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5/5/12

Horn Shoe Company

Group 5: Section FAnupam Sinha [11IT-035]Aparna Vyas [11FN-019]Ravi Dhingra [11DM-122]

Saurabh Sharma [11FN-092]Sravan Kumar [11DM-159]Sundaresh Iyer [11DM-

052]



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Problem Statement

To minimize the total cost of operation, whichincludes cost of hiring, training, salary andtermination while meeting the monthly

production target.



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Decision Variables

Novice Hired in May nh1

Experienced Hired in May eh1

Novice Terminated in May nt1

Experienced Terminated in May et1

Novice Hired in June nh2Experienced Hired in June eh2

Novice Terminated in June nt2

Apprentice Terminated in June at2

Experienced Terminated in June et2

Novice Hired in July nh3

Experienced Hired in July eh3

Novice Terminated in July nt3

Apprentice Terminated in July at3

Experienced Terminated in July et3

Novice Hired in August nh4

Experienced Hired in August eh4



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Objective function

Z min = Total hiring cost+ Total training cost +Total salary paid+ Total termination cost

Hiring cost = 1000*(nh1+nh2+nh3+nh4)+

1500*(eh1+eh2+ eh3+eh4) Training cost = 800*(∑working no of novices in each

month) + 400*(∑working no of apprentices in eachmonth)

Salary = 1600*(∑working no of novices in eachmonth) +2400*(∑working no of apprentices in eachmonth) +3000*(∑working no of experienced in eachmonth)

Termination cost = 250*(∑no of novices terminated



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Working no. of novices each

monthMay nh1

June nh2

July nh3

August nh4



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Working no. of apprentices

each monthMay Nil, as it takes one month for a novice to

become apprentice

June nh1-nt1

July nh2-nt2

August nh3-nt3



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Working no. of experienced

each monthMay eh1

June eh1-et1+eh2

July (nh1-nt1) - at2 + (eh1-et1+eh2) – et2+eh3

August (nh2-nt2) – at3 + [(nh1-nt1) - at2 +(eh1-et1+eh2) – et2 +eh3] - et3 + eh4



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Total no. of terminations

Novices nt1 + nt2 + nt3 + nh4

Apprentices at2 + at3 + (nh3 - nt3)

Experienced et1 + et2 + et3 + (nh2-nt2) – at3 +[(nh1-nt1) - at2 + (eh1-et1+eh2) – et2

+eh3] - et3 + eh4



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Production Constraints

May productionconstraint

400*nh1 + 800*eh1 = 200000

June productionconstraint

400*nh2 + 600* (nh1-nt1)+ 800* (eh1-et1+eh2) =300000

July productionconstraint

400*nh3 + 600* (nh2-nt2) + 800*[(nh1-nt1) - at2 + (eh1-et1+eh2) – et2 +eh3]

= 270000

August productionconstraint

400*nh4 + 600* (nh3-nt3) + 800*[(nh2-nt2) – at3 + [(nh1-nt1) - at2 + (eh1-et1+eh2) – et2 +eh3] - et3 + eh4] =

150000



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Termination Constraints

May terminationconstraint

(i) nt1<=nh1 (ii) et1<=eh1

June termination

constraint

(i) nt2<=nh2 (ii) at2<=(nh1-nt1)

(iii) et2<= (eh1-et1+eh2)

July terminationconstraint

(i) nt3<=nh3 (ii) at3<=(nh2-nt2)(iii) et3<= [(nh1-nt1) - at2 + (eh1-

et1+eh2) – et2 +eh3]

• In addition to these constraints, all the decisionvariables are non-negative integers.



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Thank you for your

attention !!!

Questions ???

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