 0 1 11 19 1.5 -1 0 200
1 1 0 -12 -22 -2 2 0 400
7 0 0 -4 1.6 .1 -.4 1 20
: 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
(28 40 5 2)
-λ (11 19 1.5 -1) 0,
(2.54 2.1 3.3 -2) , -2 ≤ λ≤2.1
-14+2.1=-11.9; thus, the answer is 11.9
3. before it would be profitable to manufacture it? 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. before it would be profitable to manufacture it? 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. before it would be profitable to manufacture it? 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.
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:
Standard Primal Production Form:
Standard (Equality) Primal 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.
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