DEF – ILPAn integer linear program is a linear program with the additional constraint that the variables are requiredto be integer.
min cTx (1)
Ax ≥ b (2)
x ≥ 0 (3)
x ∈ N (4)
where A ∈ Rm X n , b ∈ Rm, c ∈ Rn.
Write an example of the matrix and the vectors...
Equivalent way an writing an ILP:
min
n∑i=1
cixi (5)
n∑i=1
Aijxi ≥ bj j = 1, . . . ,m (6)
xi ≥ 0 i = 1, . . . , n (7)
xi ∈ N i = 1, . . . , n (8)
DEF – MILPAn ILP when some variables are restricted to be integer and some are not, the problem is called: mixedinteger linear program.
min cTx (9)
Ax ≥ b (10)
x ≥ 0 (11)
xj ∈ N j = 1, . . . , p (12)
DEF – Binary programsAn ILP where the integer variables are restricted to be 0 or 1 (it comes up surprisingly often). Such problemsare called pure (mixed) 0–1 linear programs or pure (mixed) binary integer linear programs.
x ∈ {0, 1}
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Modeling PotentialModeling with integer variables is useful in a wide variety of applications and in many financial applica-tion/problems:
• model logical requirements (boolean conditions)
• model managerial decisions
• select among different resources
• allocate scarce resources
• manage portfolios and projects
• model fixed costs
• many others . . .
SoftwareMany SOLVERs are available to direct tackle these kind of problems.
Commercial solvers:
• IBM ILOG CPLEX
• GUROBI
• XPRESS
• EXCEL
Open solvers:
• SCIP
• CBC
• GLPK
• LPSOLVE
• Libre Office
The downside of this power, however, is that problems with more than a few thousand variables are oftennot possible to solve unless they show a specific exploitable structure.
Exact AlgorithmsSolution techniques:
• Branch and Bound
• Cutting planes
• Branch and Cut
• Branch and Price
• A combination of the above methods
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Example 1
Suppose we wish to invest 19,000 e and we have four investment opportunities:
1 investment of 6,700 e and has a net present value of 8,000 e
2 investment of 10,000 e and has a net present value of 11,000 e
3 investment of 5,500 e and has a net present value of 6,000 e
4 investment of 3,400 e and has a net present value of 4,000 e
Into which investments should we place our money so as to maximize our total present value? Suchproblems are called capital budgeting problems.
Ignoring integrality constraints: the optimal linear programming solution is:
• the optimal linear programming solution is: x1=1, x2=0.89, x3= 0, x4 = 1
• the objective function value is 21,790 e
• this solution is not integral
• rounding x2 down to 0 gives a feasible solution with a value of 12,000 e
• the optimal solution is: x1=0, x2=1, x3= 1, x4 = 1
• the objective function value is 21,000 e
There are a number of additional constraints we might want to add:
• We can only make two investments
• If investment 2 is made, then investment 4 must also be made
• If investment 1 is made, then investment 3 cannot be made and vice-versa
• x1 + x2 + x3 + x4 ≤ 2
• x2 − x4 ≤ 0
• x1 + x3 ≤ 1
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Example 2
As the leader of an oil exploration drilling venture, you must determine the best selection of 5 out of 10possible sites.
• Label the sites s1, s2, . . . , s10
• Label the profits p1, p2, . . . , p10
Subject to the additional operational constraints:
• If site s2 is explored, then site s3 must also be explored
• Exploring sites s1 and s7 will prevent you from exploring site s8
• Exploring sites s3 or s4 will prevent you from exploring site s5
xj =
{1 then we select site j0 otherwise
max
10∑j=1
pjxj (16)
10∑j=1
xj = 5 (17)
x2 − x3 ≤ 0 (18)
x1 + x7 + x8 ≤ 2 (19)
x3 + x5 ≤ 1 (20)
x4 + x5 ≤ 1 (21)
xj ∈ {0, 1} i = 1, . . . , 10 (22)
– show how to solve a model using libre office
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1.2 Branch and bound
Branch and bound has been the most effective technique for solving MILPs in the following forty years orso. It was proposed in 1960 by Land and Dong.
It is based on dividing the problem into a number of smaller problems (branching) and evaluating theirquality based on solving the underlying linear programs (bounding).
However, in the last ten years, cutting planes have made a resurgence and are now efficiently combinedwith branch and bound into an overall procedure called branch and cut. This term was coined by Padbergand Rinaldi 1987.
All these approaches involve solving a series of linear programs:
(MILP) min cTx (23)
Ax ≥ b (24)
x ≥ 0 (25)
xj ∈ N j = 1, . . . , p (26)
Linear Programming Relaxation:
(LP) min cTx (27)
Ax ≥ b (28)
x ≥ 0 (29)
(30)
The best known (and most successful) methods for solving LPs are the interior-point and simplex methods.
