Sales price per ton Production cost per ton Purchase price per ton Alloy A $60 $
ID: 3122514 • Letter: S
Question
Sales price per ton Production cost per ton Purchase price per ton Alloy A $60 $4 Steel $50 Alloy B $70 $2.8 Brass $40 Alloy C $45 $2.8 Bronze $30 The maximum demand (in tons) for each alloy, the minimum steel and brass proportion in each alloy is detailed in the following table. max demand min steel proportion min brass proportion Alloy A 4000 0.6 0.3 Alloy B 3800 0.65 0.25 Alloy C 3500 0.3 0.4 Let xij 0 be a decision variable that denotes the number of tons of Alloy j for j {A,B,C} to be produced from Metals i {Steel, Brass, Bronze}. Formulate an LP model for the Haus that maximizes the profit, while satisfying the demand as well as the steel and brass proportion constraints. [Hints: 1. Let xij be a decision variable that indicates the number of tons of Metal i be blended into Alloy j.
Explanation / Answer
LP FORMULATION
Note: Proportion of only steel and brass are given for each of the alloys. Since total proportion must be 1, proportion of bronze is taken as: 1 - proportion of steel - proportion of brass.
The given data set is tabulated below for easy reference.
Alloy-wise details
Alloy
Sales price ($per ton)
Production cost ($per ton)
Maximum demand (tons)
A
60
4.0
4000
B
70
2.8
3800
C
45
2.8
3500
Component -wise details
Component
Purchase price
($per ton)
Proportion in
Alloy A
Alloy B
Alloy C
Steel
50
0.6
0.65
0.3
Brass
40
0.3
0.25
0.4
Bronze
30
0.1
0.1
0.3
Let xij = quantity (in ton) of ith component in jth alloy; j = 1 for A, 2 for B, 3 for C; i = 1 for steel, 2 for brass and 3 for bronze.
Then, from proportional composition of steel, brass and bronze in the three alloys, we derive the following relationships:
x21 = x11/2; x22 = 5x12/13; x23 = 4x13/3; x31 = x11/6; x32 = 2x12/13; x33 = x13.
From the above, we get the quantity of the three alloys as:
A: x11 + x21 + x31 = (5/3)x11 ……………………………………………………..(1)
B: x12 + x22 + x32 = (20/13)x12 ……………………………………………………..(2)
C: x13 + x23 + x33 = (10/3)x13 ……………………………………………………..(3)
Profit for alloy = {(Sales price – production cost) x quantity produced} – sum of {component weight x component purchase price}(summed over the three components). Thus,
Profit for alloy A = 56(5/3)x11 - (50x11 + 40x21 + 30x31)
= 56(5/3)x11 - {50x11 + 40(1/2)x11 + 30(1/6)x11} = (55/3)x11
Similarly, Profit for alloy B = 67.2(20/13)x12 - (50x12 + 40x22 + 30x32)
= 67.2(20/13)x12 - {50x12 + 40(5/13)x12 + 30(2/13)x12)} = (434/13)x12
Profit for alloy C = 42.2(10/3)x13 - (50x13 + 40x23 + 30x33)
= 42.2(10/3)x13 - {50x13 + 40(4/3)x13 + 30x13)} = (22/3)x13
Total profit, Z = (55/3)x11 + (434/13)x12 + (22/3)x13 ……………………………………(4)
Using (1), (2), (3) and (4), the LPP formulation is:
Max Z = (55/3)x11 + (434/13)x12 + (22/3)x13 [vide (4)]
Subject to (5/3)x11 4000; (20/13)x12 3800; (10/3)x13 3500 [demand constraint, vide (1), (2), (3)]
x11 , x12 , x13 0.
DONE
Alloy
Sales price ($per ton)
Production cost ($per ton)
Maximum demand (tons)
A
60
4.0
4000
B
70
2.8
3800
C
45
2.8
3500
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