Statics Exam Test Analysis - Math Problem Example

MSA 640 • Exam II • March 20, Open Book/Open s • 100 points total 20 points per question Show all work leading to your answers Andre Candess manages an office supply store. One product in the store is computer paper. Andre knows that 18,200 boxes will be sold this year at a constant rate throughout the year. The daily demand rate is 70 and the lead-time is 4 days. The cost of placing an order is $25, while the holding cost is $12 per box per year. Show each formula that you use. N.B.: See the last page of the exam for a list of essential formulas. (a) How many boxes should Andre order each time an order is placed if he wishes to minimize total inventory cost? (275) Given: D = 18,200, Co = $35, Ch = $12, L (lead time) = 4 days, d = 70 The Economic Order Quantity (EOQ), Q* is about 275 boxes. Therefore, Andre should order about 275 boxes each time an order is placed if he wishes to minimize total inventory cost (b) What is the reorder point? ROP = d*L = 70*4 = 280 The reorder point (ROP) is 280 boxes. (c) What is the average inventory? (137.69) Average Inventory = Q / 2 = 275.38/2 = 137.69 The average inventory is about 137.69 boxes. (d) What is the annual holding cost? Annual Holding Cost = The annual holding cost is about $1,652.27. (e) How many orders per year would be placed? (66) Optimal number of orders per year = D / Q* = 18200/275.38 ≈66.09 About 66 orders per year would be placed. 2. Consider the following linear programming problem: Maximize 27X + 15Y Subject to: 14X + 6Y ≤ 420 8X + 12Y ≤ 480 all variables ≥ 0 (a) Create a graphical LP solution on the attached graph paper, show the feasible region, and label the corner points. Show calculations for plotting the constraints and use simultaneous linear equations to determine corner point(s) when necessary. Use the reverse side of this sheet if you need more space for calculations. The line 14X + 6Y = 420 crosses x-axis at point (30, 0), and y-axis at point (0, 70). The line 8X + 12Y = 480 crosses x-axis at point (60, 0), and y-axis at point (0, 40). Solving simultaneous linear equations 14X + 6Y = 420 and 8X + 12Y = 480, we get solution as (18, 28). y = 28 Setting the objective function equal to an arbitrary value of 500. Therefore, an arbitrary isoprofit line is given by 27X + 15Y = 500 For getting the isoprofit line corresponding to optimal value, moving the arbitrary isoprofit line upward the region. When it gets tangent to the feasible line, the optimal solution is reached. Therefore, the isoprofit line corresponding to optimal value is given by 27X + 15Y = 906 A graph show the feasible region (shaded area) and all extreme (corner) points A(0, 0), B(30, 0), C(0, 40), and D(18, 28). The graph also show an arbitrary isoprofit line (blue), and the isoprofit line corresponding to optimal value (black). (b) Compute values for all corner points and identify the maximum possible value for the objective function. (max value 906) Corner X Y Value A 0 0 0 B 30 0 810 C 0 40 600 D 18 28 906 3. Consider the following linear programming problem: Maximize 40X + 20Y Subject to: 11X + 4Y ≤ 440 7X + 9Y ≤ 630 12X + 12Y ≥ 600 all variables ≥ 0 (a) Create a graphical LP solution on the attached graph paper, show the feasible region, and label the corner points. Show calculations for plotting the constraints and use simultaneous linear equations to determine corner point(s) when necessary. Use the reverse side of this sheet if you need more space for calculations. The line 11X + 4Y = 440 crosses x-axis at point (40, 0), and y-axis at point (0, 110). The line 7X + 9Y = 630 crosses x-axis at point (90, 0), and y-axis at point (0, 70). The line 12X + 12Y = 600 crosses x-axis at point (50, 0), and y-axis at point (0, 50). Solving simultaneous linear equations 11X + 4Y = 440 and 7X + 9Y = 630, we get solution as (20.28, 54.23). y ≈ 54.23 Solving simultaneous linear equations 11X + 4Y = 440 and 12X + 12Y = 600, we get solution as (34.29, 15.71). y ≈ 15.71 Setting the objective function equal to an arbitrary value of 3000. Therefore, an arbitrary isoprofit line is given by 40X + 20Y = 3000 For getting the isoprofit line corresponding to optimal value, moving the arbitrary isoprofit line upward the region. When it gets tangent to the feasible line, the optimal solution is reached. Therefore, the isoprofit line corresponding to optimal value is given by 40X + 20Y = 1896 A graph show the feasible region (shaded area) and all extreme (corner) points A(0, 50), B(34.29, 15.71), C(0, 70), and D(20.28, 54.23). The graph also show an arbitrary isoprofit line (blue), and the isoprofit line corresponding to optimal value (black). (b) Compute values for all corner points and identify the maximum possible value for the objective function. (max value 1,896) Corner X Y Value A 0 50 1000 B 34.29 15.71 1685.8 ≈ 1686 C 0 70 1400 D 20.28 54.23 1895.8 ≈ 1896 4. A development project has 13 major activities. The estimated times for the activities appear on the attached PERT Network. (a) Compute the ES, EF, LS, LF values and fill in the network chart. Activity Activity Time Early Start Early Finish Late Start Late Finish Slack A 4 0 4 0 4 0 B 3 0 3 6 9 6 C 2 0 2 4 6 4 D 5 4 9 4 9 0 E 3 2 5 6 9 4 F 4 9 13 9 13 0 G 7 5 12 10 17 5 H 4 13 17 13 17 0 I 5 17 22 17 22 0 J 4 17 21 21 25 4 K 3 22 25 22 25 0 L 6 25 31 25 31 0 M 5 25 30 26 31 1 (b) What is the expected completion time for this project? (completion time = 31) The expected completion time for this project is 31. (c) Find the critical path for this project by computing slack times. (A–D–F–H–I–K-L) Slack time is 0 for activities A, D, F, H, I, K, and L (Highlighted). Therefore, critical path for this project is A–D–F–H–I–K-L. (d) Assuming that the standard deviation is 4, what is the probability that the project will be finished in 33 weeks or less? Finish drawing and labeling the following normal curve to represent your solution. (0.69) Z = (X - µ)/σ = (33 – 31)/4= 0.5 Area = 0.6915, thus P(Z< (X- μ)/s)) ≈ 0.69 The probability that the project will be finished in 33 weeks or less is about 0.69. 5. Customers arrive at a corner grocery store every 5 minutes and the Poisson distribution accurately defines this rate. A single cashier works at the store, the average time to serve a customer is 4 minutes, and the exponential distribution may be used to describe the distribution of service time. Show each formula that you use. (a) What are λ and μ in this situation? (12, 15) λ = per hour µ = per hour The λ and μ in this situation are 12 per hour and 15 per hour, respectively. (b) What percentage of time is the cashier idle? (0.2) or 20% 20 percentage of time is the cashier idle. (c) What is the average length of the line? (3.2) The average length of the line is 3.2. (d) How many minutes does the average customer spend waiting in the waiting line? (16 min.) hours or 16 minutes The average customer spends 16 minutes waiting in the waiting line. (e) How many minutes typically elapse from the time the person enters the line until the person pays the cashier and leaves the system? (20 min.) hours or 20 minutes 20 minutes typically elapse from the time the person enters the line until the person pays the cashier and leaves the system. (f) Find the probability that the number of people in the system is greater than 2. (0.51) ≈ 0.51 The probability that the number of people in the system is greater than 2 is about 0.51. Read More