Monday 3 February 2014

QAT1 Quantitative Analysis WGU Task 1- 5 - Complete Course

QAT1  Quantitative Analysis WGU Task  1- 5 - Complete Course


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QAT1 Task 1

QAT1 Task 309.3.1-01 Simulation Template

QAT1 Task 2

Part A:

Nutrient constraint is a maximum constraint:
4x + 4y < 30                                        4x + 4y <30
4x + 4y = 30                                        4x + 4y = 30
4x + 4(0) = 30                                     4(0) + 4y = 30
4x + 0 = 30                                          0 + 4y = 30
4x = 30                                                4y = 30
4       4                                                 4       4

x=7.5                                                   y=7.5

Part B:

Total contribution to profit from cases produced as illustrated by profit line:
40x + 30y =Total Contribution
Point on profit line  (0,8)

Part C:
The optimal production point on the graph occurs at the intersection of:
 Flavor line (12x + 6y=72)   and  Nutrient line (4x + 4y = 30)
I determined this by moving the profit line equally outward in both directions until I reached the outmost point in the feasible region.
Part D:
40x + 30y = Total Contribution to Profit

QAT1 Task 3

TASK A:
Company A:
Demand (D) = 18,000 units per year
Ordering Costs (Co) = $38 per order
Per Unit Cost of inventory = $12
Holding cost rate = 26%
Holding cost(Ch) = $12(.26) = $3.12

TASK B:
Company B:
Demand (D) = 15,000 units per year
Production Setup Cost (Co) = $84 per order
Per Unit Cost of inventory = $19
Holding cost rate = 28%
Holding cost(Ch) = $19(.28) = $5.32

QAT  Task 4

Task A1:

Expected completion times for each of the ten project activities:

A         (2+4(3)+4)/6                18/6                 3
B         (5+4(6)+13)/6              42/6                 7
C         (3+4(4)+8)/6                27/6                 4.5
D         (10+4(11)+15)/6          69/6                 11.5
E          (4+4(5)+6)/6                30/6                 5
F          (8+4(10)+12)/6            60/6                 10
G         (4+4(6)+11)/6              39/6                 6.5
H         (8+4(10)+18)/6            66/6                 11
I           (3+4(6)+12)/6              39/6                 6.5
J           (2+4(3)+7)/6                21/6                 3.5

Task A2:

Variance for each project activity:


A         ((4-2)/                            .11
B         ((13-5)/                           1.78
C         ((8-3)/                             .69
D         ((15-10)/                          .69
E          ((6-4)/                              .11
F          ((12-8)/                           .44
G         ((11-4)/                          1.36
H         ((18-8)/                          2.78
I           ((12-3)/                                2.25
J           ((7-2)/                              .69

Task A3:

Calculation of the earliest start times are done by calculating the maximum of the earliest completion times of all of the preceding projects.  The earliest finish times are calculated by adding the completion time of the particular project being calculated to the earliest start time.

 Here are the expected start and completion times for this project:

For project Task A:
Latest Start Time is 0 minus the Earliest Start Time which is 6.5.  6.5-0=6.5  OR
Latest Finish Time is 3 minus the Earliest Finish Time which is 9.5.  9.5-3=6.5

Slack for project task A is 6.5 weeks

c.  Slack for project Task H:

Latest Start Time is 19 minus the Earliest Start Time which is 18.5.  19-18.5=.5  OR
Latest Finish Time is 30 minus the Earliest Finish Time which is 29.5.  30-29.5=.5

Slack for project task H is .5 weeks

d.  The week project task F is scheduled to start:

Project B must be completed before Project F can begin and the earliest finish time for Project B is week 7, therefore the earliest start time for Project F is also week 7.  We also know that the earliest start time for the next project, G, is week 17 and Project F takes 10 weeks, therefore the latest start time for Project F is also week 7 (17-10=7).  The earliest start time and the latest start time for Project F are both week 7, therefore we must schedule Project Task F to start on week 7.

e. The week project task I is scheduled to finish:

Project I takes 6.5 weeks to complete and the previous project, G, has to be finished before Project I can begin. The earliest finish time for Project G is week 23.5, therefore 23.6 + 6.5 = 30 weeks being the earliest finish time for Project I.  Project J is the next project and the earliest start time for Project J is week 30, therefore the latest finish time for Project I is week 30, or it will delay the start of Project J.  The earliest finish time and the latest finish time for Project I are both week 30, therefore we must schedule Project Task I to start on week 30.

QAT1 Task 4 PERT Chart


QAT1 Task 5

TASK A:

Calculate the expected value for each of the four decision branches:

Branch 1:  Develop a new product thoroughly

.4(500,000) + .4(25,000) + .2(1,000) = 200,000 + 10,000 + 200 = $210,200

TASK B:
Determine the decision alternative that has the most favorable total expected value:
The decision alternative that has the most favorable expected value is to develop a new product and develop it thoroughly.   


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