Rounding the solution of LP will not in general give the optimal solution of MILP.
Since LP is less constrained than MILP, the following are immediate:
• The optimal objective value for LP is less than or equal to the optimal objective for MILP
• If LP is infeasible, then so is MILP
• If the optimal solution x of LP satisfies xj integer for j = 1, ..., p, then x is also optimal for MILP
• it gives a bound on the optimal value of MILP
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1.2.1 Branch and bound example 1
We first explain branch and bound by solving the following pure integer linear program:
maxx1 + x2 (31)
−x1 + x2 ≤ 2 (32)
8x1 + 2x2 ≤ 19 (33)
x1, x2 ≥ 0 (34)
x1, x2 ∈ N (35)
The first step is to solve the linear programming relaxation: x1 = 1.5, x2 = 3.5 with objective value 5.
Figure 1: Attention x1 and x2 are misplaced.
How can we exclude this solution while preserving the feasible integral solutions?
One way is to branch, creating two linear programs:
• x1 ≤ 1
• x1 ≥ 2
• Clearly, any solution to the integer program must be feasible to one or the other of these two problems.
• We will solve both of these linear programs.
maxx1 + x2 (36)
−x1 + x2 ≤ 2 (37)
8x1 + 2x2 ≤ 19 (38)
x1 ≤ 1 (39)
x1, x2 ≥ 0 (40)
(41)
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The solution is x1 = 1, x2 = 3 with objective value 4. This is a feasible integral solution. So we nowhave an upper bound of 5 as well as a lower bound of 4 on the value of an optimum solution to the integerprogram.
maxx1 + x2 (42)
−x1 + x2 ≤ 2 (43)
8x1 + 2x2 ≤ 19 (44)
x1 ≥ 2 (45)
x1, x2 ≥ 0 (46)
(47)
The solution is x1 = 2, x2 = 1.5 with objective value 3.5. Because this value is worse that the lower boundof 4 that we already have, we do not need any further branching. We conclude that the feasible integralsolution of value 4 found earlier is optimum.
These problems can be arranged in a branch-and-bound tree. Each node of the tree corresponds toone of the problems that were solved:
Figure 2: B&B tree of example 1
We can stop the enumeration at a node of the branch-and-bound tree for three different reasons (whenthey occur, the node is said to be pruned).
• Pruning by integrality occurs when the corresponding linear program has an optimum solution thatis integral.
• Pruning by bounds occurs when the objective value of the linear program at that node is worse thanthe value of the best feasible solution found so far.
• Pruning by infeasibility occurs when the linear program at that node is infeasible.
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1.2.2 Branch and bound example 2
max 3x1 + x2 (48)
−x1 + x2 ≤ 2 (49)
8x1 + 2x2 ≤ 19 (50)
x1, x2 ≥ 0 (51)
x1, x2 ∈ N (52)
Figure 3: B&B tree of example 2
Point of intersection among two lines:{a×x1 + b×x2 + c = 0r×x1 + s×x2 + t = 0
where matrix A ∈ Rm × n , m-vector b ∈ Rm, n-vector c ∈ Rn and p is an integer such that 1 ≤ p ≤ n.
• The branch-and-bound algorithm keeps a list of linear programming problems obtained by relaxing theintegrality requirements on the variables and imposing constraints such as:
– xj ≤ uj– xj ≥ lj
• Each such linear program corresponds to a node of the branch-and-bound tree.
• For a node Ni , let zi denote the value of the corresponding linear program. Let L denote the list ofnodes that must still be solved. Let zU denote an upper bound on the optimum value (initially, thebound zU can be derived from a heuristic solution of MILP, or it can be set to +∞).
Step 0 Initialize L = MILP, zU = +∞, x∗ = ∅.
Step 1 Terminate? If L = ∅, the solution x∗ is optimal.
Step 2 Select node Choose and delete a problem Ni from L.
Step 3 Bound Solve Ni . If it is infeasible, go to Step 1. Else, let xi be its solution and zi its objective value.
Step 4 Prune
– If zi ≥ zU , go to Step 1.
– If xi is not feasible for MILP, go to Step 5.
– If xi is feasible for MILP, let zU = zi , x∗ = xi. Delete from L all problems with zj ≥ zU . Go toStep 1.
Step 5 Branch From Ni , construct linear programs N1i , ..., N
ki with smaller feasible regions whose union
contains all the feasible solutions of (MILP) in Ni . Add N1i , ..., N
ki to L and go to Step 1.
