Question

Question 2: Problem solving Number of crates Sea-food eatery 3 0 Profits (($x100) Sea-food eatery 2 0 5 7 12 Sea-food eatery 1 0 5 8 10 0 1 2 3 7 11 12 1. Construct the tables for 72 = 3n = 2, and n = 1. Show the calculations. (11.5 Marks) Stage 3, n = 3 (2 Marks) S3 f(s) 0 --- 1 2 3 Stage 2, n-2_(7 Marks) f(y,xy) = P(x) + f($2 - x2) 2 0 X2 1 3 f(s) S2 0 --- ... 1 2 .... 3 Stage 1,72 = 1 (2.5 Marks) 71(81%) = P1(x1) +(3 – x1) 1 2 X1 0 3 fi (5) 51 3 III. III. 2. Based on the tables you constructed write down the optimal allocation, and the maximum total profit. (3.5 Marks)

Answer 1

Solution in o Given that profits Sea tood Ea tory a Sea food Eatory 3 Sea - fool Eatery I Number of crats 0 0 7 5 5 ۱۱ Э 12 12 3 10 Let A is the no. of crates available Prilen) denotes the profit Earned when an no noot Crates allocated to store "u-n" (store is food-Eatery) where food Eatery are referred to as 1,2,3 an is the decision variable cie) no of crates to be allocated at ftgin' Emprassing on "u-n" to say that allocation has been done from east Eatery am Detine. In * (A) . Max. fn (A, an). fon (A, 10) = P(an) + font (S-an)

n=1 * S 4tra) 0 ܗ ܗ - ܘ -* 7 1) مدا 3 * no -2. fal Ane) = Pel22) + 4* (A-22) 3 424 (A) a24 0 0 - 0 0 7 o d - - 5 la 2 1) Э. 12 3 12 16 14 12 16 " 42 10,0) ta (10) 42 (1,1) P210) + 4 * (0) =0. Pa (0) + fit (1) = 0+1 =7 P2 (1) + f (o) 5+0=5 f2 (210) fa lail) f2 (212) Pa 10) + fit (Q) = 0+1 =))! Pa (1) + f * (1) 547 =12 P2 (2) + 4,410) +0=2

3. * n=3 43 (, 13) = P3 (13) + fot (A-23) 43*() 12 * sa3 o 1 2 3 3 16 17 15. 10 X3 12 t =), 234 = 1. is optional. (ie) Eatery 1 Gak Each Eatory 2 = eatury 3 = 1 profit : 174