The document describes a mathematical model for currency arbitrage. Currency arbitrage involves simultaneously buying and selling currencies in different markets to profit from temporary price differences. The model seeks to maximize final dollar holdings by determining the optimal amounts to convert between US dollars, euros, British pounds, Japanese yen, and Kuwaiti dinars, given exchange rates and transaction limits. The model is formulated as a linear program to find the currency conversion amounts that maximize profits through arbitrage opportunities across multiple currencies.

1. Currency ArbitrageCurrency Arbitrage
 In today's global economy, a multinational company must dealIn today's global economy, a multinational company must deal
with currencies of the countries in which it operates. Currencywith currencies of the countries in which it operates. Currency
arbitrage, or simultaneous purchase and sale of currencies inarbitrage, or simultaneous purchase and sale of currencies in
different markets, offers opportunities for advantageousdifferent markets, offers opportunities for advantageous
movement of money from one currency to another.movement of money from one currency to another.
 For example, converting £1000 to U.S. dollars in 2001 with anFor example, converting £1000 to U.S. dollars in 2001 with an
exchange rate of $1.60 to £1 will yield $1600. Another way ofexchange rate of $1.60 to £1 will yield $1600. Another way of
making the conversion is to first change the British pound tomaking the conversion is to first change the British pound to
Japanese yen and then convert the yen to U.S. dollars using theJapanese yen and then convert the yen to U.S. dollars using the
2001 exchange rates of £1 = ¥175 and $1 = ¥105. The resulting2001 exchange rates of £1 = ¥175 and $1 = ¥105. The resulting
dollar amount isdollar amount is
(£1,000 x ¥175)/¥105 = $1,666.67(£1,000 x ¥175)/¥105 = $1,666.67
 This example demonstrates the advantage of converting theThis example demonstrates the advantage of converting the
British money first to Japanese yen and then to dollars. ThisBritish money first to Japanese yen and then to dollars. This
section shows how the arbitrage problem involving manysection shows how the arbitrage problem involving many
currencies can be formulated and solved as a linear program.currencies can be formulated and solved as a linear program.

2. Currency Arbitrage ModelCurrency Arbitrage Model
 Suppose that a company has a total of 5 million dollars that canSuppose that a company has a total of 5 million dollars that can
be exchanged for euros (€), British pounds (£), yen (¥), andbe exchanged for euros (€), British pounds (£), yen (¥), and
Kuwaiti dinars (KD). Currency dealers set the following limitsKuwaiti dinars (KD). Currency dealers set the following limits
on the amount of any single transaction: 5 million dollars, 3on the amount of any single transaction: 5 million dollars, 3
million euros, 3.5 million pounds, 100 million yen, and 2.8million euros, 3.5 million pounds, 100 million yen, and 2.8
million Kds. The table below provides typical spot exchangemillion Kds. The table below provides typical spot exchange
rates. The bottom diagonal rates are the reciprocal of the toprates. The bottom diagonal rates are the reciprocal of the top
diagonal rates. For example,diagonal rates. For example,
rate(€ -> $) = 1/rate( $ -> €) = 1/0.769 = 1.30rate(€ -> $) = 1/rate( $ -> €) = 1/0.769 = 1.30
$ € £ ¥ KD
$ 1 0.769 0.625 105 0.342
€ 1/0.769 1 0.813 137 0.445
£ 1/0.625 1/0.813 1 169 0.543
¥ 1/105 1/137 1/169 1 0.0032
KD 1/0.342 1/445 1/543 1/0.0032 1
Is it possible to increase the dollar holdings (above the initial $5 million)
by circulating currencies through the currency market?

3. Mathematical Model:Mathematical Model:
 The situation starts with $5 million. This amount goes through a numberThe situation starts with $5 million. This amount goes through a number
of conversions to other currencies before ultimately being reconverted toof conversions to other currencies before ultimately being reconverted to
dollars. The problem thus seeks determining the amount of eachdollars. The problem thus seeks determining the amount of each
conversion that will maximize the total dollar holdings.conversion that will maximize the total dollar holdings.
 For the purpose of developing the model and simplifying the notation, theFor the purpose of developing the model and simplifying the notation, the
following numeric code is used to represent the currencies.following numeric code is used to represent the currencies.
Currency $ € £ ¥ KD
Code 1 2 3 4 5
Define:Define:
 xxijij = Amount in currency i converted to currency j, i and j = 1,2, ... ,5= Amount in currency i converted to currency j, i and j = 1,2, ... ,5
For example,For example, xx1212 is the dollar amount converted to euros andis the dollar amount converted to euros and xx5151 is the KDis the KD
amount converted to dollars. We further define two additional variablesamount converted to dollars. We further define two additional variables
representing the input and the output of the arbitrage problem:representing the input and the output of the arbitrage problem:
 II = Initial dollar amount (= $5 million)= Initial dollar amount (= $5 million)
 yy = Final dollar holdings (to be determined from the solution)= Final dollar holdings (to be determined from the solution)