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1.3 Branch-and-bound additional elements
• Formulation (so that the gap zI - zLP is small)
• Branching
• Node selection
• Heuristics (to find a good upper bound zU )
1.3.1 Branching
Binary Branching (branching on a variable) Problem Ni is a linear program. A way of dividing itsfeasible region is to impose bounds on a variable. Let xij be one of the fractional values for j = 1, . . . , p, in
the optimal solution xi of Ni.From problem Ni , we can construct two linear programs (satisfy the requirements of Step 5):
• N−ij = Ni with the additional bound xj ≤ bxijc
• N+ij = Ni with the additional bound xj ≥ dxije
The advantage of branching on a variable is that the number of constraints in the linear programs doesnot increase, since linear programming solvers treat bounds on variables implicitly.
On which variable xj should we branch, among the j = 1, . . . , p such that xi is fractional?
Branching on the most fractional variable
Strong Branching Compute for all the fractional variables:
• D+ij = increase of the objective values from Ni to N+
ij
• D−ij = increase of the objective values from Ni to N−ij
branch on the variable j = 1, . . . , p such that min(D−ij , D+ij) is the largest (thus tightening the lower bound
for pruning). Others have proposed to choose j such that (D+ij + D−ij) is the largest. Combining these two
criteria is even better, with more weight on the first.
But the computing time of doing it at each node Ni , for every fractional variable xij , may be too high.Significantly more time should be spent on these evaluations towards the top of the tree.
Pseudocost Branching Let f ij = xij − bxijc be the fractional part of xij , for j = 1, . . . , p. For an index j
such that f ij > 0, define:
• down pseudocost P−j =D−ij
fij
• up pseudocost P+j =
D+ij
1−fij
It can be observed that the pseudocosts tend to remain fairly constant throughout the branch-and-boundtree. Therefore the pseudocosts need not be computed at each node of the tree. They are estimated instead.
• A good way of initializing the pseudocosts is through strong branching at the root node or other nodesof the tree when new variables become fractional for the first time
• To update the pseudocost Pj , we average the observations over all the nodes of the tree where xj wasbranched on.
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• The estimated pseudocosts Pj and P+j are used to compute estimates of D−ij and D+
ij at node Ni
– D−ij = P−j fij
– D+ij = P+
j (1− f ij)
• Among these candidates, the branching variable xj is chosen to be the one with largest min(D−ij , D+ij)
(or other criteria such as those mentioned earlier).
1.3.2 Node selection
How does one choose among the different problems Ni available in Step 2 of the algorithm?
Two goals need to be considered:
• finding good feasible solutions (thus decreasing the upper bound zU )
• proving optimality of the current best feasible solution (by increasing the lower bound as quickly aspossible)
Best estimate criterion For the first goal, we estimate the value of the best feasible solution in each nodeNi:
Ei = zi +
p∑j=1
min(P−j fij , P
+j (1− f ij))
This corresponds to rounding the noninteger solution xi to a nearby integer solution and using thepseudocosts to estimate the degradation in objective value.
Depth-first search strategy It is an effective strategy to quickly look for a feasible solution.
Breadth-first search strategy It is an effective strategy to increase the global lower bound of the problem.
Practical considerations:
• The most successful node selection strategy may differ depending on the application.
• For this reason, most MILP solvers have several node selection strategies available as options.
• The default strategy is usually a combination of the best estimate criterion (or a variation) and depth-first search.
1.3.3 Heuristics
• Heuristics are useful for improving the bound zU , which helps in Step 4 for pruning by bounds.
• Heuristics are even more important when the branch-and-bound algorithm is too time consuming andwe need to stop the earlier the optimization
• Examples: diving heuristics, rins, solution polishing, feasibility pump and many others . . .
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Local Branching This heuristic is particularly suited for MILPs that are too large to solve to optimality,but where the linear programming relaxation can be solved in reasonable time. For simplicity, assume thatall the integer variables are 0,1 valued.
• Let x be a feasible solution of (MILP) (found by a diving heuristic, for example).
• The idea is to define a neighborhood of x as follows:
–∑pj=1 |xj − xj | ≤ k
–∑pj∈I:xj=0 xj +
∑pj∈I:xj=1(1− xj) ≤ k (linearised)
• where k is an integer chosen by the user (for example k = 20)
• Instead of getting lost in a huge enumeration tree, the search is restricted to the neighborhood of x bythis constraint
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1.4 Cutting planes
We next describe an algebraic method for generating cuts due to R. E. Gomory.
Suppose we are given an instance of ILP and begin by solving the LP-relaxation, producing an optimalbasic feasible solution xB associated with basis B.
• If the solution is integer then stop.
• Otherwise add a cutting plane such that xB :
– is not feasible
– all feasible integer points remain feasible
• Then, repeat above processes.
