FIN 203

Additional Problems (assignment 7)

Problem 1.
  a)   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)
         s.t.
             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 x1A+x2A+x3A+x4A 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 x2A+x2B+x2C etc.

b) Excel Solution


Problem 2
a) 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. 
           Max .06T + .12C + .09M +.09B
           s.t  T  + C  +  M  +  B = 10
                T       +  M        >=  3
                     C        +  B  <=  4  
                                 
                T  <= 7,  C <= 2, M <= 5,  B < = 4       
              
                T, C, M, B >  =0
b) Excel Solution
 

Problem 3.
a) Since each employee will work 6 hours a day we can divide the hours required by 6 (round to the next integer) to convert the hours to employes needed.
Let Xi be the nember of waiters starting their shift on day i (1, 2, . . .,7). The table below shows what days each shift will cover. For instance, those starting their 5-day shift on Wed (X3 of them in Column Wed) will be on their first day on Wed, 2nd day on Thu, 3rd day on Fri etc.

  Mon (X1) Tue (X2) Wed (X3) Thu (X4) Fri (X5) Sat (X6) Sun (X7) Needed
Mon
1
 
 
5
4
3
2
25
Tue
2
1
 
 
5
4
3
34
Wed
3
2
1
 
 
5
4
67
Thu
4
3
2
1
 
 
5
50
Fri
5
4
3
2
1
 
 
117
Sat
 
5
4
3
2
1
 
134
Sun
 
 
5
4
3
2
1
50

Numbers in the table indicate the number of day in a group's shift. For instance, waiters that start their shift on Wednesday will be on their 5th day on Friday, etc.


Now reason this way. On Monday People who started on Monday will be available plus people who started their shift on Thursday will be on the 5th and last day of their shift, etc.
Since the wage rate is constant and each hired worker works the same number of hours we need to minimize the total number of waiters, subject to the minimum number of waiters each day:

			Min X1 + X2 + X3 + X4 + X5 + X6 + X7
            s.t.
            X1 + X4 + X5 + X6 + X7 > = 25
            X1 + X2 + X5 + X6 + X7 > = 34
            X1 + X2 + X3 + X6 + X7 > = 67
            X1 + X2 + X3 + X4 + X7 > = 50
            X1 + X2 + X3 + X4 + X5 > = 117
            X2 + X3 + X4 + X5 + X6 > = 134
            X3 + X4 + X5 + X6 + X7 >= 50
  all
            x > =0.
b) Excel Solution

Problem 4.

a) Let xij be the number of units shipped from ith plant to the jth customer.
We want to minimize total cost subject to supply constraints that prevent shipments from any plant beyond its capacity and demand constraints. Relevant cost i sthe sum of production at a plant pluss shipping to a customer.

            Min  20x11+ 19x12 + 22x13 +24x14 + 26X21 +24x22 + 28x23 + 23x24 + 33x31 + 25x32 + 29x33 + 28x34
            s.t
            x11 + x12 + x13 + x14 < = 800
            x21 + x12 + x23 + x24 < = 600
            x31 + x32 + x33 + x34 < = 700
            x11 + x21 + x31 > = 300
            x12 + x22 + x32 > = 500
            x13 + x23 + x33 > = 400
            x14 + x24 + x34 > = 600
            all xij > = 0.
b) Excel solution