Currency arbitrage

9,753 views

Published on

Published in: Technology, Economy & Finance
0 Comments
2 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
9,753
On SlideShare
0
From Embeds
0
Number of Embeds
119
Actions
Shares
0
Downloads
244
Comments
0
Likes
2
Embeds 0
No embeds

No notes for slide

Currency arbitrage

  1. 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. 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. 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 (= $5 million)  yy = Final dollar holdings (to be determined from the solution)= Final dollar holdings (to be determined from the solution)
  4. 4. Mathematical Model:Mathematical Model:  Our goal is to determine the maximum final dollar holdings, y,Our goal is to determine the maximum final dollar holdings, y, subject to the currency flow restrictions and the maximum limitssubject to the currency flow restrictions and the maximum limits allowed for the different transactions.allowed for the different transactions. 1 ($) 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. 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. 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 = (€ $) + (€ £) + (€ ¥) + (€ KD)→ → → → = x= x2121 + x+ x2323 + x+ x2424 + x+ x2525 Thus, the constraint isThus, the constraint is xx2121 + x+ x2323 + x+ x2424 + x+ x2525 - (0.769 x- (0.769 x1212 + 1/0.813 x+ 1/0.813 x3232 + 1/137 x+ 1/137 x4242 + 1/0.445 x+ 1/0.445 x5252) =) = 00
  7. 7. Mathematical Model:Mathematical Model:  Pound (i = 3)Pound (i = 3) Total available pounds = ($ £) + (€ £) + (¥ £) + (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 = (£ $) + (£ €) + (£ ¥) + (£ KD)→ → → → = x= x3131 + x+ x3232 + x+ x3434 + x+ x3535 Thus, the constraint isThus, the constraint is xx3131 + x+ x3232 + x+ x3434 + x+ x3535 – (0.625 x– (0.625 x1313 + 0.813 x+ 0.813 x2323 + 1/169 x+ 1/169 x4343 + 1/0.543 x+ 1/0.543 x5353) = 0) = 0
  8. 8. Mathematical Model:Mathematical Model:  Yen (i = 4)Yen (i = 4) Total available yen = ($ ¥) + (€ ¥) + (£ ¥) + (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. 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. 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. 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. 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. 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.

×