Quantitative Analysis Performance - Assignment Example

DD Month YYYY QUANTITATIVE ANALYSIS Question a) North-West Corner Rule Factory Shipping Centers Supply A B C   D 5 (200) 4 3 300 E 8 4 3 150 F 9 7 5 250 Demand 200 200 300 700 The Northwest figure is 5. The supply for the figure is 300 while the demand for the figure is 200. Comparing the two supply and demand the lowest figure is 200. Therefore, 200 units will be allocated to square DA. Factory Shipping Centers Supply A B C   D 5 (200) 4 3 100 E 8 4 3 150 F 9 7 5 250 Demand 0 200 300 700 Since supply is unsatisfied, we will move to DB Factory Shipping Centers Supply A B C   D 5 (200) 4 (100) 3 0 E 8 4 3 150 F 9 7 5 250 Demand 0 100 300 700 Since the demand is unsatisfied, we will move to EB Factory Shipping Centers Supply A B C   D 5 (200) 4 (100) 3 0 E 8 4 (100) 3 50 F 9 7 5 250 Demand 0 0 300 700 Since the demand is unsatisfied, we will move to EC Factory Shipping Centers Supply A B C   D 5 (200) 4 (100) 3 0 E 8 4 (100) 3 (50) 0 F 9 7 5 250 Demand 0 0 250 700 Since the demand is still unsatisfied, we will move to FC Factory Shipping Centers Supply A B C   D 5 (200) 4 (100) 3 0 E 8 4 (100) 3 (50) 0 F 9 7 5(250) 0 Demand 0 0 0 700 The shipping schedule is as shown in brackets and is found to add up to 700 To calculate the total cost, Shipment Units Shipped Transport Cost/Unit Total Cost From To       D A 200 5 1000 D B 100 4 400 E B 100 4 400 E C 50 3 150 F C 250 5 1250 Total Cost 3200 b) Stepping Stone Method Factory Shipping Centers Supply A B C   D 5 (200) 4 (100) 3 300 E 8 4 (100) 3 (50) 150 F 9 7 5(250) 250 Demand 200 200 300 700 The closed path will be +DC , -DA, +FA, -FC = 3 – 5 + 9 – 5 = 2 The second closed path will be +EA, -EC, +FC, -FA =8 – 3 + 5 – 9= - 9 The third closed path will be +FA, -DA, +,DC FC =9 – 5+3 – 5=2 The fourth closed path will be +FB, -DB, +DC, -FC =7 – 4 + 3 – 5 = - 5 The improvement indices recorded both the negative and positive. The paths that recorded negative indices needed an improvement which is the second and the fourth paths. It can be concluded can the second feasible solution has the optimality solution. c) When the improvement index is zero then the case is regarded as degeneracy. Normally, in a feasible solution there are m+n-1 used routes. When the routes, supply, and demand are satisfied when the index is zero. When the non-optimal develops a feasible solution happens, we add the zero allocation to a path and regard it as one of the used routes where the cumulative number of used path are m+n-1 so that the algorithm can continue. Question 2 a) K Factory Shipping Centers Plant Capacities A B C   1 10 4 11 70 2 12 5 8 50 3 9 7 6 30 Project Requirement 40 50 60 150 The plant capacities will be dissatisfied then Factory Shipping Centers Plant Capacities A B C   1 10(400) 4 11 30 2 12 5 8 50 3 9 7 6 30 Project Requirement 0 50 60 150 Factory Shipping Centers Plant Capacities A B C   1 10(400) 4 (300) 11 0 2 12 5 8 50 3 9 7 6 30 Project Requirement 0 20 60 150 Factory Shipping