horn shoe company_group5
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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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