# Finance

Originally Contributed by: Arpit Bhatia

Optimization models play an increasingly important role in financial decisions. Many computational finance problems can be solved efficiently using modern optimization techniques.

In this tutorial we will discuss two such examples taken from the book Optimization Methods in Finance.

This tutorial uses the following packages

using JuMP
import GLPK

## Short Term Financing

Corporations routinely face the problem of financing short term cash commitments such as the following:

MonthJanFebMarAprMayJun
Net Cash Flow-150-100200-20050300

Net cash flow requirements are given in thousands of dollars. The company has the following sources of funds:

• A line of credit of up to $100K at an interest rate of 1% per month, • In any one of the first three months, it can issue 90-day commercial paper bearing a total interest of 2% for the 3-month period, • Excess funds can be invested at an interest rate of 0.3% per month. Our task is to find out the most economical way to use these 3 sources such that we end up with the most amount of money at the end of June. We model this problem in the following manner: We will use the following decision variables: • the amount$u_{i}$drawn from the line of credit in month$i$• the amount$v_{i}$of commercial paper issued in month$i$• the excess funds$w_{i}$in month$i$Here we have three types of constraints: 1. for every month, cash inflow = cash outflow for each month 2. upper bounds on$u_{i}$3. nonnegativity of the decision variables$u_{i}$,$v_{i}$and$w_{i}$. Our objective will be to simply maximize the company's wealth in June, which say we represent with the variable$m$. financing = Model(GLPK.Optimizer) @variables(financing, begin 0 <= u[1:5] <= 100 0 <= v[1:3] 0 <= w[1:5] m end) @objective(financing, Max, m) @constraints(financing, begin u[1] + v[1] - w[1] == 150 # January u[2] + v[2] - w[2] - 1.01u[1] + 1.003w[1] == 100 # February u[3] + v[3] - w[3] - 1.01u[2] + 1.003w[2] == -200 # March u[4] - w[4] - 1.02v[1] - 1.01u[3] + 1.003w[3] == 200 # April u[5] - w[5] - 1.02v[2] - 1.01u[4] + 1.003w[4] == -50 # May -m - 1.02v[3] - 1.01u[5] + 1.003w[5] == -300 # June end) optimize!(financing) objective_value(financing) 92.49694915254233 ## Combinatorial Auctions In many auctions, the value that a bidder has for a set of items may not be the sum of the values that he has for individual items. Examples are equity trading, electricity markets, pollution right auctions and auctions for airport landing slots. To take this into account, combinatorial auctions allow the bidders to submit bids on combinations of items. Let$M=\{1,2, \ldots, m\}$be the set of items that the auctioneer has to sell. A bid is a pair$B_{j}=\left(S_{j}, p_{j}\right)$where$S_{j} \subseteq M$is a nonempty set of items and$p_{j}$is the price offer for this set. Suppose that the auctioneer has received$n$bids$B_{1}, B_{2}, \ldots, B_{n}.$The goal of this problem is to help an auctioneer determine the winners in order to maximize his revenue. We model this problem by taking a decision variable$y_{j}$for every bid. We add a constraint that each item$i\$ is sold at most once. This gives us the following model:

\begin{aligned} \max && \sum_{i=1}^{n} p_{j} y_{j} \\ \text { s.t. } && \sum_{j : i \in S_{j}} y_{j} \leq 1 && \forall i=\{1,2 \ldots m\} \\ && y_{j} \in\{0,1\} && \forall j \in\{1,2 \ldots n\} \end{aligned}
bid_values = [6 3 12 12 8 16]
bid_items = [[1], [2], [3 4], [1 3], [2 4], [1 3 4]]

auction = Model(GLPK.Optimizer)
@variable(auction, y[1:6], Bin)
@objective(auction, Max, sum(y' .* bid_values))
for i in 1:6
@constraint(auction, sum(y[j] for j in 1:6 if i in bid_items[j]) <= 1)
end

optimize!(auction)

objective_value(auction)
21.0
value.(y)
6-element Array{Float64,1}:
1.0
1.0
1.0
0.0
0.0
0.0