ace
Project Description:
2. the ace manufacturing company produces two lines of its product, the super and the regular. resource requirements for production are given in the table. there are 1,600 hours of assembly worker hours available per week, 700 hours of paint time, and 1200 hours of inspection time. regular customers will demand at least 150 units of the regular line and at least 90 of the super line.
profit assembly paint inspection
product line contribution time (hr.) time (hr.) time (hr.)
regular 50 1.2 .8 1.5
super 75 1.6 .5 .7
a) formulate an lp model which the ace company could use to determine the optimal product mix on a weekly basis. use two decision variables (units of regular and units of super). suggest any feasible solution and explain what “feasible solution” means.
b) find the optimal solution by using the graphical solution technique. what is the value of the objective function? what are the values of all variables?
c) by how many units can the demand for the super product increase before the optimal intersection point changes? explain. for the regular product?
d) how much would it be worth to the ace company if it could obtain an additional hour of paint time? of assembly time? of inspection time? explain fully. show all calculations.
e) find the upper and lower bounds for assembly time by identifying the corner points on either end of the line and substituting these points into the assembly equation. what do these bounds mean? explain.
f) solve this problem with lindo or pom and verify that your answers are correct.
solution:
let x1 and x2 denote the number of units produced per week of the product ‘regular’
and ‘super’respectively.
maximize z =50 x1 + 75 x2
subject to
1.2x1 + 1.6x2 ≤ 1,600 or 12x1 + 16x2 ≤ 16,000 (i)
0.8 x1 +0.9 x2 ≤ 700 or 8 x1 + 9 x2 ≤ 7,000 (ii)
0.2 x1 + 0.2 x2 ≤ 300 or 2 x1 + 2 x2 ≤ 3,000 (iii)
x1 ≥ 150 (iv)
x2 ≥ 90 (v)
let
x1 = y1 + 150
x2 =y2 + 90 where y1 , y2 ≥ 0
maximize z = 50(y1+ 150) + 75 (y2 + 90) or , z = 50y1 + 75y2 + 14,250
subject to:
12(y1 + 150) + 16(y2 + 90) ≤ 16,000
8(y1 + 150) + 9(y2 + 90) ≤ 7,000
2(y1 + 150) + 2(y2 + 90) ≤ 3,000
and y1 , y2 ≥ 0
adding slack variables s1, s2 , s3 , we get
maximize z= 50y1+75y2 +14,250 subject to
12y1+ 16y2 + s1 = 12,760
8y1 + 9y2 + s2 = 4,990
2y1 + 2y2 + s3 = 2,520
table 1
50 75 0 0 0
cb
y1 y2 s1 s2 s3
0 s 1 12,760 12 16 1 0 0 12760/16
0 s 2 4,990 8 9 0 1 0 4990/9
0 s 3 2,520 2 2 0 0 1 2520/2
δj
50  75 0 0 0
table ii
cj 50 75 0 0 0
cb
y1 y 2 s 1 s 2 s 3
0 s1 3889 20/9 0 1 16/9 0
75 y2 554.44 8/9 1 0 1/9 0
0 s 3 1411 2/9 0 0 2/9 1
δj 50/3 0 0 75/9 0
since all the elements in the index row are either positive or equal to zero , table ii gives an optimum solution which is y1 = 0 and y2 = 554.44
substituting these values we get
x1 = 0+150 =150
x 2 = 90+554.44 =644.44 and the value of objective function is
z = 50 x 150 + 75 x 644.44 = rs. 55,833
profit assembly paint inspection
product line contribution time (hr.) time (hr.) time (hr.)
regular 50 1.2 .8 1.5
super 75 1.6 .5 .7
a) formulate an lp model which the ace company could use to determine the optimal product mix on a weekly basis. use two decision variables (units of regular and units of super). suggest any feasible solution and explain what “feasible solution” means.
b) find the optimal solution by using the graphical solution technique. what is the value of the objective function? what are the values of all variables?
c) by how many units can the demand for the super product increase before the optimal intersection point changes? explain. for the regular product?
d) how much would it be worth to the ace company if it could obtain an additional hour of paint time? of assembly time? of inspection time? explain fully. show all calculations.
e) find the upper and lower bounds for assembly time by identifying the corner points on either end of the line and substituting these points into the assembly equation. what do these bounds mean? explain.
f) solve this problem with lindo or pom and verify that your answers are correct.
solution:
let x1 and x2 denote the number of units produced per week of the product ‘regular’
and ‘super’respectively.
maximize z =50 x1 + 75 x2
subject to
1.2x1 + 1.6x2 ≤ 1,600 or 12x1 + 16x2 ≤ 16,000 (i)
0.8 x1 +0.9 x2 ≤ 700 or 8 x1 + 9 x2 ≤ 7,000 (ii)
0.2 x1 + 0.2 x2 ≤ 300 or 2 x1 + 2 x2 ≤ 3,000 (iii)
x1 ≥ 150 (iv)
x2 ≥ 90 (v)
let
x1 = y1 + 150
x2 =y2 + 90 where y1 , y2 ≥ 0
maximize z = 50(y1+ 150) + 75 (y2 + 90) or , z = 50y1 + 75y2 + 14,250
subject to:
12(y1 + 150) + 16(y2 + 90) ≤ 16,000
8(y1 + 150) + 9(y2 + 90) ≤ 7,000
2(y1 + 150) + 2(y2 + 90) ≤ 3,000
and y1 , y2 ≥ 0
adding slack variables s1, s2 , s3 , we get
maximize z= 50y1+75y2 +14,250 subject to
12y1+ 16y2 + s1 = 12,760
8y1 + 9y2 + s2 = 4,990
2y1 + 2y2 + s3 = 2,520
table 1
50 75 0 0 0
cb
y1 y2 s1 s2 s3
0 s 1 12,760 12 16 1 0 0 12760/16
0 s 2 4,990 8 9 0 1 0 4990/9
0 s 3 2,520 2 2 0 0 1 2520/2
δj
50  75 0 0 0
table ii
cj 50 75 0 0 0
cb
y1 y 2 s 1 s 2 s 3
0 s1 3889 20/9 0 1 16/9 0
75 y2 554.44 8/9 1 0 1/9 0
0 s 3 1411 2/9 0 0 2/9 1
δj 50/3 0 0 75/9 0
since all the elements in the index row are either positive or equal to zero , table ii gives an optimum solution which is y1 = 0 and y2 = 554.44
substituting these values we get
x1 = 0+150 =150
x 2 = 90+554.44 =644.44 and the value of objective function is
z = 50 x 150 + 75 x 644.44 = rs. 55,833
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