A construction company is planning to build a new holiday camping site. There wi
ID: 3199024 • Letter: A
Question
A construction company is planning to build a new holiday camping site. There will be three types of accommodation available here, Standard, Extra, Luxury. The capital finance available for the planning, construction and advertising costs of each type and the cost per unit are given in the table below.
Basic (€)
Extra (€)
Luxury (€)
Capital (€)
Planning
8
8
12
4600
Construction
12
24
18
12000
Advertising
6
4
6
2400
Yearly profit
€220
€320
€300
Use the simplex method and slack variables to determine how many of each type of accommodation should be built to obtain a maximum yearly profit?
Basic (€)
Extra (€)
Luxury (€)
Capital (€)
Planning
8
8
12
4600
Construction
12
24
18
12000
Advertising
6
4
6
2400
Yearly profit
€220
€320
€300
Explanation / Answer
Let the number of basic be x
Let the number of extra be y
Let the number of luxury be z
Constraints:
8x + 8y + 12z <= 4600 (Planning constraints)
12x + 24y + 18z <= 12000 (Construction constraints)
6x + 4y + 6z <= 2400 (advertising constraints)
Maximize
Z = 220x + 320y + 300z
Other constraints, the value of x,y and z all must be zero
Now solving using simplex tableau method
Tableau #1
x y z s1 s2 s3 s4 s5 s6 p
8 8 12 1 0 0 0 0 0 0 4600
12 24 18 0 1 0 0 0 0 0 12000
6 4 6 0 0 1 0 0 0 0 2400
1 0 0 0 0 0 -1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-220 -320 -300 0 0 0 0 0 0 1 0
Tableau #2
x y z s1 s2 s3 s4 s5 s6 p
8 8 12 1 0 0 0 0 0 0 4600
12 24 18 0 1 0 0 0 0 0 12000
6 4 6 0 0 1 0 0 0 0 2400
-1 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-220 -320 -300 0 0 0 0 0 0 1 0
Tableau #3
x y z s1 s2 s3 s4 s5 s6 p
8 8 12 1 0 0 0 0 0 0 4600
12 24 18 0 1 0 0 0 0 0 12000
6 4 6 0 0 1 0 0 0 0 2400
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-220 -320 -300 0 0 0 0 0 0 1 0
Tableau #4
x y z s1 s2 s3 s4 s5 s6 p
8 8 12 1 0 0 0 0 0 0 4600
12 24 18 0 1 0 0 0 0 0 12000
6 4 6 0 0 1 0 0 0 0 2400
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 -1 0 0 0 0 0 1 0 0
-220 -320 -300 0 0 0 0 0 0 1 0
Tableau #5
x y z s1 s2 s3 s4 s5 s6 p
4 0 6 1 -0.333333 0 0 0 0 0 600
0.5 1 0.75 0 0.0416667 0 0 0 0 0 500
4 0 3 0 -0.166667 1 0 0 0 0 400
-1 0 0 0 0 0 1 0 0 0 0
0.5 0 0.75 0 0.0416667 0 0 1 0 0 500
0 0 -1 0 0 0 0 0 1 0 0
-60 0 -60 0 13.3333 0 0 0 0 1 160000
Tableau #6
x y z s1 s2 s3 s4 s5 s6 p
0 0 3 1 -0.166667 -1 0 0 0 0 200
0 1 0.375 0 0.0625 -0.125 0 0 0 0 450
1 0 0.75 0 -0.0416667 0.25 0 0 0 0 100
0 0 0.75 0 -0.0416667 0.25 1 0 0 0 100
0 0 0.375 0 0.0625 -0.125 0 1 0 0 450
0 0 -1 0 0 0 0 0 1 0 0
0 0 -15 0 10.8333 15 0 0 0 1 166000
Tableau #7
x y z s1 s2 s3 s4 s5 s6 p
0 0 1 0.333333 -0.0555556 -0.333333 0 0 0 0 66.6667
0 1 0 -0.125 0.0833333 0 0 0 0 0 425
1 0 0 -0.25 0 0.5 0 0 0 0 50
0 0 0 -0.25 0 0.5 1 0 0 0 50
0 0 0 -0.125 0.0833333 0 0 1 0 0 425
0 0 0 0.333333 -0.0555556 -0.333333 0 0 1 0 66.6667
0 0 0 5 10 10 0 0 0 1 167000
Hence the maximum profit will be occuring when Optimal Solution: p = 166799; x = 50, y = 425, z = 66
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