(MILP) min cTx (57)
Ax ≥ b (58)
x ≥ 0 (59)
xj ∈ N j = 1, . . . , p (60)
Linear Programming Relaxation:
(LP) min cTx (61)
Ax ≥ b (62)
x ≥ 0 (63)
(64)
Consider any equation in the optimized final tableau, say the i-th equation:
xiB +∑j /∈B
yijxj = yi0 (1G)
for some i, 0 ≤ i ≤ m. Due to nonnegativity of x∑j /∈B
byijcxj ≤∑j /∈B
yijxj
Therefore we have:
xiB +∑j /∈B
byijcxj ≤ byi0c (2G)
Subtracting (2G) from (1G), we get the constraint:∑j /∈B
(yij − byijc)xj ≥ yi0 − byi0c
called the Gomory cut corresponding to the source or generating row i.
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Example
maxx1 + x2 (65)
−x1 + x2 ≤ 2 (66)
8x1 + 2x2 ≤ 19 (67)
x1, x2 ≥ 0 (68)
x1, x2 ∈ {0, 1} (69)
We first add slack variables x3 and x4 to turn the inequality constraints into equalities. The problembecomes:
max zx1x2 = 0 (70)
−x1 + x2 + x3 = 2 (71)
8x1 + 2x2 + x4 = 19 (72)
x1, x2, x3, x4 ≥ 0 (73)
x1, x2, x3, x4 ∈ {0, 1} (74)
Solving the linear programming relaxation by the simplex method, we get the optimal tableau:
z + 0.6x3 + 0.2x4 = 5 (75)
x2 + 0.8x3 + 0.1x4 = 3.5 (76)
x1 − 0.2x3 + 0.1x4 = 1.5 (77)
x1, x2, x3, x4 ≥ 0 (78)
The corresponding basic solution is x3 = x4 = 0, x1 = 1.5, x2 = 3.5 and z = 5. This solution is notinteger. Let us generate the Gomory mixed integer cut corresponding to the equation:
x2 + 0.8x3 + 0.1x4 = 3.5 (79)
we get the GMI cut:
8x3 + x4 ≥ 5. (80)
We can express the above Gomory cut in the space (x1 , x2 ). This yields:
An auction is a process of buying and selling goods or services by offering them up for bid, taking bids, andthen selling the item to the highest bidder. In economic theory, an auction may refer to any mechanism orset of trading rules for exchange.
• the value that a bidder has for a set of items may not be the sum of the values that he has for individualitems
• Examples are equity trading, electricity markets, pollution right auctions and auctions for airportlanding slots.
• To take this into account, combinatorial auctions allow the bidders to submit bids on combinations ofitems.
Problem data:
• let M = {1, 2, . . . ,m} be the set of items that the auctioneer has to sell
• a bid is a pair Bj = (Sj , pj ) where Sj ⊂ M is a nonempty set of items and pj is the price offer forthis set
• Suppose that the auctioneer has received n bids B1 , B2 , . . . , Bn
• How should the auctioneer determine the winners in order to maximize his revenue?
The auctioneer maximizes his revenue by solving the integer program:
xj =
{1 if bid Bj wins0 otherwise
max
n∑j=1
pjxj (82)
10∑j:i∈Sj
xj ≤ 1 i = 1, . . . ,m (83)
xj ∈ {0, 1} j = 1, . . . , n (84)
The constraints impose that each item i is sold at most once.
For example, if there are four items for sale and the following bids have been received:
• B1 = ({1}, 6)
• B2 = ({2}, 3)
• B3 = ({3, 4}, 12)
• B4 = ({1, 3}, 12)
• B5 = ({2, 4}, 8)
• B6 = ({1, 3, 4}, 16)
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The winners can be determined by the following integer program:
max 6x1 + 3x2 + 12x3 + 12x4 + 8x5 + 16x6 (85)
x1 + x4 + x6 ≤ 1 (86)
x2 + x5 ≤ 1 (87)
x3 + x4 + x6 ≤ 1 (88)
x3 + x5 + x6 ≤ 1 (89)
xj ∈ {0, 1} j = 1, . . . , 6 (90)
1.5.2 The Lockbox Problem
Consider a national firm that receives checks from all over the United States.
• Due to the vagaries of the U.S. Postal Service, as well as the banking system, there is a variable delayfrom when the check is postmarked and when the check clears
• For instance, a check mailed in Pittsburgh sent to a Pittsburgh address might clear in just 2 days. Asimilar check sent to Los Angeles might take 4 days to clear.
In order to speed up this clearing, firms open offices (called lockboxes) in different cities to handle thechecks. Which lockboxes should we open?
Example :
• We receive payments from 4 regions (West, Mid-west, East, and South)
• The average daily value from each region is as follows:
– 300,000 from the West
– 120,000 from the Midwest
– 360,000 from the East
– 180,000 from the South
• We are considering opening lock- boxes in (and/or):
– L.A.
– Cincinnati
– Boston
– Houston
• Operating a lockbox costs 90,000 per year.
The average days from mailing to clearing is given in the table:
From – To L.A. Cincinnati Boston HoustonWest 2 4 6 6Midwest 4 2 5 5East 6 5 2 5South 7 5 6 3
Table 1: Average days from mailing to clearing
First we must calculate the losses due to lost interest for each possible assignment.
