Note: A waiting line model solver computer package is needed to answer these questions. The Kolkmeyer Manufacturing Company uses a group of six identical machines, each of which operates an average of 18 hours between breakdowns. With randomly occurring breakdowns, the Poisson probability distribution is used to describe the machine breakdown arrival process. One person from the maintenance department provides the single-server repair service for the six machines. Management is now considering adding two machines to its manufacturing operation. This addition will bring the number of machines to eight. The president of Kolkmeyer asked for a study of the need to add a second employee to the repair operation. The service rate for each individual assigned to the repair operation is 0.50 machines per hour. (a) Compute the operating characteristics if the company retains the single-employee repair operation. (Round your answers to four decimal places. Report time in hours.) La = L = Wa = W =…

Use the Euclidean algorithm to find two sets of integers (a, b, c) such that 55a65b+143c: Solution = 1. By the Euclidean algorithm, we have: 143 = 2.65 + 13 and 65 = 5.13, so 13 = 143 – 2.65. - Also, 55 = 4.13+3, 13 = 4.3 + 1 and 3 = 3.1, so 1 = 13 — 4.3 = 13 — 4(55 – 4.13) = 17.13 – 4.55. Combining these, we have: 1 = 17(143 – 2.65) - 4.55 = −4.55 - 34.65 + 17.143, so we can take a = − −4, b = −34, c = 17. By carrying out the division algorithm in other ways, we obtain different solutions, such as 19.55 23.65 +7.143, so a = = 9, b -23, c = 7. = = how ? come [Note that 13.55 + 11.65 - 10.143 0, so we can obtain new solutions by adding multiples of this equation, or similar equations.]

- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.