BUS 203 Quantitative Applications

Solutions to Assignment 6

Problem 5-28
Foreign exchange Problem

5-28.

a)We need to pay close attention to how the problem variables are measured:
- W -> Wheat raised (acres)
- AR -> Alfalfa Raised (tons)
- B -> Beef raised (tons)
- AB -> Alfalfa bought (tons)
- AS -> Alfalfa sold (tons)
The objective function must express the net contribution of each activity (in the units they are measured) in dollar terms. e.g., since wheat is measured in acres we must calculate the contribution of wheat in dollars per acre, etc.
- W (wheat): 70 bushels per acre * $2 per bushel - $20 cost per acre = $120 per acre.
- AR (alfalfa raised): $28 per acre/4 tons per acre = -$7 per ton (cost).
- B (Beef raised): $500 per ton - $50 per ton = $450 per ton.
- AB (Alfalfa bought): -$28 per ton (ton)
- AS (Alfalfa sold): $21 per ton
First constraint is acreage constraint: so we must calculate acres per ton of alfalfa raised.
4 tons/acre => 1/4=.25 acres per ton.
The second constraint is in acre-feet: so we must calculate acre-feet per ton of AR.
.25 acres/ton * 3 acre-feet per acre = .75 acre-feet per ton.
b) Since the slack is 600, a total of 400 acre-feet of water is being used.
c) 8000 tons.
d) Buys 40,000 tons of alfalfa. Note that it will never be optimal to have AB and AS both positive.
e) None. The dual price is zero, or the water constraint is inactive.
f) An additional acre of land will add $3,100 to the profits. Notice also that this return for additional land holds for up to 1200 acres.
g) If the price of wheat tripled it will be 140*3 = $420. The objective function coefficient will become $420 - $20 =$400. Since the reduced cost of wheat is $2980. It is still too cheap to allocate any resources to wheat. Hence W is still zero and OV remains at $2480000.
h)$2,480,000
i) Since the increase of $1 is within the allowable range, the optimal variable values will stay the same but the OV will decrease by a dollar for every ton of Alfalfa bought (40,000 tons). Thus the OV will be $2,480,000 - 40,000 ($1) = $2,440,000.
j) Notice that as the cost of buying alfalfa decreases the OFC of this variable increases. Since the allowable increase in the OFC of AB is $6, the cost must become $22 or better before the solution changes. If that happens, Sod will probably buy more AB and increase its profits.

Foreign Exchange Problem

 
  MAX     Y1 + 1.42699 Y2 + 0.37998 Y4 + 0.1236 Y3 + 0.00444 Y5
  SUBJECT TO
         2)Y1 - 1.425 X21 - 0.1234 X31 - 0.3793 X41 - 0.00443 X51
+ X12 + X13 + X14 + X15 = 2
         3)Y2 - 0.6998 X12 - 0.08647 X32 - 0.2662 X42 -0.0031 X52
+ X21 + X23 + X24 + X25 = 5
         4)Y3 - 8.078 X13 - 11.55 X23 - 3.073 X43 - 0.03586 X53 
+ X31 + X32 + X34 + X35 = 0
         5)Y4 - 2.627 X14 - 3.754 X24 - 0.325 X34 - 0.01163 X54 
+ X41 + X42 + X43 + X45 = 3
         6) Y5 - 224.7 X15 - 320.7 X25 - 27.76 X35 - 85.51X 45
+ X51 + X52 + X53 + X54 = 0
         7)   Y3 >=  8
         8)   Y5 >=  1280
  END
 
 LP OPTIMUM FOUND AT STEP      3

        OBJECTIVE FUNCTION VALUE

        1)     10.269500    

  VARIABLE        VALUE          REDUCED COST
        Y1         2.000000           .000000
        Y2          .000000           .000590
        Y4          .000000           .000302
       X21          .000000           .002580
       X31          .000000           .000200
       X41          .000000           .000982
       X51          .000000           .000017
       X12          .000000           .000979
       X13          .000000           .001559
       X14          .000000           .000998
       X15          .000000           .000708
       X32          .000000           .000157
       X42          .000000           .000261
       X52          .000000           .000022
       X23         1.811665           .000000
       X24         3.188335           .000000
       X25          .000000           .001354
        Y3        20.924730           .000000
       X43          .000000           .000460
       X53          .000000           .000015
       X34          .000000           .000008
       X35          .000000           .000145
       X54          .000000           .000025
       X45        14.969010           .000000
        Y5      1280.000000           .000000


       ROW   SLACK OR SURPLUS     DUAL PRICES
        2)          .000000          1.000000
        3)          .000000          1.427580
        4)          .000000           .123600
        5)          .000000           .380282
        6)          .000000           .004447
        7)        12.924730           .000000
        8)          .000000          -.000007

 NO. ITERATIONS=       3

 RANGES IN WHICH THE BASIS IS UNCHANGED:
 
                           OBJ COEFFICIENT RANGES
 VARIABLE         CURRENT        ALLOWABLE        ALLOWABLE
                   COEF          INCREASE         DECREASE
       Y1        1.000000          .001621          .000708
       Y2        1.426990          .000590         INFINITY 
       Y4         .379980          .000302         INFINITY 
      X21         .000000          .002580         INFINITY 
      X31         .000000          .000200         INFINITY 
      X41         .000000          .000982         INFINITY 
      X51         .000000          .000017         INFINITY 
      X12         .000000          .000979         INFINITY 
      X13         .000000          .001559         INFINITY 
      X14         .000000          .000998         INFINITY 
      X15         .000000          .000708         INFINITY 
      X32         .000000          .000157         INFINITY 
      X42         .000000          .000261         INFINITY 
      X52         .000000          .000022         INFINITY 
      X23         .000000          .000095          .000590
      X24         .000000          .000978          .000095
      X25         .000000          .001354         INFINITY 
       Y3         .123600          .000088          .000051
      X43         .000000          .000460         INFINITY 
      X53         .000000          .000015         INFINITY 
      X34         .000000          .000008         INFINITY 
      X35         .000000          .000145         INFINITY 
      X54         .000000          .000025         INFINITY 
      X45         .000000          .000618          .000269
       Y5         .004440          .000007         INFINITY 
 
                           RIGHTHAND SIDE RANGES
      ROW       CURRENT            ALLOWABLE      ALLOWABLE
                    RHS             INCREASE       DECREASE
        2      2.000000            INFINITY        2.000000
        3      5.000000            INFINITY        1.119024
        4       .000000            INFINITY       12.924730
        5      3.000000            11.969010       4.200818
        6       .000000          1023.470000     359.211900
        7      8.000000            12.924730      INFINITY
        8   1280.000000           359.211900    1023.470000

The summary of the solution (obtaine by NONZ command which lists only the nonzero values:


        OBJECTIVE FUNCTION VALUE

        1)     10.269500    

  VARIABLE        VALUE          REDUCED COST
        Y1         2.000000           .000000
       X23         1.811665           .000000
       X24         3.188335           .000000
        Y3        20.924730           .000000
       X45        14.969010           .000000
        Y5      1280.000000           .000000


       ROW   SLACK OR SURPLUS     DUAL PRICES
        2)          .000000          1.000000
        3)          .000000          1.427580
        4)          .000000           .123600
        5)          .000000           .380282
        6)          .000000           .004447
        8)          .000000          -.000007

The optimal solution calls for exchanging 1.812M pounds into Francs, 3.188M pounds into Marks, and 14.969M Marks into Yens. The final holding will consist of 2M Dollars, 20.92M Francs and 1280M Yens. The total dollar value of this bundle is 10.2695$.