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• For example, if the West sends to Boston, then on average there will be 1,800,000 (= 6 × 300, 000) inprocess on any given day.
• Assuming an investment rate of 10%, this corresponds to a yearly loss of 180,000.
We can calculate the losses for the other possibilities in a similar fashion to get the table:
From To L.A. Cincinnati Boston HoustonWest 60 120 180 180Midwest 48 24 60 60East 216 180 72 180South 126 90 108 54
Table 2: losses
ILP formulation
• n = number of lockboxes (j = 1, . . . , n)
• m = number of regions (i = 1, . . . ,m)
• αij = annual cost of assigning region i to the lockbox j
• βj = annual cost of lockbox j
Decision variables:
yj =
{1 if lockbox j is opened0 otherwise
xij =
{1 if region i sends to lockbox j0 otherwise
min
n∑j=1
m∑i=1
αijxij +
n∑j=1
βjyj (91)
n∑j=1
xij = 1 i = 1, . . . ,m (92)
m∑i=1
xj ≤ myj j = 1, . . . , n (93)
xij ∈ {0, 1} i = 1, . . . ,m, j = 1, . . . , n (94)
yj ∈ {0, 1} j = 1, . . . , n (95)
Explanation:
• Our objective is to minimize our total yearly costs
The above is a correct 01 programming formulation of the lockbox problem. There are other formulations,however. For instance, consider the sixteen constraints of the form:
xij ≤ yj i = 1, . . . ,m, j = 1, . . . , n (112)
Different formulations can have very different properties with respect to their associated linear program.One very active research area is to take common problems and find good reformulations.
1.5.3 Constructing an Index Fund
Different possibilities of portfolio management:
• active portfolio management tries to achieve superior performance by using forecasting techniques
• aassive portfolio management avoids any forecasting techniques and rather relies on diversificationto achieve a desired performance
There are 2 types of passive management strategies:
• buy and hold: assets are selected on the basis of some fundamental criteria and there is no activeselling or buying of these stocks afterwards
• indexing: the goal is to choose a portfolio that mirrors the movements of a broad market populationor a market index. Such a portfolio is called an index fund.
Index fund – Definition Given a target population of n stocks, one selects q stocks and their weights inthe index fund, to represent the target population as closely as possible.
In the last twenty years, an increasing number of investors, both large and small, have established indexfunds. The rising popularity of index funds can be justified both theoretically and empirically:
• Market Efficiency: If the market is efficient, no superior risk-adjusted returns can be achieved bystock picking strategies since the prices reflect all the information available in the marketplace. Anindex fund captures the efficiency of the market via diversification.
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• Empirical Performance: Considerable empirical literature provides strong evidence that, on average,money managers have consistently underperformed the major indexes (luck is an explanation for goodperformance).
• Transaction Cost: Actively managed funds incur transaction costs, which reduce the overall perfor-mance of these funds. In addition, active management implies significant research costs.
Additional features:
• Here we take the point of view of a fund manager who wants to construct an index fund.
• Strategies for forming index funds involve choosing a broad market index as a proxy for an entiremarket, e.g. the Standard and Poor list of 500 stocks (S&P 500)
• A pure indexing approach consists in purchasing all the issues in the index, with the same exact weightsas in the index.
• In most instances, this approach is impractical (many small positions) and expensive (rebalancing costsmay be incurred frequently)
A Large-Scale Deterministic Model
• We propose a large-scale deterministic model for aggregating a broad market index of stocks into asmaller more manageable index fund.
• This approach will produce a portfolio that closely replicates the underlying market population.
We present a model that:
• clusters the assets into groups of similar assets
• selects one representative asset from each group to be included in the index fund portfolio
The model is based on the following data:
ρij = similarity between stock i and stock j (113)
Typical features:
• ρii = 1
• ρij ≤ 1 for i 6= j
• ρij is larger for more similar stocks
• An example of this is the correlation between the returns of stocks i and j
Decision variables:
yj =
{1 if stock j is in the index fund0 otherwise
xij =
{1 if j is the most similar stock to stock i in the index fund0 otherwise
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(IF ) max
n∑i=1
n∑j=1
ρijxij (114)
n∑j=1
yj = q (115)
n∑j=1
xij = 1 i = 1, . . . , n (116)
xij ≤ yj i = 1, . . . , n (117)
xij ∈ {0, 1} i = 1, . . . , n, j = 1, . . . , n (118)
yj ∈ {0, 1} j = 1, . . . , n (119)
• Interpret each of the constraints
• Explain why the objective of the model can be interpreted as selecting q stocks out of the populationof n stocks so that the total loss of information is minimized
Once the model has been solved and a set of q stocks has been selected for the index fund, a weight wjis calculated for each j in the fund:
wj =
n∑i=1
Vixij (120)
where Vi is the market value of stock i.