4. Mathematical Model:Mathematical Model:
 Our goal is to determine the maximum final dollar holdings, y,Our goal is to determine the maximum final dollar holdings, y,
subject to the currency flow restrictions and the maximum limitssubject to the currency flow restrictions and the maximum limits
allowed for the different transactions.allowed for the different transactions.
1
($)
3
(£)
x13
0.625 x13
X13 ≤ 5
Definition of the input/output variable, x13, between $ and £
 We start by developing the constraints of the model. FigureWe start by developing the constraints of the model. Figure
demonstrates the idea of converting dollars to pounds. The dollardemonstrates the idea of converting dollars to pounds. The dollar
amountamount xx1313 at originating currency 1 is converted toat originating currency 1 is converted to 0.625 x0.625 x1313
pounds at end currency 3. At the same time, the transacted dollarpounds at end currency 3. At the same time, the transacted dollar
amount cannot exceed the limit set by the dealer,amount cannot exceed the limit set by the dealer, xx1313 ≤ 5≤ 5..
 To conserve the flow of money from one currency to another, eachTo conserve the flow of money from one currency to another, each
currency must satisfy the following input-output equationcurrency must satisfy the following input-output equation
Total sum available of
currency (input)
== Total sum converted of
other currencies (output)

5. Mathematical Model:Mathematical Model:
 Dollar (i=1)Dollar (i=1)
Total available dollars = Initial dollar amount + dollar amount from otherTotal available dollars = Initial dollar amount + dollar amount from other
currenciescurrencies
= I= I + (€ $) + (£ $) + (¥ $) + (KD $)→ → → →+ (€ $) + (£ $) + (¥ $) + (KD $)→ → → →
= I + 1/0.769 x= I + 1/0.769 x2121 + 1/0.625 x+ 1/0.625 x3131 + 1/105 x+ 1/105 x4141 + 1/0.342 x+ 1/0.342 x5151
Total distributed dollars = Final dollar holdings + dollar amount to otherTotal distributed dollars = Final dollar holdings + dollar amount to other
currenciescurrencies
= y= y + ($ €) + ($ £) + ($ ¥) + ($ KD)→ → → →+ ($ €) + ($ £) + ($ ¥) + ($ KD)→ → → →
= y + x= y + x1212 + x+ x1414 + x+ x1414 + x+ x1515
GivenGiven II = 5, the dollar constraint thus becomes= 5, the dollar constraint thus becomes
y + xy + x1212 + x+ x1313 + x+ x1414 + x+ x1515 – (1/0.769 x– (1/0.769 x2121 + 1/0.625 x+ 1/0.625 x3131 + 1/105 x+ 1/105 x4141 + 1/0.342+ 1/0.342
xx5151) = 5) = 5

6. Mathematical Model:Mathematical Model:
 Euro (i = 2)Euro (i = 2)
Total available euros = ($ €) + (£ €) + (¥ €) + (KD €)→ → → →Total available euros = ($ €) + (£ €) + (¥ €) + (KD €)→ → → →
= 0.769 x= 0.769 x1212 + 1/0.813 x+ 1/0.813 x3232 + 1/137 x+ 1/137 x4242 + 1/0.445 x+ 1/0.445 x5252
Total distributed euros = (€ $) + (€ £) + (€ ¥) + (€ KD)→ → → →Total distributed euros = (€ $) + (€ £) + (€ ¥) + (€ KD)→ → → →
= x= x2121 + x+ x2323 + x+ x2424 + x+ x2525
Thus, the constraint isThus, the constraint is
xx2121 + x+ x2323 + x+ x2424 + x+ x2525 - (0.769 x- (0.769 x1212 + 1/0.813 x+ 1/0.813 x3232 + 1/137 x+ 1/137 x4242 + 1/0.445 x+ 1/0.445 x5252) =) =
00

7. Mathematical Model:Mathematical Model:
 Pound (i = 3)Pound (i = 3)
Total available pounds = ($ £) + (€ £) + (¥ £) + (KD £)→ → → →Total available pounds = ($ £) + (€ £) + (¥ £) + (KD £)→ → → →
= 0.625 x= 0.625 x1313 + 0.813 x+ 0.813 x2323 + 1/169 x+ 1/169 x4343 + 1/0.543 x+ 1/0.543 x5353
Total distributed pounds = (£ $) + (£ €) + (£ ¥) + (£ KD)→ → → →Total distributed pounds = (£ $) + (£ €) + (£ ¥) + (£ KD)→ → → →
= x= x3131 + x+ x3232 + x+ x3434 + x+ x3535
Thus, the constraint isThus, the constraint is
xx3131 + x+ x3232 + x+ x3434 + x+ x3535 – (0.625 x– (0.625 x1313 + 0.813 x+ 0.813 x2323 + 1/169 x+ 1/169 x4343 + 1/0.543 x+ 1/0.543 x5353) = 0) = 0

