(2) Find an optimal solution for the following transportation problem. To FROM Valley (S) Town (7) Junction S Burg (B) Supply Macon $6 $8 $7 $4 (M) 300 50 350 York $3 $10 $2 59 (1) 150 250 200 600 Perry (2) $14 $4 59 $12 250 250 Demand 300 200 250 450 1,200 The above feasible plan is obtained by applying the Northwest Corner Rule (e.g. 300 from M to V, etc.), and its total cost 300(56)+50(58)+150($10)+250($2)+200(59)+250($12)-$9.000. Using the Stepping Stone method to find an optimal plan. The improvement indices below are determined as follows: Supply (row M first, then rows Y and P) MJ: 7-2+10-8-7 350 MB: 4-9+10-8--3 To Valley Town Junction Burg FROM (D) (S) (B) Macon $6 $8 $7 $4 (M) York 300 50 $3 $10 $2 $9 (2) 150 250 200 600 $14 $4 $9 $12 250 450 Perry (P) Demand 300 200 250 YV: 3-10+8-6--5 PV: 14-12+9-10+8-6-3 PT: 4-12+9-10--9* 250 PJ: 9-12+9-2-4 1,200 the "most negative index" We focus on cell PT (index=-9). Its closed path is: PT to PB to YB to YT, and back to PT. To increase the quantity in PT, we need to adjust PB down, YB up, and YT down. Therefore, the next improved plan is: To FROM Valley (1) Town (7) Junction (S) Burg (B) Part (b) Find the indices (show steps) Supply Row M first, then rows Y and P Macon $6 $8 $7 $4 MS. (M) 300 50 350 York $3 $10 $2 $9 (1) 250 350 600 Perry $14 $4 $9 $12 (P) 150 100 250 300 200 250 450 1,200 Asterisk the "most negative index" = improved plan. To Valley Town Junction FROM (1) (7) (S) Demand Part (a) Find the new TC Part (c) Determine the next $300 (35) +50 ($7) +1250 ($1) + 350 (38) +$150 (83)+ (00 (410) (show steps). =315+ $350 $250 +$2500 +$450 $1000 $6350 Part (e) Find the indices (show steps) Row M first, then rows Y and P Burg (B) Supply Macon $6 $8 $7 $4 (M) 350 York $3 $10 $2 $9 (D) 600 Perry $14 $4 $9 $12 (P) 250 Demand 300 200 250 450 1,200 Asterisk the "most negative index" Part (d) Find the new TC= (show steps). Part (0) Is the current solution optimal? Explain (hint: see Online Module 8, p.M8-4).