• So wj is the total market value of the stocks represented by stock j in the fund
• The fraction of the index fund to be invested in stock j is proportional to the stocks weight wj
• wj∑qf=1 wf
• Note that, instead of the objective function used, one could have used an objective function that takesthe weights wj directly into account
∑ni=1
∑nj=1 Viρijxij
Solution strategies
• Branch-and-bound is a natural candidate for solving the model, however that the formulation can bevery large. Indeed, for the S&P 500, there are 250,000 variables xij and 250,000 constraints xij ≤ yj
• The linear programming relaxation needed to get upper bounds in the branch-and-bound algorithm isdifficult to solve
• Instead of the linear programming relaxation, one can use the Lagrangian relaxation
max L(u) =
n∑i=1
n∑j=1
ρijxij +
n∑i=1
ui(1−n∑j=1
xij) (121)
n∑j=1
yj = q (122)
xij ≤ yj i = 1, . . . , n (123)
xij ∈ {0, 1} i = 1, . . . , n, j = 1, . . . , n (124)
yj ∈ {0, 1} j = 1, . . . , n (125)
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where u = (u1, . . . , un) is the vector of the Lagrangian Multipliers
Property 1: L(u) ≥ Z, where Z is the maximum for the model
Proof : Let x∗ be the optimal solution of the problem IF, then x∗ is also a feasible solution of L(u) foreach u ≥ 0.
cTx∗ + u(Ax∗ − b) ≤ L(u)
the quantity (Ax∗ − b) is ≥ 0, then:
cTx∗ ≤ L(u)
proving that L(u) ≥ Z.
The objective function L(u) may be equivalently stated as:
max L(u) =
n∑i=1
n∑j=1
(ρij − ui)xij +
n∑i=1
ui (126)
Let:
(ρij − ui)+ =
{(ρij − ui) if ρij - ui ≥ 00 otherwise
and
Cj =
n∑i=1
(ρij − ui)+ (127)
Property 2:
max L(u) =
n∑j=1
Cjyj +
n∑i=1
ui (128)
n∑j=1
yj = q (129)
yj ∈ {0, 1} j = 1, . . . , n (130)
Property 3: In an optimal solution of the Lagrangian relaxation, yj is equal to 1 for the q largest valuesof Cj , and the remaining yj are equal to 0.
• Interestingly, the set of q stocks corresponding to the q largest values of Cj can also be used as aheuristic solution for the model
• Specifically, construct an index fund containing these q stocks and assign each stock i = 1, . . . , n to themost similar stock in this fund
• This solution is feasible to the model, although not necessarily optimal.
• So for any vector u, we can compute quickly both a lower bound and an upper bound on the optimumvalue of the model
How does one minimize L(u)?
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• Since L(u) is non-differentiable and convex, one can use the subgradient method
• At each iteration, a revised set of Lagrange multipliers u and an accompanying lower bound and upperbound to the model are computed
• The algorithm terminates when these two bounds match or when a maximum number of iterations isreached
• It can be proved that min L(u) is equal to the value of the linear programming relaxation of the model
• In general, this is not true and therefore it is not possible to match the upper and lower bounds
• If one wants to solve the integer program to optimality, one can use a branch-and-bound algorithm,using the upper bound min L(u) for pruning the nodes.
A Linear Programming Model
• we consider an approach which tries to rebalance the portfolio at minimum cost
• This approach assumes that we have identified important characteristics of the market index we wouldlike to achieve. Such characteristics might be:
– the fraction fi of companies with market capitalization in various range
– the fraction fi of companies that pay no dividends
– . . .
• Let us assume that there are m such characteristics that we would like our index fund to track as wellas possible
• Let aij = 1 if company j has characteristic i and 0 if it does not
• Let xj denote the optimum weight of asset j in the portfolio.
• Assume that initially, the portfolio has weights x0j .
• Let yj denote the fraction of asset j bought and zj the fraction sold.
The problem of rebalancing the portfolio at minimum cost is the following:
min
n∑j=1
(yj + zj) (131)
n∑j=1
aijxj = fi i = 1, . . . ,m (132)
n∑j=1
xj = 1 (133)
xj − x0j ≤ yj j = 1, . . . , n (134)
x0j − xj ≤ zj j = 1, . . . , n (135)
xj ≥ 0 j = 1, . . . , n (136)
yj ≥ 0 j = 1, . . . , n (137)
zj ≥ 0 j = 1, . . . , n (138)
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1.6 Portfolio Selection and Asset Allocation
• The theory of optimal selection of portfolios was developed by Harry Markowitz in the 1950s
• His work formalized the diversification principle in portfolio selection
Consider an investor who has a certain amount of money to be invested in a number of different securities(stocks, bonds, etc.) with random returns
For each security i = 1, . . . , n, we estimates:
• the expected return µi
• the variance σi
For any two securities i and j, we estimates:
• correlation coefficient ρij
If we represent the proportion of the total funds invested in security i by xi we can define a portfolio asa vector x = (x1, ..., xn), and then we can compute:
the expected return of a portfolio
E[x] = x1µ1 + · · ·+ xnµn = µTx
the variance return of a portfolio
Var[x] =∑i
∑j
ρijσiσjxixj = xTQx
where ρii = 1, Qij = ρijσiσj , and µ = (µ1, . . . , µn).