8. Mathematical Model:Mathematical Model:
 Yen (i = 4)Yen (i = 4)
Total available yen = ($ ¥) + (€ ¥) + (£ ¥) + (KD ¥)→ → → →Total available yen = ($ ¥) + (€ ¥) + (£ ¥) + (KD ¥)→ → → →
= 105 x= 105 x1414 + 137 x+ 137 x2424 + 169 x+ 169 x3434 + 1/0.0032 x+ 1/0.0032 x5454
Total distributed yen = (¥ $) + (¥ €) + (¥ £) + (¥ KD)→ → → →Total distributed yen = (¥ $) + (¥ €) + (¥ £) + (¥ KD)→ → → →
= x= x4141 + x+ x4242 + x+ x4343 + x+ x4545
Thus, the constraint isThus, the constraint is
xx4141 + x+ x4242 + x+ x4343 + x+ x4545 - (105 x- (105 x1414 + 137 x+ 137 x2424 + 169 x+ 169 x3434 + 1/0.0032 x+ 1/0.0032 x5454) = 0) = 0

9. Mathematical Model:Mathematical Model:
 KD (i = 5)KD (i = 5)
Total available KDs = (KD $) + (KD €) + (KD £) + (KD ¥)→ → → →Total available KDs = (KD $) + (KD €) + (KD £) + (KD ¥)→ → → →
= 0.342 x= 0.342 x1515 + 0.445 x+ 0.445 x2525 + 0.543 x+ 0.543 x3535 + 0.0032 x+ 0.0032 x4545
Total distributed KDs = ($ KD) + (€ KD) + (£ KD) + (¥ KD)→ → → →Total distributed KDs = ($ KD) + (€ KD) + (£ KD) + (¥ KD)→ → → →
= x= x5151 + x+ x5252 + x+ x5353 + x+ x5454
Thus, the constraint isThus, the constraint is
xx5151 + x+ x5252 + x+ x5353 + x+ x5454 - (0.342 x- (0.342 x1515 + 0.445 x+ 0.445 x2525 + 0.543 x+ 0.543 x3535 + 0.0032 x+ 0.0032 x4545) = 0) = 0

10. Mathematical Model:Mathematical Model:
 The only remaining constraints are the transaction limits, which are 5The only remaining constraints are the transaction limits, which are 5
million dollars, 3 million euros, 3.5 million pounds, 100 million yen, andmillion dollars, 3 million euros, 3.5 million pounds, 100 million yen, and
2.8 million KDs.2.8 million KDs.
 These can be translated asThese can be translated as
• xx1j1j ≤ 5, j≤ 5, j = 2, 3, 4, 5= 2, 3, 4, 5
• xx2j2j ≤ 3, j≤ 3, j = 2, 3, 4, 5= 2, 3, 4, 5
• xx3j3j ≤ 3.5, j≤ 3.5, j = 2, 3, 4, 5= 2, 3, 4, 5
• xx4j4j ≤ 100, j≤ 100, j = 2, 3, 4, 5= 2, 3, 4, 5
• xx5j5j ≤ 2.8, j≤ 2.8, j = 2, 3, 4, 5= 2, 3, 4, 5

11. Mathematical Model:Mathematical Model:
 The complete model is now gives asThe complete model is now gives as
Maximize z = yMaximize z = y
subject tosubject to
y + xy + x1212 + x+ x1313 + x+ x1414 + x+ x1515 – (1/0.769 x– (1/0.769 x2121 + 1/0.625 x+ 1/0.625 x3131 + 1/105 x+ 1/105 x4141 + 1/0.342 x+ 1/0.342 x5151)) = 5= 5
xx2121 + x+ x2323 + x+ x2424 + x+ x2525 - (0.769 x- (0.769 x1212 + 1/0.813 x+ 1/0.813 x3232 + 1/137 x+ 1/137 x4242 + 1/0.445 x+ 1/0.445 x5252)) = 0= 0
xx3131 + x+ x3232 + x+ x3434 + x+ x3535 – (0.625 x– (0.625 x1313 + 0.813 x+ 0.813 x2323 + 1/169 x+ 1/169 x4343 + 1/0.543 x+ 1/0.543 x5353)) = 0= 0
xx4141 + x+ x4242 + x+ x4343 + x+ x4545 - (105 x- (105 x1414 + 137 x+ 137 x2424 + 169 x+ 169 x3434 + 1/0.0032 x+ 1/0.0032 x5454)) = 0= 0
xx5151 + x+ x5252 + x+ x5353 + x+ x5454 - (0.342 x- (0.342 x1515 + 0.445 x+ 0.445 x2525 + 0.543 x+ 0.543 x3535 + 0.0032 x+ 0.0032 x4545)) = 0= 0
xx1j1j ≤ 5, j≤ 5, j = 2, 3, 4, 5= 2, 3, 4, 5
xx2j2j ≤ 3, j≤ 3, j = 2, 3, 4, 5= 2, 3, 4, 5
xx3j3j ≤ 3.5, j≤ 3.5, j = 2, 3, 4, 5= 2, 3, 4, 5
xx4j4j ≤ 100, j≤ 100, j = 2, 3, 4, 5= 2, 3, 4, 5
xx5j5j ≤ 2.8, j≤ 2.8, j = 2, 3, 4, 5= 2, 3, 4, 5
xxijij ≥ 0≥ 0, for all i and j, for all i and j

