Question

 

Show your work/steps clearly in order to receive credit. 1. Apply the Northwest Corner (NWC) rule to obtain a feasible plan. TO FROM Denver (D) Erie (E) Fresno (F) Griffin (G) Supply Atlanta (A) 5 3 6 7 35 Boston (B) 4 8 1 9 60 Chicago (C) 8 12 10 3 35 Demand 30 45 25 30 Total cost = __________________________________________________________________ (show calculation).

Copy your feasible plan obtained by the NWC rule to the table below. Determine the Improvement TO FROM Denver (D) Erie (E) Fresno (F) Griffin (G) Supply Indices as follows: (Row A first, then rows B and C) Atlanta (A) 5 3 6 7 35 A-F: 6–1+8–3 = 10 (10) (18) A-G: 7–3+10–1+8–3 = 18 Boston (B) 4 8 1 9 60 B-D: 4–5+3–8 = –6 (–6) (15) B-G: 9–3+10–1 = 15 Chicago (C) 8 12 10 3 35 C-D: 8–5+3–8+1–10 = –11 (–11) (–5) C-E: 12–8+1–10 = –5 Demand 30 45 25 30

Determine the next Transportation Table. Determine the Improvement TO FROM Denver (D) Erie (E) Fresno (F) Griffin (G) Supply Indices as follows: (Row A first, then rows B and C) Atlanta (A) 5 3 6 7 35 Boston (B) 4 8 1 9 60 Chicago (C) 8 12 10 3 35 Demand 30 45 25 30 Total cost = __________________________________________________________________ (show calculation).

Determine the next Transportation Table. Determine the Improvement TO FROM Denver (D) Erie (E) Fresno (F) Griffin (G) Supply Indices as follows: (Row A first, then rows B and C) Atlanta (A) 5 3 6 7 35 Boston (B) 4 8 1 9 60 Chicago (C) 8 12 10 3 35 Demand 30 45 25 30 Total cost = __________________________________________________________________ (show calculation). 2.

The stepping-stone method is being used to solve a transportation problem. There is only one empty cell having a 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 Figure 12.5). Complete the following table (similar in format to Table 12.3). Time Slack = Critical Activity (Weeks) ES EF LS LF LS ─ ES Activity? A B C D E F G H Identify the critical path(s): __________________________________________________________________ The project completion time = ________ weeks. 12-18 (p.490). State the input data:  = , 2 = ,  = 

Question: What is the probability the project will be finished in 37.5 weeks or less? _______________________ ________________________________________________________________________________

Show your steps and how you use the Normal Curve Table to obtain your probability below.

Answer 1

SOLUTION

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	[5	[3	[6	[7	35
Boston (B)	[4	[8	[1	[9	60
Chicago (C)	[8	[12	[10	[3	35
Demand	30	45	25	30	130/130

Here Both Demand and Supply are equal and it is balanced

First select extreme northwest cell and assign the value

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	30          [5	5          [3	[6	[7	35-30=5
Boston (B)	[4	[8	[1	[9	60
Chicago (C)	[8	[12	[10	[3	35
Demand	30	45-5 =40	25	30	130/130

Entire Demand can be supplied from Atlanta to Denver as demand is 30 and supply is 35, the remaining supply is 5, so we need to move to next column and check next northwest cell which is A—E and assign remaining 5 there. So, final solution for Atlanta is A---D -30 and A---E is 5

Now we need to come to next row

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	30          [5	5          [3	[6	[7	35-30=5-5 =0
Boston (B)	[4	40          [8	20          [1	[9	60-40 =20-20 =0
Chicago (C)	[8	[12	[10	[3	35
Demand	30	45-5 =40	25	30	130/130

5 units are already supplied from A---E , so remaining demand for E is 40, which can be completely supplied from B—E and then remaining 20 from B---F

Now we need to come to next row

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	30          [5	5          [3	[6	[7	35-30=5-5 =0
Boston (B)	[4	40          [8	20          [1	[9	60-40 =20-20 =0
Chicago (C)	[8	[12	5         [10	30            [3	35
Demand	30	45-5 =40	25	30	130/130

20 units of demand is already given from B---F, so remaining demand of 5 is supplied from C---F, however there is still 30 left which is given from C to G.

Entire supply and demand are now balanced, and thus this is optimal solution.

Total cost for this solution is 30*5+5*3+40*8+20*1+5*10+30*3 = 150+15+320+20+50+90 = 645

Therefore, total cost is 750

To find, optimality, first check whether the no.of occupied cells in equal to no.of rows+no.of columns-1

No,of rows = 3 and columns 4, so rows+columns-1 =6

No.of occupied cells = 6

So, we can get an optimal solution

Unoccupied cells are A—F, A---G, B—D, B—G,C---D,C---E,

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	30          [5	5          [3	[6	[7	35-30=5-5 =0
Boston (B)	[4	40          [8	20          [1	[9	60-40 =20-20 =0
Chicago (C)	[8	[12	5         [10	30            [3	35
Demand	30	45-5 =40	25	30	130/130

Opportunity costs for A---F is 10, A—G is 18, B—D is -6, B---G is 15, C---D is -11 and C-E is -5

As there are negatives, it is not optimal solution and there is scope of improvement

So, first check B—D closed path, the path is BD—AD—AE—BE

The positives are bd and ae, the minimum value is 5, so add 5 to positive cells and subtract five from negative cells

We need to go with maximum penalty cell which is CD-- -11

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	25         [5	15          [3	[6	[7	35-30=5-5 =0
Boston (B)	[4	30          [8	25          [1	[9	60-40 =20-20 =0
Chicago (C)	5          [8	[12	[10	30            [3	35
Demand	30	45-5 =40	25	30	130/130

Solution is 25*5+5*8+15*3+30*8+25*1+30*3= 75+40+45+240+25+90= 515

Unoccupied cells are BD, CE,CF,AF,AG,BG

Opportunity costs for BD is -6, CE is CE-GE—AE—BE which is 4, CF is 10-1+8-3+5-8=11, AF is 10,AG is 7-3+8-3=9.

So only negative is BD which is -6, the minimum value is 15

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	10         [5	[3	[6	[7	35-30=5-5 =0
Boston (B)	15     [4	45          [8	25          [1	[9	60-40 =20-20 =0
Chicago (C)	5          [8	[12	[10	30            [3	35
Demand	30	45-5 =40	25	30	130/130

This is still not optimal solution and solution is 10*5+15*4+5*8+45*8+25*1+30*3= 50+60+40+360+25+90=625

Still AE is negative

TO FROM	Denver (D)	Eric (E)	Fresno (F)	Griffin (G)	Supply
Atlanta (A)	[5	10        [3	[6	[7	35-30=5-5 =0
Boston (B)	25     [4	35          [8	25          [1	[9	60-40 =20-20 =0
Chicago (C)	5          [8	[12	[10	30            [3	35
Demand	30	45	25	30	130/130

This is optimal solution as all the cells are positive for opportunity costs, the solution is 25*4+5*8+10*3+35*8+25*1+30*3= 100+40+30+280+25+90=565