Haii. Is it okay if you can check if my answers are right? I'm not really sure about this. The topic is about transportation method using stepping stone and north-west corner method. Thank you very much

ACTIVITY 2.A FOR EACH OF THE FOLLOWING PROBLEMS USING AN APPROPRIATE METHOD.. WRITE YOUR ANSWERS ON SHORT BOND PAPERS, DO NOT USE THE BACK PAGE. A. Transportation Problems: 1. STA Candies manufacturing Company, Inc. wants to find the lowest cost of delivery of its products to Jabby's Supermarket(JS). JS Tacloban needs 500 boxes, JS Ormoc needs 700 boxes, And JS Catbalogan needs 400 boxes. The owner has three branches at three different places: Palo Warehouse, Carigara Warehouse and Sogod Warehouse. Palo Warehouse has 550 boxes, Carigara Warehouse has 625 boxes, and Sogod Warehouse has 425 boxes. The cost of transportation from Palo to JS Tacloban is P200, to JS Ormoc is P600, and to JS Catbalogan is P1,100. While the cost from Carigara Warehouse to JS Tacloban, JS Ormoc, and to JS Catbalogan is P300, P600, and P1300 respectively; and from Sogod Warehouse the cost of transportation is P700 to JS Tacloban, P300 to JS Ormoc, and P1500 to JS Catbalogan. Design a plan of distribution that will give a minimum cost. 2. Jabboy's Chicken House has three branches: Palo Branch, TAcloban Branch, and Tanauan Branch. These branches deliver fried chicken to three distributors: X, Y, Z. The following table gives the delivery costs, the number available at each branch, and the number needed by each distributor: From / To: X Y DEMAND SUPPLY PALO Br P30 P40 P80 PALO Br 200 X 400 TAC Br 40 70 40 TAC Br 250 Y 200 TAN Br 20 40 70 TAN Br 300 Z 150 Find an optimum solution to cut down the total delivery cost. 3. 5Js manufacturing company must ship its products from three factories to three warehouses. The weekly production of the factories are 120, 75, 55. The weekly requirements of the warehouses are 100, 90, 60. The shipping cost from each factory to the warehouse is given below: To W1 W2 W3 From F1 P 100 P120 P70 F2 50 70 80 F3 80 80 110 Design a plan of distribution that will give a minimum cost9:26 0 9 a . 4G+ Problem 1-1.pdf . . . Problem 1: Table 1: JS IS IS Supply Tacloban Ormoc Catbalogan Palo Warehouse 200 600 1100 550 (500) (50) Carigara Warehouse 300 600 1300 625 (625) Sogod Warehouse 700 300 1500 425 (25) (400) Demand 500 700 400 1600 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 500x200=100,000 PW-JSC: 1,100-1,500+300-600= -700 600x50=30,000 CW-JST: 300-600+600-200=100 600x625=375,000 CW-JSC: 1,300-1,500+300-600=-500 300x25=7,500 SW-JST: 700-300+600-200= 800 400x1500=600,000 Total: 1,112,500 Table 2: JS IS JS Supply Tacloban Ormoc Catbalogan Palo Warehouse 200 600 1100 550 (500) (50 Carigara Warehouse 300 600 1300 625 (625) Sogod Warehouse 700 300 1500 425 (75 (350) Demand 500 700 400 1600 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 200x500= 100,000 PW-JSO: 600-300+1,500-1,100= 700 600x625= 375,000 CW-JST: 300-200+1,100-1,500+300-600= -600 300x75= 22,500 CW-JSC: 1,300-1,500+300-600= -500 1100x50= 55,000 SW-JST: 700-1,500+1,100-200= 100 1500x350=525,000 TOTAL: 1,077,500 Table 3: JS JS JS Supply O9:27 P S . 