12. Mathematical Model:Mathematical Model:
 SolutionSolution
The optimum solution (using file amplEx2.3-2.txt or solverEx2.3-2.xls) is:The optimum solution (using file amplEx2.3-2.txt or solverEx2.3-2.xls) is:
SolutionSolution InterpretationInterpretation
y = 5.09032y = 5.09032 Final holdings = $5,090,320.Final holdings = $5,090,320.
Net dollar gain = $90,320, which representsNet dollar gain = $90,320, which represents
a 1.8064% rate of returna 1.8064% rate of return
xx1212 = 1.46206= 1.46206 Buy $1,462,060 worth of eurosBuy $1,462,060 worth of euros
xx1515 = 5= 5 Buy $5,000,000 worth of KDBuy $5,000,000 worth of KD
xx2525 = 3= 3 Buy €3,000,000 worth of KDBuy €3,000,000 worth of KD
xx3131 = 3.5= 3.5 Buy £3,500,000 worth of dollarsBuy £3,500,000 worth of dollars
xx3232 = 0.931495= 0.931495 Buy £931,495 worth of eurosBuy £931,495 worth of euros
xx4141 = 100= 100 Buy ¥100,000,000 worth of dollarsBuy ¥100,000,000 worth of dollars
xx4242 = 100= 100 Buy ¥100,OOO,000 worth of eurosBuy ¥100,OOO,000 worth of euros
xx4343 = 100= 100 Buy ¥100,000,000 worth of poundsBuy ¥100,000,000 worth of pounds
xx5353 = 2.085= 2.085 Buy KD2,085,000 worth of poundsBuy KD2,085,000 worth of pounds
xx5454 = .96= .96 Buy KD960,OOO worth of yenBuy KD960,OOO worth of yen

13. Mathematical Model:Mathematical Model:
 Remacks. At first it may appear that the solution is nonsensical because itRemacks. At first it may appear that the solution is nonsensical because it
calls for usingcalls for using
xx1212 + x+ x1515 = 1.46206 + 5 = 6.46206= 1.46206 + 5 = 6.46206
or $6,462,060 to buy euros and KDs when the initial dollar amount is onlyor $6,462,060 to buy euros and KDs when the initial dollar amount is only
$5,000,000. Where do the extra dollars come from? What happens in$5,000,000. Where do the extra dollars come from? What happens in
practice is that the given solution is submitted to the currency dealer aspractice is that the given solution is submitted to the currency dealer as
one order, meaning we do not wait until we accumulate enough currencyone order, meaning we do not wait until we accumulate enough currency
of a certain type before making a buy. In the end, the net result of allof a certain type before making a buy. In the end, the net result of all
these transactions is a net cost of $5,000,000 to the investor. This can bethese transactions is a net cost of $5,000,000 to the investor. This can be
seen by summing up all the dollar transactions in the solution:seen by summing up all the dollar transactions in the solution:
I = y + xI = y + x1212 + x+ x1313 + x+ x1414 + x+ x1515 – (1/0.769 x– (1/0.769 x2121 + 1/0.625 x+ 1/0.625 x3131 + 1/105 x+ 1/105 x4141 + 1/0.342+ 1/0.342
xx5151) = 5.09032 + 1.46206 + 5 – (3.5/0.625 + 100/105) = 5) = 5.09032 + 1.46206 + 5 – (3.5/0.625 + 100/105) = 5
 Notice thatNotice that xx2121, x, x3131, x, x4141 and xand x5151 are in euro, pound, yen, and KD, respectively,are in euro, pound, yen, and KD, respectively,
and hence must be converted to dollars.and hence must be converted to dollars.