FIN 203

Solutions to Assignment 8

• Problem 5
• Problem 13
• Problem 15
• Problem 21

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Problem 5
see problem 1 on the Linear Programming Formulations and Solutions in the course documents/Linear Programming Folder

Problem 13.
Let T, C, M, and B be the millions of dollars invested in Treasury Bonds, Common Stock, Money Market and Municipal Bonds respectively.

Max .08T + .06C + .12M +.09B
S.t.
     T + C + M + B = 10 ( all 10 million to be invested)

         T   <= 5            (Max on T)
         C   <= 7            (Max on C)
         M   <= 2            (Max on M)
         B   <= 4            (Max on B)
         T + C >= 3          (T + C no more that 30%, since total investment will be 10 30% of the total is 3)
         M + B <= 4          (M + B no less than 40%, again 40% of total investment is 4 million) 
     M, C, T, B >= 0     (non-negativity)

Solver sheet

[go back] [Any questions?]

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Problem 15.

     Let x ij be the gallons of vintage i (i=1,2,. .4)   mixed in blend j (j=A,B,C)
     Max 80(x1A + x2A + x3A + x4A)+
         50(x1B + x2B + x3B + x4B)+
         35(x1C + x2C + x3C + x4C)      
     st
          x1A + x1B + x1C <= 130
          x2A + x2B + x2C <= 200
          x3A + x3B + x3C <= 150
          x4A + x4B + x4C <= 350

          x2A + x3A >= .75(x1A + x2A + x3A + x4A)
          x4A       >= .08(x1A + x2A + x3A + x4A)
          x2B       >= .10(x1B + x2B + x3B + x4B)
          x4B       <= .35(x1B + x2B + x3B + x4B)
          x2C + x3C >= .35(x1C + x2C + x3C + x4C) 
               xij   >= 0 for all i and j.

Note:
Total of blend is the sum of the vintages used in that blend. For instance, total of blend A (in gallons) is given by x_1A+x_2A+x_3A+x_4A etc. Likewise, total of any vintage used is the sum of that vintage mixed in all the blends. For instance, total vintage 2 used is given by x_2A+x_2B+x_2C etc.
Solver sheet

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Problem 21

Let T, C, M and B be the millions of dollars placed in the four investmnets.
Substituting Rf = .06 and Rm = .12 in the CAPM model, the returns for T, C, M and B are found to be .06, .12,.08 and .09 respectively. Therfore the problem is the same as 13 except with the new objective function.

    Max .06T + .12C + .08M +.09B
          s.t   T  + C  +  M  +  B  = 10
                T       +  M        >= 3
                     C        +  B  <= 4
                 T, C, M, B  >=0 




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