Essay About Objective Function And 0S1

Callaway Golf
Callaway Golf
GCO 2802 Computer models for business decision
Assignment 1
Question 1
Denim
Corduroy
Total
Cotton(pounds)
Profit($ per yard)
$2.25
$3.10
Demand(yards per month)
Linear Programming Model:
Problem definition
Resource Availability: 6500 pounds of cotton
3000hours processing time
Decision Variables: X1 = number of yards of denim to produce per day
X2 = number of yards of corduroy to produce per day
Objective Function: Maximize Z= $2.25X1+ $3.10X2
Where Z = profit per day
Resource Constraints: 5X1+ 7.5X2ЎЬ 6500
3X1+ 3.2X2ЎЬ 3000
Non-Negativity: X1, X2ЎЭ 0
Complete Linear programming model
Maximize Z= $2.25X1+ $3.10X2
Subject to: 5X1+ 7.5X2ЎЬ 6500
3X1+ 3.2X2ЎЬ 3200
X2ЎЬ 510
X1, X2ЎЭ 0
Standard Form
Max Z= $2.25X1+ $3.10X2+ 0S1+ 0S2
Subject to: 5X1+ 7.5X2+ S1ЎЬ 6500
3X1+ 3.2X2+ S2ЎЬ 3200
X2ЎЬ 510
X1, X2, S1, S2ЎЭ 0
Where X1 = number of yards of denim to produce per day
X2 = number of yards of corduroy to produce per day
S1, S2 are slack variables
Optimal solution
ZA= $2.25* 0+ $3.1* 510= $1581
ZB= $2.25*456+ $3.1*510= $2607(highest)
ZC= $2.25* 1000+ $3.1*0= $2250
So point B is the optimal solution
Cotton used= 5X1+ 7.5X2= 5*456+ 7.5* 510= 6105 so 395 pounds of cotton was left over.
Corduroy used= 3X1+ 3.2X2= 3*456+ 3.2*510= 3200 so no corduroy was left over.
The number of yards of corduroy is 510 so the demand is meeted.
$2.25—$3.00
Denim
Corduroy
Total
Cotton(pounds)
Profit($ per yard)
$3.00
$3.10
Demand(yards per month)
ZA= $3* 0+ $3.1* 510= $1581
ZB= $3*456+ $3.1*510= $2607(highest)
ZC= $3* 1000+ $3.1*0= $3000
Point C is the optimal solution
(e) 6500—6000
Denim
Corduroy
Total
Cotton(pounds)
Profit($ per yard)
$3.00
$3.10
Demand(yards per month)
Complete Linear programming model
Maximize Z= $2.25X1+ $3.10X2
Subject to: 5X1+ 7.5X2ЎЬ 6000
3X1+ 3.2X2ЎЬ 3200
ZA= $2.25* 0+ $3.1* 510= $1581
ZB= $2.25*435+ $3.1*510= $2559.75