Assignment4Solution

# Assignment4Solution - Fall 2010 Optimization I(ORIE...

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Fall 2010 Optimization I (ORIE 3300/5300) Assignment 4 Solution Problem 2 (AMPL book 2-3) (a) The AMPL model that minimizes the cost of supply while producing a blend that contains 52 tons of cane sugar, 56 tons of corn sugar, and 59 tons of beet sugar is as follows: Model file: set SUGAR; set SUPPLIER; param percent {SUPPLIER, SUGAR} >=0; param cost {SUPPLIER} > 0; param blend_amt {SUGAR} >= 0; var Buy {i in SUPPLIER} >= 0; minimize Total_Cost: sum{i in SUPPLIER} cost[i]*Buy[i]; subject to BlendAmt {j in SUGAR}: sum {i in SUPPLIER} percent[i,j]*Buy[i] = blend_amt[j]; #meet blending requirements for each sugar type Data file: set SUGAR := Cane Corn Beet; set SUPPLIER := A B C D E F G; param percent (tr): A B C D E F G := Cane .1 .1 .2 .3 .4 .2 .6 Corn .3 .4 .4 .2 .6 .7 .1 Beet .6 .5 .4 .5 0 .1 .3; param: cost := A 10 B 11 C 12 D 13 E 14 F 12 G 15; param: blend_amt := 1

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Fall 2010 Optimization I (ORIE 3300/5300) Cane 52 Corn 56 Beet 59; With the following result: MINOS 5.5: optimal solution found. 2 iterations, objective 2068.5 ampl: display Buy; Buy [*] := A 60 B 0 C 0 D 0 E 0 F 45.5 G 61.5 ; (b) If the manufacturer requires that it is necessary to buy at least 10 tons from each manufac- turer, we must simply set a lower bound on the Buy variable. This is done by adding the following constraint on the decision variable: var Buy {i in SUPPLIER} >= 10; Now, the solution is: MINOS 5.5: optimal solution found. 2 iterations, objective 2093.954545 ampl: display Buy; Buy [*] := A 43.6364 B 10 C 10 D 10 E 10 F 30.9545 G 52.4091 ; (c) Now, we need to formulate an alternative model that finds the lowest-cost way to blend one ton of supplies so that the amount of each sugar is between 30 and 37 percent of the total. This is a blending problem. The formulation is as follows: 2
Fall 2010 Optimization I (ORIE 3300/5300) Model file: set SUGAR; set SUPPLIER; param percent {SUPPLIER, SUGAR} >= 0; param cost {SUPPLIER} > 0; param blend_min {SUGAR} >= 0; param blend_max {SUGAR} >= 0; var Buy {i in SUPPLIER} >= 0; minimize Total_Cost: sum{i in SUPPLIER} cost[i]*Buy[i]; subject to BlendAmt_LB {j in SUGAR}: sum {i in SUPPLIER} (percent[i,j]- blend_min[j])*Buy[i] >= 0 ; #Blend meets min requirement subject to BlendAmt_UB {j in SUGAR}: sum {i in SUPPLIER} (percent[i,j]- blend_max[j])*Buy[i] <= 0 ; #Blend meets max requirement subject to OneTon: sum {i in SUPPLIER} Buy[i] =1 ; Data file: set SUGAR := Cane Corn Beet; set SUPPLIER := A B C D E F G; param percent (tr): A B C D E F G := Cane .1 .1 .2 .3 .4 .2 .6 Corn .3 .4 .4 .2 .6 .7 .1 Beet .6 .5 .4 .5 0 .1 .3; param: cost := A 10 B 11 C 12 D 13 E 14 F 12 G 15; param: blend_min blend_max := Cane .30 .37 Corn .30 .37 Beet .30 .37; 3

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Fall 2010 Optimization I (ORIE 3300/5300) With the following result: MINOS 5.5: optimal solution found. 5 iterations, objective 12.25 ampl: display Buy; Buy [*] := A 0.4 B 0 C 0 D 0 E 0 F 0.25 G 0.35 ; Problem 3 (a) Since ¯ c 2 and ¯ c 7 are both positive, x 2 and x 7 could enter the basis. If x 2 enters, then ¯ t = min ¯ b 3 ¯ a 32 , ¯ b 6 ¯ a 62 = min 2 1 , 6 3 = 2 .
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