4G+ Problem 1-1.pdf . . . Demand 500 700 400 1600 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 200x500= 100,000 PW-JSO: 600-300+1,500-1,100= 700 600x625= 375,000 CW-JST: 300-200+1,100-1,500+300-600= -600 300x75= 22,500 CW-JSC: 1,300-1,500+300-600= -500 1100x50= 55,000 SW-JST: 700-1,500+1,100-200= 100 1500x350=525,000 TOTAL: 1,077,500 Table 3: JS IS JS Supply Tacloban Ormoc Catbalogan Palo Warehouse 200 600 1100 550 (150) (400 Carigara Warehouse 300 600 1300 625 (350 (275) Sogod Warehouse 700 300 1500 425 (425 Demand 500 700 400 1600 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 200x150= 30,000 PW-JSO: 600-200+300-600=100 300x350= 105,000 CW-JSC: 1,300-300+200-1,100=100 600x275= 165,000 SW-JST: 700-300+600-300=700 300x425= 127,500 SW-JSC: 1,500-1,100+200-300+600-300=600 1100x400= 440,000 TOTAL: 867,500 O9:27 P . 4G+ Problem 1-1.pdf . . . Problem 2: = Table 1: X Y Z Supply PALO Br. 30 40 80 400 (200) (200) TAC Br. 40 70 40 200 50) (150) TAN Br. 20 40 70 150 (150) Demand 200 250 300 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 200x30=6,000 PALO Br.- Z: 80-40+70-40=70 200x40=8,000 TAC Br.- X: 40-70+40-30=-20 70x50=3,500 TAN Br.- X: 20-70+40-70+40-30= -70 150x40=6,000 TAN Br-Y: 40-70+40-70= -60 150x70-10,500 Total:34,000 3 of 7 O9:27 P . 4G+ Problem 1-1.pdf . . . Table 2: X Y Z Supply PALO Br. 30 40 80 400 (150 (250) TAC Br. 40 70 40 200 (200) TAN Br. 20 40 70 150 (50) (100) Demand 200 250 300 COST: IMPROVEMENT: (from the occupied cells) from the vacant cells) 150x30= 4,500 PALO Br.- Z: 80-40+70-40=70 50x20= 1,000 TAC Br.- X: 40-20+70-40=50 250x40=10,000 TAC Br. - Y: 70-40+30-20+70-40= 70 200x40= 8,000 TAN Br-Y: 40-20+30-40=10 100x70= 7,000 Total:30,500 O9227 Ii 0') - 6 Problem 1-1.pdf N 1? .l\" i / Problem 3: Table 1: W1 W2 W3 Supply F1 100 120 70 120 (100) (20) F2 50 70 80 75 (70) (5) F3 80 80 110 55 (55) Demand 100 90 60 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 100x100: 10,000 F1W3: 70-80+70120= -60 120x20: 2,400 F2W1:50100+12070=0 70x70: 4,900 F3W1: 80-110+8070+120100=0 80x5: 400 F3-W2: 80-110+8070= -20 110x55: 6,050 TOTAL: 23,750 Table 2: W1 W2 W3 Supply F1 100 120 70 120 (100) (15) (5) F2 50 70 80 75 (75) F3 80 80 110 55 (55) Demand 100 90 60 9:27 P S . 4G+ Problem 1-1.pdf . . . Demand 100 90 60 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 100X100=10,000 F2-W1: 50-100+120-70=0 120X15= 1,800 F2-W3: 80-70+120-70=60 70X75= 5,250 F3-W1: 80-110+70-100=-60 70X5= 350 F3-W2: 80-120+70-110= -80 110X55= 6,050 TOTAL: 23,450 Table 3: W1 W2 W3 Supply F1 100 120 70 120 (100) (20) F2 50 70 80 75 (75) F3 80 80 110 55 15) (40) Demand 100 90 60 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 100X100= 10,000 F1-W2: 120-80+110-70=80 70X75= 5,250 F2-W1: 50-100+70-110+80-70=-80 80X15= 1,200 F2-W3: 80-110+80-70=-20 70X20= 1,400 F3-W1: 80-110+70-100=-60 110X40= 4,400 TOTAL: 22,250 6 of 7 O9:27 P . 4G+ Problem 1-1.pdf . . . TOTAL: 22,250 Table 4: W1 W2 W3 Supply F1 100 120 70 120 (60) (60) F2 50 70 80 75 (40) (35) F3 80 80 110 55 55 Demand 100 90 60 COST: IMPROVEMENT: (from the occupied cells) (from the vacant cells) 100X60= 6,000 F1-W2: 120-100+50-70=0 50X40= 2,000 F2-W3: 80-50+100-70=60 70X35= 2,450 F3-W1: 80-50+70-80=20 80X55= 4,400 F3-W3: 110-80+70-50+100-70=50 70X60= 4,200 TOTAL: 19,050 7 of 7 O