Callaway Golf

GCO 2802 Computer models for business decision

Assignment 1

Question 1

Denim Corduroy Total

Cotton(pounds) 5 7.5 6500

Hour 3.0 3.2 3000

Profit($ per yard) $2.25 $3.10

Demand(yards per month) ЎЮ 510

(a) 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

(b) 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

(c) 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.

(d) $2.25---$3.00

Denim Corduroy Total

Cotton(pounds) 5 7.5 6500

Hour 3.0 3.2 3000

Profit($ per yard) $3.00 $3.10

Demand(yards per month) ЎЮ 510

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) 5 7.5 6000

Hour 3.0 3.2 3000

Profit($ per yard) $3.00 $3.10

Demand(yards per month) ЎЮ 510

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

ZC= $2.25*507.69+ $3.1*461.54= $2573.08(highest)

Zd= $2.25* 1000+ $3.1*0= $2250

Point C is the optimal solution, so the optimal solution is changed

(e) Cotton