Portfolio features:
• The portfolio vector x must satisfy∑i xi = 1
• There may or may not be additional feasibility constraints
• A feasible portfolio x is called efficient if:
– it has the maximal expected return among all portfolios with the same variance
– it has the minimum variance among all portfolios that have at least a certain expected return
• The collection of efficient portfolios form the efficient frontier of the portfolio universe
1.6.1 Mean-variance optimization (MVO) problem
• it is also called Markowitz portfolio optimization problem
• it can be formulated in three but equivalent ways
• one formulation results in the problem of finding a minimum variance portfolio of the securities 1 to nthat yields at least a target value R of expected return.
Convex quadratic programming problem:
min xTQx (139)
eTx = 1 (140)
µTx ≥ R (141)
x ≥ 0 (142)
(143)
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• where e is an n-dimensional vector all of which components are equal to 1
• The first constraint indicates that the proportions xi should sum to 1
• The second constraint indicates that the expected return is no less than the target value R
• the objective function corresponds to the total variance of the portfolio
• Nonnegativity constraints on xi are introduced to rule out short sales
• Note that the matrix Q is positive semidefinite
1.6.2 Portfolio Optimization with Minimum Transaction Levels
When solving the classical Markowitz model, the optimal portfolio often contains xi that are too small. Inpractice, one would like a solution of:
min xTQx (144)
eTx = 1 (145)
µTx ≥ R (146)
Cx ≥ d (147)
x ≥ 0 (148)
(149)
With the property that:
xj ≥ 0⇒ xj ≥ lj (A)
where lj are given minimum transaction levels.
• This constraint states that, if an investment is made in a stock, then it must be large enough, forexample at least 100 shares
• We are interested in portfolio optimization problem where there is an upper bound on the number ofpositive variables
• Because the constraint (11.2) is not a simple linear constraint, it cannot be handled directly by quadraticprogramming.
Branch-and-bound algorithm
• Let x∗ be the optimal solution found (QP with interior point methods). If no minimum transactionlevel constraint is violated by x∗ , then x∗ is also optimum and we can stop.
• Otherwise, let j be an index for which Requirement (A) is violated by x∗ and form two subproblems,obtained from by adding:
(i) xj = 0
(ii) xj ≥ lj
• Both are quadratic programs that can be solved again using interior points methods.
• A branch-and-bound tree can be constructed in this way
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2 Models: Value-at-Risk and Conditional Value-at-Risk
• Financial activities involve risk
• Financial and other institutions must manage risk using sophisticated mathematical techniques
• necessity of quantitative risk measures that adequately reflect the vulnerabilities of a company
• Value-at-Risk and Conditional Value-at-Risk are widely used measure of risk in finance
• They can be used instead of the variance of a portfolio as in the Markowitz model
• This is acheived through stochastic programming
• The random events are modeled by a large but finite set of scenarios, leading to a linear programmingequivalent of the original stochastic program.
2.1 Value-at-Risk – VaR
Definition – Take a random variable X that represents for example loss from an investment portfolio overa fixed period of time (a negative value for X indicates gains). Given a probability level α, α-VaR of therandom variable X is given by the following relation:
V aRα(X) = min{γ : P (X ≤ γ) ≥ α}
VaR: represents the predicted maximum loss with a specified probability level (e.g., 95%) over a certainperiod of time (e.g.,one day).
Figure 4: 0.95-VaR on a portfolio loss distribution plot
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Features:
• VaR is widely used by people in the financial industry and VaR calcu- lators are common features inmost financial software
• VaR can be estimated either parametrically (for example, variance-covariance VaR or delta-gammaVaR) or nonparametrically (for examples, historical simulation VaR or resampled VaR)
• 85% of large banks were using historical simulation. The other 15% used Monte Carlo methods
Problems:
• it lacks of sub-additivity
• a sub-additive function f(x1 + x2) ≤ f(x1) + f(x1) ∀x1, x2
• a good Risk measures should respect the principle that diversification reduces risk
• When VaR is computed by generating scenarios, it turns out to be a non-smooth and non-convexfunction of the positions in the investment port- folio.
• Therefore, when one tries to optimize VaR computed in this manner, the optimization is difficult (globaloptimization algorithms)
• Another criticism on VaR is that it pays no attention to the magnitude of losses beyond the VaR value.
