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Examples A.2 · Example 3

Q.Interpret the model in Example 2, in case
[!FORMULA] A=(1015610200),B=(3407935127)A = \begin{pmatrix} 10 & 15 & 6 \\ 10 & 20 & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 3 & 4 & 0 \\ 7 & 9 & 3 \\ 5 & 12 & 7 \end{pmatrix}
and the available raw materials are 330330 units of R1R_1, 455455 units of R2R_2 and 140140 units of R3R_3.

Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★est
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✓ Free question

Multiplying ABAB gives (1652478717022060)\begin{pmatrix} 165 & 247 & 87 \\ 170 & 220 & 60 \end{pmatrix}. The two orders together demand 335335 units of R1R_1, 467467 of R2R_2 and 147147 of R3R_3 — more than the 330, 455, 140330,\,455,\,140 in stock — so the orders cannot be met as placed. Reducing the orders (e.g. to A1A_1) brings the demand within stock.

Step 1 — Identify. Substitute the given numbers into the Example 2 model to see whether the firm's stock can fill both orders.

Step 2 — Set up. Here row FiF_i of AA is client ii's order in products P1,P2,P3P_1,P_2,P_3; BB gives the raw material R1,R2,R3R_1,R_2,R_3 per unit of each product; stock is R1=330, R2=455, R3=140R_1=330,\ R_2=455,\ R_3=140.

Step 3 — Formulate. The raw material required is ABAB (clients ×\times raw materials).

Step 4 — Solve. Compute the product row by row.

AB=(1015610200)(3407935127).AB = \begin{pmatrix} 10 & 15 & 6 \\ 10 & 20 & 0 \end{pmatrix}\begin{pmatrix} 3 & 4 & 0 \\ 7 & 9 & 3 \\ 5 & 12 & 7 \end{pmatrix}.

Row F1F_1: R1=10⋅3+15⋅7+6⋅5=30+105+30=165R_1 = 10\cdot3+15\cdot7+6\cdot5 = 30+105+30 = 165; R2=10⋅4+15⋅9+6⋅12=40+135+72=247R_2 = 10\cdot4+15\cdot9+6\cdot12 = 40+135+72 = 247; R3=10⋅0+15⋅3+6⋅7=0+45+42=87R_3 = 10\cdot0+15\cdot3+6\cdot7 = 0+45+42 = 87.

Row F2F_2: R1=10⋅3+20⋅7+0=170R_1 = 10\cdot3+20\cdot7+0 = 170; R2=10⋅4+20⋅9+0=220R_2 = 10\cdot4+20\cdot9+0 = 220; R3=10⋅0+20⋅3+0=60R_3 = 10\cdot0+20\cdot3+0 = 60. Hence

AB=(1652478717022060)(rows F1,F2; columns R1,R2,R3).AB = \begin{pmatrix} 165 & 247 & 87 \\ 170 & 220 & 60 \end{pmatrix}\quad\text{(rows } F_1,F_2;\ \text{columns } R_1,R_2,R_3).

Total raw material for both orders (column sums):

R1=165+170=335,R2=247+220=467,R3=87+60=147.R_1 = 165+170 = 335,\quad R_2 = 247+220 = 467,\quad R_3 = 87+60 = 147.

Step 5 — Interpret and validate. The demand 335, 467, 147335,\,467,\,147 exceeds the stock 330, 455, 140330,\,455,\,140 in every raw material, so both orders cannot be filled as they stand. Since the recipe BB is fixed, the firm must either buy more raw material or have the clients cut their orders. For instance, replacing AA by the reduced orders

A1=(912610200)A_1 = \begin{pmatrix} 9 & 12 & 6 \\ 10 & 20 & 0 \end{pmatrix}

gives A1B=(1412167817022060)A_1 B = \begin{pmatrix} 141 & 216 & 78 \\ 170 & 220 & 60 \end{pmatrix}, whose column sums 311, 436, 138311,\,436,\,138 lie comfortably below stock — so the trimmed orders can be supplied.

✓Final answer

AB=(1652478717022060)AB = \begin{pmatrix} 165 & 247 & 87 \\ 170 & 220 & 60 \end{pmatrix}; the orders need 335335 units of R1R_1, 467467 of R2R_2 and 147147 of R3R_3, which exceed the available 330, 455, 140330,\,455,\,140, so they cannot be met as placed. With reduced orders A1A_1 the requirement drops to 311, 436, 138311,\,436,\,138 — within stock.

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