Question

PLEASE formulate and solve all the problems below by using BOTH EXCEL SOLVER AND GUROBI.

Weenies is a food processing plant which manufactures hot dogs and hot dog buns. They grind their own flour for the hot dog buns at a maximum rate of 200 pounds per week. Each hot dog bun requires 0.1 pound of flour. They currently have a contract with Pigland, Inc, which specifies that a delivery of 800 pounds of pork product is delivered every Monday. Each hot dog requires 4 pound of port product All the other ingredients in the hot dogs and hot dog buns are in plentiful supply. Finally, the labor force at Weenies consists of 5 employees working full time (40 hours per week each). Each hot dog requires 3 minutes of labor, and each hot dog bun requires 2 minutes of labor. Each hot dog yields a profit of $0.80, and each bun yields a profit of $0.30. Weenies would like to know how many hot dogs and how many hot dog buns they should produce each week so as to achieve the highest possible profit. Formulate a linear programming model and solve it by using both Excel Solver and Gurobi

Answer 1

let x1 = hot dogs they should produce each week

x2 = hot dog buns they should produce each week

Objective is maximize the profit so; Max Z = 0.80 x1 + 0.30 x2

Flour constraint: 0,1 x2 ≤ 200

Pork product constraints: x1/4 ≤ 800

0,25 x1 ≤ 800

Labor constraint: 3x1 +2x2 ≤ 12000

(40 hours = 2400 minutes, 2400 minutes per worker, so for 5worker= 2400*5 =12000 minutes)

Non-negativity; x1, x2≥0

MAX z = 0.8 x1 + 0.3 x2

subject to

0.1 x2 ≤ 200 0.25 x1 ≤ 800 3 x1 + 2 x2 ≤ 12000

and x1,x2≥0;

1. To draw constraint 0.1x2 ≤ 200

x1	0	1
x2	2000	2000

2. To draw constraint 0.25x1 ≤ 800

x1	3200	3200
x2	0	1

3. To draw constraint 3x1+2x2 ≤ 12000

x1	0	4000
x2	6000	0

The value of the objective function at each of these extreme points is as follows:

Extreme Point Coordinates (x1,x2)	Lines through Extreme Point	Objective function value z=0.8x1+0.3x2
O(0,0)	4→x1≥0 5→x2≥0	0.8(0)+0.3(0)=0
A(3200,0)	2→0.25x1≤800 5→x2≥0	0.8(3200)+0.3(0)=2560
B(3200,1200)	2→0.25x1≤800 3→3x1+2x2≤12000	0.8(3200)+0.3(1200)=2920
C(2666.67,2000)	1→0.1x2≤200 3→3x1+2x2≤12000	0.8(2666.67)+0.3(2000)=2733.33
D(0,2000)	1→0.1x2≤200 4→x1≥0	0.8(0)+0.3(2000)=600

The maximum value of the objective function z = 2920 occurs at the extreme point (3200,1200).

Hence, the optimal solution to the given LP problem is :

x1 = 3200

x2 = 1200

max z = 2920

So, the no. of hot dogs to be produced are 3200

and no. of hot dog buns to be produced are 1200

with max profit of $2920

  

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