Example of application
• take a generic portfolio
• you know its current market value, at the beginning of the day while it is not known its market valueat the end of the day
• The investment bank that holds this portfolio may declare that its portfolio has a 1-day VaR of 10million with a confidence level of 95%
• It means that the bank expects that, with a probability of 95%, the portfolio’s value will not be lessthan 10 million during the day
– This implies that the bank expects that the market value of its portfolio at the end of the day willbe less than 10 million, with a probability of 5%
– So the bank expects that 5 times out of 100, the portfolio will be below the threshold of 10 million,while above this threshold will be 95 times out of 100
2.2 Conditional Value-at-Risk – CVaR
• We consider a portfolio of assets with random returns
• We denote the portfolio choice vector with x and the random events by the vector y
• Let f (x, y) denote the loss function when we choose the portfolio x from a set X of feasible portfoliosand y is the realization of the random events
• We assume that the random vector y has a probability density function denoted by p(y)
For a fixed decision vector x, we compute the cumulative distribution function of the loss associated withthat vector x:
Ψ(x, y) =
∫f(x,y)<γ
p(y) dy
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Then, for a given confidence level α, the α-VaR associated with portfolio x is given as:
V aRα(x) = min{γ ∈ R : Ψ(x, y) ≥ α}
We define the α-CVaR associated with portfolio x as:
CV aRα(x) =1
1− α
∫f(x,y)≥V aRα(x)
f(x, y)p(y) dy
For a discrete probability distribution (where event yj occurs with probability pj , for j = (1, . . . , n), theabove definition of CVaR becomes:
CV aRα(x) =1
1− α∑
j:f(x,yj)≥V aRα(x)
pjf(x, yj)
Features:
• CVaR of a portfolio is always at least as big as its VaR
• Consequently, portfolios with small CVaR also have small VaR
• However, in general minimizing CVaR and VaR are not equivalent
Example:
• Suppose we are given the loss function f (x, y) for a given decision x as:
• f (x, y) = y where y = 75 j with probability 1% for j = 0, . . . , 99
• We would like to determine the maximum loss incurred with 95% probability
• This is the Value-at-Risk VaRα (x) for = 95%
• We have VaR95% (x) = 20 since the loss is less than 20 with probability 95%
• To compute the Conditional Value-at-Risk, we use the above formula:
• CVaR95% (x) =1
0.05(20 + 21 + 22 + 23 + 24) × 1% = 22
Since the definition of CVaR involves the VaR function explicitly, it is difficult to work with and optimizethis function. Instead, we consider the following simpler auxiliary function:
Fα(x, γ) = γ +1
1− α
∫f(x,y)≥γ
(f(x, y)− γ)p(y) dy
Alternatively, we can write
Fα(x, γ) = γ +1
1− α
∫(f(x, y)− γ)+p(y) dy
This function, viewed as a function of γ, has the following important properties that makes it useful forthe computation of VaR and CVaR:
• Fα(x, γ) is a convex function of γ
• VaRα (x) is a minimizer over γ of Fα(x, γ)
• The minimum value over γ of the function Fα(x, γ) is CVaRα (x)
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As a consequence of the listed properties, we immediately deduce that, in order to minimize CVaRα (x)over x, we need to minimize the function Fα(x, γ) with respect to x and γ simultaneously
minx∈X
CV aRα(x) = minx∈X,γ
Fα(x, γ)
• Consequently, we can optimize CVaR directly, without needing to compute VaR first
• If the loss function f (x, y) is a convex (linear) function of the portfolio variables x, then Fα(x, γ) isalso a convex (linear) function of x
• In this case, provided the feasible portfolio set X is also convex, the optimization problems above aresmooth convex optimization problems that can be solved using well known optimization techniques forsuch problems.
Scenarios
• Often it is not possible or desirable to compute/determine the joint density function p(y) of the randomevents in our formulation
• Instead, we may have a number of scenarios, say ys for s = 1, . . . , S, which may represent some historicalvalues of the random events or some values obtained via computer simulation.
In this case, we obtain the following approximation to the function:
˜Fα(x, γ) = γ +1
(1− α)S
S∑s=1
(f(x, ys)− γ)
Now, the problem:
minx∈X
CV aRα(x) = minx∈X,γ
γ +1
(1− α)S
S∑s=1
(f(x, ys)− γ)
To solve this optimization problem, we introduce artificial variables zs to replace (f(x, ys) − γ), thenobtaining:
minx,z,γ
γ +1
(1− α)S
S∑s=1
zs (150)
zs ≥ 0 s = 1, . . . , S (151)
zs ≥ f(x, ys)− γ s = 1, . . . , S (152)
x ∈ X (153)
(154)
In the case that f (x, y) is linear in x, all the constraints are linear constraints and therefore the problemis a linear programming problem that can be solved using the simplex method or alternative LP algorithms.
