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EMGT 501 Fall 2005 Midterm Exam SOLUTIONS

EMGT 501 Fall 2005 Midterm Exam SOLUTIONS. 1. a. M 1 = units of component 1 manufactured M 2 = units of component 2 manufactured M 3 = units of component 3 manufactured P 1 = units of component 1 purchased P 2 = units of component 2 purchased P 3 = units of component 3 purchased.

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EMGT 501 Fall 2005 Midterm Exam SOLUTIONS

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  1. EMGT 501 Fall 2005 Midterm Exam SOLUTIONS

  2. 1. a M1 = units of component 1 manufactured M2 = units of component 2 manufactured M3 = units of component 3 manufactured P1 = units of component 1 purchased P2 = units of component 2 purchased P3 = units of component 3 purchased Min M1, M2, M3, P1, P2, P30

  3. b Component 2 4000 0 Component 1 2000 4000 Component 3 1400 2100 Source Manufacture Purchase Total Cost: $73,550

  4. c Max all

  5. d. u1* = 0.9063, u2* = 0, u3* = 0.1250, u4* = 6.5 u5* = 7.9688 and u6* = 7.000 If a dual variable is positive on optimality, then its corresponding constraint in primal formulation becomes binding (=). Similarly, if a primal variable is positive on optimality, then its corresponding constraint in dual formulation becomes binding (=).

  6. 2. a b

  7. c The objective is changed from 9 to 11 NOTE:

  8. 3. a Min

  9. b. Optimal solution: u1 = 3/10, u2 = 0, u3 = 54/30 The (z-c)j values for the four surplus variables of the dual show x1 = 0, x2 = 25, x3 = 125, and x4 = 0. c. Since u1 = 3/10, u2 = 0, and u3 = 54/30, machines A and C (uj > 0) are operating at capacity. Machine C is the priority machine since each hour is worth 54/30.

  10. 4. a

  11. b

  12. c Critical Path: A-D-H-I d E(T)= tA + tD + tH + tI = 4.83 + 8.83 + 8 + 4 = 25.66 days e Using the normal distribution, From Appendix, area for z = -0.65 is 0.2422. Probability of 25 days or less = 0.5000 - 0.2422 = 0.2578

  13. 5. a b Number of production runs = D / Q* = 7200 / 1078.12 = 6.68 [days] C

  14. d Production run length = days e Maximum Inventory

  15. f Holding Cost Ordering cost Total Cost = $2,003.48 g

  16. 6. a C = current cost per unit C ' = 1.23 C new cost per unit Let Q' = new optimal production lot size

  17. Q' = 0.9017(Q*) = 0.9017(5000) = 4509

  18. Queueing Theory

  19. The Basic Structure of Queueing Models The Basic Queueing Process: Customers are generated over time by an input source. The customers enter a queueing system. A required service is performed in the service mechanism.

  20. Input source Queueing system Served customers Service mechanism Queue Customers

  21. Input Source (Calling Population): The size of Input Source (Calling Population) is assumed infinite because the calculations are far easier. The pattern by which customers are generated is assumed to be a Poisson process.

  22. The probability distribution of the time between consecutive arrivals is an exponential distribution. The time between consecutive arrivals is referred to as the interarrival time.

  23. Queue: The queue is where customers wait before being served. A queue is characterized by the maximum permissible number of customers that it can contain. The assumption of an infinite queue is the standard one for most queueing models.

  24. Queue Discipline: The queue discipline refers to the order in which members of the queue are selected for service. For example, (a) First-come-first-served (b) Random

  25. Service Mechanism: The service mechanism consists of one or more service facilities, each of which contains one or more parallel service channels, called servers. The time at a service facility is referred to as the service time. The service-time is assumed to be the exponential distribution.