Centers Plant Capacities A B C   1 10(40) 4 (30) 11 0 2 12 5(20) 8 30 3 9 7 6 30 Project Requirement 0 0 60 150 Factory Shipping Centers Plant Capacities A B C   1 10(40) 4 (30) 11 0 2 12 5(20) 8 (30) 0 3 9 7 6 30 Project Requirement 0 0 30 150 Factory Shipping Centers Plant Capacities A B C   1 10(40) 4 (30) 11 0 2 12 5(20) 8 (30) 0 3 9 7 6(30) 0 Project Requirement 0 0 0 150 Finding the improvement indices 1C, 1A, 3A, 3C 11 – 10 + 9 – 6= 4 2A, 2C, 1C, 1A 12 – 8 + 11 – 10 =5 3A, 3C, 1C, 1A 9 – 6 + 11 – 10 = 4 3B, 3C, 2C, 2B 7 – 6 + 8 – 5 = 4 The solution is optimal because it has no negative improvement indices. b) There are more than one optimal solution to this problem, because all the indices are positive. For this reason no improvement is needed. Question 3 Factory Shipping Centers Plant Capacities A B C   1 10 4 11 70 2 12 5 8 50 3 18 14 12 60 Project Requirement 50 60 70 180 Factory Shipping Centers Plant Capacities A B C   1 10 (50) 4 11 20 2 12 5 8 50 3 18 14 12 60 Project Requirement 0 60 70 180 Factory Shipping Centers Plant Capacities A B C   1 10 (50) 4 (20) 11 0 2 12 5 8 50 3 18 14 12 60 Project Requirement 0 40 70 180 Factory Shipping Centers Plant Capacities A B C   1 10 (50) 4 (20) 11 0 2 12 5(40) 8 10 3 18 14 12 60 Project Requirement 0 0 70 180 Factory Shipping Centers Plant Capacities A B C   1 10 (50) 4 (20) 11 0 2 12 5(40) 8(10) 0 3 18 14 12 60 Project Requirement 0 0 60 180 Factory Shipping Centers Plant Capacities A B C   1 10 (50) 4 (20) 11 0 2 12 5(40) 8(10) 0 3 18 14 12 (60) 0 Project Requirement 0 0 0 180 Stepping Stone Method 1C, 1A, 3A, 3C 11 – 10 + 18 – 12 =7 2A, 2C, 3C, 3A 12 – 8 + 12 – 18 = - 2 3A, 3C, 1C, 1A 18 – 12 + 11 – 10 = 7 3B, 3C, 1C, 1B 14 – 12 + 11 – 4 =9 The third plant capacity altered the indices for the second path to register a negative index. When the improvement index is zero then the case is regarded as degeneracy. Normally, in a feasible solution there are m+n-1 used routes. When the routes, supply, and demand are satisfied when the index is zero (Hsieh, 1280). When the non-optimal develops a feasible solution happens, we add the zero allocation to a path and regard it as one of the used routes where the cumulative number of used path are m+n-1 so that the algorithm can continue. Question 4 Factory Shipping Centers Mill Capacity (Tons) Supply House 1 Supply House 2 Supply House 3   Pineville 3 3 2 25 Oak Ridge 4 2 3 40 Mapletown 3 2 3 30 Supply House Demanded 30 30 35 95 NorthWest Rule Factory Shipping Centers Mill Capacity (Tons) Supply House 1 Supply House 2 Supply House 3   Pineville 3 (25) 3 2 0 Oak Ridge 4 2 3 40 Mapletown 3 2 3 30 Supply House Demanded 5 30 35 95 Factory Shipping Centers Mill Capacity (Tons) Supply House 1 Supply House 2 Supply House 3   Pineville 3 (25) 3 2 0 Oak Ridge 4(5) 2 3 35 Mapletown 3 2 3 30 Supply House Demanded 0 30 35 95 Factory Shipping Centers Mill Capacity (Tons) Supply House 1 Supply House 2 Supply House 3   Pineville 3 (25) 3 2 0 Oak Ridge 4(5) 2 (30) 3 5 Mapletown 3 2 3 30 Supply