Solution to Foreign Exchange Case

1. Let x_ij be the currency i exchanged into currency j (measured in denomination of i), and y_i be the final holding of currency i (measured in currency's own denomination)

     Max  1.0y1  + 1.6785y2 + .1840y3 + .6211y4 + .0085y5
st
     y1 = 2.0 + 1.665x21 + .1823x31 + .6149x41 + .00847x51
              - x12 - x13 - x14 -x15
     y2 = 5.0 + .591x12 + .1095x32 + .3694x42 + .005093x52
              - x21 - x23 - x24 - x25
     y3 = 0.0 + 5.385x13 + 9.12x23 + 3.351x43 + .0465x53  
              - x31 - x32 - x34 - x35
     y4 = 3.0 + 1.594x14 + 2.607x24 + .2965x34 + .01379x54
              - x41 - x42 - x43 - x45               
     Y5 = 0.0 + 116.3x15 + 193.1x25 + 21.11x35 + 72.14x45
              - x51 - x52 - x53 - x54
     y1 >= 0.0
     y2 >= 0.0
     y3 >= 7.0
     y4 >= 0.0
     y5 >= 1040.0

Notes on the formulation
• The objective function is based on the dollar value of each currency calculated as the average of bid and ask prices.
• Final holding of each currency, say pounds, is beginning excess pounds plus the sum of all other currencies exchanged into pounds (converted into pounds multiplying by the exchange rate) minus amounts of pounds used to buy other currencies. etc.
• Finally we require that the final holdings of each currency satisfy the deficiencies of various currencies.

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2. The statement is correct. OV₁ represents the best combination of the currencies to hold if no particular cash needs were present. Therefore maximizing OV₂ (the value of the best combination after meeting the requirements) is tantamount to deviating from OV₁ as little as possible, or minimizing OV₁-OV₂.



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3. a) If you could trade Francs to Marks and back to Francs and end up with more Francs, you coul do this cycle using the increased Franc holding again, ending up with even more Francs. This cycle could be repeated again and again without bound,  increasing the franc holding at each step.  In currency exchage jargon this is called arbitrage. The solution of the linear program would be unbounded.

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   b) If they stood pat, their current holding of 2M Dollars plus 5M Pounds and plus 3M Marks would be worth 12.2558M Dollars, while their best attainable value is 12.261M Dollars as given in Exhibit 4. Therefore the statement is false.



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c)This is also false. Compared to the worst solution which has a value of 10.164M Dollars, the best solution (OV=12.184M) represents a 2M Dollar difference. Furthermore even when the % differences are small, they may represent substantial amounts-- a mere 1% of 12 million Dollars is $120,000!


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d) the falsity of this statement is obvious from the comparison of Exhibit 1 (representing a value of 12.2558) with exhibit 4 representing a value of 12.261. Currencies may be exchanged profitably regardless of the cash needs of various denominations. Thus the entire set of opportunities must be kept open.. [Any questions?]

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