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A job production unit has four jobs P, Q, R, and S which can be manufactured on each of the four machines I, II, III, and IV. The processing cost of each job for each machine is given in the following table:
| Job | Machines (Processing cost in ₹) |
|||
| I | II | III | IV | |
| P | 31 | 25 | 33 | 29 |
| Q | 25 | 24 | 23 | 21 |
| R | 19 | 21 | 23 | 24 |
| S | 38 | 36 | 34 | 40 |
Find the optimal assignment to minimize the total processing cost.
Concept: Assignment Problem
Five jobs are performed first on machine M1 and then on machine M2. Time taken in hours by each job on each machine is given below:
| Machines↓\Jobs→ | 1 | 2 | 3 | 4 | 5 |
| M1 | 6 | 8 | 4 | 5 | 7 |
| M2 | 3 | 7 | 6 | 4 | 16 |
Determine the optimal sequence of jobs and total elapsed time. Also, find the idle time for two machines.
Concept: Finding an Optimal Sequence
Three new machines M1, M2, M3 are to be installed in a machine shop. There are four vacant places A, B, C, D. Due to limited space, machine M2 can not be placed at B. The cost matrix (in hundred rupees) is as follows:
| Machines | Places | |||
| A | B | C | D | |
| M1 | 13 | 10 | 12 | 11 |
| M2 | 15 | - | 13 | 20 |
| M3 | 5 | 7 | 10 | 6 |
Determine the optimum assignment schedule and find the minimum cost.
Concept: Special Cases of Assignment Problem
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e−3 = 0.0497
Concept: Poisson Distribution
The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.
Concept: Poisson Distribution
Choose the correct alternative:
A distance random variable X is said to have the Poisson distribution with parameter m if its p.m.f. is given by P(x) = `("e"^(-"m")"m"^"x")/("x"!)` the condition for m is ______
Concept: Poisson Distribution
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, using the following activity find the value of m.
Solution: X : Follows Poisson distribution
∴ P(X) = `("e"^-"m" "m"^x)/(x!)`, P(X = 1) = 0.4 and P(X = 2) = 0.2
∴ P(X = 1) = `square` P(X = 2).
`("e"^-"m" "m"^x)/(1!) = square ("e"^-"m" "m"^2)/(2!)`,
`"e"^-"m" = square "e"^-"m" "m"/2`, m ≠ 0
∴ m = `square`
Concept: Poisson Distribution
State whether the following statement is true or false:
lf X ∼ P(m) with P(X = 1) = P(X = 2) then m = 1.
Concept: Poisson Distribution
If X ∼ P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2).
Given e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
∴ `("e"^square"m"^1)/(1!) = ("e"^"-m""m"^2)/square`
∴ m = `square`
∴ P(X = 2) = `("e"^-2. "m"^2)/(2!)` = `square`
Concept: Poisson Distribution