  26. Elementary Queueing Process Served customers Queueing system Queue C C C C S S Service S facility S Customers C C C C C C C Served customers

  27. Distribution of service times Number of servers Distribution of interarrival times Where M = exponential distribution (Markovian) D = degenerate distribution (constant times) = Erlang distribution (shape parameter = k) G = general distribution(any arbitrary distribution allowed)

  28. Both interarrival and service times have an exponential distribution. The number of servers is s . Interarrival time is an exponential distribution. No restriction on service time. The number of servers is exactly 1.

  29. Terminology and Notation State of system = # of customers in queueing system. Queue length = # of customers waiting for service to begin. N(t) = # of customers in queueing system at time t (t 0) = probability of exactly n customers in queueing system at time t.

  30. # of servers in queueing system. A mean arrival rate (the expected number of arrivals per unit time) of new customers when n customers are in system. A mean service rate for overall system (the expected number of customers completing service per unit time) when n customers are in system. Note: represents a combined rate at which all busy servers (those serving customers) achieve service completions.

  31. When is a constant for all n, it is expressed by When the mean service rate per busy server is a constant for all n 1, this constant is denoted by . Under these circumstances, and are the expected interarrival time and the expected service time. is the utilization factor for the service facility.

  32. The state of the system will be greatly affected by the initial state and by the time that has since elapsed. The system is said to be in a transient condition. After sufficient time has elapsed, the state of the system becomes essentially independent of the initial state and the elapsed time. The system has reached a steady-state condition, where the probability distribution of the state of the system remains the same over time.

  33. The probability of exactly n customers in queueing system. The expected number of customers in queueing system The expected queue length (excludes customers being served)

  34. A waiting time in system (includes service time) for each individual customer. A waiting time in queue (excludes service time) for each individual customer.

  35. Relationships between and Assume that is a constant for all n. In a steady-state queueing process, Assume that the mean service time is a constant, for all It follows that,

  36. The Role of the Exponential Distribution

  37. An exponential distribution has the following probability density function:

  38. Relationship to the Poisson distribution Suppose that the time between consecutive arrivals has an exponential distribution with parameter . Let X(t) be the number of occurrences by time t (t 0) The number of arrivals follows for n = 0, 1, 2, …;

  39. The Birth-and-Death Process Most elementary queueing models assume that the inputs and outputs of the queueing system occur according to the birth-and-death process. In the context of queueing theory, the term birth refers to the arrival of a new customer into the queueing system, and death refers to the departure of a served customer.

  40. The assumptions of the birth-and-death process are the following: Assumption 1. Given N(t) = n, the current probability distribution of the remaining time until the next birth is exponential. Assumption 2. Given N(t) = n, the current probability distribution of the remaining time until the next death is exponential

  41. Assumption 3. The random variable of assumption 1 (the remaining time until the next birth) and the random variable variable of assumption 2 (the remaining time until the next death) are mutually independent. The next transition in the state of the process is either (a single birth) (a single death), depending on whether the former or latter random variable is smaller.

  42. The birth-and-death process is a special type of continuous time Markov chain. State: 0 1 2 3 n-2 n-1 n n+1 and are mean rates.

  43. Starting at time 0, suppose that a count is made of the number of the times that the process enters this state and the number of times it leaves this state, as demoted below: the number of times that process enters state n by time t. the number of times that process leaves state n by time t.

  44. Rate In = Rate Out Principle. For any state of the system n (n = 0,1,2,…), average entering rate = average leaving rate. The equation expressing this principle is called the balance equation for state n.

  45. Rate In = Rate Out State 0 1 2 n – 1 n

  46. State: 0: 1: 2: To simplify notation, let for n = 1,2,…

  47. and then define for n = 0. Thus, the steady-state probabilities are for n = 0,1,2,… The requirement that so that implies that

  48. The definitions of L and specify that is the average arrival rate. is the mean arrival rate while the system is in state n. is the proportion of time for state n,

  49. The M/M/s Model A M/M/s model assumes that all interarrival times are independently and identically distributed according to an exponential distribution, that all service times are independent and identically distributed according to another exponential distribution, and that the number of service is s (any positive integer).

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