Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 7 Assignment Problem and Sequencing Ex 7.2 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Ex 7.2
Solution & Step-by-Step Answer:
Let turning to be A and threading be B. ∴ Observe Min{A, B} = 1 for job 6 on B. Then the problem reduces to ∴ Now Min {A, B} = 2 for job 4 on A Then the problem reduce to Now Min {A, B} = 3 for job 1 on A and job 5 on B Then the problem reduces to Now Min {A, B) = 5 for job 3 on A Only job 2 is left so the optimal sequence is Worktable is given by Total elapsed time = 43 minutes Idle time for A (turning) = 43 – 42 = 1 min Idle time for B (threshing) = 2 + 4 = 6 min











Solution & Step-by-Step Answer:
Observe Min {A, B} = 3 for job II on B. Then the problem is reduced to Now Min {A, B} = 4 for job III at B Now only job I in left ∴ the optimal sequence is given by The work table is Total elapsed time = 23 days Idle time for A = 23 – 19 = 4 days Idle time for B = 8 days






Solution & Step-by-Step Answer:
Let Data entry be A and filing be B. So Observe min {A, B} = 90 for policy 1 at A Then the problem reduces to Observe min {A, B} = 100 for policy 3 at B Now only policy 2 is left ∴ The optimal sequence is Worktable So Total elapsed time = 490 min Idle time for A (data entry) = 490 – 390 = 100 min Idle time for B (filing) = 140 min.







Solution & Step-by-Step Answer:
Observe min {x, y} = 2 for job B on x The problem reduces to Now min [x, y] = 4 for job A on x The problem reduces to Now min [x, y] = 6 for job D on x The problem reduces to Now min [x, y] = 8 for job E on y Now only job C in left ∴ The optimal sequence is Worktable Total elapsed time = 60 hrs Idle time for X = 60 – 56 = 4 hrs Idle time for Y = 6 hrs










Solution & Step-by-Step Answer:
Observe min {A, B} = 5 for job VI for B and job VII for A The problem reduces to Now min {A, B] = 7 for job I on A The problem reduces to Now min {A, B] = 10 for job IV on A and B so we have two options. Or we take the 1st one. The problem reduces to Now min {A, B} = 14 for job V on A and job II and III for job B. ∴ We have Or We take the optimal sequence as. VII – I – IV – V – III – II – VI Worktable Total elapsed time = 91 units Idle time for A = 91 – 86 = 5 units Idle time for B = 13 units











Solution & Step-by-Step Answer:
(i) Min A = 5, Max B = 5 As min A ≥ max B. The problem can be converted into two machine problems. Let G and H be two fictitious machines such that G = A + B and H = B + C we get Now min {G, H} = 7 for job III & V for G and job I for H ∴ We have two options Or We take the first one The problem reduces to Min {G, H} = 9 for job IV on H The problem reduces to Now min {G, H} = 10 for job II for G and job VII for H Now job VI is left ∴ The optimal sequence is The work table is Total elapsed time = 61 hrs Idle time for A = 61 – 54 = 7 hrs Idle time for B = 35 + [61 – 58] = 38 hrs Idle time for C = 15 hrs











(ii) Min A = 5, Max B = 5
Min A ≥ Max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C we get
Now min {G, H} = 5 for job 1 for H.
The problem reduces to
Now min {G, H} = 8 for job 2 for G and job H also job 5 for G
∴ We have two options
Or
We take the first one
The problem reduces to
Now min {G, H} = 9 for job 3 for H
Now only job 4 is left
∴ The optimal sequence is
Worktable
Total elapsed time = 40 hrs
Idle time for A = 40 – 32 = 8 hrs
Idle time for B = 19 + [40 – 34] = 25 hrs
Idle time for C = 12 hrs









Solution & Step-by-Step Answer:
Min R = 6, Max Q = 6 As min R ≥ max Q. The problem can be converted into a two-machine problem. Let G and H be two fictitious machines such that G = P + Q and H = Q + R we get Min {G, H} = 9 for books A, D, E for G. ∴ We have more than one option we take The problem reduces to Min {G, H} = 8 for book C on H Now only B is left. So the optimal sequence is Worktable Total elapsed time = 51 units Idle time for P = 51 – 32 = 19 units Idle time for Q = 14 + [51 – 34] = 31 units Idle time for R = 9 units






