Question

copy your feasible plan and determine next transportation table

Copy your feasible plan obtained by the NWC rule to the table be To Denver (D) Erie(E) Fresno (n Griffin (G) Row A first, then rows B and C A-F:6-1+8-3-10 B-D:4-5+38-6 C-D: 8-5+3-8+1-10-1 6 Atlanta (4 Boston (B Chicage ( (18 35A-G:7-3+10-1+8-3- 18 15 60 B-G:9-3+10-1-15 C-E: 12-8+1-105 45 Determine the nest Transportation T To Denver(D) Erie (E) Fresno () Griffin (G Supply (Row A first, then rows B and O Atlanta (A Boston (B) Chicago ( 45 Total cost- (show calculation). Determine the next Transportation Table. To Denver (D) Erie (E) Fresne () Griffin (G) Row A first, then rows B and C Atlanta (A Boston (B) Chicage (O 35 60 35 10 45 Total cost show calculation). 2. The stepping-stone method is being used to solve a transpontation problem. There is only one empty cell havinga negative improvement index, and this index is-7. The stepping-stone path for this cell indicates that the smallest quantity for the cells with minus signs is 30 units. If the total cost for the current solution is $1,000, what will the total cost be for the improved solution? 12-17 (p. 489) Construct a network with t, ES, EF. LS, and LF(similar in format to Eigure 12.5 Complete the following table (similar in format to Table 12.3 Time Slack Critical LF ILS-ESI Activity Activity Weeks) ES EF LS Identify the critical path The project completion time- 12-18 (p.490). State the input data: μ-- f Question: What is the probability the project will be finished in 37.5 weks or less? Show your steps and how you use the Normal Curve Tabkk to obtain your probability below

Answer 1

Using NWC rule, start with the Northwest Corner cell and allocate the maximum quantity possible and then move on to either the next cell to the right or below, depending upon the remaining demand or supply in that particular row or column. The process is repeated until all the demand is satisfied.

Initial feasibe plan by the MWC rule is following:

Copy your fasibk plan obtained by theNWC rule tothe table below Erie (E) Fresno (F) |Griffin (G) T) Denver (D) Supply (Row A first, then rows B and C A-F:6-1+8-3-1 5 A-G:7-3+10-1+8-3-18 B-D:4-5+3-8-6 ROM Atlanta (4) Beston (B Chkage ( 30 5 l0 20 5 25 40 15 60 B-G:9-3+101-15 10 C-D:8-543 8+1-10-1 30 30 C-E: 12-8-1-10- s$ 30

Total cost of inital solution = 30*5+5*3+40*8+20*1+5*10+30*3 = 645

Improvement index of cell C-D is the most negative. Therefore, it is the next basic cell.

Using the close loop for this cell, allocate the minimum value of the non-empty cell with negative sign (as used in calculation of improvement index). That cell is C-F and the value is 5. So 5 is subtracted from C-F, added to B-F, subtracted from B-E, added to A-E, subtracted from A-D and added to C-D. Resulting next transportation is following:

Determine the pest Transportation Table To Denver (D Erie (E) Fresno (D) Griffin(G Supply (Row A first, then rows B and Atlanta (A Boston (B Chkago (O Demand 25 10 35 35 25 60 10 5 0 30 30 45 25*5+10 3+35 8+25*1+5 8+0*10+30*3590 Total cost (show calculation)

Improvement indices are following:

A-F: 6-1+8-3 = 10

A-G: 7-3+8-5 = 7

B-D: 4-5+3-8 = (-6)

B-G: 9-3+8-5+3-8 = 4

C-E: 12-8+5-3 = 6

C-F: 10-8+5-3+8-1 = 11

Improvement index of B-D is the most negative. Therefore, it is the bext basic cell.

Using the close loop for this cell, allocate the minimum value of the non-empty cell with negative sign (as used in calculation of improvement index). That cell is A-D and the value is 25. So 25 is subtracted from A-D, added to A-E, subtracted from B-E, and added to B-D. Resulting next transportation is following:

Determine the pest Transportation Table. To Denver (D) Erie (E) Fresno () Griffin (G) Supply Row A first, then rows B ando Atlanta (4 Boston (B Chikage (O 35 35 60 35 25 10 25 10 5 30 30 45 25 30 35*3+25 4+10*8+25*1+5 8+30*3440 Total cost (show calculation)

Improvement indices are following:

A-D: 5-3+8-4 = 6

A-F: 6-1+8-3 = 10

A-G: 7-3+8-4+8-3 = 13

B-G: 9-3+8-4 = 10

C-E: 12-8+4-8 = 0

C-F: 10-8+4-1 = 5

Improvement indices of all empty cells are negative.

Therefore, optimal solution is reached.