Solutions to Assignment 2

• Problem 2-12
• Problem 2-17
• Problem 2-18
• Problem 2-20
• Problem 2-25
• Foreign Exchange

• 2-12

    Let S, W and C be the number of different models
    Min 12S +   15W +    24C
    St
      6000S + 8000W + 11000C >= 12,000,000
          S                  >= 100
                  W          >= 200 
                           C >= 300

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Note
Although the problem does state that "...she wants to make sure that total contribution margin equals total cost ..." writing the constraint as a greater-than-or-equal-to makes more sense. For otherwise if the demand for the models had been much higher with that constraint written as a strict equality, it would have been impossible to satisfy all demand and not make a profit (namely total contribution margin = fixed costs).
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• 2-17.

     Let xij be the amount shipped from location i (i=1,2)
     to wholesaler j (j=1,2,.. 5).

     Min 5.31x11 + 5.29x12 + 5.37x13 + 5.34x14 + 5.30x15
       + 5.85x21 + 5.79x22 + 5.75x23 + 5.78x24 + 5.78x25
     st
     x11 + x12 + x13 + x14 + x15 <= 20,000
     x21 + x22 + x23 + x24 + x25 <=12,000
     x11 + x21 >=4,000
     x12 + x22 >=6,000
     x13 + x23 >=2,000
     x14 + x24 >=10,000
     x15 + x25>=8,000
          xij>=0 for all i,j

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Note
Notice that the relevant costs are the sum of packaging and shipping costs since total capacity is greater than total demand. in this case we need to let the model decide which of the two capacities will not be fully utilized on the basis of minimization of this relevant cost. If, on the other hand, total capacity were equal to total demand, we would achieve the optimal cost by simply minimizing the shipping costs with no attention paid to packaging costs. The total packaging costs would be the same (since we have to use all the capacity) no matter how the shipping is arranged. the total optimal cost would then be equal to the minimum shipping cost plus the constant packaging cost.
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• 2-18.

     Let xt be the number of trained employees (before 
     layoffs) at the beginning of period t;
         yt be the number of new hires at the start of t
         zt be the number of lay-offs at the start of period t

    Min  600 (y1+ y2 +y3 ..) + 1200x1 + 1200x2 + 1300x3....

    St       80yt + Pt <=  165(xt -zt)          t = 1,2, . .,6
               zt <= .15xt                     t = 1,2, . .,6
               xt = x(t-1) - z(t-1) + y(t-1)   t = 2,3, . .,6
               x1 = 40 

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Notes

1. The objective function includes the wages of those laid off by summing the wage rate times xt (number before the layoffs)
2. The requirement constraints simply insure that total time needed (for training, 80 yt plus the pit time, Pt) will be covered by the available xt - zt workers each providing 165 hours.
3. In any given period t, the number of trained employees is equal to the number of employees during period (t-1) (which is xt - zt) plus the number hired at the beginning of (t-1) (which is yt). People hired in (t-1) completed their training during t-1 and are now available in t.
4. You may have noticed that with 40 beginning employees, it will be impossible to meet the first period's demand. Again, we need to separate the modeling from the solution process.

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• 2-20.

     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.

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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.
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• 2-25.

     Let x1 and x2 be the hours processes 1 and 2 used.

     Max p1 x1 + p2 x2
st
     15x1 +  9x2 >=600
      6x1 + 24x2 >=225
      3x1 + 12x2 <=300
      9x1 +  6x2 <=450
          x1, x2 >=0 

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Note
Each output and each input can be written solely in terms of hours of operation of the two processes
• 3x1 + 12x2 (Kerosene)
• 9x1 + 6x2 (Benzene)
• 15x1 + 9x2 (St. fluid)
• 6x1 + 24x2 (Lighter fluid)

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• Foreign Exchange Problem

     Let xij be the currency i exchanged into currency j
(measured in denomination of i)
         yi be the final holding of currency i (measured in
currency's denomination)

     Max  y1  + 1.42699y2 + .1236y3 + .37998y4 + .00444y5
st
     y1 = 2.0 + 1.425x21 + .1234x31 + .3793x41 + .00443x51
              - x12 - x13 - x14 -x15
     y2 = 5.0 + .6998x12 + .08647x32 + .2662x42 + .0031x52
              - x21 - x23 - x24 - x25
     y3 = 0.0 + 8.078x13 + 11.55x23 + 3.073x43 + .03586x53  
              - x31 - x32 - x34 - x35
     y4 = 3.0 + 2.627x14 + 3.754x24 + .325x34 + .01163x54
              - x41 - x42 - x43 - x45               
     Y5 = 0.0 + 224.7x15 + 320.7x25 + 27.76x35 + 85.51x45
              - x51 - x52 - x53 - x54
     y1 >= 0.0
     y2 >= 0.0
     y3 >= 8.0
     y4 >= 0.0
     y5 >= 1280.0

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Notes

1. The objective function is based on the dollar value of each currency calculated as the average of bid and ask prices.
   
2. Final holding of each currency, say pounds, is beginning excess plus the sum of all other currencies exchanged into pounds, minus amounts of pounds used to buy other currencies.
   
3. Finally we require that the final holdings of each currency satisfy the deficiencies of various currencies.

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