House Demanded 0 0 35 95 Factory Shipping Centers Mill Capacity (Tons) Supply House 1 Supply House 2 Supply House 3   Pineville 3 (25) 3 2 0 Oak Ridge 4(5) 2 (30) 3 (5) 0 Mapletown 3 2 3 30 Supply House Demanded 0 0 30 95 Factory Shipping Centers Mill Capacity (Tons) Supply House 1 Supply House 2 Supply House 3   Pineville 3 (25) 3 2 0 Oak Ridge 4(5) 2 (30) 3 (5) 0 Mapletown 3 2 3(30) 0 Supply House Demanded 0 0 0 95 Stepping Stone Methods 3 – 3 + 3 – 2 =1 2 – 3 + 3 – 3 = -1 3 – 3 + 2 – 3 = -1 2 – 3 + 2 – 3 = - 2 The best transportation path is the first route. Question 5 Minimize Costs= 20X1 + 28 X2 Subject to 5 X1 + 4 X2 ≤ 1000---------1 X1 ≥ 50--------------------------2 X1 + X2 ≥ 200----------------------3 X2 ≥ 100-------------------------------4 X1 , X2 ≥ 0--------------------------5 For equation 1 x1 0 250 x2 250 0 For equation 5 x1 0 200 x2 200 0 250 200 150 100 50 0 50 100 150 200 250 300 350 Some of the feasible points are (50, 150) , ( 50, 200) , (100, 100) , (150, 100) The optimal points are attained when x1 = 150 and X2 = 100 Question 6 Maximize Monthly Earnings = 50 X1 + 25 X2 Subject to 4 X1 + 8 X2 ≤ 800 (advertising budget available)----------------------------1 100 X1 + 50 X2 ≤ 8000 (square footage available)----------------------------2 X1 ≤ 50 (rental limit expected)------------------------------------------3 X1 , X2 ≥ For equation 1 x1 0 100 x2 100 50 For equation 2 x1 0 80 x2 160 0 160 140 120 100 80 60 40 20 0 20 40 60 80 100 120 140 160 180 200 The feasible points include (0, 100), (50, 80) , (50, 0), (0, 0) The company will build 50 rooms of X1, and 80 rooms of X2 Question 7 Factory Shipping Centers Supply Dubai Abu Dhabi Umm Alquin   Sharja 7 15 4 300 Ajman 8 18 4 500 Al-Ain 11 10 12 550 Demand 550 350 450 1350 Factory Shipping Centers Supply Dubai Abu Dhabi Umm Alquin   Sharja 7(300) 15 4 0 Ajman 8 18 4 500 Al-Ain 11 10 12 550 Demand 250 350 450 1350 Factory Shipping Centers Supply Dubai Abu Dhabi Umm Alquin   Sharja 7(300) 15 4 0 Ajman 8 (250) 18 4 250 Al-Ain 11 10 12 550 Demand 0 350 450 1350 Factory Shipping Centers Supply Dubai Abu Dhabi Umm Alquin   Sharja 7(300) 15 4 0 Ajman 8 (250) 18 (250) 4 0 Al-Ain 11 10 12 550 Demand 0 100 450 1350 Factory Shipping Centers Supply Dubai Abu Dhabi Umm Alquin   Sharja 7(300) 15 4 0 Ajman 8 (250) 18 (250) 4 0 Al-Ain 11 10(100) 12 450 Demand 0 0 450 1350 Factory Shipping Centers Supply Dubai Abu Dhabi Umm Alquin   Sharja 7(300) 15 4 0 Ajman 8 (250) 18 (250) 4 0 Al-Ain 11 10(100) 12(450) 0 Demand 0 0 0 1350 From the above table the shipping routes are Sharia –Dubai Ajman – Dubai Ajman – Abu Dhabi Al-Ain – Abu Dhabi Al – Ain – Um Alquin The number of beds are indicated in brackets Sharia –Dubai (300) Ajman – Dubai (250) Ajman – Abu Dhabi (250) Al-Ain – Abu Dhabi (100) Al – Ain – Um Alquin (450) The transportation cost =300* 7 + 250*8 + 250*18 + 100*10 + 450*12 = $15000 References Hsieh, H.-F. Three Approaches To Qualitative Content Analysis. Qualitative Health Research 15.9 (2005): 1277-1288. Web. Read More