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AMPL Stochastic Programming: It is late summer in a northern city and the direct

ID: 3773543 • Letter: A

Question

AMPL Stochastic Programming:

It is late summer in a northern city and the director of streets is making decisions regarding the purchase of fuel and salt to be used in snow removal in the approaching winter. If insufficient supplies are purchased before a severe winter, substantially higher prices must be paid to purchase these supplies quickly during the winter. There are two methods of snow removal: plowing or salting. Salting is generally cheaper. However, excess salt after a warm winter must be carried until next winter. On the other hand, excess fuel is readily usable by other city departments, so the penalty for stocking too much fuel is lower. The director of streets classifies a winter as either warm or cold and attaches probabilities 0.4 and 0.6, respectively, to these two possible states of nature.

Daily cost of operating a truck, exclusive of fuel and salt, is $110 in a warm winter and $120 in a cold winter.

Truck fleet has a capacity of 5,000 truck-days during the winter season.

If only plowing is used, then a warm winter will require 3,500 truck-days of snow removal; whereas, a cold winter will require 5,100.

Salting is a more effective use of trucks than plowing. o In a warm winter, one truck salting is equivalent to 1.2 trucks plowing o In a cold winter, one truck salting is equivalent to 1.1 trucks plowing.

Note that in a cold winter, the limited truck capacity implies some salting will be necessary.

A. Develop an AMPL model and Data file

I need what I would put in a .mod file and a .dat file into the program AMPL. If you don't know what AMPL is please do not answer the question.

Explanation / Answer

BF1    = truck-days of fuel bought in period 1
BS1    = truck-days of salt bought in period 1
BFW = fuel bought in period 2, warm winter
BSW = salt bought in period 2, warm winter
XFW = excess fuel at end of warm winter
XSW = excess salt at end of warm winter
PW     = truck-days plowing, warm winter
SW     = truck-days spreading salt, warm winter
KW    = costs incurred during a warm winter in dollars
BFC   = fuel bought in period 2, cold winter
BSC   = salt bought in period 2, cold winter
XFC   = excess fuel at end of a cold winter
XSC   = excess salt at end of a cold winter
PC      = truck-days plowing, cold winter
SC      = truck-days spreading salt, cold winter
KC     = costs incurred during a cold winter in dollars

The Warm Winter Solution
If the winter is known to be warm, then the relevant LP is:
MIN = 70 * BF1 + 20 * BS1 + KW;
-BF1 - BFW + XFW + PW + SW = 0;      !(Fuel Usage);
-BS1 - BSW + XSW + SW = 0;           !(Salt Usage);
PW + SW < 5000;                      !(Truck Usage);
PW + 1.2 * SW > 3500;                !(Snow Removal);
KW - 73*BFW - 30*BSW + 65*XFW + 15*XSC !(Cost);
- 110 * PW - 110 * SW = 0;

When solved, the solution is:
Objective value:   583333.3
Variable        Value      Reduced Cost
BF1           2916.667        0.0000000
BS1           2916.667        0.0000000
KW           320833.3        0.0000000   
BFW          0.0000000         3.000000
XFW          0.0000000         5.000000
PW          0.0000000         13.33334
SW           2916.667        0.0000000   
BSW          0.0000000         10.00000
XSW          0.0000000          5.00000
Row      Slack or Surplus    Dual Price
1             583333.3         1.000000
2            0.0000000         70.00000
3            0.0000000         20.00000
4             2083.333        0.0000000   
5            0.0000000        -166.6667
6            0.0000000        -1.000000

The Cold Winter Solution
The corresponding LP, if we knew the winter would b
e cold, is:
MIN = 70 * BF1 + 20 * BS1 + KC;
-BF1 - BFC + XFC + PC + SC = 0;
-BS1 + SC - BSC + XSC = 0;
PC + SC <= 5000;
PC + 1.1 * SC >= 5100;
KC - 73 * BFC + 65 * XFC - 120 * PC - 120 * SC - 32 * BSC + 15 *
XSC=0;

Optimal solution found at step:         6
Objective value:                 970000.0
Variable           Value        Reduced Cost
     BF1        5000.000           0.0000000
     BS1        1000.000           0.0000000
      KC        600000.0           0.0000000
     BFC       0.0000000            3.000000
     XFC       0.0000000            5.000000
      PC        4000.000           0.0000000
      SC        1000.000           0.0000000
     BSC       0.0000000            12.00000
     XSC       0.0000000            5.000000
     Row    Slack or Surplus      Dual Price
       1        970000.0            1.000000
       2       0.0000000            70.00000
       3       0.0000000            20.00000
       4       0.0000000            10.00000
       5       0.0000000           -200.0000
       6       0.0000000           -1.000000
     
     
       The solution to the complete problem is:
       Optimal solution found at step:     12
       Objective value:                 819888.3
       Variable           Value        Reduced Cost
            BF1        2916.667           0.0000000
            BS1        2916.667           0.0000000
             KW        320833.3           0.0000000
             KC        715091.7           0.0000000
            BFW       0.0000000            3.000000
            XFW       0.0000000           0.2000004
             PW       0.0000000            4.683331
             SW        2916.667           0.0000000
            BSW       0.0000000            3.580000
            XSW       0.0000000            8.420000
            BFC        1891.667           0.0000000
            XFC       0.0000000            4.799998
             PC        1891.667           0.0000000
             SC        2916.667           0.0000000
            BSC       0.0000000            7.620001
            XSC       0.0000000            2.580000
            Row    Slack or Surplus      Dual Price
              1        819888.3            1.000000
              2       0.0000000            26.20000
              3       0.0000000            8.420000
              4        2083.333           0.0000000
              5       0.0000000           -65.51667
              6       0.0000000          -0.4000000
              7       0.0000000            43.80000
              8       0.0000000            11.58000
              9        191.6667           0.0000000
             10       0.0000000           -115.8000
             11       0.0000000          -0.6000000
           
            

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