C.The configuration of a byte addressable computer is given as 16-bit memoryaddresses,4K-byte cache which is direct-mapped, each line is of 32 bytes and a clock of1GHz . i.Find Tag, Block, and Word. in bits. ii.Find the total number of bits in the cache. iii.CPU issues the following addresses 125,126,127,0,127,126,125 all decimal .Compute the number of misses.?

Please check if part ii. and iii. are correct in my solution

m = 16 Cache size - 4k a uk = 2x2 10 = 212 32= f = 1 GHz a 109 H 2 23 . ton No. of lines : 2223 = 2 27 L=7 t=16-5-7 = 4 4 5 Total mo. of bito in coche Tag bits & no. of lives & cache six in bito 4 x 27+ 8 x 212 512 at 32768 133280 17 11 125 - 125 - 0000 OOOO OLO! = OOOOOH (live me) 126 0000 0111 TIO = 000001! 127 - Couo 0000 OOOOOO 0 0000 0000 0000 SOOOOOOO 25 T 27 M o 127 126 125 + 14 No. of misses & 6 iv) Avg access time [3x0.95+ (22xoros)] 109 3.95 ms